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MP Board Class 11th Maths Important Questions Chapter 12 Introduction to Three Dimensional Geometry

Introduction to Three Dimensional Geometry Important Questions

Introduction to Three Dimensional Geometry Objective Type Questions

(A) Choose the correct option :

Question 1.
Distance of a point (3,4, 5) from origin:
(a) 3
(b) 4
(c) 5
(d) 5\(\sqrt {2}\)
Answer:
(d) 5\(\sqrt {2}\)

Question 2.
The perpendicular distance of the point P (x, y, z) from X – axis is:
(a) \(\sqrt { { x }^{ 2 } + { z }^{ 2 } }\)
(b) \(\sqrt { { y }^{ 2 } + { z }^{ 2 } }\)
(c) \(\sqrt { { x }^{ 2 } + { y }^{ 2 } }\)
(d) \(\sqrt { { x }^{ 2 } + { y }^{ 2 } + { z }^{ 2 } }\)
Answer:
(b) \(\sqrt { { y }^{ 2 } + { z }^{ 2 } }\)

Question 3.
Distance of a point (3, 2, 5) from X – axis is:
(a) \(\sqrt {28}\)
(b)\(\sqrt {30}\)
(c) \(\sqrt {29}\)
(d) 3
Answer:
(c) \(\sqrt {29}\)

Question 4.
The YZ – plane divides the line segment joining the points (2, 3, 4) and (3, 5, – 4) in the ratio:
(a) 2 : 3 internally
(b) 1 : 2
(c) 2 : 3 externally
(d) 1 : 3
Answer:
(c) 2 : 3 externally

Question 5.
Distance between the points (1, 2,3) and (1,3, -2) is :
(a) \(\sqrt {- 26}\)
(b) \(\sqrt {26}\)
(c) 26
(d) ± \(\sqrt {24}\)
Answer:
(b) \(\sqrt {26}\)

Question 6.
Coordinate of a point on Z – axis, equidistant from the points (1, 5, 7) and (5, 1, – 4) is :
(a) (0, 0, \(\frac { 3 }{ 2 }\)
(b) (o, \(\frac { 3 }{ 2 }\), 0)
(c) (\(\frac { 3 }{ 2 }\), 0, 0)
(d) (4, – 4, – 11)
Answer:
(a) (0, 0, \(\frac { 3 }{ 2 }\)

Question 7.
Distance of a point (2, 1, 4) from Y – axis is:
(a) \(\sqrt {20}\)
(b) 1
(c) \(\sqrt {12}\)
(d) \(\sqrt {10}\)
Answer:
(a) \(\sqrt {20}\)

(B) Match the following :

Answer:

1. (e)
2. (d)
3. (b)
4. (a)
5. (c)

(C) Fill in the blanks :

1. If the points (- 1, 3, 2), (- 4, 2, – 2) and (5, 5, λ ) are collinear, then value of λ is …………….
2. The perpendicular distance of point P(x, y, z) from YZ – plane is …………….
3. XY – plane divides the line segment joining the points (- 3, 4, 8) and (5, – 6, 4) in the ratio …………….
4. Distance between the points (1, – 3, 4) and (3, 11, – 6) is …………….
5. Perpendicular distance of a point P (3,4, 5) from YZ – plane is …………….

Answer:

1. 10
2. x
3. 2 : 1
4. 10\(\sqrt {3}\)
5. 3

(D) Write true / false :

1. Distance of a point (4, 3, 5) from Y – axis is \(\sqrt {40}\).
2. Distance of a point (5, 12, 13) from YZ – axis is \(\sqrt {313}\).
3. The coordinate of a point where YZ – plane divides the join of the points (3, 5, – 7) and (- 2, 1, 8) are (0, \(\frac { 13 }{ 5 }\), 2).
4. Coordinate of mid point of the line segment joining the points (- 3, 4, – 8) and (5, – 6, 4) are (1, – 1, 2).
5. Points A(1, 2, 3), B(4, 0, 4) and C (- 2, 4, 2) are collinear.

Answer:

1. False
2. True
3. True
4. False
5. True

(E) Write answer in one word / sentence :

1. A point R lies on the line segment joining the points P (2, – 3, 4) and Q(8, 0, 10) whose x coordinate is 4, then the coordinate of R will be.
2. In what ratio does the plane 2x + y – z = 3 divide the line segment joining the points A (2, 1, 3) and 5(4, – 2, 5)?
3. If the origin is the centroid of the triangle ABC with vertices A (a, 1, 3), B (- 2, b, – 5), C (4, 7, c) then write the value of a, b and c.
4. Find the locus of the point, which is equidistant to the points (3, 4, – 5) and (- 2, 1, 4).
5. In which ratio the XY – plane divides the line joining the points (2, 4, 2) and (2, 5, – 4)?

Answer:

1. (4, – 2, 6)
2. 1 : 2 externally
3. – 2, – 8, – 2
4. 10x + 6y – 18z – 29 = 0
5. 1 : 2 internally.

Introduction to Three Dimensional Geometry Very Short Answer Type Questions

Question 1.
A point is on X – axis. What are its y – co – ordinate and z – co – ordinate?
Answer:
y – co – ordinate and z – co – ordinate of point on X – axis is 0, 0 is (x, 0, 0).

Question 2.
A point is in the xz – plane. What can you say about is y – co – ordinate?
Answer:
On xz – plane the y – co – ordinate will be zero.

Question 3.
The X – axis and Y – axis taken together to form a plane, name of the plane is.
Answer:
The plane is known as xy – plane.

Question 4.
Co – ordinate of any point on xy – plane is.
Answer:
The co – ordinate of any point on xy – plane is (x, y, 0).

Question 5.
Length of perpendicular of point P(x, y, z) from X – axis is.
Answer:
Length of perpendicular = \(\sqrt { { y }^{ 2 } + { z }^{ 2 } }\)

Question 6.
Length of perpendicular of point P(x, y, z) from yz – plane is.
Answer:
Length of perpendicular of point P(x, y, z) from yz – plane is x.

Question 7.
Find the distance of point (3, 4, 5) from xz – plane.
Answer:
Distance of point (3, 4, 5) from xz – plane M(3, 0, 5) = \(\sqrt {0 + 16 + 0}\) = 4.

Question 8.
What is the centroid of ∆ABC whose vertices A(x₁, y₁, z₁), B(x₂, y₂, Z₂) and C(x₃, y₃, z₃)?
Answer:
Centroid of ∆ABC is

Question 9.
Find distance between points (2,3, 5) and (4,3,1).
Solution:
Distance between two points A(x₁, y₁, z₁) and B(x₂, y₂, Z₂):

Question 10.
Find the distance between points (- 3, 7, 2) and (2, 4, – 1).
Solution:

Question 11.
Find the distance of the point (3, 2, 5) from X – axis.
Solution:
Point on X – axis (3, 0, 0)

Question 12.
Find the distance of the point (2, 1, 4) from Y – axis.
Solution:
Point on Y – axis (0, 1, 0)

Question 13.
The three dimensional planes divides the space in how many octants?
Answer:
Three dimensional plane divides the space in eight octants.

Introduction to Three Dimensional Geometry Short Answer Type Questions

Question 1.
Prove that the points (- 2, 3, 5) (1, 2, 3) and (7, 0, – 1) are collinear. (NCERT)
Solution:
Given point are A (- 2, 3, 5), B(1, 2, 3) and C (7, 0, – 1).

Hence points A, B, Care collinear.

Question 2.
Find the co – ordinates of the point which divides the line segment joining the points (- 2, 3, 5) and (1, – 4, 6) in the ratio 2 : 3 internally. (NCERT)
Solution:
Required co – ordinates are :

Question 3.
Given that P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, – 10) are collinear. Find the ratio in which Q divides PR. (NCERT)
Solution:

⇒ 5k + 5 = 9k + 3
⇒ 4k = 2
⇒ k = \(\frac { 1 }{ 2 }\)
Hence, point Q divides PR internally in ratio 1 : 2.

Question 4.
Find the ratio in which the YZ – plane divides the line segment formed by joining the points (- 2, 4, 7) and (3, – 5, 8).
Solution:
Let the point R (0, y, z) on YZ – plane divides the points P (- 2, 4, 7) and Q (3, – 5, 8) in the ratio m : n.
x = \(\frac{m x_{2}+n x_{1}}{m+n}\)
⇒ 0 = \(\frac { m × 3 + n( – 2) }{ m + n }\)
⇒ 3m – 2n = 0
⇒ 3m = 2n
⇒ \(\frac { m }{ n }\) = \(\frac { 2 }{ 3 }\)
⇒ m : n = 2 : 3
YZ – plane divides the line PQ internally in ratio 2:3.

Question 5.
If the origin is the centroid of the ∆PQR with vertices P(2a, 2, 6), Q(- 4, 3b, – 10) and R (8, 14, 2c) then find the value of a, b and c. (NCERT)
Solution:
Centroid of APQR is :
x = \(\frac{x_{1} + x_{2} + x_{3}}{3}\), y = \(\frac{y_{1} + y_{2} + y_{3}}{3}\), z = \(\frac{z_{1} + z_{2} + z_{3}}{3}\)
Co – ordinate of centroid are (0, 0, 0),

Question 6.
Find the co – ordinates of a point on Y – axis which are at a distance of 5\(\sqrt {2}\) from the point P(3, – 2, 5).
Solution:
Let A (0, y, 0) be any point on Y – axis.
Given : PA = 5\(\sqrt {2}\)
PA² = 50
⇒ (3 – 0)² + (- 2 – y)² + (5 – 0)² = 50
⇒ 9 + 4 + y² + 4y + 25 = 50
⇒ y² + 4y + 38 – 50 = 0
⇒ y² + 4y – 12 = 0
⇒ y² + 6y – 2y – 12 = 0
⇒ y(y + 6) – 2(y + 6) = 0
⇒ (y – 2)(y + 6) = 0
⇒ y = 2, – 6
The co – ordinate on Y – axis is (0, 2, 0) or (0, – 6, 0).

Question 7.
A point R with x co – ordinate 4 lies on the line segment joining the points P (2, – 3, 4) and Q (8, 0, 10). Find the co – ordinate of the point R.
Solution:
Let the point R (x, y, z) divides PQ in ratio m : n.

Co – ordinate of R (4, – 2, 6).

Question 8.
Find the equation of set of points which are equidistant from the points (1, 2, 3) and (3, 2, – 1). (NCERT)
Solution:
Let point P (x, y, z) be equidistant from points A (1, 2, 3) and B (3, 2, – 1).
∴ PA = PB

⇒ PA² = PB²
⇒ (x – 1)² + (y – 2)² + (z – 3)² = (x – 3)² + (y – 2)² + (z +1)²
⇒ x² – 2x + 1 + z² – 6z + 9
– 2x – 6z + 10 = – 6x + 2z + 10
⇒ 4x – 8z = 0
⇒ x – 2z = 0.
This is the required equation.

Question 9.
In which ratio the point (1, 1, 1) divides the line joining the points (3, – 2, 4) and (- 1, 4, 2)?
Solution:
Let point R (1,1,1), divides the line joining the points P (3, – 2, 4) and Q(- 1, 4, 2) in ratio λ : 1.
By formula:
x = \(\frac{m x_{2}+n x_{1}}{m+n}\)
Here x₁ = 3, x₂ = – 1, x = 1, m = λ and n = 1.
∴ 1 = \(\frac { λ(- 1) + 1 × 3 }{ λ + 1}\)
⇒ λ + 1 = – λ + 3
⇒ 2λ = 2
∴ λ = 1
Hence, required ratio is 1 : 1.

MP Board Class 11th Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 11 Conic Sections

Conic Sections Important Questions

Conic Sections Objective Type Questions

(A) Choose the correct option :

Question 1.
Coordinates of the focus of the parabola y = 2x² + x are:
(a) (0, 0)
(b) (\(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 4 }\))
(c) (- \(\frac { 1 }{ 4 }\), 0)
(d) ( – \(\frac { 1 }{ 4 }\), \(\frac { 1 }{ 8 }\))
Answer:
(c) (- \(\frac { 1 }{ 4 }\), 0)

Question 2.
In a ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 a > b, the relation between a, b an eccentricity e is:
(a) b² = a²(1 – e²)
(b) b² = a²(e² – 1)
(c) a² = b²(1 – e²)
(d) a² = b²(e² – 1)
Answer:
(a) b² = a²(1 – e²)

Question 3.
The length of latus rectum of ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, represent a circle then its eccentricity will be:
(a) \(\frac { { 2a }^{ 2 } }{ b }\)
(b) \(\frac { { 2b }^{ 2 } }{ a }\)
(c) \(\frac { { a }^{ 2 } }{ b }\)
(d) \(\frac { { b }^{ 2 } }{ a }\)
Answer:
(b) \(\frac { { 2b }^{ 2 } }{ a }\)

Question 4.
The eccentricity of the parabola is:
(a) Less than 1
(b) Greater than 1
(c) 0
(d) 1
Answer:
(d) 1

Question 5.
The eccentricity of the ellipse is:
(a) Less than 1
(b) Greater than 1
(c) 0
(d) 1
Answer:
(a) Less than 1

Question 6.
The eccentricity of the hyperbola is:
(a) Less than 1
(b) Greater than 1
(c) 0
(d) 1
Answer:
(b) Greater than 1

Question 7.
In a ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, represent a circle then its eccentricity will be:
(a) Less than 1
(b) Greater than 1
(c) 0
(d) 1
Answer:
(c) 0

Question 8.
The sum of focal distances from any point on the ellipse is:
(a) Equal to major axis
(b) Equal to minor axis
(c) The distance between two foci
(d) Equal to latus rectum.
Answer:
(a) Equal to major axis

Question 9.
The differecne of the focal distances from any point on the hyperbola is:
(a) Equal to its conjugate axis
(b) Equal to its transverse axis
(c) The distance between two foci
(d) Equal to its latus rectum.
Answer:
(b) Equal to its transverse axis

Question 10.
The value of the eccentricity of ellipse 25x² + 16y² = 400 is:
(a) \(\frac { 3 }{ 5 }\)
(b) \(\frac { 1 }{ 3 }\)
(c) \(\frac { 2 }{ 5 }\)
(d) \(\frac { 1 }{ 5}\)
Answer:
(a) \(\frac { 3 }{ 5 }\)

Question 11.
Equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represent a circle if:
(a) a = b, c = 0
(b) f = g, h = 0
(c) a = b, h = 0
(d) f = g, c = 0
Answer:
(a) a = b, c = 0

Question 12.
Area of triangle whose centre (1,2) and which is passes through the point (4,6) will be:
(a) 5π
(b) 10π
(c) 25π
(d) 25π²
Answer:
(c) 25π

Question 13.
The circle passing through (1, – 2) and touching the X – axis at (3,0), also passes through the point:
(a) (2, – 5)
(b) (5, – 2)
(c) (- 2, 5)
(d) (- 5, 2)
Answer:
(a) (2, – 5)

Question 14.
The length of the diameter of the circle which touches the X – axis at the point (1,0) and passes through the point (2,3) is:
(a) \(\frac { 10 }{ 3 }\)
(b) \(\frac { 3 }{ 5 }\)
(c) \(\frac { 6 }{ 5 }\)
(d) \(\frac { 5 }{ 3 }\)
Answer:
(a) \(\frac { 10 }{ 3 }\)

Question 15.
Eccentricity of the hyperbola 3x² – y² = 4 :
(a) 1
(b) 2
(c) – 2
(d) \(\sqrt {2}\)
Answer:
(b) 2

(B) Match the following :

Answer:

1. (c)
2. (e)
3. (b)
4. (a)
5. (d)
6. (i)
7. (h)
8. (f)
9. (j)
10. (g)

(C) Fill in the blanks :

1. The length of the latus rectum of the parabola y² = 4ax is ……………
2. The centre of the ellipse \(\frac { { (x-1) }^{ 2 } }{ 9 } +\frac { { (y-2) }^{ 2 } }{ 4 }\) = 1 will be ……………
3. The vertex of the parabola (y – 2)² = 4a(x -1) is ……………
4. The lines \(\frac { x }{ a }\) – \(\frac { y }{ b }\) = m and \(\frac { x }{ a }\) + \(\frac { y }{ b }\) = \(\frac { 1 }{ m }\) meets always at ……………
5. If an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, a > b and its eccentricity is e, then the foci will be
6. Standard form of equation of parabola is ……………
7. Parametric equation of a circle x² + y² = 4 is ……………
8. A line y = x + a\(\sqrt {2}\) touches the circle x² + y² = a² point ……………
9. A line y = mx + c touches the circle x² + y² = a² if c = ……………
10. Vertex of the parabola 3y² + 6y – 4x + 11 = 0 is ……………
11. Equation 2x² + 2y² – 12x – 16y + 4 = 0 represent a point circle if k = ……………
12. Radius of circle 3x² + 3y² – 5x – 6y + 4 = 0 is ……………

Answer:

1. 4a
2. (1, 2)
3. (1, 2)
4. Hyperbola
5. (± ae, 0)
6. y² = 4ax
7. x = 2cosθ
8. (- \(\frac { a }{ \sqrt { 2 } }\), \(\frac { a }{ \sqrt { 2 } }\) )
9. ±a\(\sqrt { 1+{ m }^{ 2 } }\)
10. (5, 1)
11. 50, 12
12. \(\sqrt {61}\)

(D) Write true / false :

1. Conic section is a locus of the point whose the ratio between the distance from the fixed point and distance from the fixed line, this ratio is called eccentricity of the conic section.
2. The ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 has two directrics, the equation of directrics are x = ± \(\frac { a }{ e }\); Where a > b and y = ± \(\frac { b }{ e }\) ; where b > a.
3. The foci of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 are (0, ± be) where a < b.
4. A circle drawn by taking major axis of the ellipse as diameter is called auxiliary circle of the ellipse.
5. The locus of intersection point of the lines bx + ay = abt and bx – ay = \(\frac { ab}{ t }\) will be a ellipse.
6. The focus of a parabola x² = – 16y will be (0, – 4).
7. Equation x² + y² – 6x + 8y + 50 = 0 represent a circle.
8. Circle x² + y² = 9 and x² + y² + 8y + c = 0 touches externally if c = 15 .
9. Eccentricity of hyperbola is 1.
10. Minimum distance between line y – x = 1 and curve x = y² is \(\frac { 3\sqrt { 2 } }{ 8 }\)

Answer:

1. True
2. True
3. True
4. True
5. False
6. True
7. False
8. True
9. False
10. True

(E) Write answer in one word / sentence :

1. If the circle x² + y² + 2ax + 8y +16 = 0, touches X – axis, then the value of α will be.
2. Coordinate of focus of parabola x² = – 10y will be.
3. Write the equation of a circle whose centre is (2,2) and passes through the point (4, 5).
4. The centre of a circle is (5, 7) and touches Y – axis, then its radius will be.
5. If the radius of a circle x² + y² – 6x + ky – 25 = 0 is \(\sqrt {38}\) the value of k will be.
6. Vertex of the parabola y = x² – 2x + 3 will be.
7. Equation of a parabola whose vertex (0, 0) and focus (0, 3) will be.
8. Length of major axis of ellipse 9x² + 16y² = 144 will be.
9. Eccentricity of an ellipse whose latus rectum in half of its minor axis will be.
10. Equation of hyperbola whose one focus in (4, 0 ) and corresponding equation of directrix x = 1 will be.

Answer:

1. ± 4
2. (0, \(\frac { – 5 }{ 2 }\))
3. x² + y² – 4x – 4y – 5 = 0
4. 7, 5
5. ± 4
6. (1, 2)
7. x² = 12y
8. 6
9. \(\frac { \sqrt { 3 } }{ 2 }\)
10. \(\frac { x^{ 2 } }{ 4 } -\frac { y^{ 2 } }{ 12 } \) = 1

Conic Sections Long Answer Type Questions

Question 1.
Find the equation of circle which touches the X – axis at a distance of 4 units in the negative direction and makes intercept of 6 units on positive direction of Y – axis.
Solution:
Here OA = CM = 4, BD = 6.
Length of perpendicular drawn from centre C on BD.
Then, BM = MD = 3
In right angled ∆ CMB,
CB² = CM² + BM²
= 4² + 3²
= 16 + 9 =25
⇒ CB = 5

∴ CA = Radius of circle = CB = 5
∴ Centre of circle (- 4, 5) and radius = 5
Hence, equation of circle :
(x + 4)² + (y – 5)² = 5²
⇒ x² + 8x + 16 + y² – 10y + 25 = 25
⇒ x² + y² + 8x – 10y + 16 = 0

Question 2.
Find the equation of circle which touches Y – axis at a distance of 4 units and makes intercept of 6 units on Y – axis?
Solution:
Given : OP = 4, AB = 6, PC = AC = radius.
CM ⊥ AB ∴ AM = BM = \(\frac { 6 }{ 2 }\) = 3
OP = CM = 4
In right angled ∆ AMC,

AC² = AM² + CM²
= (3)² + (4)² = 9 + 16 = 25
∴ AC = 5
From figure PC – OM= 5 = radius
Centre of circle is (5, 4) and radius = 5.
Hence, required equation of circle :
(x – 5)² + (y – 4)² = (5)²
x² – 10x + 25 + y² – 8y + 16 = 25
x² + y² – 10x – 8y + 16 = 0.

Question 3.
ABCD is a square. Supposing AB and AD as the coordinate axes. Find the equation of the circle circumscribing the square if each side of square is of length l.
Solution:
Taking AB and AD as X – axis and Y – axis respectively
Given : AB = BC = CD = DA = 1
M is mid point of AB.
N is mid point of AD.
AM = \(\frac { l }{ 2 }\), AN = \(\frac { l }{ 2 }\) = OM
In ∆OAM,
OA² = AM² + OM²
= \(\frac { l }{ 2 }\)² + \(\frac { l }{ 2 }\)²
= \(\frac{l^{2}}{4}+\frac{l^{2}}{4} = \frac{l^{2}}{2}\)
∴ Radius = OA = \(\frac{l}{\sqrt{2}}\)
Centre of circle (AM, OM) = ( \(\frac { l }{ 2 }\), \(\frac { l }{ 2 }\) )

Required equation of circle is :
(x – \(\frac { l }{ 2 }\) )² + (y – \(\frac { l }{ 2 }\) )² = \(\frac{l^{2}}{2}\)
Centre of circle (AM, OM) = (\(\frac { l }{ 2 }\), \(\frac { l }{ 2 }\))
Required equation of circle is :
(x – \(\frac { l }{ 2 }\))² + (y – \(\frac { l }{ 2 }\))² = \(\frac{l^{2}}{2}\)
⇒ x² – lx + \(\frac{l^{2}}{4}\) + y² – ly + \(\frac{l^{2}}{4}\) = \(\frac{l^{2}}{2}\)
⇒ x² + y² – l(x + y) = 0
⇒ x² + y² = l(x + y)

Question 4.
Find the equation of the circle passing through the points (4, 1) and (6, 5). Whose centre lies on line 4x + y = 16. (NCERT)
Solution:
Let the equation of circle be
x² + y² + 2gx + 2 fy + c = 0 …. (1)
It passes through points (4, 1) and (6, 5).
∴ 8g + 2f + c + 17 = 0 …. (2)
and 12g + 10f + c + 61 = 0 …. (3)
Centre of circle (1) is (- g, – f) which lies on line 4x + y = 16.
∴ – 4g – f – 16 = 0
⇒ 4g + f + 16 – 0 …. (4)
Subtracting equation (2) from equation (3), we get
4g + 8f + 44 = 0
⇒ g + 2f + 11 = 0 …. (5)
On solving equation (4) and (5), g = – 3, f = – 4
Put g = – 3 and f = – 4 in equation (2),
– 24 – 8 + C + 17 = 0
⇒ c = 15
Put values of g, f and c in equation (1), then required equation of circle is :
x² + y² – 6x – 8y + 15 = 0.

Question 5.
Find the equation of the circle which passes through the points (2, 3) and (- 1, 1) whose centre lies on line x – 3y – 11 = 0. (NCERT)
Solution:
Let the equation of circle is :
x² + y² + 2gx + 2fy + c = 0 …. (1)
∵Points (2, 3) and (- 1, 1) lies on equation (1),
∴ (2)² + (3)² + 2g(2) + 2f(3) + c = 0
⇒ 4 + 9 + 4g + 6f + c + 13 = 0
4g + 6f + c + 13 = 0 …. (2)
and (-1)² + (l)² – 2g + 2f + c = 0
⇒ 1 + 1 – 2g + 2f + c = 0
⇒ – 2g + 2f + c + 2 = 0 …. (3)

Putting the value of g, f and c in equation (1), then required equation of circle will be :
x² + y² + 2(\(\frac { -7 }{ 2 }\))x + 2(\(\frac { 5 }{ 2 }\))y + c = 0 …. (1)
x² + y² – 7x + 5y -14 = 0.

Question 6.
Find the equation of circle whose radius is 5, centre is on Y – axis and which passes through point (2, 3).
Solution:
Centre of circle is on X – axis, so k = 0.
Let the equation of circle be :
(x – h)² + (y – k)² = a²
Here a = 5
(x – h)² + (y – 0)² = (5)²
(x – h)² + y² = 25
Circle (1) passes through point (2, 3),
∴ (2 – h)² + (3)² = 25
⇒ (2 – h)² = 25 – 9 = 16 = (4)²
⇒ 2 – h = ± 4

Question 7.
y = mx is a chord of the circle whose radius is ‘a’ and its diameter is X – axis. Origin is one of the limiting points of the chord. Show that the equation to a circle whose diameter is the given chord is given by the equation (1 + m²) (x² + y² ) – 2a (x + my) = 0. Solution:
Equation of circle whose radius is a and centre is (a, 0) will be
(x – a)² + y² = a²
⇒ x² – 2ax + y² + a² = a²
⇒ x² – 2ax + y² = 0 …. (1)
Equation of given line is :
y = mx …. (2)
Now, equation of circle passing through the intersection of eqns. (1) and (2) will be :
x² + y² – 2ax + λ(y – mx) = 0 … (3)
Centre of co – ordinate of circle (3) are (\(\frac { λm + 2a }{ 2 }\), \(\frac { λ }{ 2 }\))
∵ Centre lies on line y = mx.
∴ – \(\frac { λ }{ 2 }\) = m\(\frac { λm + 2a }{ 2 }\)
⇒ λ = \(\frac{-2 a m}{1+m^{2}}\)

Put the value of λ in Equation (3),
x² + y² – 2ax + \(\frac{-2 a m}{1+m^{2}}\) (y – mx) = 0
⇒ (l + m²)(x² + y²) = 2ax + 2am²x + 2amy – 2am²x
⇒ (l + m²)(x² + y²) = 2a(x + my)
⇒ (l + m²)(x² + y²) – 2a(x + my) = 0
Which is required equation of circle.

Question 8.
If the straight line x cos α + y sin α = p cuts a circle x² + y² = a² in two points M and N, then show that the equation of the circle whose diameter is MN will be x² + y² – a² = 2p(x cos α + y sin α – p).
Solution:
Given : Equation of line is :
x cos α + y sin α = p …. (1)
and Equation of circle is
x² + y² = a² …. (2)
Now, equation of circle passing through the intersection of line (1) and circle (2) at points M and N is :
x² + y² – a² + λ(x cos α + y sin α – p) = 0 …. (3)
If MN is diameter of above circle then centre is :
(- \(\frac { λ }{ 2 }\)cos α, – \(\frac { λ }{ 2 }\)sin α)
Which is lies on line x cos α + y sin α = p.
– ( \(\frac { λ }{ 2 }\)cos α )cos α + (- \(\frac { λ }{ 2 }\)sin α)sin α = p
⇒ – \(\frac { λ }{ 2 }\)[cos² α + sin² α] = p
⇒ λ = – 2p
Put the value of λ in equation (3), then required equation of circle is
x² + y² – a² – 2p(x cos α + y sin α – p) = 0
⇒ x² + y² – a² = 2p(x cos α + y sin α – p)

Question 9.
Find the following equation of parabola : (i) co – ordinates of focus, (ii) axis, (iii) equation of directrix, (iv) length of Iatus rectum. (NCERT)
(A) y² = 12x
Solution:
Equation of parabola : y² = 12x
Comparing with y² = 4 ax,
4a = 12 ⇒ a = 3
∴Co – ordinates of focus (a, 0) = (3, 0).
Axis of parabola = X – axis.
Equation of directrix is x = – a ⇒ x = – 3.
Length of latus rectum = 4a = 4 x 3 = 12.

(B) x² = 6y.
Solution:
Equation of parabola: x² = 6y
Comparing with x² = 4ay
4a = 6 a ⇒ 3/2
∴ Co – ordinate of focus (0, a) = (0, 3/2).
Axis of parabola = Y – axis.
Equation of directrix is y = – a ⇒ y = – 3/2.

(C) y² = – 8x
Solution:
Equation of parabola : y² = – 8x
Comparing with y² = – 4ax
– 4a – = – 8 ⇒ a = 2
∴ Co – ordinates of focus (- a, 0) = (- 2, 0).
Axis of parabola = X – axis.
Equation of directrix is x = a ⇒ x = 2.
Length of latus rectum 4a = 4 x 2 = 8.

(D) x² = – 16y
Solution:
Equation of parabola : x² = – 16y
Comparing with x² = – 4ay
– 4a = – 16 ⇒ a = 4
∴ Co – ordinate of focus (0, – a) = (0, – 4)
Axis of parabola = Y – axis
Equation of directrix is y = a ⇒ y = 4
Length of latus rectum = 4a = 16.

Question 10.
An equilateral triangle inscribed in the parabola y² = 4ax, where one vertex is at the vertex of parabola. Find the length of the side of triangle. (NCERT)
Solution:
Let the equation of parabola is y² = 4ax.
Let APQ be the equilateral triangle whose vertex A(0, 0), P(h, k) and Q(h, – k).
AP² = (h – 0)² + (k – 0)²
= h² + k²
⇒ AP = \(\sqrt{h^{2}+k^{2}}\)
Similarly, AQ = \(\sqrt{h^{2}+k^{2}}\)

Again, PQ = \(\sqrt{(h-k)^{2}+(k+h)^{2}}\)
= \(\sqrt{(2k)^{2}}\) = 2k
∴ AP = PQ
⇒ \(\sqrt{h^{2}+k^{2}}\) = 2k
⇒ h² + k² = 4k²
⇒ h² = 3k²
⇒ h = \(\sqrt {3}\).k
∵ Point P(h, k) lies on parabola y² = 4ax.
k² = 4ah = 4a.\(\sqrt {3}\)k
⇒ k = 4a\(\sqrt {3}\), [∵ k ≠ 0]
Hence, length of side PQ = 2k = 2.(4a\(\sqrt {3}\)) = 8a\(\sqrt {3}\).

Question 11.
If a parabola reflector is 20 cm in diameter and 5 cm deep. Find the focus. (NCERT)
Solution:
Taking vertex of parabola reflector at origin and X – axis along the axis of parabola.

Equation of parabola y² = 4ax …. (1)
Given : OS = 5 cm, AB = 20 cm, AS = 10 cm
∴ Co – ordinate of A will be (5, 10).
∴ (10)² = 4a x 5
⇒ 100 = 20a
⇒ a = 5
∴ OS = 5 cm
Co – ordinates of focus S will be (5, 0).

Question 12.
An arch is in the form of a parabola with its vertical axis. The arch is 10 m high and 5 m wide at the base. How wide it is 2 m vertex of the parabola. (NCERT)
Solution:
Let the equation of parabola is :
x² = 4 ay …. (1)
Given : AB = 5 metre
AF = BF = \(\frac { 5 }{ 2 }\) metre
OE = 2 metre
OF = 10 metre
Co – ordinate of A will be (\(\frac { 5 }{ 2 }\), 10)
This point lies on parabola, hence it will be satisfy equation (1),
∴ \(\frac { 5 }{ 2 }\)² = 4a x 10
⇒ \(\frac { 25 }{ 4 }\) = 4 x a x 10
⇒ a = \(\frac { 25 }{ 4 × 4 × 10 }\)
⇒ a = \(\frac { 5 }{ 32 }\)

Put the value of a in Equation (1),
∴ x² = 4 x \(\frac { 5 }{ 32 }\)y
⇒ x² = \(\frac { 5 }{ 8 }\)y
Let EC = k
OE = 2
Co – ordinate of C will be (k, 2) and it will satisfy equation of parabola.
We get k² =\(\frac { 5 }{ 8 }\) x 2
⇒ k² = \(\frac { 5 }{ 4 }\)
⇒ k = \(\frac{\sqrt{5}}{2}\)
DE = 2EC
2 x \(\frac{\sqrt{5}}{2}\) = \(\sqrt {5}\)
= 2.23 metre (approx.)

Question 13.
In each of the following ellipse. Find the co – ordinates of the foci and vertices, the length of major axis and minor axis, the eccentricity and the length of latus rectum of the ellipse. (NCERT)
(A) = \(\frac{x^{2}}{36}+\frac{y^{2}}{16}\) = 1.
Solution:
Comparing with standard form of ellipse, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
We get, a² = 36 ⇒ a = 6, b² = 16 ⇒ b = 4
Here a > b.
∴ b² = a²(1 – e²)
⇒ 16 = 36(1 – e²)
⇒ 1 – e² = \(\frac { 16 }{ 36 }\) = \(\frac { 4 }{ 9 }\)
⇒ e² = 1 – \(\frac { 4}{9 }\) = \(\frac { 5 }{ 9 }\)
∴ Eccentricity e = \(\frac{\sqrt{5}}{3}\)
Foci (± ae, o) = (± 6 × \(\frac{\sqrt{5}}{3}\), o)
= (± 2\(\sqrt {5}\), 0)
Vertices (± a, 0) = (± 6, 0)
Length of major axis = 2a = 2 x 6 = 12.
Length of minor axis = 2b = 2 x 4 = 8.
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac { 2 × 16 }{ 6 }\) = \(\frac { 16 }{ 3 }\).

(B) \(\frac{x^{2}}{4}+\frac{y^{2}}{25}\) = 1.
Solution:
Comparing with standard form of ellipse, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
We get, a² = 36 ⇒ a = 6, b² = 16 ⇒ b = 4
Here a < b.
∴ a² = b²(1 – e²)
⇒ 4 = 25(1 – e²)
⇒ 1 – e² = \(\frac { 4 }{ 25 }\)
⇒ e² = 1 – \(\frac { 4}{25 }\) = \(\frac { 21 }{ 25 }\)
∴ Eccentricity e = \(\frac{\sqrt{21}}{5}\)
Foci (0, ± b) = (0, ± 5 × \(\frac{\sqrt{21}}{5}\))
= (0, ± \(\sqrt {21}\))
Vertices (0, ± b) = (0, ± 5)
Length of major axis = 2b = 2 x 5 = 12.
Length of minor axis = 2a = 2 x 2 = 8.
Length of latus rectum = \(\frac{2 a^{2}}{b}\) = \(\frac{2 \times 2^{2}}{5}\) = \(\frac { 2 × 4 }{ 5 }\) = \(\frac { 8 }{ 5 }\).

Question 14.
Find the equation of hyperbola whose foci is (± 4, 0) and length of latus rectum is 12. (NCERT)
Solution:
Foci of hyperbola (± 4, 0).
Hence equation of hyperbola will be :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
Foci (±ae, 0) = (± 4, 0)
∴ ae = 4
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = 12
⇒ b² = 6a
We know, b² = a²(e² – 1)
⇒ 6a = a²e² – a²
⇒ 6a = 4² – a²
⇒ 6a = 16 – a²
⇒ 6a = 16 – a²
⇒ a² + 6a – 16 = 0
⇒ a² + 8a – 2a – 16 = 0
⇒ a(a – 2)(a + 8) = 0
⇒ a = 2, a = – 8, (∵ a cannot be negative)
∴ a = 2
b² = 6a = 6 x 2 = 12
⇒ b = \(\sqrt {12}\)
Putting values of a and b in equation (1), then required equation of hyperbola will be :
\(\frac{x^{2}}{2^{2}}-\frac{y^{2}}{(\sqrt{12})^{2}}\) = 1
⇒ \(\frac{x^{2}}{4}+\frac{y^{2}}{12}\) = 1
⇒ 3x² – y² = 12.

Question 15.
Find the axis, foci, directrix, eccentricity and the latus rectum of the ellipse 9x² + 4y² = 36.
Solution:
Given equation of ellipse
9x² + 4y² = 36
⇒ \(\frac{x^{2}}{4}+\frac{y^{2}}{9}\) = 1
Here, b² > a² or b > a
∴ Major axis = 2.3 = 6
Minor axis = 2.2 = 4
Now,
a² = b²(1 – e²)
(2)² = (3)²(1 – e²)
⇒ 4 = 9(1 – e²)
⇒ \(\frac { 4 }{ 9 }\) = 1 – e²
⇒ e² = 1 – \(\frac { 4 }{ 9 }\) = \(\frac { 9 – 4 }{ 9 }\) = \(\frac { 5 }{ 9 }\)
∴ e = \(\frac{\sqrt{5}}{3}\)
Co – ordinate of foci = (0, ± be )
= (0, ± 3. \(\frac{\sqrt{5}}{3}\))
= (0, ± \(\sqrt {5}\)).
Co – ordinate of vertex = (0, ± b )
= (0 ± 3 ).
Length of latus rectum = \(\frac{2 a^{2}}{b}\)
= \(\frac { 2.4 }{ 3 }\) = \(\frac { 8 }{ 3 }\).

Question 16.
(A) Find the equation of ellipse whose vertices are (± 5, 0) and foci (± 4, 0).
Solution:
Given : Vertices are (± 5, 0) and foci are (± 4, 0)
∴ a = 5
and ae = 4
⇒ 5e = 4
⇒ e = \(\frac { 4 }{ 5 }\)
Let the equation of ellipse is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, where a > b …. (1)
∴ From b² = a²(1 – e²),
b² = 5²(1 – (\(\frac { 4 }{ 5 }\))²)
b² = 25(1 – \(\frac { 16 }{ 25 }\))
b² = 25 x \(\frac { 9 }{ 25 }\) = 9
Putting values of a and b in equation (1), the required equation of ellipse will be :
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1
⇒ 9x² + 25y² = 225

(B) Find the equation of ellipse whose vertices are (0, ± 13) and foci is (0, ± 5).
Solution:
Let the equation of ellipse is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, where a < b …. (1)
Vertices of ellipse = (0, ± b) = (0, ± be)
∴ b = 13
Foci = (0, ± 5) = (0, ± be)
∴ be = 15
⇒ 13 x 3 = 5
⇒ e = \(\frac { 5 }{ 13 }\)
Now, a² = b²(1 – e²)
⇒ a² = 13²[1 – ( \(\frac { 5 }{ 13 }\))² ]
⇒ a² = 169[1 – \(\frac { 25 }{ 169 }\) ]
⇒ a² = 169\(\frac { 169 – 25 }{ 169 }\)
⇒ a² = 144
⇒ a = 12.
Putting values of a and b in equation (1), then required equation of ellipse will be :
\(\frac{x^{2}}{(12)^{2}}-\frac{y^{2}}{(13)^{2}}\) = 1
⇒ \(\frac{x^{2}}{144}+\frac{y^{2}}{169}\) = 1

Question 17.
Find the equation of ellipse whose centre is at (0, 0), major axis on the Y – axis passing through the points (3, 2) and (1, 6).
Solution:
Let the equation of ellipse is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, where a < b …. (1)
∵ Equation (1) passes through points(3, 2) and (1, 6)
∴ \(\frac{9}{a^{2}}+\frac{4}{b^{2}}\) = 1 …. (2)
and \(\frac{1}{a^{2}}+\frac{36}{b^{2}}\) = 1 …. (3)
Multiply equation (2) by 9, we get

Putting value of a² in equation (2), we get
\(\frac { 9 }{ 10 }\) + \(\frac{4}{b^{2}}\) = 1
⇒ \(\frac{4}{b^{2}}\) 1 – \(\frac { 9 }{ 10 }\)
⇒ \(\frac{4}{b^{2}}\) = \(\frac { 1}{ 10 }\)
⇒ b² = 40
Putting values of a² and b² in equation (1), then required equation of ellipse will be :
\(\frac{x^{2}}{10}+\frac{y^{2}}{40}\) = 1

Question 18.
Find the equation of ellipse whose major axis on the X – axis which passes through the points (4, 3) and (6, 2).
Solution:
Let the equation of ellipse is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 …. (1)
∵ It passes through points(4, 3) and (6, 2)
∴\(\frac{36}{a^{2}}+\frac{4}{b^{2}}\) = 1 …. (2)
and \(\frac{16}{a^{2}}+\frac{9}{b^{2}}\) = 1 …. (3)
Subtracting equation (3) from equation (2), we get
\(\frac{20}{a^{2}}-\frac{5}{b^{2}}\) = 1
⇒ \(\frac{4}{a^{2}}-\frac{1}{b^{2}}\) = 1
⇒ \(\frac{4}{a^{2}} = \frac{1}{b^{2}}\)
⇒ a² = 4b²
Putting value of a² in equation (2), we get
\(\frac{36}{4b^{2}}+\frac{4}{b^{2}}\) = 1
⇒ \(\frac{9}{b^{2}}+\frac{4}{b^{2}}\) = 1
⇒ 9 + 4 = b²
⇒ b² = 13
Putting values of a² and b² in equation (1), hence
Required equation of ellipse \(\frac{x^{2}}{52}+\frac{y^{2}}{13}\) = 1

Question 19.
An arch is the form of a semi ellipse. It is 8 m wide and 2 m high of the centre. Find the height of the arch at a point 1-5 m from one end. (NCERT)
Solution:
Let the equation of ellipse is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 …. (1)
Here 2a = 8 ⇒ a = 4, b = 2
Putting the values of a and b, we get

\(\frac{x^{2}}{4^{2}}-\frac{y^{2}}{2^{2}}\) = 1 …. (1)
\(\frac{x^{2}}{16}+\frac{y^{2}}{4}\) = 1
Given :
AP = 1.5m, OA = \(\frac { 8 }{ 2 }\) = 4m,
OP = OA – AP = 4 – 1.5 = 2.5m.
Let PQ = k
∴ Co – ordinate of Q will be (2.5, k), which will satisfy ellipse’s equation.
Hence,
\(\frac{(2.5)^{2}}{16}+\frac{k^{2}}{4}\) = 1
⇒ \(\frac { 6.25}{ 16 }\) + \(\frac{k^{2}}{4}\) = 1
⇒ \(\frac{k^{2}}{4}\) = 1 – \(\frac { 6.25 }{ 16 }\)
⇒ \(\frac{k^{2}}{4}\) = \(\frac { 16 – 6.25 }{ 16 }\)
⇒ k² = \(\frac { 9.75 }{ 4 }\)
⇒ k² = 2.437
⇒ k = 1.56metre (approx).

Question 20.
A rod of length 12 cm moves with its ends always touching the co – ordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the X – axis. (NCERT)
Solution:
Let AB be the rod of length 12 cm which make an angle θ with X – axis.
∴ ∠BAO = θ
AB = 12 cm
AP = 3 cm, then PB = 9 cm

In ∆PNA,
sin θ = \(\frac {PN }{ PA }\) = \(\frac { y }{ 3 }\)
In ∆PMB,
cos θ = \(\frac { PM }{ PB }\) = \(\frac { x }{ 9 }\)
sin²θ + cos²θ = \(\frac { y }{ 3 }\)² + \(\frac { x }{ 9 }\)²
⇒ 1 = \(\frac{y^{2}}{9}+\frac{x^{2}}{81}\)
Hence required equation is :
\(\frac{x^{2}}{81}+\frac{y^{2}}{9}\) = 1

Question 21.
Find the eccentricity, co – ordinate of foci, equation of directrix and length of Iatus rectum of ellipse 4x² + y² – 8x + 2y + 1 = 0.
Solution:
4x² + y² – 8x + 2y + 1 = 0
⇒ 4x² – 8x + y² + 2y +1 = 0
⇒ 4x² – 8x + (y + 1)² = 0
⇒ 4(x² – 2x) + (y + 1)² = 0
⇒ 4(x² – 2x + 1) + (y + 1)² = 4
⇒ 4(x² – 2x + 1) + (y + 1)² = 4
⇒ \(\frac{(x-1)^{2}}{1}+\frac{(y+1)^{2}}{4}\) or \(\frac{X^{2}}{1}+\frac{Y^{2}}{4}\) = 1
Here, b > a
a² = b²(1 – e²)
⇒ 1 = (1 – e²)
⇒ \(\frac { 1 }{ 4}\) = 1 – e²
⇒ e² = 1 – \(\frac { 1 }{ 4}\) = \(\frac { 3 }{ 4}\)
⇒ Eccentricity e = \(\frac{\sqrt{3}}{2}\)
Co – ordinate of foci (0, ± be)
= (0, ±2. \(\frac{\sqrt{3}}{2}\)
= (0, ± \(\sqrt {3}\) )
Here X = 0, Y = ± \(\sqrt {3}\)
∴ x – 1 = 0, y + 1 = ± \(\sqrt {3}\)
⇒ x = 1, y = – 1 ± \(\sqrt {3}\)
foci = (1 ± \(\sqrt {3}\) – 1)
Equation of directrix Y = ± \(\frac { b }{ e}\)
⇒ Y = ± \(\frac{2}{\sqrt{3}}\).2
⇒ Y = ± \(\frac{4}{\sqrt{3}}\)
Y + 1 = \(\frac{4}{\sqrt{3}}\), (∵ Y = y+1)
⇒ y = ± \(\frac{4}{\sqrt{3}}\) – 1
Length of latus rectum = \(\frac{2 a^{2}}{b}\)
= 2. \(\frac { 1 }{ 2 }\) = 1.

Question 22.
Find the vertices, co-ordinate of foci, eccentricity and length of latus rectum of hyperbola :
(A) 9y² – 4x² = 36.
Solution:
Given : 9y² – 4x² = 36
⇒ \(\frac{9 y^{2}}{36}-\frac{4 x^{2}}{36}\) = 1
⇒ \(\frac{y^{2}}{4}-\frac{x^{2}}{9}\) = 1
Comparing the above equation with the standard form of hyperbola
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1 …. (1)
b² = 4 ⇒ b = 2, a² = 9 ⇒ a = 3
Let e is the eccentricity of hyperbola.
Then, a² = b²(e² – 1)
⇒ 9 = (e² – 1)
⇒ \(\frac { 9 }{ 4 }\) = e² – 1
⇒ e² = \(\frac { 9 }{ 4 }\) + 1 = \(\frac { 13 }{ 4 }\)
∴ Eccentricity e = \(\frac{\sqrt{3}}{2}\)
Vertices = (0, ± b) = (0, ± 2)
Foci = (0, ± be) = (0, ± 2 x \(\frac{\sqrt{3}}{2}\)) = (0, ±\(\sqrt {3}\))
Length of latus rectum = \(\frac{2 a^{2}}{b}\) = \(\frac { 2 x 9 }{ 2 }\) = 9

(B) 16x² – 9y² = 576.
Solution:
Given : 16x² – 9y² = 576
⇒ \(\frac{16 x^{2}}{576}-\frac{9 y^{2}}{576}\) = 1
⇒ \(\frac{x^{2}}{36}-\frac{y^{2}}{64}\) = 1
Comparing the above equation with the standard form of hyperbola
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 …. (1)
Here, a² = 36 ⇒ a = 6, b² = 64 ⇒ b = 3
We Know that, b² = a²(e² – 1)
64 = 36(e² – 1)
⇒ \(\frac { 64 }{ 36 }\) = e² – 1
⇒ \(\frac { 16 }{ 9 }\) = e² – 1
⇒ e² = \(\frac { 16 }{ 9 }\) + 1 = \(\frac { 25 }{ 9 }\)
∴ Eccentricity e = \(\frac{5}{3}\)
Vertices = (± a, 0) = (± 6, 0)
Foci = (± ae, 0) = (± 6 x \(\frac { 5 }{ 3 }\)) = (± 10, 0)
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac { 2 x 64 }{ 6 }\) = \(\frac { 64 }{ 3 }\).

(c) 5y² – 9x² = 36.
Solution:
Given : 5y² – 9x² = 36
⇒ \(\frac{5 y^{2}}{36}-\frac{9 x^{2}}{36}\) = 1
\(\frac{y^{2}}{\frac{36}{5}}-\frac{x^{2}}{4}\) = 1
Comparing the above equation with the standard form of hyperbola
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1 …. (1)
Here b² = \(\frac { 36 }{ 5 }\) ⇒ b = \(\frac{\sqrt{6}}{5}\), a² = 4 ⇒ a = 2
We know that, a² = b²(e² – 1)
⇒ 4 = \(\frac{\sqrt{36}}{5}\)(e² – 1)
⇒ e² – 1 = \(\frac { 20 }{ 36 }\) = \(\frac { 5 }{ 9 }\)
⇒ e² = 1 + \(\frac { 5 }{ 9 }\) = \(\frac { 14 }{ 9 }\)

Question 23.
Find the equation of hyperbola whose foci are (0, ± \(\sqrt {10}\)) and which passes through point (2,3).
Solution:
Foci of hyperbola are (0, ±\(\sqrt {10}\)).
∴Form of hyperbola is :
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1 …. (1)
Foci (0, ± be) = (0, ± \(\sqrt {10}\))
be = \(\sqrt {10}\)
Equation (1) passes through point (2, 3).
∴ \(\frac{9}{b^{2}}-\frac{4}{a^{2}}\)
⇒ 9a² – 4b² = a²b²
We know that, a² = b² (e² – 1)
⇒ a² = b²e² – b²
a² = ( \(\sqrt {10}\))² – b²
⇒ a² = 10 – b²
⇒ b² = 10 – a²
Putting value of b² in equation (2),
9a² – 4(10 – a²) = a² (10 – a²)
⇒ 9a² – 40 + 4a² = 10a² – a⁴
⇒ 13a² – 40 = 10a² – a⁴
⇒ a⁴ + 13a² – 10a² – 40 = 0
⇒ a⁴ + 3a² – 40 = 0
⇒ a⁴ + 8a² – 5a² – 40 = 0
⇒ a²(a² + 8) – 5(a² + 8) = 0
⇒ (a² – 5)(a² + 8) = 0
a² = 5, a² = – 8
∵ The value of a cannot be negative.
∴ a² = 5
b² = 10 – a²
⇒ b²  = 10 – 5
⇒  b² = 5
Putting values of a² and b² in equation (1), then required equation of hyperbola will be :
\(\frac{y^{2}}{5}-\frac{x^{2}}{5}\) = 1
⇒ y² – x² = 5.

Question 24.
Find the equation of hyperbola in which the distance between foci is 8 and distance between directrix is 6.
Solution:
Let the equation of hyperbola is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) … (1)
and Eccentricity of hyperbola is e and foci are (ae, 0) and (- ae, 0) and latus rectum
are x = \(\frac { a }{ e }\) and x = – \(\frac { a }{ e }\)
Distance between foci = 2ae
Distance between latus rectum = \(\frac { 2a }{ e }\)
According to question, 2ae = 8
and \(\frac { 2a }{ e }\) = 6
Multiplying equation (2) and (3),
4a² = 48 ⇒ a² = 12
⇒ a = 2\(\sqrt {3}\)
Putting value of a in equation (2),
2.2\(\sqrt {3}\)e = 8 ⇒ e = \(\frac{2}{\sqrt{3}}\)
b² = a²(e² – 1)
= 12( \(\frac { 4 }{ 3 }\) – 1) = 4
Putting values of a² and b² in equation (1), the required equation of hyperbola is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\)
⇒ x² – 3y² = 12.

Question 25.
Find the centre, eccentricity, foci and length of latus rectum of hyperbola 9x² – 16y² + 18x + 32y – 151 = 0.
Solution:
Equation of hyperbola is :
9x² – 16y² + 18x + 32y – 151 = 0
⇒ 9x² + 18x – 16y² + 32y = 151
⇒ 9(x² + 2x) – 16(y² – 2y) = 151
⇒ 9(x + 1)² – 16(y – 1)² = 151 – 16 + 9
⇒ 9(x + 1)² – 16(y – 1)² = 144
⇒ \(\frac{9(x+1)^{2}}{144}-\frac{16(y-1)^{2}}{144}\)
⇒ \(\frac{(x+1)^{2}}{16}-\frac{(y-1)^{2}}{9}\) = 1
Let x + 1 = X and y – 1 = Y, then equation of hyperbola is
\(\frac{X^{2}}{16} – \frac{Y^{2}}{9}\) = 1
∴ a² = 16 ⇒ a = 4 and b² = 9 ⇒ b = 3.
∴ Centre is (- 1, 1).
For eccentricity = e
b² = a²(e² – 1)
⇒ 9 = 16(e² – 1)
⇒ \(\frac { 9 }{ 16 }\) = e² – 1
⇒ e² = 1 + \(\frac { 9 }{ 16 }\) = \(\frac { 25 }{ 16 }\)
⇒ e = \(\frac { 5 }{ 4 }\)
For foci, X = ± ae, Y = 0
⇒ x +1 = ± 4 x \(\frac { 5 }{ 4 }\), y – 1 = 0
⇒ x + 1 = ± 5, y = 1
⇒ x = 4, – 6, y = 1
∴ Foci are (4, 1) and (6, 1).
Equation of directrix is X = ± \(\frac { a }{ e }\)
⇒ x +1 = ± \(\frac { 4 }{ 5/4 }\)
⇒ x = ± \(\frac { 16 }{ 5 }\) – 1
⇒ x = ± \(\frac { 11 }{ 5 }\) and x = – \(\frac { 21 }{ 5 }\)
⇒ 5x = 11 and 5x + 21 = 0.

Question 26.
If e and ex are the eccentricity of hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) and \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1, then prove that: \(\frac{1}{e^{2}}+\frac{1}{e_{1}^{2}}\) = 1.
Solution:
Equation of hyperbola is
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
and
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1
For eccentricity e of equation (1),
b² = a²(e² – 1)
⇒ \(\frac{b^{2}}{a^{2}}\) = (e² – 1)
⇒ e ² = 1 + \(\frac{b^{2}}{a^{2}}\) = \(\frac{a^{2}+b^{2}}{b^{2}}\)
⇒ \(\frac{1}{e_{1}^{2}}\) = \(\frac{a^{2}}{a^{2}+b^{2}}\)
Again for eccentricity e of equation (2),

Question 27.
On a level plain the crack of the rifle and the thud of the ball striking the target are heard at the same instant, prove that the locus of the hearer is a hyperbola.
Solution:
Let P be the situation of hearer and T be the situation of the rifle and S is target. Let the velocity of the ball be v₁ and the velocity of sound be v₂.

Then,
Time to reach the ball from T to S = \(\frac{T S}{v_{1}}\)
Time to reach the sound from S to P = \(\frac{S P}{v_{2}}\)
and Time to reach the sound from T to P = \(\frac{T P}{v_{2}}\)
∴ The crack of the rifle and the thud of the ball are heard at the same instant.
∴ \(\frac{T S}{v_{1}}\) + \(\frac{S P}{v_{2}}\) = \(\frac{T P}{v_{2}}\)
⇒ \(\frac{T P}{v_{2}}\) – \(\frac{S P}{v_{2}}\) = \(\frac{T S}{v_{1}}\)
⇒ TP – SP = \(\frac{v_{2}}{v_{1}}\)
⇒ PT – PS = A constant (∵ v₂, v₂, TS are constant)
Hence locus of point P is hyperbola whose foci is T and S.

MP Board Class 11th Maths Important Questions

MP Board Class 12th Maths Important Questions Chapter 10 सदिश बीजगणित

सदिश बीजगणित Important Questions

सदिश बीजगणित वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प चुनकर लिखिए –

प्रश्न 1.
सदिशों 2\(\bar { i } \) + 4\(\bar { j } \) – 5\(\bar { k } \) तथा \(\bar { i } \) + 2\(\bar { j } \) + 3\(\bar { k } \) के परिणामी सदिश के समान्तर एकांक सदिश है –

उत्तर:

प्रश्न 2.
यदि \(\bar { OA } \) = a, \(\bar { OB } \) = -b तथा C, AB पर एक ऐसा बिन्दु है कि \(\bar { AC } \) = 3AB, तो सदिश \(\bar { OC } \) बराबर है –
(a) 3\(\bar { a } \) – 2\(\bar { b } \)
(b) 3\(\bar { b } \) – 2\(\bar { a } \)
(c) 3\(\bar { a } \) – \(\bar { b } \)
(d) 3\(\bar { b } \) – \(\bar { a } \).
उत्तर:
(b) 3\(\bar { b } \) – 2\(\bar { a } \)

प्रश्न 3.
दो सदिश \(\bar { a } \) और \(\bar { b } \) इस प्रकार है कि |\(\bar { a } \)| = 2, |b| = 1 तथा \(\bar { a } \).\(\bar { b } \) = \(\sqrt{3}\) हो तो उनके बीच का कोण होगा –
(a) \(\frac { \pi }{ 2 } \)
(b) \(\frac { \pi }{ 4 } \)
(c) \(\frac { \pi }{ 6 } \)
(d) \(\frac { \pi }{ 7 } \)
उत्तर:
(c) \(\frac { \pi }{ 6 } \)

प्रश्न 4.
समान्तर चतुर्भुज का क्षेत्रफल जिसकी संलग्न भुजाएँ \(\bar { i } \) – 2\(\bar { j } \) + 3\(\bar { k } \) तथा 2\(\bar { i } \) + \(\bar { j } \) – 4\(\bar { k } \) है, हैं –
(a) 3\(\sqrt{6}\)
(b) 4\(\sqrt{6}\)
(c) 5\(\sqrt{6}\)
(d) 6\(\sqrt{6}\)
उत्तर:
(c) 5\(\sqrt{6}\)

प्रश्न 5.
यदि \(\bar { a } \) = \(\bar { b } \) + \(\bar { c } \) तो \(\bar { a } \).( \(\bar { b } \) × \(\bar { c } \) ) बराबर है –
(a) 2\(\bar { a } \).( \(\bar { b } \) + \(\bar { c } \) )
(b) 0
(c) \(\bar { b } \).( \(\bar { a } \) + \(\bar { c } \) )
(d) इनमें से कोई नहीं।
उत्तर:
(b) 0

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिये –

1. दो सदिशों का योग या अन्तर सदैव एक ……………………… होता है।
2. सदिशों का योग ……………………….. का पालन करता है।
3. ( \(\vec { a } \) + \(\vec { b } \) ) + \(\vec { c } \) = \(\vec { a } \) + …………………………..
4. दो सदिशों का योग ………………………. से प्राप्त किया जा सकता है।
5. बिन्दु (1, 2, 3) का मूलबिन्दु के सापेक्ष स्थिति सदिश ……………………………. होगा।
6. यदि \(\vec { A } \)C = 3 \(\vec { A } \)B हो, तो बिन्दु A, B, C …………….. होंगे।
7. यदि \(\vec { a } \) और \(\vec { b } \)समान्तर हों, तो \(\vec { a } \) × \(\vec { b } \) = ……………………. होगा।
8. सदिश \(\vec { a } \) की दिशा में एकांक सदिश ……………………………. होगा।
9. सदिश \(\vec { b } \) का \(\vec { a } \) की दिशा में प्रक्षेप …………………………. होगा।
10. यदि सदिश 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) + 2\(\hat { j } \) – 3\(\hat { k } \) तथा 3\(\hat { i } \) + p\(\hat { j } \) + 5\(\hat { k } \) समतलीय हों, तो p का मान …………………………….. होगा।
11. एक बल 2\(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \), एक बिन्दु A जिसका स्थिति सदिश 2\(\hat { i } \) – \(\hat { j } \) है, पर कार्य करता है। बल का मूलबिन्दु के सापेक्ष आघूर्ण ……………………………… होगा।
12. एक समान्तर चतुर्भुज का क्षेत्रफल …………………. होगा, जिसके विकर्ण 3\(\hat { i } \) + \(\hat { j } \) – 2\(\hat { k } \) तथा \(\hat { i } \)
    – 3\(\hat { j } \) + 4\(\hat { k } \) हैं।

उत्तर:

1. नया सदिश
2. क्रम विनिमेय नियम और साहचर्य नियम
3. ( \(\vec { b } \) + \(\vec { c } \) )
4. सदिश योग के त्रिभुज – नियम
5. \(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \)
6. समरेख
7. \(\vec { O } \)
8. \(\frac { \vec { a } }{ |\vec { a } | } \)
9. 
10. -4
11. \(\hat { i } \) + 2\(\hat { j } \) + 4\(\hat { k } \)
12. 5\(\sqrt{3}\) वर्ग इकाई।

प्रश्न 3.
निम्न कथनों में सत्य/असत्य बताइए –

1. एक त्रिभुज की भुजाओं द्वारा क्रमानुसार निरूपित सदिशों का योग शून्य होता है।
2. यदि \(\vec { a } \) व \(\vec { b } \) दो असंरेख सदिश हैं तो |\(\vec { a } \) + \(\vec { b } \)| ≥ |\(\vec { a } \) + \(\vec { b } \)
3. एक सदिश जिसके आदि और अंतिम बिंदु संपातो होते हैं एकांक सदिश कहलाता है।
4. यदि बिन्दुओं P और Q के स्थिति सदिश क्रमशः \(\hat { i } \) + 3\(\hat { j } \) – 7\(\hat { k } \) और 5\(\hat { i } \) – 2\(\hat { j } \) + 4\(\hat { k } \) हों, तो |\(\vec { P } \)Q| का मान 9\(\sqrt{2}\) होगा!
5. यदि |\(\vec { a } \) + \(\vec { b } \)| = |\(\vec { a } \) – \(\vec { b } \)|, तो \(\vec { a } \) + \(\vec { b } \) = 0
6. \(\vec { a } \). ( \(\vec { a } \) × \(\vec { b } \) ) का मान शून्य होता है।
7. सदिश \(\hat { i } \) – λ\(\hat { j } \) + \(\hat { k } \) और \(\hat { i } \) – \(\hat { j } \) + 5\(\hat { k } \) परस्पर लम्ब हों तो λ का मान 6 होगा।

उत्तर:

1. सत्य
2. असत्य
3. असत्य
4. सत्य
5. असत्य
6. सत्य
7. असत्य।

प्रश्न 4.
सही जोड़ी बनाइए –

उत्तर:
(a) (iv)
(b) (v)
(c) (i)
(d) (ii)
(e) (iii)
(f) (vii)
(g) (vi).

प्रश्न 5.
एक शब्द/वाक्य में उत्तर दीजिए –

1. यदि \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) किसी ∆ABC के शीर्षों के स्थिति सदिश हों, तो ∆ABC के क्षेत्रफल का सूत्र लिखिए।
2. यदि \(\vec { a } \) = \(\hat { i } \) – 2\(\hat { j } \) + 3\(\hat { k } \), \(\vec { b } \) = 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \) तथा \(\vec { c } \) = \(\hat { j } \) + \(\hat { k } \) तथा [ \(\vec { a } \) \(\vec { b } \) \(\vec { c } \) ] का मान ज्ञात कीजिए।
3. दो सदिशों 3\(\hat { i } \) – 2\(\hat { j } \) + 4\(\hat { k } \) और \(\hat { i } \) – \(\hat { j } \) + 5\(\hat { k } \) के बीच का कोण ज्ञात कीजिए।
4. \(\hat { i } \) × ( \(\hat { j } \) + \(\hat { k } \) ) + \(\hat { j } \) × ( \(\hat { k } \) + \(\hat { i } \) + \(\hat { k } \) × ( \(\hat { i } \) + \(\hat { j } \) ) का मान ज्ञात कीजिए।
5. \(\vec { a } \) का \(\vec { b } \) की दिशा में प्रक्षेप ज्ञात कीजिए।
6. यदि \(\vec { a } \) और \(\vec { b } \) परस्पर लम्ब सदिश हों तो ( \(\vec { a } \) + \(\vec { b } \) )² का मान ज्ञात कीजिए।

उत्तर:

1. \(\frac{1}{2}\) |\(\vec { a } \) × \(\vec { b } \) + \(\vec { b } \) × \(\vec { c } \) + \(\vec { c } \) × \(\vec { a } \)|
2. 12
3. cos^-1 \(\frac { 25 }{ \sqrt { 783 } } \)
4. 0
5. 
6. |\(\vec { a } \)|² + |\(\vec { b } \)|²

सदिश बीजगणित अति लघु उत्तरीय प्रश्न

प्रश्न 1.
सदिश \(\vec { a } \) = \(\hat { i } \) – 2\(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = -2\(\hat { i } \) + 4\(\hat { j } \) + 5\(\hat { k } \) और \(\vec { c } \) = \(\hat { i } \) – 6\(\hat { j } \) – 7\(\hat { k } \) तो |\(\vec { a } \) + \(\vec { b } \) + \(\vec { c } \)| का मान ज्ञात कीजिए। (NCERT, CBSE 2012)
हल:
\(\vec { a } \) + \(\vec { b } \) + \(\vec { c } \) = \(\hat { i } \) – 2\(\hat { j } \) + \(\hat { k } \) – 2\(\hat { i } \) + 4\(\hat { j } \) + 5\(\hat { k } \) + \(\hat { i } \) – 6\(\hat { j } \) – 7\(\hat { k } \)
⇒ \(\vec { a } \) + \(\vec { b } \) + \(\vec { c } \) = -4\(\hat { j } \) – \(\hat { k } \)
⇒ |\(\vec { a } \) + \(\vec { b } \) + \(\vec { c } \)| = \(\sqrt { 0+16+1 } \)
= \(\sqrt{17}\)

प्रश्न 2.
सदिशों \(\vec { a } \) = 2\(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) और \(\vec { b } \) = – \(\hat { i } \) + \(\hat { j } \) + 3\(\hat { k } \) के योगफल के अनुदिश मात्रक सदिश ज्ञात कीजिये।
हल:
\(\vec { a } \) = 2\(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \), \(\vec { b } \) = –\(\hat { i } \) + \(\hat { j } \) + 3\(\hat { k } \)
⇒ \(\vec { a } \) + \(\vec { b } \) = (2 – 1)\(\hat { i } \) + (-1 + 1)\(\hat { j } \) + (2 + 3)\(\hat { k } \)
= \(\hat { i } \) + 0\(\hat { j } \) + 5\(\hat { k } \)
माना \(\vec { a } \) + \(\vec { b } \) = \(\vec { r } \)

प्रश्न 3.
सदिश \(\vec { a } \) = \(\hat { i } \) – 2\(\hat { j } \) के अनुदिश एक ऐसा सदिश ज्ञात कीजिए, जिसका परिमाण 7 इकाई है। (NCERT)
हल:
दिया है:
\(\vec { a } \) = \(\hat { i } \) – 2\(\hat { j } \)
\(\vec { a } \) के अनुदिश मात्रक सदिश
\(\hat { a } \) = \(\frac { \vec { a } }{ |\vec { a } | } \)
|\(\hat { a } \)| = |\(\hat { i } \) – 2\(\hat { j } \)|
⇒ |\(\hat { a } \)| = \(\sqrt { (1)^{ 2 }+(-2)^{ 2 } } \) = \(\sqrt{5}\)

\(\hat { a } \) के अनुदिश 7 परिमाण वाला सदिश

प्रश्न 4.
दर्शाइये कि सदिश 2\(\hat { i } \) – 3\(\hat { j } \) + 4\(\hat { k } \) और -4\(\hat { i } \) + 6\(\hat { j } \) – 8\(\hat { k } \) संरेख हैं।
हल:
\(\vec { a } \) = 2\(\hat { i } \) – 3\(\hat { j } \) + 4\(\hat { k } \), \(\hat { b } \) = -4\(\hat { i } \) + 6\(\hat { j } \) – 8\(\hat { k } \)
अब \(\vec { b } \) = -2(2\(\hat { i } \) – 3\(\hat { j } \) + 4\(\hat { k } \) )
⇒ \(\vec { b } \) = -2\(\vec { a } \)
∵ \(\vec { b } \) = λ\(\vec { a } \) (λ = स्थिरांक)
सदिश \(\vec { a } \), \(\vec { b } \) संरेख हैं।

प्रश्न 5.
सदिश \(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) की दिक् कोज्यायें ज्ञात कीजिये। (NCERT)
हल:
माना
\(\vec { r } \) = \(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \)
|\(\vec { r } \)| = \(\sqrt { 1+4+9 } \) = \(\sqrt { 14 } \)

= \(\frac { 1 }{ \sqrt { 14 } } \) \(\hat { i } \) + \(\frac { 2 }{ \sqrt { 14 } } \) \(\hat { j } \) + \(\frac { 3 }{ \sqrt { 14 } } \) \(\hat { k } \)
दिक् कोज्यायें = \(\frac { 1 }{ \sqrt { 14 } } \), \(\frac { 2 }{ \sqrt { 14 } } \), \(\frac { 3 }{ \sqrt { 14 } } \)

प्रश्न 6.
यदि \(\vec { a } \) = 2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \) तथा \(\vec { b } \) = \(\hat { i } \) + \(\hat { j } \) – 2\(\hat { k } \) है, तो \(\vec { a } \) – \(\vec { b } \) का मान ज्ञात कीजिये।
हल:

प्रश्न 7.
यदि \(\vec { a } \) = \(\hat { i } \) + \(\hat { j } \) + 2\(\hat { k } \) और \(\vec { b } \) = 3\(\hat { i } \) + 2\(\hat { j } \) – \(\hat { k } \) हो, तो |2\(\vec { a } \) – \(\vec { b } \)| का मान ज्ञात कीजिये।
हल:

प्रश्न 8.
यदि \(\vec { a } \) = 2\(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \) और \(\vec { b } \) = \(\hat { i } \) – 4\(\hat { j } \) + λ\(\hat { k } \) हो, तो λ का मान ज्ञात कीजिये।
हल:
दिये गये सदिश परस्पर लम्ब होंगे यदि उनका अदिश गुणन शून्य है।
अर्थात् \(\vec { a } \).\(\vec { b } \) = 0

प्रश्न 9.
सिद्ध कीजिये कि सदिश 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) और –\(\hat { i } \) + 3\(\hat { j } \) + 5\(\hat { k } \) परस्पर लम्बवत् हैं।
हल:
माना
\(\vec { a } \) = 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = –\(\hat { i } \) + 3\(\hat { j } \) + 5\(\hat { k } \) ) = 0
L.H.S = – 2 – 3 + 5
= 0 = R.H.S यही सिद्ध करना था।

प्रश्न 10.
(A) सिद्ध कीजिये कि सदिश 3\(\hat { i } \) – 2\(\hat { j } \) + \(\hat { k } \) और 2\(\hat { i } \) + \(\hat { j } \) – 4\(\hat { k } \) परस्पर लम्बवत् हैं।
हल:
प्रश्न क्र. 9 की भाँति हल करें।

(B)
यदि \(\vec { a } \) = 4\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) और \(\vec { b } \) = P\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) परस्पर लम्बवत् हों, तो p का मान ज्ञात कीजिये।
हल:
\(\vec { a } \) = 4\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = P\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \)
\(\vec { a } \) तथा \(\vec { b } \) परस्पर लम्बवत् होंगे, यदि \(\vec { a } \).\(\vec { b } \) = 0
4p – 2 + 3 = 0
⇒ 4p = -1
⇒ p = – \(\frac{1}{4}\)

प्रश्न 11.
(A) सदिशों (2\(\hat { i } \) + 3\(\hat { j } \) – 4\(\hat { k } \) ) और (3\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) ) के बीच का कोण ज्ञात कीजिये।
हल:
माना \(\vec { a } \) = 2\(\hat { i } \) + 3\(\hat { j } \) – 4\(\hat { k } \), \(\vec { b } \) = 3\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \)
यदि इनके बीच का कोण θ हो, तो

∴ θ = 90°.

(B) सदिशों \(\vec { a } \) = 2\(\hat { i } \) – 2\(\hat { j } \) – \(\hat { k } \) और \(\vec { b } \) = 6\(\hat { i } \) – 3\(\hat { j } \) + 2\(\hat { k } \) के बीच के कोण की कोज्या ज्ञात कीजिए।
हल:
प्रश्न क्र. 11 (A) की भाँति हल कीजिये।

(C) \(\vec { a } \) = 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) तथा \(\vec { b } \) = 3\(\hat { i } \) – 4\(\hat { j } \) – 4\(\hat { k } \) का अदिश गुणनफल तथा उनके बीच का कोण ज्ञात कीजिए।
हल:

यदि इनके बीच का कोण θ हो, तो

∴ θ = cos^-1 \(\sqrt { \frac { 6 }{ 41 } } \)

प्रश्न 12.
यदि |\(\bar { a } \)| = 10, |\(\bar { b } \)| = 2 तथा \(\bar { a } \).\(\bar { b } \) = 12 हो, तो |\(\bar { a } \) × \(\bar { b } \) का मान ज्ञात कीजिए।
हल:
दिया है:
|\(\bar { a } \)| = 10, |\(\bar { b } \)| = 2 तथा \(\bar { a } \).\(\bar { b } \) = 12
हम जानते हैं कि |\(\bar { a } \) × \(\bar { b } \)|² = |a|² |b|² – ( \(\bar { a } \).\(\bar { b } \) )²
= 100 × 4 – 144 = 400 – 144 = 256
∴ |\(\bar { a } \) × \(\bar { b } \)| = \(\sqrt{256}\) = 16

प्रश्न 13.
यदि \(\vec { a } \) = \(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \) तथा \(\vec { b } \) = \(\hat { i } \) – \(\hat { j } \) – \(\hat { k } \) हो, तो \(\vec { a } \) × \(\vec { b } \) ज्ञात कीजिए।
हल:
दिया है:
\(\vec { a } \) = \(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = \(\hat { i } \) – \(\hat { j } \) – \(\hat { k } \)

प्रश्न 14.
विस्थापन \(\vec { d } \) = – \(\hat { i } \) – 3 \(\hat { j } \) + 5 \(\hat { k } \) के अनुदिश कार्य करने वाले बल \(\vec { F } \) = 4\(\hat { i } \) – 3\(\hat { j } \) + 2\(\hat { k } \) द्वारा किया गया कार्य ज्ञात कीजिए।
हल:
दिया है:
\(\vec { d } \) = – \(\hat { i } \) – 3 \(\hat { j } \) + 5 \(\hat { k } \), \(\vec { F } \) = 4\(\hat { i } \) – 3\(\hat { j } \) + 2\(\hat { k } \) द्वारा किया गया कार्य ज्ञात कीजिये।
हल:
दिया है:
\(\vec { d } \) = – \(\hat { i } \) – 3 \(\hat { j } \) + 5 \(\hat { k } \), \(\vec { F } \) = 4\(\hat { i } \) – 3\(\hat { j } \) + 2\(\hat { k } \)
∴ बल द्वारा किया गया करध
W = \(\vec { F } \).\(\vec { d } \)
= (4\(\hat { i } \) – 3\(\hat { j } \) + 2\(\hat { k } \) ).(-\(\hat { i } \) – 3\(\hat { j } \) + 5\(\hat { k } \) )
= -4 + 9 + 10 = 15 इकाई।

प्रश्न 15.
यदि |\(\vec { a } \) + \(\vec { b } \)| = |\(\vec { a } \) – \(\vec { b } \)| हो, तो सिद्ध कीजिए कि \(\vec { a } \) और \(\vec { b } \) परस्पर लम्बवत् हैं।
हल:
|\(\vec { a } \) + \(\vec { b } \)| = |\(\vec { a } \) – \(\vec { b } \)|

इस प्रकार, अदिश गुणनफल शून्य है। अतः सदिश \(\vec { a } \) और \(\vec { b } \) परस्पर लम्बवत् होंगे। यही सिद्ध करना था।

प्रश्न 16.
दो सदिश \(\vec { a } \) और \(\vec { b } \) के इस प्रकार हैं कि |\(\vec { a } \)| = 2, |\(\vec { b } \)| = 3 तथा \(\vec { a } \).\(\vec { b } \) = 3, सदिशों \(\vec { a } \) के और \(\vec { b } \) के बीच का कोण ज्ञात कीजिये।
हल:
प्रश्न क्र. 17 की भाँति हल कीजिये।

प्रश्न 17.
यदि |\(\vec { a } \)| = 4, |\(\vec { b } \)| = 4 तथा \(\vec { a } \).\(\vec { b } \) = 6 हो, तो \(\vec { a } \) सदिशों है और \(\vec { b } \) के बीच का कोण ज्ञात कीजिये।
हल:
\(\vec { a } \).\(\vec { b } \) = |\(\vec { a } \)| |\(\vec { b } \)|
⇒ 6 = 4.4 cos θ
⇒ 6 = 16 cos θ
⇒ cos θ = \(\frac{6}{16}\) = \(\frac{3}{8}\)
∴ θ = cos^-1 \(\frac{3}{8}\)

प्रश्न 18.
दो सदिश \(\vec { a } \) और \(\vec { b } \) इस प्रकार हैं कि |\(\vec { a } \)| = 2, |\(\vec { b } \)| = 7 तथा \(\vec { a } \) × \(\vec { b } \) के बीच का कोण ज्ञात कीजिये।।
हल:
माना \(\vec { a } \) और \(\vec { b } \) के बीच का कोण θ है।

प्रश्न 19.
सदिश 2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \) और \(\hat { i } \) + \(\hat { j } \) – 2\(\hat { k } \) के बीच के कोण की कोज्या ज्ञात कीजिए।
हल:
माना \(\vec { a } \) = 2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = \(\hat { i } \) + \(\hat { j } \) – 2\(\hat { k } \)

प्रश्न 20.
उस समान्तर षटफलक का आयतन ज्ञात कीजिये जिसकी कोरेंनिम्न सदिशों से निरूपित हैं –
2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \), 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \)?
हल:
माना \(\vec { a } \) = 2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = \(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) तथा \(\vec { c } \) = 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \)
समान्तर षट्फलक का आयतन = [a b c]

= 2(1 – 2) + 3(-1 -4) + 1(1 + 2)
= – 2 – 15 + 3
= -14 घन इकाई।

प्रश्न 21.
सिद्ध कीजिये –
\(\hat { i } \).( \(\hat { j } \) × \(\hat { k } \) ) + ( \(\hat { i } \) × \(\hat { k } \) ).\(\hat { j } \) = 0।
हल:
\(\hat { i } \).( \(\hat { j } \) × \(\hat { k } \) ). \(\hat { j } \) = \(\hat { i } \).\(\hat { i } \) + (-\(\hat { j } \) ).\(\hat { j } \)
= 1 – 1
= 0. यही सिद्ध करना था।

प्रश्न 22.
यदि सदिश \(\vec { a } \) और \(\vec { b } \) परस्पर लम्बवत् हैं तो सिद्ध कीजिए कि |\(\vec { a } \) + \(\vec { b } \)|² = |\(\vec { a } \)|² + |\(\vec { b } \)|²
हल:
हम जानते हैं कि
|\(\vec { a } \) + \(\vec { b } \)|² = |\(\vec { a } \)|² + |\(\vec { b } \)|² + 2|\(\vec { a } \) |.|\(\vec { b } \)|
सदिश \(\vec { a } \) और \(\vec { b } \) परस्पर लम्बवत् हैं, तब
\(\vec { a } \) .\(\vec { b } \) = 0
⇒ |\(\vec { a } \) + \(\vec { b } \)|² = |\(\vec { a } \)|² + |\(\vec { b } \)|² यही सिद्ध करना था।

प्रश्न 23. सिद्ध कीजिए कि
\(\vec { a } \) × ( \(\vec { b } \) + \(\vec { c } \) ) + \(\vec { b } \) × ( \(\vec { c } \) + \(\vec { a } \) ) + \(\vec { c } \) × ( \(\vec { a } \) + \(\vec { b } \) ) = 0
हल:

प्रश्न 24.
विस्थापन \(\vec { d } \) = 3\(\hat { i } \) + 2\(\hat { j } \) + 5\(\hat { k } \) के अनुदिश कार्य करने वाले बल \(\vec { F } \) = 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) द्वारा किया गया कार्य ज्ञात कीजिये।
हल:
कार्य = \(\vec { F } \).\(\vec { d } \)
= (2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) ).(3\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) )
= 6 – 2 + 3 = 7 इकाई।

प्रश्न 25.
दो सदिशों \(\vec { d } \) तथा \(\vec { b } \) जिनके परिमाण समान हैं में से प्रत्येक का परिणाम ज्ञात कीजिये जबकि उनके बीच का कोण 60° है तथा उनका अदिश गुणनफल \(\frac{9}{2}\) हैं।
हल:
अदिश गुणन की परिभाषा से,

प्रश्न 26.
समांतर चतुर्भुज का क्षेत्रफल ज्ञात कीजिये जिसकी दो आसन्न भुजायें सदिशों \(\vec { a } \) = 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) और
\(\vec { b } \) = 3\(\hat { i } \) + 4\(\hat { j } \) – \(\hat { k } \)
समांतर चतुर्भुज का क्षेत्रफल = \(\frac{1}{2}\) |\(\vec { a } \) × \(\vec { b } \)|

अभीष्ट क्षेत्रफल = \(\frac{1}{2}\) \(\sqrt{155}\).

प्रश्न 27.
यदि \(\vec { a } \) = 4\(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \) और \(\vec { b } \) = \(\hat { i } \) – 2\(\hat { k } \) हो, तो |2\(\vec { b } \) × \(\vec { a } \)| का मान ज्ञात कीजिये।
हल:

प्रश्न 28.
यदि \(\vec { a } \) = 4\(\hat { i } \) + 3\(\hat { j } \) + 2\(\hat { k } \) और \(\vec { b } \) = 3\(\hat { i } \) + 2\(\hat { k } \) हो, तो |\(\vec { b } \) × 2\(\vec { a } \)| का मान ज्ञात कीजिए।
हल:

प्रश्न 29.
यदि \(\vec { a } \) = 2\(\hat { i } \) + \(\hat { j } \) + 2\(\hat { k } \) और \(\vec { b } \) = 5\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \) तो \(\vec { b } \) का \(\vec { a } \) पर अदिश प्रक्षेप ज्ञात कीजिये।
हल:
सदिश \(\vec { b } \) का \(\vec { a } \) की दिशा में अदिश प्रक्षेप

= \(\frac{10-3+2}{3}\)
= \(\frac{9}{3}\) = 3.

प्रश्न 30.
यदि \(\vec { a } \) = \(\hat { i } \) + 3\(\hat { j } \) – 2\(\hat { k } \), \(\vec { b } \) = – \(\hat { j } \) + 3\(\hat { k } \) हो, तो |\(\vec { a } \) × \(\vec { b } \)| का मान ज्ञात कीजिए।
हल:

सदिश बीजगणित दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
दर्शाइये कि बिन्दु A(-2\(\hat { i } \) + 3\(\hat { j } \) + 5\(\hat { k } \) ), B( \(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) ) और C(7\(\hat { i } \) + 0\(\hat { j } \) – \(\hat { k } \) ) संरेख हैं। (NCERT)
हल:
माना O मूलबिन्दु है, तब बिन्दुओं A, B और C के स्थिति सदिश हैं –

अतः सदिश \(\overrightarrow{A B}\) और \(\overrightarrow{B C}\) समांतर हैं परन्तु B, \(\overrightarrow{A B}\), और \(\overrightarrow{B C}\) का उभयनिष्ठ बिन्दु हैं। अतएव दिए हुए बिन्दु A, B और C संरेख हैं। यही सिद्ध करना था।

प्रश्न 2.
यदि बिन्दुओं A, B, C और D के स्थिति सदिश क्रमश: 2\(\hat { i } \) + 4\(\hat { k } \), 5\(\hat { i } \) + 3\(\sqrt{3}\)\(\hat { j } \) + 4\(\hat { k } \), -2\(\sqrt{3}\)\(\hat { j } \) + \(\hat { k } \) और 2\(\hat { i } \) + \(\hat { k } \) हैं, तो सिद्ध कीजिए –
CD||AB और CD = \(\frac{2}{3}\) \(\vec { A } \)B
हल:
माना O मूलबिन्दु है, तब

⇒ \(\overrightarrow{C D}\), \(\overrightarrow{A B}\) के समांतर है और \(\overrightarrow{C D}\) = \(\frac{2}{3}\) × \(\overrightarrow{A B}\) यही सिद्ध करना था।

प्रश्न 3.
यदि त्रिभुज ABC का केन्द्रक G हो, तो सिद्ध कीजिए कि –
\(\overrightarrow{G A}\) + \(\overrightarrow{G B}\) + \(\overrightarrow{G C}\) = \(\vec { 0 } \).
हल:
माना त्रिभुज ABC के शीर्षों A, B और C के स्थिति सदिश क्रमशः \(\vec { a } \), \(\vec { b } \) और \(\vec { c } \) हैं।
∴ केन्द्रक G का स्थिति सदिश

अब \(\vec { G } \)A + \(\vec { G } \)B + \(\vec { G } \)C = (A का स्थिति सदिश – G का स्थिति सदिश) + (B का स्थिति सदिश – G का स्थिति सदिश) + (C का स्थिति सदिश – G का स्थिति सदिश)

अतः \(\vec { G } \)A + \(\vec { G } \)B + \(\vec { G } \)C = \(\vec { 0 } \). यही सिद्ध करना था।

प्रश्न 4.
सदिशों का प्रयोग करते हुए सिद्ध कीजिये कि त्रिभुज की माध्यिकाएँ संगामी होती हैं।
हल:
ABC त्रिभुज की माध्यिकाएँ AD, BE तथा CF हैं।
माना बिन्दुओं A, B और C के स्थिति सदिश क्रमशः \(\vec { a } \), \(\vec { b } \) और \(\vec { c } \) हैं।


अब माध्यिका AD को 2:1 के अनुपात में विभाजित करने (6) वाले बिन्दु का स्थिति सदिश

माध्यिका BE को 2:1 के अनुपात में विभाजित करने वाले बिन्दु का स्थिति सदिश

माध्यिका CF को 2:1 के अनुपात में विभाजित करने वाले बिन्दु का स्थिति सदिश

अतः त्रिभुज की माध्यिकाएँ एक बिन्दु G पर मिलती है अर्थात् संगामी हैं जिसका स्थिति सदिश \(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\) है। बिन्दु G को त्रिभुज का केन्द्रक कहते हैं। यही सिद्ध करना था।

प्रश्न 5.
एक सदिश \(\overrightarrow{O P}\), OX के साथ 45° और OY के साथ 60° कोण बनाती हैं। \(\overrightarrow{O P}\) के द्वारा OZ के साथ बनाया गया कोण ज्ञात कीजिए।
हल:
माना सदिश \(\overrightarrow{O P}\), अक्षों OX, OY और OZ के साथ क्रमशः α, β, γ कोण बनाती हैं, तब
α = 45°, β = 60°
∴ l = cos α = cos 45° = \(\frac { 1 }{ \sqrt { 2 } } \)
m = cos β = cos 60° = \(\frac{1}{2}\)
और n = cos γ
हम जानते हैं कि
l² + m² + n² = 1

cos γ = \(\frac{1}{2}\) था cos γ = – \(\frac{1}{2}\)
γ = 60° था 120°

प्रश्न 6.
परिमाण 5\(\sqrt{2}\), की एक सदिश \(\vec { a } \) ज्ञात कीजिए जो X – अक्ष के साथ \(\frac { \pi }{ 4 } \), Y – अक्ष के साथ \(\frac { \pi }{ 2 } \) और Z – अक्ष के साथ न्यूनकोण θ बनाती है।
हल:
दिया गया है –
α = \(\frac { \pi }{ 4 } \), β = \(\frac { \pi }{ 2 } \), γ = θ
I = cos \(\frac { \pi }{ 4 } \) = \(\frac { 1 }{ \sqrt { 2 } } \), m = cos \(\frac { \pi }{ 2 } \) = 0, n = cos θ
हम जानते हैं कि
l² + m² + n² = 1
⇒ ( \(\frac { 1 }{ \sqrt { 2 } } \) )² + 0 + n² = 1
⇒ \(\frac{1}{2}\) + n² = 1
∴ n² = \(\frac{1}{2}\)
cos² θ = \(\frac{1}{2}\)
cos θ = \(\frac { 1 }{ \sqrt { 2 } } \)
∴ θ = \(\frac { \pi }{ 4 } \)
सदिश की दिक् कोज्यायें \(\frac { 1 }{ \sqrt { 2 } } \), 0, \(\frac { 1 }{ \sqrt { 2 } } \)

प्रश्न 7.
सिद्ध कीजिए कि ( \(\vec { a } \) × \(\vec { b } \) )² = a²b² – ( \(\vec { a } \).\(\vec { b } \) )²।
हल:
L.H.S. = ( \(\vec { a } \) × \(\vec { b } \) )² = ( \(\vec { a } \) × \(\vec { b } \) )
= (absin θ\(\hat { n } \) ).(ab sinθ \(\hat { n } \) ) = a²b² sin² θ \(\hat { n } \).\(\hat { n } \)
= a²b² sin² θ
= a²b² – a²b² cos ² θ
= a²b² – (ab cos θ)²
= a²b² – ( \(\vec { a } \).\(\vec { b } \) )² = R.H.S यही सिद्ध करना था।

प्रश्न 8.
यदि \(\vec { a } \) = 2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = \(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) हो तो \(\vec { a } \) × ( \(\vec { b } \) × \(\vec { c } \) ) का मान ज्ञात कीजिये।
हल:
दिया है:
\(\vec { a } \) = 2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = \(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \), \(\vec { c } \) = 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \)

प्रश्न 9.
उस समान्तर षटफलक का आयतन ज्ञात कीजिये जिसकी तीन कोरें निम्न सदिशों से निरूपित हैं –
\(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \)
हल:
माना
\(\vec { a } \) = \(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \)

= 1(1 – 2) – 1(-1 -2) + 1(1 + 1)
= -1 + 3 + 2
= 4 इकाई।

प्रश्न 10.
यदि \(\vec { a } \) = 2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = \(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) और \(\vec { c } \) = 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \) हो, तो [\(\vec { a } \) \(\vec { b } \) \(\vec { c } \)] का मान ज्ञात कीजिए।
हल:
दिया है:
\(\vec { a } \) = 2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = \(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \), \(\vec { c } \) = 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \)

= 2(1 – 2) – 1(3 – 1) + 2(- 6 + 1)
= – 2 – 2 – 10 = – 14.

प्रश्न 11.
यदि \(\vec { a } \) = 3\(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \), \(\vec { b } \) = 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \) और \(\vec { c } \) = \(\hat { i } \) – 2\(\hat { j } \) + 2\(\hat { k } \) हो, तो \(\vec { a } \) का मान ज्ञात कीजिए।
हल:
प्रश्न क्र. 8 की भाँति हल कीजिये।

प्रश्न 12.
यदि a = \(\vec { a } \) = \(\hat { i } \) – 2\(\hat { j } \) + 3\(\hat { k } \), \(\vec { b } \) = -2\(\hat { i } \) + 3\(\hat { j } \) – 4\(\hat { k } \) तथा \(\vec { c } \) = \(\hat { i } \) – 3\(\hat { j } \) + 5\(\hat { k } \) हो, तो सिद्ध कीजिए कि सदिश \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) समतलीय हैं।
हल:
यदि \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) समतलीय हैं, तो उनके अदिश त्रिक गुणन शून्य होना चाहिए अर्थात्
[ \(\vec { a } \) \(\vec { b } \) \(\vec { c } \) ] = 0

= 1(15 – 12) + 2(-10 + 4) + 3(6 – 3)
= 1(3) + 2(-6) + 3(3) = 12 – 12 = 0

प्रश्न 13.
सिद्ध कीजिये 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) + 2\(\hat { j } \) – 3\(\hat { k } \) तथा 3\(\hat { i } \) – 4\(\hat { j } \) + 5\(\hat { k } \)
दिये गये सदिश समतलीय हैं तब [ \(\bar { a } \) \(\bar { b } \) \(\bar { c } \) ] = 0

L.H.S = 2(10 – 12) -1(-5 + 4) + 3(3 – 2)
= -4 + 1 + 3
= 0 यही सिद्ध करना था।
अतः \(\bar { a } \) \(\bar { b } \) \(\bar { c } \) समतलीय हैं।

प्रश्न 14.
(A) λ का मान ज्ञात कीजिये यदि सदिश λ\(\hat { i } \) + 3\(\hat { j } \) + 2\(\hat { k } \), 2\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) और 2\(\hat { i } \) + 3\(\hat { j } \) + 4\(\hat { k } \) समतलीय हैं।
हल:
माना \(\vec { a } \) = λ\(\hat { i } \) + 3\(\hat { j } \) + 2\(\hat { k } \), \(\vec { b } \) = 2\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \), \(\vec { c } \) = 2\(\hat { i } \) + 3\(\hat { j } \) + 4\(\hat { k } \)
दिये गये सदिश समतलीय होंगे यदि
[ \(\vec { a } \) \(\vec { b } \) \(\vec { c } \) ] = 0

⇒ λ(8 – 9) – 2 (12 – 6) + 2(9 – 4) = 0
⇒ -λ – 12 + 10 = 0
⇒ λ = -2

(B)
λ का मान ज्ञात कीजिये यदि निम्न सदिश समतलीय हैं –
\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \), λ\(\hat { i } \) – \(\hat { j } \) + λ\(\hat { k } \)।
हल:
प्रश्न क्र. 13 की भाँति हल कीजिये।
[उत्तर: λ = – \(\frac{18}{5}\) ]

(C)
λ का मान ज्ञात कीजिये जबकि सदिश 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) + 2\(\hat { j } \) – 3\(\hat { k } \) तथा 3\(\hat { i } \) + λ\(\hat { j } \) + 5\(\hat { k } \) समतलीय हैं।
हल:
प्रश्न क्र. 14 (A) की भाँति हल कीजिये।
[उत्तर: λ = – \(\frac{18}{5}\) ]

प्रश्न 15.
यदि दो इकाई सदिशों \(\bar { a } \) तथा \(\bar { b } \) के बीच का कोण हो, तो सिद्ध कीजिए कि
cos \(\frac { \theta }{ 2 } \) = \(\frac{1}{2}\) |\(\bar { a } \) + \(\bar { b } \)|.
हल:

प्रश्न 16.
दो इकाई सदिश \(\vec { a } \) तथा \(\vec { b } \) के बीच का कोण θ हो, तो सिद्ध कीजिए कि
sin \(\frac { \theta }{ 2 } \) = \(\frac{1}{2}\) |\(\vec { a } \) – \(\vec { b } \)|.
हल:

प्रश्न 17.
किसी त्रिभुज ABC में सिद्ध कीजिए कि
(A) ac cos B – bc cos A = a² – b²
(B) 2(bc cos A + ca cos B + ab cos C) = a² + b² + c².
हल:
माना की \(\overrightarrow{B C}\) = \(\vec{a}\), \(\overrightarrow{C A}\) = \(\vec{b}\), \(\overrightarrow{A B}\) = \(\vec{c}\) तथा \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) के मापांक क्रमश: a, b, c हैं।

(A) ac cos B – bc cos A = -ac cos (π – B) + bc cos (π – A)

यही सिद्ध करना था।

(B) 2 (bc cos A + ca cos B + ab cos C)

यही सिद्ध करना था।

प्रश्न 18.
∆ABC में सदिश विधि से सिद्ध कीजिए कि c = a cos B + bcos A.
हल:
∆ ABC में त्रिभुज के योग नियम से,
\(\vec { a } \) + \(\vec { b } \) + \(\vec { c } \) = \(\vec { 0 } \)

⇒ c² = ac cos B + bc cos A
⇒ c² = c(a cos B + b cos A)
⇒ c = a cos B + b cos A

प्रश्न 19.
∆ABC में सदिश विधि से सिद्ध कीजिए कि
b² = a² + c² – 2ac cos B.
हल:
हम जानते हैं कि ∆ABC में,


⇒ b² = a² + c² + 2ac cos (π – B)
∴ b² = a² + c² – 2ac cos B. यही सिद्ध करना था।

प्रश्न 20.
∆ABC में सदिश विधि से सिद्ध कीजिए –
(A) a² = b² + c² – 2bc cos A
(B) c² = a² + b² – 2ab cos C.
हल:
प्रश्न क्र. 19 की भाँति स्वयं हल कीजिये।

प्रश्न 21.
(A) सदिशों \(\vec { a } \) = 2\(\hat { i } \) + 2\(\hat { j } \) + \(\hat { k } \) और \(\vec { b } \) = 4\(\hat { i } \) + 4\(\hat { j } \) – 7\(\hat { k } \) में से प्रत्येक पर लम्ब मात्रक सदिश ज्ञात कीजिए।
हल:
दिया है:
\(\vec { a } \) = 2\(\hat { i } \) + 2\(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = 4\(\hat { i } \) + 4\(\hat { j } \) – 7\(\hat { k } \)

(B)
सदिशों \(\vec { a } \) = \(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) और \(\vec { b } \) = \(\hat { i } \) + 2\(\hat { j } \) – \(\hat { k } \) में से प्रत्येक पर लम्ब मात्रक सदिश ज्ञात कीजिए।
हल:
प्रश्न क्र. 21 (A) की भाँति हल कीजिये।

(C)
सदिशों \(\vec { a } \) = 3\(\hat { i } \) + \(\hat { j } \) – 2\(\hat { k } \) और \(\vec { b } \) = 2\(\hat { i } \) + 3\(\hat { j } \) – \(\hat { k } \) में से प्रत्येक पर लम्ब मात्रक सदिश ज्ञात कीजिए।
हल:
प्रश्न क्र. 21 (A) की भाँति हल कीजिये।
उत्तर:
\(\bar { n } \) = ± \(\frac{1}{5 \sqrt{3}}\) ( 5\(\hat { i } \) – \(\hat { j } \) + 7\(\hat { k } \) ).

प्रश्न 22.
सदिशों \(\vec { a } \) = 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) और \(\vec { b } \) = 3\(\hat { i } \) – 4\(\hat { j } \) – \(\hat { k } \) में से प्रत्येक पर लम्ब मात्रक सदिश ज्ञात कीजिए।
हल:
प्रश्न क्र. 21 (A) की भाँति हल कीजिये।
उत्तर:
\(\bar { n } \) = ± \(\frac{1}{\sqrt{3}}\) ( \(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \) )

प्रश्न 23.
उस समानान्तर चतुर्भुज का क्षेत्रफल ज्ञात कीजिए जिसके विकर्ण 3\(\hat { i } \) + \(\hat { j } \) – 2\(\hat { k } \) और \(\hat { i } \) – 3\(\hat { j } \) + 4\(\hat { k } \) हैं।
हल:
ABCD एक समानान्तर चतुर्भुज है जिसके विकर्ण \(\overline{A C}=\bar{d}_{1}\) तथा \(\overline{B D}=\bar{d}_{2}\) हैं।
दिया है:



∴ समानान्तर चतुर्भुज का क्षेत्रफल

प्रश्न 24.
सदिश विधि से सिद्ध कीजिए कि समकोण त्रिभुज में कर्ण का वर्ग शेष दो भुजाओं के वर्गों के योग के बराबर होता है।
हल:
माना OAB एक समकोण त्रिभुज है जिसमें सिद्ध करना है कि AB² – OA² + OB²
माना कि मूलबिन्दु O के सापेक्ष A और B के स्थिति सदिश \(\vec { a } \) और \(\vec { b } \) हैं।
∵ OA और OB परस्पर लम्बवत् हैं,
∴ \(\vec { a } \).\(\vec { b } \) = 0
अब \(\vec { AB } \) = B का स्थिति सदिश – A का स्थिति सदिश

प्रश्न 25.
5\(\hat { i } \) + \(\hat { k } \) से निरूपित बल बिन्दु 9\(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) पर लगा हुआ है, बिन्दु 3\(\hat { i } \) + 2\(\hat { j } \) + \(\hat { k } \) के परितः सदिश आघूर्ण ज्ञात कीजिये।
हल:
\(\vec { F } \) = 5\(\hat { i } \) + \(\hat { k } \) = 5\(\hat { i } \) + 0\(\hat { j } \) + 1\(\hat { k } \)
O का स्थिति सदिश = 3\(\hat { i } \) + 2\(\hat { j } \) + k
P का स्थिति सदिश = 9\(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \)
\(\vec { O } \)P = P का स्थिति सदिश – O का स्थिति सदिश
= 9\(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) – (3\(\hat { i } \) + 2\(\hat { j } \) + \(\hat { k } \) )


बल \(\vec { F } \) का बिंदु P के परितः आधूर्ण = \(\vec { r } \) × \(\vec { F } \)

प्रश्न 26.
(A) सिद्ध कीजिये कि
\(\hat { i } \) × ( \(\vec { a } \) × \(\hat { i } \) ) + \(\hat { j } \) × ( \(\vec { a } \) × \(\hat { j } \) ) + \(\hat { k } \) × ( \(\vec { a } \) × \(\vec { k } \) ) = 2\(\vec { a } \).
हल:

(B) सिद्ध कीजिये कि
\(\vec { a } \) × ( \(\vec { b } \) × \(\vec { c } \) ) + \(\vec { b } \) × ( \(\vec { c } \) × \(\vec { a } \) ) + \(\vec { c } \) × ( \(\vec { a } \) × \(\vec { b } \) ) = \(\vec { 0 } \).
हल:
चूँकि

प्रत्येक को जोड़ने पर,

प्रश्न 27.
सिद्ध कीजिये कि [ \(\vec { a } \) + \(\vec { b } \) \(\vec { b } \) × ( \(\vec { c } \) × \(\vec { a } \) ) + \(\vec { c } \) × ( \(\vec { a } \) × \(\vec { b } \) )
हल:

प्रश्न 28.
दो बल \(\vec { P } \) = 4\(\hat { i } \) + \(\hat { j } \) – 3\(\hat { k } \) और \(\vec { Q } \) = 3\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \) एक कण पर क्रिया करते हैं, जिससे कण बिन्दु A (1, 2, 3) से बिन्दु B (5, 4, 1) तक विस्थापित होता है। बलों द्वारा किया गया कार्य ज्ञात कीजिए।
हल:

प्रश्न 29.
एक कण पर दो स्थिर बल 4\(\hat { i } \) + 3\(\hat { j } \)) व 3\(\hat { i } \) + 2\(\hat { j } \) कार्य करते हैं। वह कण बिन्दु \(\hat { i } \) + 2\(\hat { j } \) से बिन्दु 5\(\hat { i } \) + 4\(\hat { j } \) तक विस्थापित हो जाता है। बलों द्वारा किया गया कार्य ज्ञात कीजिए।
हल:

प्रश्न 30.
6 इकाई का बल जो सदिश 2\(\hat { i } \) – 2\(\hat { j } \) + \(\hat { k } \) के समान्तर कार्य करता है एवं एक कण को बिन्दु \(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) से 5\(\hat { i } \) + 3\(\hat { j } \) + 7\(\hat { k } \) तक विस्थापित कर देता है। बल के द्वारा किया गया कार्य ज्ञात कीजिये।
हल:
सदिश 2\(\hat { i } \) – 2\(\hat { j } \) + \(\hat { k } \) समान्तर इकाई सदिश

प्रश्न 31.
सिद्ध कीजिए कि [ \(\vec { a } \) – \(\vec { b } \) \(\vec { b } \) – \(\vec { c } \) \(\vec { c } \) – \(\vec { a } \) ] = 0।
हल:
अदिश त्रिक गुणन की परिभाषा से,

MP Board Class 12th Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 8 Binomial Theorem

Binomial Theorem Important Questions

Binomial Theorem Objective Type Questions

(A) Choose the correct option :

Question 1.
The total number of terms in the expansion of
(a) 7
(b) 12
(c) 13
(d) 6.
Answer:
(c) 13

Question 2.
If y = 3x + 6x² + 10x³ + …………… ∞, then the correct relation will be :
(a) x = 1 – (1 + y)^{–\(\frac { 1 }{ 3 }\)}
(b) x = (1 + y)^{–\(\frac { 1 }{ 3 }\)}
(c) y = 1 – (1 – x)^-3
(d) x = 1 + (1 + y)^{–\(\frac { 1 }{ 3 }\)}
Answer:
(a) x = 1 – (1 + y)^{–\(\frac { 1 }{ 3 }\)}

Question 3.
The total number of terms in the expansion of (1+ x)^-1 will be :
(a) 0
(b) ∞
(c) 2
(d) It can not be expand
Answer:
(b) ∞

Question 4.
The mid – term in the expansion of (x – \(\frac { 1 }{ x }\))¹⁰ will be :
(a) – ¹⁰C₅
(b) ¹⁰C₅
(c) 251
(d) 252
Answer:
(a) – ¹⁰C₅

Question 5.
For all positive integer of n, n(n – 1) is :
(a) Integer
(b) Natural number
(c) Even positive integer
(d) Odd positive integer.
Answer:
(c) Even positive integer

Question 6.
Expansion of (a + x)ⁿ is :
(a) aⁿ + ⁿC₁a^n-1x + ⁿC₂a^n-2x² + ……….. + ⁿC_ra^n-rx^r + ………… + aⁿ
(b) xⁿ + ⁿC₁x^n-1a +ⁿC₂x^n-2a² + ……….. + ⁿC_rx^n-ra^r + ………… + aⁿ
(c) aⁿ – ⁿC₁a^n-1x + ⁿC₂a^n-2x² + ……….. + (-1)^rnC_ra^n-rx^r + ………… + (-1)ⁿaⁿ
(d) xⁿ – ⁿC₁a^n-1a + ⁿC₂a^n-2a² + ……….. + (-1)^rnC_ra^n-rx^r + ………… + (-1)ⁿxⁿ
Answer:
(a) aⁿ + ⁿC₁a^n-1x +ⁿC₂a^n-2x² + ……….. + ⁿC_ra^n-rx^r + ………… + aⁿ

Question 7.
The fifth term in the expansion of ( x – \(\frac { 1 }{ x }\) )¹⁰ from the end, will be :
(a) \(\frac { ^{ 10 }{ C }_{ 6 } }{ x }\)
(b) \(\frac { 105 }{ 32{ x }^{ 2 } }\)
(c) \(\frac { ^{ 10 }{ C }_{ 6 } }{ { x }^{ 2 } }\)
(d) \(\frac { ^{ 10 }{ C }_{ 6 } }{ { x }^{ 10 } }\)
Answer:
(a) \(\frac { ^{ 10 }{ C }_{ 6 } }{ x }\)
[Hint: The total number of terms be 11 in the expansion of it.
∵ The 5^th term from end = (11 – 5)^th = 7^th term from the beginning.]

Question 8.
The number of mid – terms in the expansion of ( x – \(\frac { 1 }{ x }\) )¹⁰
(a) 1
(b) 2
(c) – \(\frac { ^{ 13 }{ C }_{ 7 } }{ x }\)
(d) 1716x
Answer:
(b) 2

Question 9.
The value of ⁿC₀ + ⁿC₁ + ⁿC₂ + …………….. + ⁿC_n :
(a) 2ⁿ + 1
(b) 2^{n – 1}
(c) 2ⁿ – 1
(d) 2ⁿ
Answer:
(d) 2ⁿ

Question 10.
The value of ⁿC₀ + ⁿC₂ + ⁿC₄ + …………….. = ⁿC₁ + ⁿC₃ + ……………. will be :
(a) 2ⁿ + 1
(b) 2^{n – 1}
(c) 2ⁿ – 1
(d) 2ⁿ
Answer:
(b) 2^{n – 1}

Question 11.
The total number of terms in the expansion of (a + b + c + d)ⁿ will be :
(a) \(\frac { (n+1)(n+2) }{ 2 }\)
(b) \(\frac { n(n+1) }{ 2 }\)
(c) \(\frac { (n+1)(n+2)(n+3) }{ 6 }\)
(d) \(\frac { (n+1)(n+2) }{ 6 }\)
Answer:
(c) \(\frac { (n+1)(n+2)(n+3) }{ 6 }\)

Question 12.
The necessary condition for expansion of (1 + x)^-1 is :
(a) | x | < 1
(b) | x | > 1
(c) | x | = 1
(d) | x | = – 1.
Answer:
(a) | x | < 1

Question 13.
The general term in the expansion of (x + a)ⁿ will be :
(a) r^th
(b) (r+1)^thterm
(c) (r-1)^th
(d) (r+2)^thterm
Answer:
(b) (r+1)^thterm

Question 14.
In the expansion of ( 2x + \(\frac { 1 }{ { 3x }^{ 2 } }\) )⁹, then term independent of x will be :
(a) \(\frac { 8 }{ 127 }\)
(b) \(\frac { 124 }{ 81 }\)
(c) \(\frac { 1792 }{ 9 }\)
(d) \(\frac { 256 }{ 243 }\)
Answer:
(c) \(\frac { 1792 }{ 9 }\)

Question 15.
The coefficient of x3 in the expansion of ( x – \(\frac { 1 }{ x }\) )¹⁵ is :
(a) 14
(b) 21
(c) 28
(d) 35
Answer:
(b) 21

Question 16.
The independent term in the expansion of (x² – \(\frac { 2 }{ { x }^{ 3 } }\))¹⁵ is :
(a) 5^th
(b) 6^th
(c) 7^th
(d) 8^th
Answer:
(c) 7^th

Question 17.
The value of ⁿC₀ + ⁿC₁ + ⁿC₂ + …………….. = ⁿC_n the expansion of (l + x)ⁿ is :
(a) 2ⁿ – 1
(b) 2ⁿ – 2
(c) 2ⁿ
(d) 2^n-1
Answer:
(c) 2ⁿ

Question 18.
The value of ¹⁵C₀ + ¹⁵C₂ + ¹⁵C₄ + ¹⁵C₆ + …………….. = ¹⁵C₁₄ is :
(a) 2¹⁴
(b) 2¹⁵
(c) 2¹⁵ – 1
(d) None of these
Answer:
(a) 2¹⁴

(B) Match the following :

Answer:

1. (d)
2. (a)
3. (e)
4. (c)
5. (b)
6. (g)
7. (f)
8. (i)
9. (b)

(C) Fill in the blanks :

1. By binomial theorem the value of (102)⁴ is ……………..
2. The value of second term in the expansion of (1 – x)^-3/2 is ……………..
3. The 5^th term from the end in the expansion of (x – \(\frac { 1 }{ 2x }\) )¹⁰ is ……………..
4. The constant term will be …………….. in the expansion of ( x² – 2 + \(\frac { 1 }{ { x }^{ 2 } }\) )⁶
5. The value of C₁ + 2C₂ + 3C₃ + ………….. + nC_n will be ……………..
6. The value of ⁿC₀ – ⁿC₁ + ⁿC₂ – ⁿC₃ + ………….. will be ……………..
7. The value of is ……………..
8. The value of is ……………..
9. The value of is ……………..
10. (2x + 3y)⁵ = …………….. up to three terms.
11. The coefficient of x⁷ in the expansion of (x² + \(\frac { 1 }{ x }\) )¹¹ will be …………….
12. In the expansion of (1 – x)¹⁰ the value of middle term is ……………..
13. Third (3^rd) term in the expansion of e^-3x will be ……………..
14. If n is odd, in the expansion of (x + a)ⁿ, then number of middle terms are ……………..
15. The middle term in the expansion of (\(\frac { x }{ a }\) + \(\frac { a }{ x }\) )¹⁰
16. The coefficient of xⁿ in the expansion of (1 + x) (1 – x)ⁿ will be ……………..

Answer:

1. 08243216
2. \(\frac { { x }^{ 3 } }{ 16 }\)
3. \(\frac { 105 }{ 32{ x }^{ 2 } }\)
4. 924
5. n.2^n-1
6. 0
7. \(\frac { n(n+1) }{ 2 }\)
8. \(\frac { n(n+1)(2n+1) }{ 6 }\)
9. [ \(\frac { n(n+1) }{ 2 }\) ]²
10. 32x⁵ + 240x⁴y + 720x³y²
11. 462
12. – 252 x^-5
13. \(\frac { 1 }{ 2 }\)
14. Two
15. 252
16. (- 1)ⁿ(1 – n)

(D) Write true / false :

1. The expansion of (1 + x)^-3 is 1 – 3x + 6x² – 10x³ + ………….. + \(\frac { (- 1)(r + 1)(r +2) }{ 2! }\)x^r + ……………..
2. The expansion of (1 – x)^-3 is 1 + 3x + 6x² + 10x³ + ………….. + \(\frac { (- 1)^r(r + 1)(r +2) }{ 2! }\)x^r + ……………..
3. The expansion of (1 – x)^-2 is 1 + 2x + 3x² + (r + 1) x^r+ ………….. +
4. The (r + 1 )^th term in the expansion of (1 – x)^-2 is (- 1)^r(r + 1) x^r+ ………….. +
5. The (r + 1 )^th term in the expansion of (1 – x)ⁿ will be x^r
6. The total number of terms in the expansion of (a + b + c)ⁿ is \(\frac { (n + 1)(n + 1) }{ 2 }\)
7. In the expansion of ( 3x – \(\frac { { x }^{ 3 } }{ 9 } \) )⁹, No . of terms is 9.
8. The number of term in the expansion of ( 3x – \(\frac { { x }^{ 3 } }{ 9 } \) )⁹ is 8.
9. In the expansion of (x + a)ⁿ then sum of powers of x and α in any term is n.
10. The coefficient of x in the expansion of (1 – 2x)^-3 is 6.
11. The second term in the expansion of (2x + 3y)⁵ is 240x⁴y.
12. The value of second term in the expansion of (1 – x)^-3/2 is \(\frac { 3 }{ 2 }\)x.

Answer:

1. True
2. True
3. True
4. True
5. False
6. True
7. False
8. False
9. True
10. False
11. True
12. True.

(E) Write answer in one word / sentence :

1. Find the value of 999³ by the binomial theorem
2. Find the middle term in expansion of (x² – \(\frac { 1 }{ x }\))⁶
3. General term in the expansion of (x + a)ⁿ will be.
4. If in the expansion of (1+x)⁵¹ the coefficient of x^r and x^{r – 5} are equal, then the value of r will be.
5. Find the coefficient of xⁿ in the expansion of (1 + x + x² + …………… ∞)², if | x | < 1.
6. In the expansion of (\(\frac { x }{ 3 }\) – \(\frac { 2 }{ { x }^{ 2 } }\))¹⁰, x⁴ comes in r^th term, then the value of r will be.
7. The 5^th term in the expansion of (1 – 2x)^{– 1} will be.

Answer:

1. 997002999
2. – 20 x⁵
3. ⁿC_r x^{n – r} a^r
4. 28
5. (n + 1)
6. 3
7. 16x²

Binomial Theorem Long Answer Type Questions

Question 1.
Expand : (\(\frac { 2 }{ x }\) – \(\frac { x }{ 2 }\))⁵ (NCERT)
Solution:

Question 2.
Expand : (2x – 3)⁶ (NCERT)
Solution:

Question 3.
Expand : (\(\frac { x }{ 3 }\) + \(\frac { 1 }{ x }\))⁵ (NCERT)
Solution:

Question 4.
Expand : (x + \(\frac { 1 }{ x }\))⁶. (NCERT)
Solution:

Question 5.
find 13^th term in the expansion of ( 9x – \(\frac { 1 }{ 3\sqrt { x } }\) )¹⁸. (NCERT)
Solution:

Question 6.
Find the middle term of (3 – \(\frac { { x }^{ 3 } }{ 6 }\))⁷
Solution:

Question 7.
Find the middle term in the expansion of (\(\frac { x }{ 3 }\) + 9y)¹⁰
Solution:
Here n =10
Total number of terms = n + 1 = 10 + 1 = 11 (odd)
Here, the term will be middle term.

Question 8.
If coefficient of x² and x³ in the expansion of (3 + ax)⁹ are equal, the value of a.
Solution:

Question 9.
Find the coefficient of x⁵ in the expansion of (x + 3)⁸
Solution:
Suppose x⁵ appears in (r + 1)^th term T_r+1 = nC₁x^n-ra^r
Here n = 8, x = x, a = 3
T_r+1 = ⁸C_r(x)^{8 – r}(3)^r
For the coefficient of x⁵,
8 – r = 5
=> r = 3
T₃₊₁ = ⁸C₃(3)³
= \(\frac { 8 × 7 × 6 }{ 3 × 2 × 1}\) × 3 × 3 × 3 × x⁵
= 1512 × x⁵
Hence coefficient of x⁵ is 1512.

Question 10.
Find the coefficient of a⁵b⁷ in the expansion of (a – 2b)¹².
Solution:

Question 11.
If the 17^th and 18^th terms in the expansion of (2 + a)⁵⁰ are equal, then find the value of a. (NCERT)
Solution:
In the expansion of (x + a)ⁿ
T_r+1 = ⁿC_r x^{n – r} a^r
Here n = 50, x = 2, a = a
T₁₇ = T_{16 + 1} = ⁵⁰C₁₆ (2)^{50 – 16} (a)¹⁶
⇒ T₁₇ = ⁵⁰C₁₆ (2)³⁴ (a)¹⁶
and T₁₈ = T_{17 + 1} = ⁵⁰C₁₇ (2)^{50 – 17} (a)¹⁷
= ⁵⁰C₁₇ (2)³³ (a)¹⁷

Question 12.
Prove that the value of the middle term in the expansion of (1+x)²ⁿ is \(\frac { { 1.3.5 …….. (2n – 1)} }{ n! }\).2ⁿ xⁿ
Solution:

Question 13.
In the expansion of (x + 1)ⁿ, the coefficient of the (r – 1)^th, r^th and (r + 1)^th terms are in the ratio 1 : 3 : 5, then find the value of n and r.
Solution:
In the expansion of (x + 1)ⁿ,
T_{r + 1} = ⁿC_rx^{n – r}(1)^r
T_{r – 1} = T_{r – 2 + 1} = ⁿC_{r – 2}(x)^{n – (r – 2)}(1)^{r – 2}
Coefficient of T_{r – 1}^th term = ⁿC_{r – 2}
T_r = T_{r – 1 + 1} = ⁿC_{r – 1}(x)^{n – (r – 1)}(1)^{r – 1}
Coefficient of T_r^thterm = ⁿC_{r – 1}
T_{r + 1} = ⁿC_r x^{n – r} (1)^r
Coefficient of T_r+1^th term = ⁿC_r

Put n = 4r – 5 from equation (1) in equation (2),
3(4r – 5) – 8r = – 3
⇒ 12r – 15 – 8r = – 3
⇒ 4r = 12
∴ r = 3
Put r = 3 in equation (2),
n – 4 x 3 = – 5
⇒ n = 12 – 5
⇒ n = l
n = 7, r = 3

Question 14.
Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ of in the expansion of (1 + x)^{2n – 1}.
Solution:
In the expansion of (x + a)ⁿ
T_r+1 = ⁿC_r x^{n – r} a^r
Here x = 1, a = x, n = 2n
T_r+1 = ²ⁿC_r(1)^{2n – r}(x)^r
For the coefficient of xⁿ, put r = n,
T_n+1 = ²ⁿC_n(a)^{2n – n}(x)ⁿ
and T₁₈ = T_{17 + 1} = ⁵⁰C₁₇ (2)^{50 – 17} (a)¹⁷
= (²ⁿC_n) xⁿ
∴ In the expansion of (1 + x)²ⁿ, the coefficient of xⁿ = ²ⁿC_n …. (1)
and in the expansion of (1 + x)^{2n – 1}, x = 1, a = x, n = 2n – 1
∴ T_r+1 = ²ⁿC_r (1)^{2n – 1  -r} (x)ⁿ
For the coefficient of xⁿ, put r = n, ‘
We get T_n+1 = ^{2n – 1}C_n xⁿ
The coefficient of xⁿ in the expansion of (1 + x)^{2n – 1} = ^{n – 1}C_n

∴ The coefficient of xⁿ in the expansion of (1 + x)²ⁿ
= 2 x The coefficient of xⁿ in the expansion of (1 + x)²ⁿ, [from equation (1) and (2)]

Question 15.
Find the constant term in the expansion of (\(\frac { 3 }{ 2 }\)x² – \(\frac { 1 }{ 3x }\))⁶
Solution:

MP Board Class 11th Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 7 Permutations and Combinations

Permutations and Combinations Important Questions

Permutations and Combinations Objective Type Questions

(A) Choose the correct option :

Question 1.
The correct relation between ⁿP_r and ⁿC_r is :
(a) ⁿC_r = \(\frac{[r}{^{n} P_{r}}\)
(b) ⁿC_r.ⁿP_r= 1
(c) ⁿC_r = \(^{n} P_{r} \times\lfloor r\)
(d) \(^{n} C_{r} \times\left\lfloor r=^{n} P_{r}\right.\)
Answer:
(d) \(^{n} C_{r} \times\left\lfloor r=^{n} P_{r}\right.\)

Question 2.
If ⁿC₂:ⁿC₄ = 2 : 1, then n = …………..
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
(c) 5

Question 3.
How many words can be made by the letter of the word ‘COMMITEE’:

Answer:

Question 4.
If ⁿP₄ = 20 x ⁿP₃, then n =
(a) 20
(b) 21
(c) 18
(d) 23
Answer:
(d) 23

Question 5.
The number of different words that can be formed from the letter of the word ‘TRIANGLE’:
(a) 7200
(b) 36000
(c) 144000
(d) 1240.
Answer:
(b) 36000

Question 6.
In how many ways three person can be seated in 4 chairs :
(a) 42
(b) 20
(c) 24
(d) 12.
Answer:
(c) 24

Question 7.
If ¹⁵C_r = ¹⁵C_{r – 3} , then the value of ^rC₆ will be :
(a) 84
(b) 83
(c) 82
(d) 80.
Answer:
(a) 84

Question 8.
⁵C₀ + ⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅ = ……………..
(a) 30
(b) 32
(c) 23
(d) 42
Answer:
(b) 32

Question 9.
The value of : \(\frac{1}{\lfloor- 4}\) will be :
(a) 0
(b) ∞
(c) 4
(d) 1.
Answer:
(a) 0

Question 10.
If ¹⁰P_r = 720, then the value of r will be :
(a) 4
(b) 5
(c) 6
(d) 3
Answer:
(d) 3

(B) Match the following :

Answer:

1. (d)
2. (a)
3. (f)
4. (b)
5. (c)
6. (e)

(C) Fill in the blanks :

1. The number of words can be made by the letters of the word ‘INDIA’ ……………
2. In order to invite 6 friends, the number of ways to send 3 servant is ……………
3. If ²⁰C_r = ²⁰C_y; x ≠ y, then the value of x + y will be ……………
4. The numbers of ways greater than 4000 can be made by the digits 2, 4, 5, 7 is ……………
5. If ¹²P_r = 1320, then r will be ……………
6. The number of ways can 8 friends be seated in a circular table is ……………
7. If ⁿC₈ = ⁿC₁₂, then n will be ……………

Answer:

1. 60
2. 4⁶
3. 20
4. 48
5. 3
6. 5040
7. 20

(D) Write true / false :

1. If ⁿC_r + ⁿC_{r + 1} = ^{n + 1}C_x, then x will be 6.
2. 5 letters be posted in 4 letter boxes in 5⁴ ways.
3. Fruit can be chosen from 5 guavas, 3 apples and 4 mangoes in 60 ways.
4. The number of ways can 8 friends stands in a row is 8.
5. The value of ⁿC_r ÷ ^{n – 1}C_ris \(\frac { n }{ r – 1 }\)

Answer:

1. True
2. False
3. False
4. True
5. False.

(E) Write answer in one word / sentence :

1. How many four digit numbers can be formed with the digits 2, 3, 4, 5 in these number divisible by 2?
2. How many ways can a garland be prepared with 6 flowers?
3. If 2.ⁿC₅ = 9.^{n – 2}C₅, then the value of n will be
4. Five points out of 15 points are collinear. If these points be joined, then how many triangles can be formed?
5. The value of will be?

Answer:

1. 12
2. 60
3. 10
4. 30
5. ⁵⁶C₄

Permutations and Combinations Short Answer Type Questions

Question 1.
If \(\frac{1}{\lfloor6}\) + \(\frac{1}{\lfloor7}\) = \(\frac{x}{\lfloor8}\), then find the value of x.
Solution:

Question 2.
Find the value of

(i) n = 6, r = 2 (ii) n = 9, r = 5
Solution:

Question 3.
If ^n-1P₃ : ⁿP₄ = 1 : 9, then find the value of n.
Solution:

Question 4.
IF ⁿP₅ : 42.ⁿP₃, then find the value of n.
Solution:
IF ⁿP₅ = 42.ⁿP₃, then find the value of n.
Solution:
Given:  ⁿP₅ = 42.ⁿP₃
⇒ n(n – 1)(n – 2)(n – 3)(n – 4) = 42n(n – 1)(n – 2)
⇒ (n – 3)(n – 4) = 42
⇒ n² – 7n – 30 = 0
⇒ n² – 10n + 3n – 30 = 00
⇒ (n – 10) (n + 3) = 0
n = 10, n ≠ – 3.

Question 5.
IF \(\frac{^{n} P_{4}}{^{n-1} P_{4}}\) = \(\frac { 5 }{ 3 }\), then find the value of n.
Solution:
Given : \(\frac{^{n} P_{4}}{^{n-1} P_{4}}\) = \(\frac { 5 }{ 3 }\)
⇒ 3.ⁿP₃ = 5.^n-1P₄
⇒ 3n(n – 1)(n – 2)(n – 3) = 5 (n – 1)(n – 2)(n – 3)(n – 4)
⇒ 3n = 5 (n – 4)
⇒ n = 10.

Question 6.
If ⁿC₉ = ⁿC₈, then find the value of ⁿC₁₇.
Solution:

Question 7.
If ⁿP_n-2 = 60, then find the value of n.
Solution:

Question 8.
If ¹²P_r = 1320, then find the value of r.
Solution:

Question 9.
If ⁿC₁₄ = ⁿC₁₆, then find the value of ³²C_n.
Solution:
Given : ⁿC₁₄ = ⁿC₁₆
⇒ ⁿC₁₄ = ⁿC_16-1
⇒ 14 = n – 16
⇒ n = 30
∴ ³²C_n = ³²C₃₀ = ³²C₂
= \(\frac { 32.31 }{ 2 }\) = 496.

Question 10.
If ²⁰C_x = ²⁰C_y, then find the value of x + y.
Solution:
Given :
⇒ ²⁰C_x = ²⁰C_y
⇒ ²⁰C_x = ²⁰C_20-y
⇒ x = 20 – y
⇒ x + y = 20

Question 11.
In how many ways can 8 friends sit around a round table?
Solution:
Total number of ways of sitting 8 friends around table = \(\lfloor 8 – 1\) = \(\lfloor 7\)
= 7 x 6 x 5 x 4 x 3 x 2 x 1
= 5040.

Question 12.
In how many ways can a garland be prepared with 6 flowers?
Solution:
Ways of preparing the garlands with 6 flowers = \(\frac { 1 }{ 2 }\) \(\lfloor 6 – 1\)
= \(\frac { 1 }{ 2 }\) \(\lfloor 5\)
= \(\frac { 1 }{ 2 }\).5.4.3.2.1 = 60.

Question 13.
How many numbers between 100 and 1000 can be formed with the digits 2, 3, 0, 4, 8, 9 if repetition of digits is not allowed in the same number?
Solution:
The numbers between 100 and 1000 are of three digits.
∴Number of three digits made from 6 digits will be ⁶P₃.
In this number which begins with 0 will be ⁵P₂.
∴Required number = ²⁰C_y – ⁵P₂ = 120 – 20 = 100.

Question 14.
How many words, with or without meaning can be formed using all the letters of the word EQUATION? (NCERT)
Solution:

Question 15.
How many chords can be drawn through 21 points on a circle? (NCERT)
Solution:
By joining two points we get one chord.
∴ No. of chords = ²¹C₂
= \(\frac { 21 x 20 }{ 2 x 1 }\)
= 210.

Question 16.
In how many ways can a team of 3 boys and 3 girls can be selected from 5 boys and 4 girls? (NCERT)
Solution:
Number of ways of selecting 3 boys from 5 boys and 3 girls from 4 girls will be
= ⁵C₃ × ⁴C₃
= \(\frac { 5 × 4 × 3 }{ 3 × 2 × 1}\) × \(\frac { 4 × 3 × 2 }{ 3 × 2 × 1 }\)
= 10 x 4 = 40.

Permutations and Combinations Long Answers Type Questions

Question 1.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour. (NCERT)
Solution:
Since, 3 red balls out of 6 red balls can be selected
= ⁶C₃ = \(\frac { 6 × 5 × 4 }{ 3 × 2 × 1}\) = 20
3 white balls out of 5 white balls = ⁵C₃ = \(\frac { 5 × 4 × 3 }{ 3 × 2 × 1}\) = 10
and 3 blue balls out of 5 blue balls = ⁵C₃ = \(\frac { 5 × 4 × 3 }{ 3 × 2 × 1}\) = 10
Total number of ways of selecting 9 balls = 20 x 10 x 10

Question 2.
Determine the number of 5 cards combinations out of a deck of 52 cards, if there is exactly one ace in each combination (NCERT)
Solution:
Number of ways of selecting one ace from 4 ace = ⁴C₁
Number of selecting 4 cards from 48 cards = ⁴⁸C₄
= \(\frac { 48 × 47 × 46 × 45 }{ 4 × 3 × 2 × 1}\)
Total number of ways = ⁴C₁ x ⁴⁸C₄
= 4 × = \(\frac { 48 × 47 × 46 × 45 }{ 4 × 3 × 2 × 1}\)
= 7, 78, 320.

Question 3.
In how many ways can one select a cricket team of eleven players from 17 players. In which only 5 players can bowl, if each cricket team of 11 must include exactly 4 bowlers? (NCERT)
Solution:
Out of 17 players, to select 11 players in which 5 bowlers and 12 batsman.
Out of 5 bowlers to select 4 bowlers = ⁵C₄ = 5
Out of 12 batsman to select 7 batsman = ¹²C₇
= \(\frac { 12 × 11 × 10 × 9 × 8 × 7 × 6 }{ 7 × 6 × 5 × 4 × 3 × 2 × 1 }\)
= 11 × 9 × 8 = 792
∴ Number of ways selecting 11 players = 5 x 792 = 3960.

Question 4.
A bag contains 5 black and 6 red balls, determine the number of ways in which 2 black and 3 red balls can be selected. (NCERT)
Solution:
There are 5 black and 6 red balls.
Out of 5 black balls, 2 black balls can be =⁵C₂ = \(\frac { 5 × 4 }{ 2 × 1 }\) = 10
Out of 6 red balls, 3 red balls can be = ⁶C₃ = \(\frac { 6 × 5 × 4 }{ 3 × 2 × 1 }\) = 20
∴ Total No. of ways = 10 x 20 = 200.

Question 5.
In how many ways can a student choose a program of 5 courses, if 9 courses are available and 2 specific courses are compulsory for each students? (NCERT)
Solution:
Number of available courses = 9.
Number of courses to choose = 5.
Compulsory courses = 2.
∴ Required number of ways = ^9-2C_5-2 = ⁷C₃
= \(\frac { 7 × 6 × 5 }{ 3 × 2 × 1 }\) = 35.

Question 6.
How many words, with or without meaning each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER? (NCERT)
Solution:
There are 3 vowels and 5 consonants in DAUGHTER.
2 vowels out of 3 vowels = ³C₂ = \(\frac { 3! }{ 1!2! }\) = \(\frac { 3 × 2 }{ 2 × 1 }\)
3 consonants out of 5 consonats = ⁵C₃ = \(\frac { 5! }{ 2!3! }\) = \(\frac { 5 × 4 × 3 }{ 3 × 2 × 1 }\) = 10
Arrangement of 2 vowels and 3 consonants = 5!
= 5 × 4 × 3 × 2 × 1 = 120

Question 7.
How many words, with or without menning can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Solution:
Given word is EQUATION.
Vowels – E, U, A, I, O
Consonants – T, Q, N.
Arrangement of 5 vowels = 5! = 5 x 4 x 3 x 2 x 1 = 120
Arrangement of 3 consonants = 3! = 3 x 2 x 1 = 6
Arrangements of vowels and consonants = 2!
∴ Total No. of ways = 120 x 6 x 2! = 720 x 2 = 1440.

Question 8.
A committee of 7 members has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (NCERT)

1. Exactly 3 girls
2. At least 3 girls
3. At most 3 girls.

Solution:
Total members in committee = 7
1. Selection of 3 girls and 4 boys in committee
= ⁴C₃ x ⁹C₄
= \(\frac { 4 × 3 × 2 × 1 }{ 3 × 2 × 1 }\) x \(\frac { 9 × 8 × 7 × 6 }{ 4 × 3 × 2 × 1 }\)
= 9 x 8 x 7 = 504.

2. Selection of minimum 3 girls in committee
= ⁴C₃ x ⁹C₄ + ⁴C₄ x ⁹C₃
= 504 + 1 x \(\frac { 9 × 8 × 7 }{ 3 × 2 × 1 }\) = 504 + 84 = 588.

3. Selection of maximum 3 girls in a committee

Question 9.
How many different words can be formed by using the letters of the word ALLAHABAD, out of which in how many words vowels will come in even places? (NCERT)
Solution:
Total number of letters in ALLAHABAD = 9.
In which A = 4 times, L = 2 times.

= 9.8.7.3.5 = 7560.
Number of even places are = 4 and odd places = 5. Hence we can put A only in 4 places and rest 5 letters L, L, H, B, D in 5 odd places inways.

Question 10.
In how many ways can a cricket team of eleven can be formed from 14 players? Which win always have :

1. Captain is always taken.
2. When two fast bowlers are always taken?

Solution:
1. When captain is always taken (11 – 1) from (14 – 1) players.
= ¹³C₁₀ = ¹³C₃
= \(\frac { 13.12.11 }{ 3.2.1 }\) = 286.

2. When two fast bowlers are always taken = ¹²C₉ = ¹²C₃
= \(\frac { 12.11.10 }{ 3.2.1}\)

Question 11.
In how many ways committee of 5 members can be formed out of 6 men and 4 women ? If the committee has : (i) Only one woman, (ii) At least one woman.
Solution:
(i) Only one woman out of 4 women and 4 men out of 6 men.
No. of ways = ⁶C₄ x ⁴C₁
= ⁶C₂ x 4
= \(\frac { 6 × 5 × 4 }{ 2.1 }\) = 60

(ii) Atleast one woman :

Hence, total No. of ways = 60 + 120 + 60 + 6 = 246.

Question 12.
If ²ⁿC₃ : ⁿC₂ = 12 : 1,then find the value of it. (NCERT)
Solution:

Question 13.
If ²ⁿC₃ : ⁿC₃ = 11 : 1, then find the value of it. (NCERT)
Solution:

Question 14.
Prove that:

Solution:

Question 15.
Prove that:
²ⁿP_n = 2ⁿ{1.3.5 …………. (2n – 1)}.
Solution:

MP Board Class 11th Maths Important Questions

MP Board Class 12th Maths Important Questions Chapter 9 अवकल समीकरण

अवकल समीकरण Important Questions

अवकल समीकरण वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प चुनकर लिखिए –

प्रश्न 1.
अवकल समीकरण \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + x² \(\frac{dy}{dx}\) = e^x की घात है –
(a) 1
(b) 2
(c) 3
(d) अस्तित्व नहीं है।
उत्तर:
(c) 3

प्रश्न 2.
अवकल समीकरण (1 + x) y dx + (1 – y) x dy = 0 का हल होगा –
(a) log xy + x + y = c
(b) log y + x – y = c
(c) log xy – x – y = c
(d) log xy – x + y = c.
उत्तर:
(b) log y + x – y = c

प्रश्न 3.
उन सभी वृत्तों का अवकल समीकरण जो मूलबिन्दू से गुजरते हैं तथा जिनके केन्द्र X – अक्ष पर स्थित है –
(a) x² = y² + xy \(\frac{dy}{dx}\)
(b) x² = y² + 3xy \(\frac{dy}{dx}\)
(c) y² = x² + 2xy \(\frac{dy}{dx}\)
(d) y² = x² – 2xy \(\frac{dy}{dx}\)
उत्तर:
(c) y² = x² + 2xy \(\frac{dy}{dx}\)

प्रश्न 4.
अवकल समीकरण \(\frac{dy}{dx}\) + y = e^-x, y(0) = 0 का हल होगा –
(a) y = e^-x (x -1)
(b) y = xe^x
(c) y = xe^-x + 1
(d) y = xe^-x
उत्तर:
(d) y = xe^-x

प्रश्न 5.
सरल रेखा जो अवकल समीकरण \(\frac{dy}{dx}\) = m को संतुष्ट करती हो तथा Y – अक्ष पर धनात्मक दिशा में 3 अन्तःखण्ड काटती हो, है –
(a) y = mx + c
(b) = mx + 3
(c) y = mx – 3
(d) y = – mx + 3
उत्तर:
(b) = mx + 3

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिये –

1. समीकरण x² + y² = a² के संगत अवकल समीकरण ……………………….. है।
2. वक्र y = e^cx से संबंधित अवकल समीकरण …………………………….. है, जहाँ c स्वेच्छ अचर है।
3. रेखीय अवकल समीकरण \(\frac{dy}{dx}\) + Py = Q में समाकलन गुणांक …………………………. है।
4. रेखीय अवकल समीकरण \(\frac{dy}{dx}\) + Py = Q में P और …………………………… हैं।
5. अवकल समीकरण (x + y + 1) dy = dx ………………………… रूप का है।
6. अवकल समीकरण e^-x+y \(\frac{dy}{dx}\) = 1 का हल हो ……………………… है।

उत्तर:

1. y \(\frac{dy}{dx}\) + x = 0
2. x\(\frac{dy}{dx}\) = y log y
3. e^∫pdx
4. अचर
5. रेखीय अवकल समीकरण

प्रश्न 3.
निम्न कथनों में सत्य/असत्य बताइए –

1. अवकल समीकरण y = x ( \(\frac{dy}{dx}\) )² + \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) की कोटि 2 है।
2. अवकल समीकरण ( \(\frac { d^{ 3 }y }{ dx^{ 3 } } \) )^4/5 – 2 ( \(\frac{dy}{dx}\) ) ( \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) )² = 0 की घात 5 है।
3. अवकल समीकरण x \(\frac{dy}{dx}\) – y = 2x² का समाकलन गुणांक e^-x है।
4. अवकल समीकरण dy = sin x dx का हल y + cos x – c = 0 है।
5. अवकल समीकरण ydx + (x – y³) dy = 0 का हल xy = \(\frac { y^{ 4 } }{ 4 } \) + c है।

उत्तर:

1. सत्य
2. असत्य
3. असत्य
4. सत्य
5. सत्य।

प्रश्न 4.
एक शब्द/वाक्य में उत्तर दीजिए –

1. अवकल समीकरण (1 + y²) + (2xy – cot y) \(\frac{dy}{dx}\) = 0 का समाकल गुणांक लिखिए।
2. अवकल समीकरण (1 + x²)dy = (1 + y²) dx का हल ज्ञात कीजिए।
3. अवकल समीकरण y = x( \(\frac{dy}{dx}\) )² + \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) की कोटि व घात का योग लिखिए।
4. अवकल समीकरण dy = sin x dx का हल y + cos x – c = 0 है।
5. अवकल समीकरण \(\frac{dy}{dx}\) + \(\frac{1}{x}\) = \(\frac { e^{ y } }{ x^{ 2 } } \) का हल ज्ञात कीजिए।

उत्तर:

1. 1 + y²
2. x – y = c (1 + xy)
3. 3
4. log x
5. 2xe^-y = cx² + 1.

अवकल समीकरण अति लघु उत्तरीय प्रश्न

प्रश्न 1.
अवकल समीकरण \(\frac{dy}{dx}\) + y = e^-x की कोटि तथा घात ज्ञात कीजिये।
उत्तर:
1, 1

प्रश्न 2.
अवकल समीकरण ( \(\frac{dy}{dx}\) )³ = \(\sqrt { 1+(\frac { dy }{ dx } )^{ 2 } } \) की कोटि तथा घात ज्ञात कीजिये।
उत्तर:
1, 6

प्रश्न 3.
अवकल समीकरण \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + \(\sqrt { 1+(\frac { dy }{ dx } )^{ 3 } } \) = 0 की कोटि तथा घात ज्ञात कीजिये।
उत्तर:
2, 2

प्रश्न 4.
वृत्त के समीकरण x² + y² = a² के संगत अवकल समीकरण क्या होगा?
उत्तर:
y \(\frac{dy}{dx}\) + x = 0

प्रश्न 5.
सरल रेखा y = mx + c के लिये अवकल समीकरण बनाइये।
उत्तर:
\(\frac{dy}{dx}\) = m

प्रश्न 6.
अवकल समीकरण \(\frac{dy}{dx}\) = 4y को हल कीजिये।
उत्तर:
y = c.e^4x

प्रश्न 7.
x² \(\frac{dy}{dx}\) = 2 का व्यापक हल ज्ञात कीजिये।
उत्तर:
y = c – \(\frac{2}{x}\)

प्रश्न 8.
अवकल समीकरण dy = sin x dx का हल ज्ञात कीजिये।
उत्तर:
y + cos x = c

प्रश्न 9.
\(\frac{dy}{dx}\) + Px = Q रूप वाले अवकल समीकरण का व्यापक हल ज्ञात कीजिये।
उत्तर:
xe^∫pdy = ∫Q.e^∫pdy.dy + c

प्रश्न 10.
अवकल समीकरण (1 – y²) \(\frac{dy}{dx}\) + yx = ay का समाकल गुणांक ज्ञात कीजिये।
उत्तर:
\(\frac { 1 }{ \sqrt { 1-y^{ 2 } } } \)

अवकल समीकरण लघु उत्तरीय प्रश्न

प्रश्न 1.
अवकल समीकरण x log x dy – y dx = 0?
हल:
दिया है:
⇒ xlog x dy = y dx
⇒ \(\frac{1}{y}\) \(\frac{dy}{dx}\) = \(\frac{1}{xlogx}\) dx
⇒ ∫\(\frac{1}{y}\) \( dy = ∫[latex]\frac{1}{xlogx}\) dx
⇒ log y = ∫\(\frac{1}{t}\) dt (माना log x = t, \(\frac{1}{x}\) \( dx = dt)
⇒ log y = log t + log c
⇒ log y = log log x + log c

प्रश्न 2.
अवकल समीकरण dy/dx = e^x-y + x.e^-y को हल कीजिए।
हल:
दिया है:
[latex]\frac{dy}{dx}\) = e^x-y + x.e^-y
⇒ \(\frac{dy}{dx}\) = e^-y(e^x + x)
⇒ e^y dy = (e^x + x) dx
दोनों पक्षों का समाकलन करने पर,
∫e^y dy = ∫(e^x + x) dx
e^y = e^x + \(\frac { x^{ 2 } }{ 2 } \) + c

प्रश्न 3.
सिद्ध कीजिए कि y = 4 sin 3x अवकल समीकरण \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + 9y = 0 का एक हल है।
हल:
दिया है:
y = 4 sin 3x ……………… (1)
x के सापेक्ष अवकलन करने पर,
∴ \(\frac{dy}{dx}\) = 12 cos 3x
पुनः x के सापेक्ष अवकलन करने पर,
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = -36 sin 3x = -9 × 4 sin 3x
⇒ \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = -9y, [समी. (1) से]
⇒ \(\frac { d^{ 2 }y }{ dx^{ 2 } } \)+ 9y = 0 यही सिद्ध करना था।

प्रश्न 4.
अवकल समीकरण \(\frac{dy}{dx}\) = sec x (sec x + tan x) का हल ज्ञात कीजिए।
हल:
दिया गया अवकल समीकरण है:
\(\frac{dy}{dx}\) = sec x (sec x + tan x)
⇒ dy = (sec²x + sec x tan x) dx
⇒ ∫dy = ∫sec²x dx + ∫sec x tan x dx
∴ y = tan x + sec x + c

प्रश्न 5.
अवकल समीकरण \(\frac{dy}{dx}\) = sec² x + 3x² को हल कीजिए।
हल:
\(\frac{dy}{dx}\) = sec² x + 3x²
⇒ dy = (sec² x + 3x²) dx
⇒ ∫dy = ∫sec² x dx + 3∫x² dx
⇒ y = tan x + \(\frac { 3x^{ 3 } }{ 3 } \) + c
⇒ y = tan x + x³ + c

प्रश्न 6.
अवकल समीकरण \(\frac{dy}{dx}\) = sec² x + 2x का हल ज्ञात कीजिए।
हल:
प्रश्न क्र. 5 की भाँति हल करें।

प्रश्न 7.
अवकल समीकरण \(\frac{dy}{dx}\) = (3x² + 2) को हल कीजिए।
हल:
दिया है:
\(\frac{dy}{dx}\) = (3x² + 2)
⇒ dy = (3x² + 2) dx
⇒ ∫dy = ∫(3x² + 2) dx
⇒ y = 3 × \(\frac { x^{ 3 } }{ 3 } \) + 2x + c = x³ + 2x + c

प्रश्न 8.
अवकल समीकरण x² \(\frac{dy}{dx}\) = 2 को हल कीजिए।
उत्तर:
x² \(\frac{dy}{dx}\) = 2
⇒ dy = 2.x^-2 dx
⇒ ∫dy = 2∫x^-2 dx
⇒ y = 2( \(\frac{-1}{x}\) ) + c

प्रश्न 9.
अवकल समीकर \(\frac{dy}{dx}\) = x³ + sin 4x का हल ज्ञात कीजिये।
दिया है:
\(\frac{dy}{dx}\) = x³ + sin 4x
⇒ dy = (x³ + sin 4x) dx
⇒ ∫dy = ∫x³ dx + ∫sin 4x dx
⇒ y = \(\frac { x^{ 4 } }{ 4 } \) + ( \(\frac { -cos4x }{ 4 } \) ) + c
⇒ y = \(\frac { x^{ 4 } }{ 4 } \) – \(\frac{cos4x}{4}\) + c

प्रश्न 10.
अवकल समीकरण \(\frac{dy}{dx}\) + 2x = e^3x का हल ज्ञात कीजिये।
हल:
दिया है:
\(\frac{dy}{dx}\) + 2x = e^3x
⇒ \(\frac{dy}{dx}\) = e^3x – 2x
⇒ dy = (e^3x – 2x) dx
⇒ ∫dy = ∫e^3x dx – 2∫x dx
⇒ y = e^3x. \(\frac{1}{3}\) – \(\frac { 2x^{ 2 } }{ 2 } \) + c = \(\frac{1}{3}\) e^3x – x² + c

प्रश्न 11.
अवकल समीकरण \(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \) का हल ज्ञात कीजिये।
हल:
\(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \)
⇒ \(\frac { 1 }{ cos^{ 2 }y } \) dy = \(\frac { 1 }{ sin^{ 2 }x } \) dx
⇒ sec² ydy = cosec² xdx
⇒ ∫sec² ydy = ∫cosec² xdx
⇒ tan y = – cot x + c

प्रश्न 12.
अवकल समीकरण (x² + 1) \(\frac{dy}{dx}\) = 1 को हल कीजिये।
हल:
दिया है:
(x² + 1) \(\frac{dy}{dx}\) = 1
⇒ \(\frac{dy}{dx}\) = \(\frac { 1 }{ (1+x^{ 2 }) } \)
⇒ dy = \(\frac { 1 }{ (1+x^{ 2 }) } \) dx
⇒ ∫dy = ∫\(\frac { 1 }{ (1+x^{ 2 }) } \) dx
⇒ y = tan^-1 x + c

प्रश्न 13.
समीकरण \(\frac{dy}{dx}\) = sin x sin y को हल कीजिये।
हल:
\(\frac{dy}{dx}\) = sin x sin y
⇒ cosec y dy = sin x dx
समाकलन करने पर,
-log_e(cosec y + cot y) = – cos x + c
⇒ cos x – log_e(cosec y + cot y) = c

प्रश्न 14.
अवकल समीकरण \(\frac{dy}{dx}\) = y sin x को हल कीजिये।
हल:
\(\frac{dy}{dx}\) = y sin x
⇒ \(\frac{1}{y}\) \(\frac{dy}{dx}\) = sin x
⇒ ∫\(\frac{1}{y}\) dy = ∫sin x dx
⇒ log y = – cos x + c

प्रश्न 15.
अवकल समीकरण \(\frac{dy}{dx}\) = x cos x को हल कीजिये।
हल:
दिया है:
\(\frac{dy}{dx}\) = x cos x
⇒ dy = x cos x dx
⇒ ∫dy = ∫x cos x dx
⇒ y = x sin x – ∫1. sin x dx + c
⇒ y = x sin x + cos x + c

प्रश्न 16.
अवकल समीकरण \(\frac{dy}{dx}\) = 1 – x + y – xy को हल कीजिए।
हल:
दिया गया अवकल समीकरण है –
\(\frac{dy}{dx}\) = 1 – x + y – xy
⇒ \(\frac{dy}{dx}\) = (1 – x) + y(1 – x)
⇒ \(\frac{dy}{dx}\) = (1 – x) (1 + y)
⇒ \(\frac{dy}{1+y}\) = (1 – x)dx
⇒ ∫\(\frac{dy}{1+y}\) = ∫(1 – x)dx
⇒ log_e(1 + y) = x – \(\frac { x^{ 2 } }{ 2 } \) + c

प्रश्न 17.
अवकल समीकरण \(\frac{dy}{dx}\) = (1 + x)(1 + y²) को हल कीजिए।
हल:
दिया गया अवकल समीकरण है:
\(\frac{dy}{dx}\) = (1 + x)(1 + y²)
⇒ \(\frac { 1 }{ 1+y^{ 2 } } \) dy = (1 + x) dx
समाकलन करने पर,
⇒ tan^-1 y = x + \(\frac { x^{ 2 } }{ 2 } \) + c

प्रश्न 18.
अवकल समीकरण हल कीजिए –
\(\frac{dy}{dx}\) = cot²x
⇒ dy = cot² x dx
⇒ ∫dy = ∫cot² x dx
⇒ y = ∫(cosec² x – 1) dx
⇒ y = – cot x – x + c

अवकल समीकरण दीर्घ उत्तरीय प्रश्न-I

प्रश्न 1.
(A) अवकल समीकरण \(\frac{dy}{dx}\) + y tan x = sec x को हल कीजिए।
हल:
दिया है:
\(\frac{dy}{dx}\) + y tan x = sec x ………… (1)
यह एक रैखिक अवकल समीकरण है, इसकी तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
P = tan x, Q = sec x
∴ I.F. = e^{∫p dx} = e^{∫tan x dx} = e^{log secx}
⇒ I.F. = sec x
अतः अवकल समी. (1) का अभीष्ट हल है:
y × (I.F.) = ∫Q × (I.F). dx + c
⇒ y × (sec x) = ∫sec x × (sec x) dx + c
= ∫sec² xdx + c
⇒ y sec x = tan x + c

(B) अवकल समीकरण \(\frac{dy}{dx}\) + y tan x = sin x को हल कीजिए।
हल:
प्रश्न क्रमांक 1 (A) की भाँति हल करें।

प्रश्न 2.
अवकल समीकरण \(\frac { dy }{ dx } \) = \(\frac { \sqrt { 1-y^{ 2 } } }{ \sqrt { 1-x^{ 2 } } } \) को हल कीजिए।
हल:
\(\frac{dy}{dx}\) = \(\frac { \sqrt { 1-y^{ 2 } } }{ \sqrt { 1-x^{ 2 } } } \)

प्रश्न 3.
अवकल समीकरण 3x²dy = (3xy + y²)dx को हल कीजिए।
हल:
दिया गया अवकल समीकरण है:
3x² dy = (3xy + y²) dx
⇒ \(\frac{dy}{dx}\) = \(\frac { 3xy+y^{ 2 } }{ 3x^{ 2 } } \) ……….. (1)
माना y = vx
⇒ \(\frac{dy}{dx}\) = v + x \(\frac{dv}{dx}\)
समी. (1) में मान रखने पर,

प्रश्न 4.
अवकल समीकरण हल कीजिए –
(1 + x)² \(\frac{dy}{dx}\) + 2xy = 4x²
हल:
दिया गया अवकल समीकरण है:
(1 + x²) \(\frac{dy}{dx}\) + 2xy = 4x²
⇒ \(\frac{dy}{dx}\) + \(\frac { 2xy }{ 1+x^{ 2 } } \) = \(\frac { 4x^{ 2 } }{ 1+x^{ 2 } } \)
इसकी तुलना रैखिक अवकल समीकरण \(\frac{dy}{dx}\) + Py = Q से करने पर,

अतः अभीष्ट हल होगा:

प्रश्न 5.
अवकल समीकरण (1 + x²) \(\frac{dy}{dx}\) + 2xy = cos x को हल कीजिए।
हल:
दिया है:
(1 + x²) \(\frac{dy}{dx}\) + 2xy = cos x
⇒ \(\frac{dy}{dx}\) + \(\frac { 2x }{ (1+x^{ 2 }) } \). y = \(\frac { cosx }{ 1+x^{ 2 } } \) …………. (1)
P = \(\frac { 2x }{ (1+x^{ 2 }) } \), Q = \(\frac { cosx }{ 1+x^{ 2 } } \)

अतः अवकल समी. (1) का अभीष्ट हल है
y.(I.F.) = ∫Q.(I.F.) dx + c
⇒ y(1 + x²) = ∫\(\frac { cosx }{ 1+x^{ 2 } } \) dx + c
⇒ y(1 + x²) = ∫cos x dx + c
⇒ y(1 + x²) = sin x + c

प्रश्न 6.
किसी वस्तु के बनाने का सीमांत लागत मूल्य c'(x) = \(\frac{dc}{dx}\) = 2 + 0.15 x समीकरण से दिया जाता है। इस वस्तु के बनाने पर कुल लागत मूल्य c(x) ज्ञात कीजिए। (दिया है: c(0) = 100)
हल:
दिया है:
समीकरण c'(x) = \(\frac{dc}{dx}\) = 2 + 0.15 x
समाकलन करने पर,
∫c'(x) dx = ∫(2 + 0.15 x) dx
c(x) = 2x + 0.15 \(\frac { x^{ 2 } }{ 2 } \) + A ………. (1)
अब यदि x = 0 तो
c(0) = 2 × 0 + \(\frac{0.15}{2}\) × o² + A
⇒ c(0) = A [∵c(0) = 100]
∴ A = 100,
समी. (1) में मान रखने पर,
c(x) = 2x + 0.075 x² + 100

प्रश्न 7.
अवकल समीकरण x\(\sqrt { 1+y^{ 2 } } \) dx + y \(\sqrt { 1+x^{ 2 } } \) dy = 0 को हल कीजिए।
हल:
दिया गया अवकल समीकरण है:
x\(\sqrt { 1+y^{ 2 } } \) dx + y \(\sqrt { 1+x^{ 2 } } \) dy = 0
⇒ \(\frac { y }{ \sqrt { 1+y^{ 2 } } } \) dy = – \(\frac { x }{ \sqrt { 1+x^{ 2 } } } \)
दोनों पक्षों का समाकलन करने पर,
∫\(\frac { y }{ \sqrt { 1+y^{ 2 } } } \) dy = -∫\(\frac { x }{ \sqrt { 1+x^{ 2 } } } \) dx + c
⇒ \(\sqrt { 1+y^{ 2 } } \) = – \(\sqrt { 1+x^{ 2 } } \) + c
⇒ \(\sqrt { 1+x^{ 2 } } \) + \(\sqrt { 1+y^{ 2 } } \) = c

प्रश्न 8.
अवकल समीकरण हल कीजिए –
(x + y + 1) \(\frac{dy}{dx}\) = 1?
हल:
दिया गया अवकल समीकरण है:
(x + y + 1) \(\frac{dy}{dx}\) = 1
⇒ \(\frac{dx}{dy}\) = x + y + 1
⇒ \(\frac{dx}{dy}\) – x = y + 1
यह y के सापेक्ष x का अवकल समीकरण है।
इसकी तुलना रैखिक अवकल समीकरण \(\frac{dx}{dy}\) + Px = Q से करने पर,
P = – 1 तथा Q = y + 1

अत: अभीष्ट हल होगा:

प्रश्न 9.
अवकल समीकरण sec² x tan y dx + sec² y tan xdy = 0 को हल कीजिए।
हल:
दिया है:
sec² xtan ydx + sec² ytan xdy = 0
⇒ sec² tan x dy = -sec² tan ydx
⇒ \(\frac { sec^{ 2 }y }{ tany } \) dy = \(\frac { sec^{ 2 }x }{ tanx } \) dx
⇒ ∫\(\frac { sec^{ 2 }y }{ tany } \) dy = -∫\(\frac { sec^{ 2 }x }{ tanx } \) dx + c
⇒ log y = – log x + log c
⇒ log x + log y = log c
⇒ log xy = log c
⇒ xy = c

प्रश्न 10.
अवकल समीकरण हल कीजिए।
\(\frac{dy}{dx}\) = y tan x – 2 sin x?
हल:
दिया गया अवकल समीकरण
\(\frac{dy}{dx}\) – y tan x = -2 sin x
इसकी तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
P = – tan x, Q = – 2 sin x
अभीष्ट हल y.(I.F.) = ∫Q.I.F.dx + c
⇒ y cos x = -2∫sin x cos x dx + c
⇒ y cos x= -∫sin 2x dx + c
⇒ y cos x = \(\frac{cos 2x}{2}\) + c

प्रश्न 11.
अवकल समीकरण \(\frac{dy}{dx}\) + 2y = 4x को हल कीजिए।
हल:
\(\frac{dy}{dx}\) + 2y = 4x
इसकी तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
P = 2,Q = 4x

⇒ y = 2x – 1 + c.e^-2x

प्रश्न 12.
अवकल समीकरण cos²x \(\frac{dy}{dx}\) + y = 2 को हल कीजिए।
हल:
cos² x \(\frac{dy}{dx}\) + y = 2
⇒ \(\frac{dy}{dx}\) + sec² x.y = 2 sec² x
इसकी तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
P = sec² x, Q = 2 sec² x

प्रश्न 13.
अवकल समीकरण cos x \(\frac { dy }{ dx } \) + y = sin x को हल कीजिये।
हल:
cos x \(\frac{dy}{dx}\) + y = sin x
⇒ \(\frac{dy}{dx}\) + secx. y = tan x
इसकी तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
P = sec x, Q = tan x

प्रश्न 14.
अवकल समीकरण (1 + y²) dx = (tan^-1 y – x) dy को हल कीजिए। (CBSE 2015)
हल:
दिया गया अवकल समीकरण है –
(1 + y²) dx = (tan^-1y – x) dy
\(\frac{dx}{dy}\) + \(\frac { x }{ 1+y^{ 2 } } \) = \(\frac { tan^{ -1 }y }{ 1+y^{ 2 } } \)

प्रश्न 15.
अवकल समीकरण (1 + y²) + (x – e^{tan-1 y}) \(\frac{dy}{dx}\) = 0 को हल कीजिए। (CBSE 2016)
हल:
दिया गया अवकल समीकरण है –
(1 + y²) + (x – e^{tan-1 y}) \(\frac{dy}{dx}\) = 0

समी (1) की तुलना \(\frac{dx}{dy}\) + Px = Q से करने पर,

अतः समीकरण का हल होगा –
x.I.F = ∫I.F. × Qdy

tan^-1y = t रखने पर,
\(\frac{d}{dy}\) tan^-1 y = \(\frac{dt}{dy}\)
⇒ \(\frac { 1 }{ 1+y^{ 2 } } \) dy = dt
∴ x.e^{tan-1 y} = ∫e^t.e^t dt = ∫e^2t dt
⇒ x.e^{tan-1 y} = \(\frac{1}{2}\) e^2t + c
⇒ x.e^{tan-1 y} = \(\frac{1}{2}\) e^{2 tan-1 y} + c

प्रश्न 16.
अवकल समीकरण (1 + x²) \(\frac{dy}{dx}\) + 2xy = \(\frac { 1 }{ 1+x^{ 2 } } \) को हल कीजिए, जहाँ y = 0 तथा x = 1(NCERT)
हल:
दिया गया अवकल समीकरण है –
(1 + x²) \(\frac{dy}{dx}\) + 2xy = \(\frac { 1 }{ 1+x^{ 2 } } \)
⇒ \(\frac{dy}{dx}\) + \(\frac { 2xy }{ 1+x^{ 2 } } \) = \(\frac { 1 }{ 1+x^{ 2 } } \) )²
समी. (1) की तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
यहाँ P = \(\frac { 2x }{ 1+x^{ 2 } } \), Q = \(\frac { 1 }{ 1+x^{ 2 } } \) )²

अतः समीकरण का हल होगा –
y.I.F. = ∫I.F. × Qdx

y = 0 तथा x = 1 रखने पर,
0(1+1)² = tan^-1 + c
⇒ 0 = \(\frac { \pi }{ 4 } \) + c
⇒ c = – \(\frac { \pi }{ 4 } \)
c का मान समी. (2) में रखने पर, अवकल समीकरण का हल होगा
y.(1 + x²) = tan^-1 x – \(\frac { \pi }{ 4 } \)

प्रश्न 17.
अवकल समीकरण \(\frac{dy}{dx}\) + cot x = 4x cosec x का एक विशिष्ट हल ज्ञात कीजिए। दिया गया है कि – y = 0 तथा x = \(\frac { \pi }{ 2 } \) (NCERT; CBSE 2012)
हल:
दिया गया अवकल समीकरण है –
\(\frac{dy}{dx}\) + y cot x = 4x cosec x
समी. (1) की तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
P = cot x, Q = 4x cosecx

अतः समीकरण का हल होगा –
y.I.F. = ∫1.F. × Qdx
⇒ y.sin x = ∫sin x × 4x cosec x dx
⇒ y sin x = 4∫\(\frac { xsinx }{ sinx } \) dx
⇒ y sin x = 4∫xdx
⇒ y sin x = \(\frac { 4x^{ 2 } }{ 2 } \) + c
⇒ y sin x = 2x² + c
x = \(\frac { \pi }{ 2 } \) तथा y = 0 रखने पर,
0(sin \(\frac { \pi }{ 2 } \) ) = 2( \(\frac { \pi }{ 2 } \) )² + c
⇒ 0 = \(\frac { 2\pi ^{ 2 } }{ 4 } \) + c
∴ c = \(\frac { -\pi ^{ 2 } }{ 2 } \)
c का मान समी. (2) में रखने पर, अवकल समीकरण का अभीष्ट हल होगा –
y.sinx = 2x² – \(\frac { \pi ^{ 2 } }{ 2 } \)

MP Board Class 12 Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 6 Linear Inequalities

Linear Inequalities Important Questions

Linear Inequalities Objective Type Questions

(A) Choose the correct option :

Question 1.
If x is a real number, then the solution of inequality 3x + 1 < 5x + 7 is :
(a) (- ∞, 3)
(b) (- 3,∞)
(c) (3, ∞)
(d) None of these.
Answer:
(b) (- 3,∞)

Question 2.
If x is a real number, then the solution of inequality \(\frac { 1 }{ 2 }\) (3x – 1) ≥ \(\frac { 1 }{ 3 }\)(4x + 3) – 1 is :
(a) (- ∞, 3)
(b) (3, ∞)
(c) (3, ∞)
(d) None of these.
Answer:
(c) (3, ∞)

Question 3.
The graph of x ≤ 2 and y ≥ 2 is situated in the :
(a) First and second quadrant
(b) Second and Third quadrant
(c) First and third quadrant
(d) None of these.
Answer:
(a) First and second quadrant

Question 4.
In linear programming problems, the objective function will be :
(a) One of the constraints
(b) A linear function which gives optimum solution
(c) Relation between varivable
(d) None of these.
Answer:
(b) A linear function which gives optimum solution

Question 5.
In linear constraints the maximum value of objective function will be :
(a) At the centre of feasible region
(b) At (0, 0)
(c) At one of the vertices of the feasible region
(d) At the vertex which is situated at maximum distance from (0, 0)
Answer:
(c) At one of the vertices of the feasible region

Question 6.
Which word is not used in linear programming problems :
(a) Slack variable
(b) Objective function
(c) Concave region
(d) Feasible solution.
Answer:
(a) Slack variable

Question 7.
The minimum value of p = 6x + 16y when constraints are x ≤ 40 and y ≥ 20 and x, y ≥ 0 is :
(a) 240
(b) 320
(c) 0
(d) None of these.
Answer:
(b) 320

Question 8.
At which point the value of 3x + 2y is maximum under the constraints x + y ≤ 2, x ≥ 0 y ≥ 0 :
(a) (0, 0)
(b) (1.5, 1.5)
(c) (2, 0)
(d) (0, 2).
Answer:
(c) (2, 0)

Question 9.
Under constraints x ≥ 0, y ≥ 0, x + y ≤ 4, maximum value of P = 3x + y is :
(a) 8
(b) 12
(c) 6
(d) 10.
Answer:
(b) 12

Question 10.
Under constraints x – 2y ≥ 6, x + 2y ≥ 0, x ≤ 6 , maximum value of P = x + 3y is :
(a) 16
(b) 17
(c) 18
(d) 19.
Answer:
(c) 18

Question 11.
Feasible solution of linear programming is in :
(a) Second quadrant only
(b) First and third quadrant
(c) First and second quadrant
(d) First quadrant only.
Answer:
(d) First quadrant only.

Question 12.
If x is a real number and |x| < 4, then :
(a) x ≥ 4
(b) – 4 < x < 4
(c) x ≤ – 4
(d) – 4 ≤ x ≤ 4
Answer:
(b) – 4 < x < 4

Question 13.
Solution of inequality 3x – 2 < 0 will be :
(a) [3, ∞]
(b) [- ∞, \(\frac { 2 }{ 3 }\)]
(c) [3, 2]
(d) [2, 3]
Answer:
(b) [- ∞, \(\frac { 2 }{ 3 }\)]

Question 14.
The set of solutions of inequality 4x – 12 ≥ 0 is :
(a) (4, 2)
(b) [4, 12]
(c) [3, ∞]
(d) [3, ∞]
Answer:
(c) [3, ∞]

Question 15.
The solution of inequality |4x – 3| < 27 is :
(a) (- 6, \(\frac { 15 }{ 2 }\))
(b) [ – 6, \(\frac { 15 }{ 2 }\) ]
(c) [ – 6, \(\frac { 15 }{ 2 }\) )
(d) ( – 6, \(\frac { 15 }{ 2 }\) ]
Answer:
(a) (- 6, \(\frac { 15 }{ 2 }\))

Question 16.
Solution of inequality x > 2 is :

Answer:

Question 17.
Sawant has a space to store at most 13 boxes of apples and oranges. If he bought x boxes of oranges and y boxes of apples, then correct inequality will be :
(a) x + y < 13
(b) x + y > 13
(c) x + y < 13
(d) x + y = 13.
Answer:
(c) x + y < 13

Question 18.
In which point the maximum of objective function P = 10x + 6y in the following points:
(a) (0, 0)
(b) (8, 4)
(c) (10, 0)
(d) (0, 0).
Answer:
(b) (8, 4)

Question 19.
The graph of inequality x +y > 2 is :

Answer:

Question 20.
The solution of inequality 3x – 7 ≥ x + 1 is :
(a) [4, ∞]
(b) (4, ∞]
(c) (4, ∞)
(d) [4, ∞).
Answer:
(d) [4, ∞).

Question 21.
The solution of inequality 3x – 6 > 0; 2x – 6 > 0 is :
(a) [3, ∞]
(b) (3, ∞)
(c) [2, ∞)
(d) (2, ∞).
Answer:
(b) (3, ∞)

Question 22.
The solution of inequality 2x + 6 = 0; 4x – 8 < 0 is :
(a) [- 3, ∞)
(b) (3, 2)
(c) [- 3, 2)
(d)[- 3, 2].
Answer:
(c) [- 3, 2)

Question 23.
The set of solution of inequality x ≥ 0 is :
(a) All points of X – axis and first, fourth quadrants
(b) All points of first and fourth quadrants
(c) All points of first and second quadrants
(d) All points of Y – axis and first, fourth quadrants.
Answer:
(d) All points of Y – axis and first, fourth quadrants.

Question 24.
Variables of the objective function of linear Programming problem are :
(a) Negative
(b) Zero or Negative
(c) Zero
(d) Zero or Positive.
Answer:
(d) Zero or Positive.

(B) Match the following :

Answer:

1. (c)
2. (d)
3. (a)
4. (f)
5. (e)

(C) Fill in the blanks :

1. The maximum or minimum value of objective function is called ……………….
2. The graph of x ≥ 0 is situated in the ………………. quadrant.
3. The graph of y ≤ 0 is situated in the ………………. quadrant.
4. Solution of inequality 3x + 1 < 5x + 7 is (where x is an integer) ……………….
5. Solution of inequality 3x – 4 < 5 is (where x is an integer) ……………….
6. Solution of inequality 4x + 3 > – 13 is (where x is a real number) ……………….
7. Solution of inequality 7x – 2 < 5x + 4 is (where x is a real number) ……………….
8. Solution of inequality 20x < 90 is (where x is a natural number) ……………….
9. Solution of inequality 3x – 2 < x + 4 is ……………….
10. The inequality of one variable is ……………….
11. The set of solutions of y > 0 are in the ………………. and ………………. quadrant.
12. x ≥ 2 and y ≥ 2 will be situated in the graph ……………….
13. The domain of inequality 3x – 15 ≤ 0 is ……………….
14. A function whose maximum and minimum value is to be found subject to the given constrains is known as ……………….

Answer:

1. Optimum value
2. First and fourth
3. Third and fourth
4. (- 3, ∞)
5. {… – 3, – 2, -1, 0, 1, 2}
6. (- 4, ∞)
7. (- ∞, 3)
8. {1, 2, 3, 4}
9. (- ∞, 3)
10. ax + b ≥ c or ax + b ≤ c; where a ≠ 0
11. First and second
12. First and second quadrant
13. (- ∞, \(\frac { 15 }{ 3 }\))
14. Objective function

(D) Write true / false :

1. If the variable x is such that its value lies between two fixed point a and b, then {x : a < x < b} is called a closed interval.
2. The function whose maximum or minimum value is to be found is called objective function.
3. The set of values of the variable satisfying all constraint is called feasible solution of the problem.
4. The process of doing certain specified steps in a given order is called programming.
5. The set {x : a < x < b} which consists of both a and b is called open interval.
6. The solution of inequality 6x – 30 ≥ 0 will be x ≤ 5
7. The solution of inequality – 2x + 7 < – 13 will be (10, ∞).
8. The linear inequality of one variable is ax + b = 0; a ≠ 0, ∀a, b ∈
9. R. The line inequality of two variables is ax + by + c = 0; a ≠ 0, b ≠ 0, ∀a, b, c ∈ R.
10. The graph of inequality x > 0 is :
    
11. The graph of inequality y ≤ 0 is :
    
12. The solution of inequality 5x – 30 ≤ 0 is :
    

Answer:

1. False
2. True
3. True
4. True
5. False
6. False
7. True
8. True
9. True
10. True
11. True
12. False.

(E) Write answer in one word / sentence :

1. In which quadrant lies the solution of x ≥ 0 and y ≥ 0?
2. In which quadrant lies the solution of x ≤ 2 and y ≥ 2?
3. Write the solution of 3(2 – x) ≥ 2(1 – x) .
4. Write the solution of \(\frac { x }{ 3 }\) > \(\frac { x }{ 2 }\) + 1.
5. Write the solution of \(\frac { x – 4 }{ x + 2 }\) ≤ 2.

Answer:

1. First quadrant
2. First and second quadrant
3. (- ∞, 4]
4. (- ∞, – 6)
5. (- ∞, – 8]∪(- 2, ∞)

Linear Inequalities Long Answer Type Questions

Question 1.
Draw the graph of inequality 2x + y ≥ 6 in two – dimensional plane. (NCERT)
Solution:
Given inequality is 2 x + y ≥ 6
Consider this as an equation 2x + y = 6.
For this equation following values of x and y are given in the table :

Plot the graph using the above table. Take a point (0, 0) and put in given inequality,
2x + y ≥ 6
⇒ 2 x 0 + 0 ≥ 6
⇒ 0 ≥ 6 Which is false.
Hence, shaded portion will be opposite of origin.

∴ The shaded region represents the inequality 2x + y ≥ 6.

Question 2.
Draw the graph of inequality 3x + 4y ≤ 12 in two – dimensional plane. (NCERT)
Solution:
Given inequality : 3x + 4y = 12
Consider this as an equation 3x + 4y = 12.
For this equation the following values of x and y are given in the table :

Plot the above points in a graph paper and join them.
Put (0, 0) in given inequality,
⇒ 3x + 4y ≤ 12
⇒ 3 x 0 + 4 x 0 ≤ 12
⇒ 0 ≤ 12
Which is true.
Hence, shaded portion will be towards origin as shown in figure.

∴ The shaded region represents the inequality 3x + 4y ≤ 12

Question 3.
Draw the graph of inequality y + 8 ≥ 2x in two – dimensional plane. (NCERT)
Solution:
Given inequality : y+8 > 2x Consider this as an equation 2x – y = 8
For this equation following values of x and y are given in the table :

Plot the above points in graph paper and join them.
Put (0, 0) in the given inequality,
y + 8 ≥ 2x
⇒ 0 + 8 ≥ 2 x 0
⇒ 8 ≥ 0 Which is true.
Hence, the shaded portion will be towards origin.

∴ Shaded portion represents the inequality y + 8 ≥ 2x.

Draw the graph of following inequality in two – dimensional plane :

Question 4.
2x – 3y > 6.
Solution:
Given inequality is :
2x – 3y > 6
Consider this as an equation 2x – 3y = 6
Following table is prepared for different values of x and y :

Plot the above points on xy – plane and join them.
Put x = 0, y = 0 in given inequality,
2x – 3y > 6
⇒ 2 x 0 – 3 x 0 > 6
⇒ 0 > 6 Which is false.
Hence, the shaded portion will be opposite of origin.

∴The shaded region represents the inequality 2x – 3y > 6

Question 5.
– 3x + 2y ≥ – 6.
Solution:
Given inequality is : – 3x + 2y ≥ – 6.
Consider this as an equation – 3x + 2y = – 6.
Following table is prepared for different values of x and y :

Plotting the above points on xy – plane and join them. Put x = 0, y = 0 in given inequality,
– 3x + 2y ≥ – 6
⇒ – 3(0) + 2(0) ≥ – 6
⇒ 0 ≥ – 6 Which is true.
∴ Shaded portion will be towards origin.

Hence, the shaded portion represents the inequality – 3x + 2y ≥ – 6.

Question 6.
3y – 5x < 30.
Solution:
Given inequality: y – 5x < 30.
Consider this as an equation 3y – 5x = 30.
Following table is prepared for different values of x and y :

Ploting the above points on xy – plane and join them.
Put x = 0, y = 0 in given inequality,
3y – 5x < 30
⇒ 3(0) – 5(0) < 30
⇒ 0 < 30
Which is true.
Hence, the shaded region will be towards origin.

∴ Shaded region represents the inequality 3y – 5x < 30

Question 7.
2x + y ≥ 6, 3x + 4y ≤ 12. (NCERT)
Solution:
Given inequalities are :
2x + y ≥ 6
and 3x + 4y ≤ 12
For the equation 2x+y = 6.
Following table is prepared for different values of x and y :

Plotting the above points on xy – plane to get straight line.
Put x = 0, y = 0 in the inequality,
2x + y ≥ 6
⇒ 2 x 0 + 0 ≥ 6
⇒ 0 ≥ 6
Which is false.
∴ The shaded region will be opposite of origin.
For the equation 3x + 4y = 12.
Following table is prepared for different values of x and y :

Plotting the above points on xy – plane to get straight line.
Put x = 0, y = 0 in given inequality,
3x + 4y ≤ 12
⇒ 3 x (0) + 4(0) ≤ 12
⇒ 0 ≤ 12
Which is true.
Hence, the shaded portion will be towards origin.

∴ The common shaded region is the required solution of the given inequalities.

Question 8.
Solve the following inequalities graphically in two – dimensional plane : x + y ≥ 4, 2x – y >0. (NCERT)
Solution:
Given inequalities are :
x + y ≥ 4
and 2x – y > 0
For the equation x + y = 4, following table is prepared for different values of x and y:

Plotting the above points on xy – plane to get straight line.
Put x = 0, y = 0 in the inequality,
x + y ≥ 4
⇒ 0 + 0 ≥ 4
⇒ 0 ≥ 4
Which is false.
Hence, the shaded region will be opposite of origin.
Preparing table for different value of x and y of the equation 2x – y = 0 :

Plotting the above points in xy – plane to get straight line.
Put x = 1 and y = 0 in given inequality,
2x – y > 0
⇒ 2(1) – 0 > 0
⇒ 2 > 0
Which is true.
Hence, the shaded region will be towards of point (1,0).

The common shaded region is the required solution of the given inequalities.

Question 9.
Solve the following inequalities graphically in two – dimensional plane :
5x + 4y ≤ 20, x ≥ 1, y ≥ 2 (NCERT)
Solution:
The given inequalities are :
5x + 4y ≤ 20
x ≥ 1
y ≥ 2
Prepared table for different values of x and y of the equation 5x +4y = 20 :

Plotting the points of the above table on xy – plane to get straight line.
Then, put x = 0, y = 0 in the inequality,
5x + 4y ≤ 20
⇒ 0 + 0 ≤ 20
⇒ 0 < 20
Which is true.
∴Shaded region will be towards origin.
For equation x + O.y = 1 to get different values of x and y are in the table :

Plotting the points from table on xy plane to get straight line.
Putting x = 0, y = 0 in the inequality,
x ≥ 1
⇒ 0 ≥ 1
Which is false.
Hence, the shaded region will be opposite side of origin.
For equation 0.x + y = 2, the different values of x and y are given in the table :

Plotting the points on xy plane to get straight line.
Then, put x = 0, y = 0 (0,0) in the inequality,
y ≥ 2
⇒ 0 ≥ 2
Which is false.
Hence, the shaded region will be opposite side of origin.

∴ The common shaded region is the required solution of the given inequalities.

Question 10.
Solve the following inequalities graphically :
3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0. (NCERT)
Solution:
The given inequalities are :
3x + 4y ≤ 60
x + 3y ≤ 30
x ≥ 0
y ≥ 0
Prepared table for values of x and y of equation 3x + 4y = 60 :

Plotting the above points on xy – plane to get straight line.
Then, put x = 0, y = 0 in the inequality,
3x + 4y ≤ 60
⇒ 0 + 0 ≤ 60
⇒ 0 ≤ 60
Which is true.
Hence, the shaded portion will be towards origin.
Prepared table for values of x and y of equation x + 3y = 30:

Plotting the above points on xy – plane to get straight line.
Then, put x = 0, y = 0 in the inequality,
x + 3y ≤ 30
⇒ 0 + 0 ≤ 30
⇒ 0 ≤ 30
Which is true.
Hence, the shaded region will be towards origin.
The graph of x = 0 is y – axis and graph of y = 0 is x – axis.

The common shaded region is the required solution set of the given inequalities.

MP Board Class 11th Maths Important Questions

MP Board Class 12th Maths Important Questions Chapter 8 समाकलनों के अनुप्रयोग

समाकलनों के अनुप्रयोग Important Questions

समाकलनों के अनुप्रयोग वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प चुनकर लिखिए –

(a) π
(b) \(\frac { \pi }{ 2 } \)
(c) \(\frac { \pi }{ 4 } \)
(d) \(\frac { -\pi }{ 4 } \)
उत्तर:
(c) \(\frac { \pi }{ 4 } \)

प्रश्न 2.
दीर्घवृत्त \(\frac { x^{ 2 } }{ y^{ 2 } } \) + \(\frac { y^{ 2 } }{ b^{ 2 } } \) = 1 के प्रथम चतुर्थांश का क्षेत्रफल होगा –
(a) πab
(b) \(\frac{1}{2}\) πab
(c) \(\frac{1}{4}\) πab
(d) \(\frac{1}{8}\) πab
उत्तर:
(c) \(\frac{1}{4}\) πab

प्रश्न 3.
\(\int_{2}^{4} x^{3} d x\) का मान है –
(a) 60
(b) 50
(c) 70
(d) 256.
उत्तर:
(a) 60

प्रश्न 4.
\(\int_{0}^{2 a} f(x) d x=0\) होगा यदि –
(a) f (2a – x) = f (x)
(b) f (2a – x) = – f (x)
(c) f (x) सम फलन हो
(d) f (x) विषम फलन हो
उत्तर:
(b) f (2a – x) = – f (x)

प्रश्न 5.
\(\int_{0}^{2 \pi}|\sin x| d x\) बराबर है –
(a) 2
(b) \(\sqrt{3}\)
(c) 4
(d) 0.
उत्तर:
(c) 4

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिये –

उत्तर:

1. 0
2. 0
3. 17
4. 0
5. \(\frac { \pi }{ 4 } \)
6. πa²

प्रश्न 3.
निम्न कथनों में सत्य/असत्य बताइए –

उत्तर:

1. सत्य
2. सत्य
3. असत्य
4. असत्य
5. सत्य
6. सत्य

प्रश्न 4.
सही जोड़ी बनाइए –

उत्तर:

1. (b)
2. (a)
3. (d)
4. (c)
5. (f)
6. (e)

प्रश्न 5.
एक शब्द/वाक्य में उत्तर दीजिए –

उत्तर:

1. 1
2. \(\frac { -\pi }{ 2 } \) log 2
3. 2
4. \(\frac { \pi }{ 2 } \)
5. -4
6. \(\frac { \pi }{ 12 } \)

समाकलनों के अनुप्रयोग दीर्घ उत्तरीय प्रश्न – II

प्रश्न 1.
दो वक्रों x² = 8y तथा y² = 8x के बीच घिरे क्षेत्र का क्षेत्रफल समाकल विधि से ज्ञात कीजिए।
हल:
दिये हुए परवलयों के समीकरण हैं:
x² = 8y ………… (1)
तथा y² = 8x ……………….. (2)
समी. (1) और (2) को हल करने पर,
( \(\frac { x^{ 2 } }{ 8 } \) )² = 8x
⇒ \(\frac { x^{ 4 } }{ 64 } \) = 8x
⇒ x⁴ = 64 × 8x = 8 × 8 × 8 × x
⇒ x³ = (8)³ या x = 8
समी. (1) से,

(8)² = 8y ⇒ y = 8
A के निर्देशांक (8, 8) हैं।
परवलय बिन्दु तथा A पर मिलते हैं।
∴ उभयनिष्ठ क्षेत्रफल = क्षेत्रफल OBALO – क्षेत्रफल OCALO

प्रश्न 2.
चक्र y = cosx, X – अक्ष और x = 0 तथा x = 2π से घिरे क्षेत्र का क्षेत्रफल समाकलन विधि से ज्ञात कीजिए।
हल:
y = f (x) = cos x
जब x ∈ [0, \(\frac { \pi }{ 2 } \) ], cos x ≥ 0;
जब x ∈ [ \(\frac { \pi }{ 2 } \), \(\frac { 3\pi }{ 2 } \) ], cos x ≤ 0
जब x ∈ [ \(\frac { 3\pi }{ 2 } \), 2π], cos x ≥ 0

अभीष्ट क्षेत्रफल = OAB का क्षेत्रफल + BCD का क्षेत्रफल + DEF का क्षेत्रफल

प्रश्न 3.
वक्र y² = 4x तथा रेखा y = 2x के बीच का क्षेत्रफल ज्ञात कीजिए। (समाकलन विधि से)
हल:
दिया गया वक्र और रेखा है –
y² = 4x और y = 2x
y² = 4x में y = 2x रखने पर,

(2x)² = 4x
⇒ 4x² = 4x
⇒ x² – x = 0
⇒ x(x – 1) = 0
∴ x = 0 या x = 1
तब y = 2x = 0 या y = 2x = 2 × 1 = 2
अत: वक्र y² = 4x तथा रेखा y = 2x के प्रतिच्छेद बिन्दु O(0, 0) एवं (1, 2) होंगे।
अत: अभीष्ट छायांकित क्षेत्र
= क्षेत्र (OPAMO) – क्षेत्र (OQAMO)

जहाँ y_c = वक्र y² = 4x तथा y_s = रेखा y = 2x के लिए,

= \(\frac{4}{3}\) [1 – 0] – \(\frac{2}{2}\) [1 – 0]
= \(\frac{4}{3}\) – 1 = \(\frac{1}{3}\) वर्ग इकाई।

प्रश्न 4.
समाकलन विधि द्वारा वक्रों x² = 4y तथा x = 4y – 2 के अंतर्गत क्षेत्रफल ज्ञात कीजिए।
हल:
प्रश्न क्रमांक 3 की भाँति हल करें।

प्रश्न 5.
समाकलन विधि से रेखाओं |x| + |y| = a से घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए।
हल:
दी गयी रेखाओं |x| + |y| = a द्वारा निरूपित रेखायें निम्न होंगी –
x + y = a ……………… (1)
– x – y = a ………………. (2)
x – y = a …………….. (3)
– x + y = a …………….. (4)
⇒ \(\frac{x}{a}\) + \(\frac{y}{a}\) = 1, \(\frac{x}{-a}\) + \(\frac{y}{-a}\) = 1
⇒ \(\frac{x}{a}\) + \(\frac{y}{-a}\) = 1, \(\frac{x}{-a}\) + \(\frac{y}{a}\) = 1

उक्त रेखाओं के ग्राफ से स्पष्ट है, कि उक्त रेखाएँ क्रमशः PQ, RS, PS और QR रेखाओं द्वारा निरूपित हैं।
अतः उक्त रेखाओं द्वारा घिरा क्षेत्रफल
= 4 × ∆OPQ का क्षेत्रफल

= 4 × \(\frac { a^{ 2 } }{ 2 } \) = 2a² वर्ग इकाई।

प्रश्न 6.
वृत्त x² + y² का क्षेत्रफल समाकलन विधि से ज्ञात कीजिए।
हल:
दिया है:
वृत्त x² + y² = a²
⇒ y² = a² – x²
⇒ y = \(\sqrt { a^{ 2 }-x^{ 2 } } \)
वृत्त X – अक्ष तथा Y – अक्ष के सापेक्ष सममित आकृति है अतः
वृत्त का क्षेत्रफल = 4 × ∆ABC का क्षेत्रफल

प्रश्न 7.
वृत्त x² + y² = 25 का क्षेत्रफल समाकलन विधि से ज्ञात कीजिए।
हल:
प्रश्न क्रमांक 6 की भाँति हल करें। [संकेत – a = 5 रखें।]

प्रश्न 8.
वक्रों y² = 4ax तथा x² = 4ay, a > 0 के अन्तर्गत क्षेत्रफल ज्ञात कीजिए। (समाकलन विधि द्वारा)
हल:
दिये हुए परवलयों के समीकरण हैं:
y² = 4ax ………….. (1)
तथा x² = 4ay ……………. (2)
समी. (1) और (2) को हल करने पर,

\(\frac { x^{ 4 } }{ 16a^{ 2 } } \) = 4ax
⇒ x⁴ – 64a³x = 0
⇒ x(x³ – 64a³) = 0
अतः या तो x = 0 अथवा x³ – 64a³ = 0
जब x³ – 64a³ = 0
जब x³ – 64a³ = 0 तो x = 4a
जब x = 0 तो y = 0
और जब x = 4a तो y = 4a
अत: दोनों परवलय बिन्दुओं (0, 0), (4a, 4a) पर मिलते हैं।
∴ उभयनिष्ठ क्षेत्रफल = क्षेत्रफल OBALO – क्षेत्रफल OCALO

प्रश्न 9.
वक्रों y² = 4x और x² = 4y के बीच का क्षेत्रफल समाकलन विधि से ज्ञात कीजिए।
हल:
दिए हुए वक्रों के समीकरण हैं:
y² = 4x ……………… (1)
xc = 4y ………………. (2)
समी. (2) से y का मान समी. (1) में रखने पर,

⇒ \(\frac { x^{ 4 } }{ 16 } \) = 4x
⇒ x⁴ – 64x = 0
⇒ x(x³ – 64) = 0
⇒ x = 0 या x³ = 64
⇒ x = 0 या x = 4
∴ बिन्दु 0 पर x = 0 तथा बिन्दु M पर x = 4
अतः अभीष्ट क्षेत्रफल = क्षेत्र. OAPBO
= क्षेत्र. OAPMO – क्षेत्र. OBPMO

प्रश्न 10.
वक्र y² = x और x² = y के बीच का क्षेत्रफल समाकलन विधि से ज्ञात कीजिए।
हल:
प्रश्न क्रमांक 9 की भाँति हल करें।

प्रश्न 11.
दीर्घवृत्त \(\frac { x^{ 2 } }{ a^{ 2 } } \) + \(\frac { y^{ 2 } }{ b^{ 2 } } \) = 1 से घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए। (समाकलन विधि द्वारा)
हल:
दिया हुआ वक्र X – अक्ष और Y – अक्ष के सापेक्ष सममित है। अतएव सम्पूर्ण क्षेत्रफल क्षेत्र OAB का 4 गुना होगा।
यहाँ y = \(\frac{b}{a}\) \(\sqrt { a^{ 2 }-x^{ 2 } } \)
वक्र पर बिन्दु P लिया। तब पट्टी PM का क्षेत्रफल ydx होगा।
x का मान O पर शून्य तथा A पर x = a है।

प्रश्न 12.
परवलय y² = 4ax तथा इसकी नाभिलम्ब जीवा से घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए। (समाकलन विधि द्वारा)
हल:
परवलय का समीकरण है:
y² = 4ax
⇒ y = ± 2\(\sqrt{ax}\)
LSL’ नाभिलम्ब है तथा S(a, 0) और 0(0, 0) है।
अतः अभीष्ट क्षेत्रफल = 2 × क्षेत्रफल OSL

प्रश्न 13.
परवलय y² = 4ax और रेखा y = mx के बीच घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए।
हल:
परवलय का समी. y² = 4ax ……………. (1)
रेखा का समी. y = mx …………….. (2)
मूलबिन्दु O(0, 0) है। P परवलय और रेखा का प्रतिच्छेद बिन्दु है। समी. (1) और (2) को हल करने पर,
y² = 4ax
⇒ (mx)² = 4ax
⇒ m²x² – 4ax = 0
⇒ x(m²x – 4a) = 0
∴ x = 0, x = \(\frac { 4a }{ m^{ 2 } } \)

प्रश्न 14.
प्रथम चतुर्थांश में x² = 4y, y = 2, y = 4 एवं Y – अक्ष से घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए। (NCERT)
हल:
दिए गए वक्र का समीकरण है:
x² = 4y
⇒ x = 2\(\sqrt{y}\)
∴ अभीष्ट क्षेत्रफल = ABCD का क्षेत्रफल

प्रश्न 15.
छेदक रेखा x = \(\frac { a }{ \sqrt { 2 } } \) द्वारा वृत्त x² + y² = a² के छोटे भाग का क्षेत्रफल ज्ञात कीजिए। (NCERT)
हल:
वृत्त का समीकरण है:
x² + y² = a²
y² = a² – x²
y = \(\sqrt { a^{ 2 }-x^{ 2 } } \)
रेखा का समीकरण है –
x = \(\frac { a }{ \sqrt { 2 } } \)
समी. (1) और (2) को हल करने पर,

( \(\frac { a }{ \sqrt { 2 } } \) )² + y² = a²
y² = a² – \(\frac { a^{ 2 } }{ 2 } \) = \(\frac { a^{ 2 } }{ 2 } \)
y = \(\frac { a }{ \sqrt { 2 } } \)
वृत्त और रेखा का प्रतिच्छेद बिन्दु A( \(\frac { a }{ \sqrt { 2 } } \), \(\frac { a }{ \sqrt { 2 } } \) ) है।
D के निर्देशांक (a, 0) होंगे।
∴ अभीष्ट क्षेत्रफल = ABD का क्षेत्रफल = 2 × ACD का क्षेत्रफल

प्रश्न 16.
यदि वक्र x = y² एवं रेखा x = 4 से घिरा हुआ क्षेत्रफल x = a द्वारा दो बराबर भागों में विभाजित होता है, तो a का मान ज्ञात कीजिए। (NCERT)
हल:
परवलय का समीकरण है:
x = y²
⇒ y = \(\sqrt{x}\) ………… (1)
परवलय तथा रेखा x = 4 से घिरे क्षेत्रफल को रेखा x = a दो बराबर भागों में विभाजित करती है।
क्षेत्रफल OEC = क्षेत्रफल EFCB

प्रश्न 17.
प्रथम चतुर्थांश में वृत्त x² + y² = 4 एवं रेखाओं x = 0, x = 2 से घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए। (NCERT)
हल:
वृत्त का समीकरण है:
x² + y² = 4
⇒ y² = 4 – x²
⇒ y = \(\sqrt { (2)^{ 2 }-x^{ 2 } } \)

प्रश्न 18.
प्रथम चतुर्थांश में वृत्त x² + y² = 32 रेखा y = x एवं X – अक्ष द्वारा घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए। (NCERT; CBSE 2018)
हल:
दिए गए वृत्त का समीकरण है:
x² + y² = 32
y² = 32 – x²
वृत्त का समीकरण y = \(\sqrt { 32-x^{ 2 } } \)
रेखा का समीकरण है:
समी. (1) और (2) को हल करने पर,
x² + x² = 32
2x² = 32
x² = 16
x = 4

x का मान समी. (2) में रखने पर,
y = 4
रेखा और वृत्त का प्रतिच्छेद बिन्दु O(0,0) तथा A(4, 4) है। B के निर्देशांक (4, 0) तथा C के निर्देशांक (4\(\sqrt{2}\),0) हैं।
अभीष्ट क्षेत्रफल = क्षेत्रफल OACBO
= क्षेत्रफल OAB + क्षेत्रफल ABC

प्रश्न 19.
उस त्रिभुज का क्षेत्रफल समाकलन द्वारा ज्ञात कीजिए जिसकी भुजाएँ y = 2x + 1, y = 3x + 1 तथा x = 4 हैं। (NCERT)
हल:
माना कि ∆ABC की भुजाओं AB, AC तथा BC के समीकरण क्रमशः हैं –
y = 2x + 1 ……….. (1)
y = 3x + 1 ……….. (2)
और x = 4 ……….. (3)
समी. (1) और (2) को हल करने पर, बिन्दु A (0, 1)
समी. (1) और (3) को हल करने पर, बिन्दु B (4, 9) और
समी. (2) और (3) को हल करने पर, बिन्दु C (4, 13) हैं।
अत: अभीष्ट क्षेत्रफल

MP Board Class 12 Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 5 Complex Numbers and Quadratic Equations

Relations and Functions Important Questions

Relations and Functions Objective Type Questions

(A) Choose the correct option :

Question 1.
If x + iy = 2 + 3i, then (x, y) will be :
(a) (3, 2)
(b) (2, 3)
(c) (- 2,- 3)
(d) (3, 3)
Answer:
(b) (2, 3)

Question 2.
The simplest form of \(\frac { 1 + i }{ 1 – i }\) :
(a) – i
(b) i
(c) ± i
(d) None of these.
Answer:
(b) i

Question 3.
If z is a complex number, then z + z is :
(a) Real number
(b) Imaginary number
(c) Nothing can be said
(d) None of these.
Answer:
(a) Real number

Question 4.
The value of \(\sqrt {i}\) = ……………..
(a) ± \(\frac { 1 – i }{ \sqrt { 2 } }\)
(b) ± \(\frac { 1 + i }{ \sqrt { 2 } }\)
(c) ± (1 + i)
(d) ± (1 – i)
Answer:
(b) ± \(\frac { 1 + i }{ \sqrt { 2 } }\)

Question 5.
Number of solution of the equation z + z = 0 is :
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 6.
Argument of complex number 0 is :
(a) 0
(b) π
(c) 2π
(d) Not defined
Answer:
(d) Not defined

Question 7.
Conjugate of \(\frac { 1 – i }{ 1 + i }\) is :
(a) \(\frac { 1 – i }{ 1 + i }\)
(b) \(\frac { 1 – i }{ 1 – i }\)
(c) i
(d) – i
Answer:
(c) i

Question 8.
Argument (z₁.z₂) = ………………
(a) arg (z₁) – arg (z₂)
(b) arg (z₁) + arg (z₂)
(c) arg (z₁). arg (z₂)
(d) None of these
Answer:
(b) arg (z₁) + arg (z₂)

Question 9.
The least positive integer n for which \(\frac { 1 – i }{ 1 + i }\)ⁿ = 1 is :
(a) 2
(b) 4
(c) 8
(d) 12
Answer:
(b) 4

Question 10.
If |z – \(\frac { 4}{ z}\)| = 2, then |z| has maximum value:
(a) \(\sqrt {3}\) + 1
(b) \(\sqrt {5}\) + 1
(c) 2
(d) 2 + \(\sqrt {2}\)
Answer:
(b) \(\sqrt {5}\) + 1

(B) Match the following :

Answer:

1. (d)
2. (a)
3. (f)
4. (b)
5. (c)
6. (e)

(c) Fill in the blanks :

1. The value of (1 + i)⁴ (1 + \(\frac {1}{i}\))⁴ is ……………….
2. The value of \(\frac {1}{i}\) is ……………….
3. The value of z – \(\bar { z }\) is ……………….
4. Argument of – 1 – i is ……………….
5. polar form of i³ is ……………….
6. Argument of complex number is \(\frac { 1+\sqrt { 3i } }{ \sqrt { 3 } +i }\) is ……………….
7. If |\(\frac { z – i }{ z + i }\)| = 1, then locus of z will be ……………….
8. If x + \(\frac {1}{x}\) = 2cosθ, then the value of xⁿ = \(\frac { 1 }{ { x }^{ n } }\) will be ……………….
9. If z = 2 – 3i, then the value of z. \(\bar { z }\) is ……………….
10. If 2 + (x + yi) = 3 – i, the x = ……………… and y = ……………….

Answer:

1. 16
2. – i
3. Imaginary number
4. \(\frac {5π}{4}\)
5. cos\(\frac {π}{2}\) – i sin\(\frac {π}{2}\)
6. \(\frac {π}{6}\)
7. X – axis
8. 2cos nθ
9. 13
10. 1, – 1

(D) Write true / false :

1. Additive inverse of complex number – 2 + 5i is 2 – 5 i.
2. If | z² – 1 |= z² + 1, then z lies on a circle.
3. If | z₁ | = 12 and | z₂ – 3 – 4i | = 5, then the least value of | z₁ – z₂ | is 7.
4. If arg(z) < 0, then arg(- z) – arg(z) = π.
5. Polar form of 1 + \(\sqrt {- 1}\) is \(\sqrt {2}\)(cos\(\frac {π}{4}\) + isin\(\frac {π}{4}\)).
6. If z is a complex number such that | z | ≥ 2, then the minimum value of |z + \(\frac { 1 }{ 2 }\) | equal \(\frac { 5 }{ 2 }\)

Answer:

1. True
2. False
3. False
4. True
5. True
6. False.

(E) Write answer in one word / sentence :

1. If Z₁ = 2 – i, z₂ = 1 + i, them find the value \(\frac { { { z }_{ 1 }+{ z }_{ 2 }+1 } }{ { z }_{ 1 }-{ z }_{ 2 }+1 }\)
2. Modulus of \(\frac {1 – i }{ 1 + i }\) – \(\frac { 1 – i }{ 1 + i }\) will be.
3. Polar form of complex number \(\sqrt {3}\) + i will be.
4. If Z₁ = 2 – i and z₂ = – 2 + i, then Re(\(\frac { { { z }_{ 1 }{ z }_{ 2 } } }{ { z }_{ 1 } }\)) will be
5. If the number \(\frac { z – 1 }{ z + 1 }\) is purly imaginary, then the value of |z| will be.

Answer:

1. \(\sqrt {2}\)
2. 2
3. 2(cos\(\frac {π}{6}\) + isin\(\frac {π}{6}\))
4. \(\frac { – 2 }{ 5 }\)
5. 1

Complex Numbers and Quadratic Equations Very Short Answer Type Questions

Question 1.
If 1, ω, ω² are the cube root of unity then find the value of ω³ⁿ.
Solution:
Since ω³ = 1
ω³ⁿ = (ω³)ⁿ = (1)ⁿ = 1.

Question 2.
Find the value of i⁴ⁿ.
Solution:
i⁴ⁿ = (i⁴)ⁿ = (i².i²)ⁿ = [(- 1)(- 1)]ⁿ = (1)ⁿ = 1.

Question 3.
Find the condition of two complex numbers x + iy and a + ib are compaired.
Solution:
If x + iy = a + ib
Then, x = a, y = b.

Question 4.
Write conjugate of a complex number z = a – ib.
Solution:
z = conjugate of a – ib = a – (- ib)
= a + ib.

Question 5.
Write the complex number – 6₃ in ordered form.
Solution:
Ordered form of z = a + ib is (a, b).
Given: z = 0 – 6i
Ordered pair (0, – 6).

Question 6.
Write the complex number 3 – \(\sqrt { 7i }\) in ordered form.
Solution:
Ordered pair of z = a + ib is (a, b).
∴Ordered pair of 3 – \(\sqrt { 7i }\) = (3, – \(\sqrt { 7i }\))

Question 7.
Find the value of i + \(\frac { 1 }{ i }\).
Solution:
i + \(\frac { 1 }{ i }\) = \(\frac { { i }^{ 2 }+1 }{ i }\) = \(\frac { – 1 + i }{ i }\) = 0.

Question 8.
Find the value of i^-19.
Solution:
i^-19 = \(\frac { 1 }{ { i }^{ 19 } }\) = \(\frac { 1 }{ { i }^{ 18 }i } \) = \(\frac { 1 }{ (i^{ 2 })^{ 9 }.i }\)
= \(\frac { 1 }{ -1.i }\) = \(\frac { – 1 }{ i }\)
= \(\frac { – 1.i }{ { i }^{ 2 } }\) = i.

Question 9.
Write (1 – i)⁴ in form of a + ib.
Solution:
(1 – i)⁴ = [ (1 – i)² ]² (NCERT)
= [1 + i² – 2i]² = [1 – 1 – 2i]², [∵ i² = – 1]
= (- 2i)² = 4i² = 4(- 1) = – 4
= – 4 + 0i.

Question 10.
Find the modulus of complex number z = – 1 – i \(\sqrt {3}\) .
Solution:
Z = – 1 – i\(\sqrt {3}\)
a = – 1, b = \(\sqrt {3}\)
Modulus = \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\) = \(\sqrt {1 + 3}\) = \(\sqrt {2}\)

Question 11.
Find the modulus of complex number 1 – i.
Solution:
z = 1 – i
a = 1, b = – 1
Modulus = \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\) = \(\sqrt {1 + 1}\) = \(\sqrt {2}\)

Question 12.
Find the sum of 2 – 3i and its conjugate.
Solution:
z = 2 – 3i and \(\bar { z }\) = 2 + 3i
z + \(\bar { z }\) = 2 – 3i + 2 + 3i = 4.

Question 13.
Solve the quadratic equation x² + 2 = 0. (NCERT)
Solution:
x² + 2 = 0
⇒ x² = – 2
⇒ x² = – 1 x 2
= 2i² [∵ i² = – 1]
x = ± \(\sqrt {2i}\) or + \(\sqrt {2}\)i, – \(\sqrt {2}\)i

Question 14.
Solve the quadratic equation x² + 3 = 0. (NCERT)
Solution:
x² + 3 = 0
⇒ x² = – 3 = – 1 x 3
⇒ x² = 3i², [∵ i² = – 1]
x² = ± \(\sqrt {3i}\) or + \(\sqrt {3}\)i, – \(\sqrt {3}\)i

Question 15.
Write the triangle inequality for the two complex number Z₁ and z₂.
Solution:
| Z₁| and | z₂| are the two sides of any triangle
Then, | z₀ + z₂| ≤ | z₁| +| z₂|

Question 16.
If T and 9 are the modulus and argument of the complex number a + ib, then write its polar form.
Solution:
Polar form of a + ib is r(cosθ + i sinθ).

Question 17.
If z = cosθ + i sinθ where |z| = 1, then find \(\frac { 1 }{ z }\).
Solution:
\(\frac { 1 }{ z }\) = cosθ – isinθ.

Question 18.
Write the statement of De – moivre’s theorem.
Solution:
If polar form of z = a + ib is r(cosθ + isinθ), then De – moivre’s theorem is
(cosθ + isinθ)n = cos nθ + i sin nθ.

Question 19.
Write polar form of i.
Solution:
a + ib = 0 + i. 1
r = \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\) =1, θ = tan^-1\(\frac { b }{ a}\) = tan^-1∞ = \(\frac { π }{ 2}\)
Polar form of i = (cos\(\frac { π }{ 2}\) + i sin \(\frac { π }{ 2}\)).

Question 20.
Find the polar form of (- 2 + 2i).
Solution:

Complex Numbers and Quadratic Equations Long Answer Type Questions

Question 1.
Convert (\(\frac { 1 }{ 3}\) + 3i)³ in form of a + ib. (NCERT)
Solution:

Question 2.
Write ( – 2 – \(\frac { 1 }{ 3}\)i)³ in form of a + ib. (NCERT)
Solution:

Question 3.
Write (5 – 3i)³ in the form of a + ib.
Solution:
(5 – 3i)³ = (5)³ – (3i)³ – 3(5².3i) + 3.5(3i)²
= 125 – 27.i³ – 75.3i + 15.9i²
= 125 – 27i.i² – 225i + 15(- 9)
= 125 + 27i – 225i – 135
= – 10 – 198i.

Find the multiplicative inverse of following complex number :

Question 4.
4 – 3i
Solution:
Let the multiplicative inverse of 4 – 3i is a + ib.
Then, (4 – 3i) x (a + ib) = 1

Question 5.
\(\sqrt {5}\) + 3i
Solution:
Let z = \(\sqrt {5}\) + 3i
Be multiplicative inverse

Question 6.
Find the polar form of the following :
(i) \(\frac { 1 + 7i }{ { 2 – i }^{ 2 } }\), (ii) \(\frac { 1 + 3i }{ 1 – 2i }\)
Solution:

Now, let – 1 + i = r(cosθ + i sinθ),
Then, r cosθ = – 1 …. (1)
and r sinθ = 1 …. (2)
Squaring and adding equation (1) and (2),
r² = 1 + 1 = 2
⇒ r = \(\sqrt {2}\)
Put value of r in equation (1) and (2),

Comparing the real and imaginary parts,
r cosθ = – 1
r sinθ = 1

Squaring and adding equation (1) and (2),

∵ Real part is negative and imaginary part is positive of z, hence θ will lie in 4^th quadrant

Question 7.
Find modulus and argument of complex number z = – 1 – i\(\sqrt {3}\) (NCERT)
Solution:
Let z = – 1 – i\(\sqrt {3}\) = rcosθ + ir sinθ

Comparing the real and imaginary parts,
rcosθ = – 1 …. (1)
rsinθ = – \(\sqrt {3}\) …. (2)
Squaring and adding equation (1) and (2),

Since both the real and imaginary part of z are negative.
∴ θ lies in third quadrant

Question 8.
Find the modulus and argument of z = – \(\sqrt {3}\)+ i.
Solution:
Let z = – \(\sqrt {3}\)+ i = rcosθ + ir sinθ

Comparing the real and imaginary parts,
r cosθ = – \(\sqrt {3}\)
rsinθ = 1
Squaring and adding eqns. (1) and (2),
rcosθ = – \(\sqrt {3}\) …. (1)
rsinθ = 1 …. (2)
Squaring and adding equation (1) and (2),

Since both the real and imaginary part of z are negative and positive respectively.
∴ θ lies in 2^2nd quadrant

Question 9.
Find the modulus and argument of 1 – i.
Solution:
Let z = 1 – i = rcosθ + ir smθ
Comparing the real and imaginary parts,
1 = r cosθ
and – 1 = rsinθ
Squaring and adding equation (1) and (2),
(1)² + (- 1)² = r² cos²0 + r² sin²0
⇒ r²(cos²0 + sin²0) = 1 + 1,
⇒ r²= 2 ⇒ r = \(\sqrt {2}\) (modulus)
Dividing equation (2) by equation (1),

Question 10.
Find square root of complex number 3 + 4i.
Solution:
Let \(\sqrt {3 + 4i}\) = x + iy
On squaring,
3 + 4 i = (x + iy)²
⇒ 3 + 4i = x² + 2ixy + i²y²
⇒ 3 + 4i = x² + 2ixy – y²
⇒ 3 + 4i = (x² – y²) + i(2xy)
x² – y² = 3 …. (1)
2xy = 4 …. (2)
(x² + y²)² = (x² – y²)² + 4x²y²
= (3)² + (4)²
= 25
∴ x² + y² = 5 …. (3)
Now adding eqns. (1) and (3),
2x² = 8
⇒ x² = 4
⇒ x = ± 2
From equation (2), 2(± 2)y = 4
∴ y = ± 1
\(\sqrt { x + iy }\) = ± 2 ± i.

Question 11.
Find square root of complex number 8 – 6i.
Solution:
Let \(\sqrt { 8 – 6i }\) = x + iy
On squaring,
8 – 6i = (x + iy)²
⇒ 8 – 6i = x² + 2ixy + i²y² .
⇒ 8 – 6i = (x² – y²) + 2ixy
∴ 8 = x² – y² …. (1)
and – 6 = 2 xy …. (2)
(x² + y²)² = (x² – y²)² + 4x²y²
(x²+y²)² = (8)² + (- 6)²
= 64 + 36 = 100
∴ x² + y² = 10 …. (3)
Now adding equation (1) and (3),
2x² = 18
⇒ x² = 9
⇒ x = ±3
From equation (3), y² = 10 – 9 = 1
∴ y = ± 1
\(\sqrt { 8 – 6i}\) = ± 3 ± i.

Question 12.
x² + 3x + 9 = 0.
Solution:
Given : x² + 3x + 9 = 0
Comparing the above equation by ax² + bx + c = 0,
a = 1, b = 3, c = 9
Now, D = b² – 4ac
= 3² – 4 x 1 x 9 = 9 – 36 = – 27 < 0

Question 13.
– x² + x – 2 = 0.
Solution:
Given : – x² + x – 2 = 0
Comparing the above equation by ax² + bx + c = 0,

Question 14.
x² + 3x + 5 = 0.
Solution:
Given : x² + 3x + 5 = 0
Comparing the above equation by ax² + bx+c = 0,
a = 1, b = 3, c = 5

Question 15.
x² – x + 2 = 0
Solution:
Given : x² – x + 2 = 0
Comparing the above equation by ax² + bx + c = 0,
a = 1, b = – 1, c = 2
Now, D = b² – 4ac
= (- 1)² – 4 x 1 x 2 = 1 – 8 = – 7 < 0

Question 16.
If (\(\frac { 1 + i }{ 1 – i }\)) = 1, then And the minimum positive integer number of m. (NCERT)
Solution:

MP Board Class 11th Maths Important Questions

MP Board Class 12th Maths Important Questions Chapter 7B निशिचत समाकलन

निशिचत समाकलन Important Questions

निशिचत समाकलन दीर्घ उत्तरीय प्रश्न – II

प्रश्न 1.

हल:
माना

समी. (1) और (2) को जोड़ने पर,

प्रश्न 2.
सिद्ध कीजिए कि

हल:
माना


समी. (1) और (2) को जोड़ने पर,

प्रश्न 3.

हल:

प्रश्न 4.

हल:
माना

प्रश्न 5.
सिद्ध कीजिए कि

हल:
माना

समी. (1) और (2) को जोड़ने पर,


माना cos x = t ⇒ – sin x dx = dt ⇒ sin x dx = – dt
[जब x = 0, t = cos 0 = 1, जब x = π, t = cos π = -1]

प्रश्न 6.
सिद्ध कीजिए कि

हल:
माना

समी. (1) और (2) को जोड़ने पर,

प्रश्न 7.
सिद्ध कीजिए –

हल:
माना

समी. (1) और (2) को जोड़ने पर,

प्रश्न 8.
सिद्ध कीजिए कि

हल:
माना

प्रश्न 9.

समी. (1) और (2) को जोड़ने पर,

प्रश्न 10.
सिद्ध कीजिए –

हल:
माना


समी. (1) और (2) को जोड़ने पर,

प्रश्न 11.

हल:
माना

मान tan \(\frac { \pi }{ 2 } \) = t
∴ \(\frac{1}{2}\) sec² \(\frac { \pi }{ 2 } \) dx = dt
जब x = π, तब t = tan 0 = 0
तथा जब x = π, तब t = tan \(\frac { \pi }{ 2 } \) = ∞

प्रश्न 12.
सिद्ध कीजिए –

हल:
माना

समी. (1) और (2) को जोड़ने पर,

प्रश्न 13.
सिद्ध कीजिए –

हल:
माना


समी. (1) और (2) को जोड़ने पर,

प्रश्न 14.
सिद्ध कीजिए –

हल:
माना

प्रश्न 15.

हल:
माना

समी. (1) और (2) को जोड़ने पर,

माना cos x = t, तब – sin x dx = dt
जब x = 0, तब t = 1 तथा जब x = π, तब t = -1

प्रश्न 16.

हल:
माना

MP Board Class 12 Maths Important Questions