## MP Board Class 12th Maths Important Questions Chapter 1 Relations and Functions

### Relations and Functions Important Questions

**Relations and Functions Objective Type Questions:**

Question 1.

Choose the correct answer:

Question 1.

Let f : R → R te a function defined as f(x) = 8x:

(a) f is one – one onto (c) / is one-one but not onto

(b) f is many – one onto

(c) f is one – one but not onto

(d) f is neither one – one nor onto

Answer:

(a) f is one – one onto (c) / is one-one but not onto

Question 2.

Let f : R → R be a function is defined as f (x) = \(\frac { e^{ x^{ 2 } }-e^{ -x^{ 2 } } }{ e^{ x^{ 2 } }+e^{ -x^{ 2 } } } \):

(a) f is one – one but not onto

(b) f is one – one onto

(c) f is neither one – one nor onto

(d) f is many one onto

Answer:

(a) f is one – one but not onto

Question 3.

If f : R → R be given by f(x) = \((3-x^{ 3 })^{ 1/3 }\) then (fof) x is:

(a) x \(x^{ 1/3 }\)

(b) x\(x^{ 3 }\)

(c) x

(d) \((3-x^{ 3 })\)

Answer:

(c) x

Question 4.

Consider a binary operation * on N defined as a*b = \(a^{ 3 }\) + \(b^{ 3 }\):

(a) Is * both associative and commulative

(b) Is * commutative but not associative

(c) Is * associative but not commutative

(d) Is * neither commutative nor associative

Answer:

(b) Is * commutative but not associative

Question 5.

Number of binary operations on the set {a,b} are:

(a) 10

(b) 16

(c) 20

(d) 8

Answer:

(b) 16

Question 2.

Fill in the blanks:

- Number of equivalance relation in the set {1, 2, 3} containing (1,2) is …………………………….
- If f : R → R, when f(x) = 5x – 7,∀x∈R, then value of \(f^{ -1 }\) (7) is ……………………….
- If f : R → R, be defined by f(x) = x
^{2}– 3x + 2, then f {f(x)} …………………………… - If f : R → R, be defined by f(x) = 2x + 5, then \(f^{ -1 }\) (y) ………………………………
- If f : R → R, be defined by f(x) = x
^{3}the f is ………………………………….

Answer:

- 2
- 0
- x
^{4}– 6x^{3}+ 10x^{2}– 3x - \(\frac{1}{2}\) (y – 5)
- Is one – one

Question 3.

Write True/False:

- Let R = {(1, 3), (3, 1), (3, 3)} be a relation defined on A = {1, 2, 3}. Then R is symmetric, transitive but not reflexive.
- If f : A → B is bijective function, then the inverse of \(f^{ -1 }\) of f is unique.
- The composition of funtion is commutative
- Every function is invertible
- Let a binary operation * on the set Q
^{+}of positive rational numbers be defined by a*b = \(\frac{ab}{3}\), ∀a,b ∈ Q^{+}. Then the inverse of 4*6 is \(\frac{9}{8}\)

Answer:

- False
- True
- False
- False
- True

Question 4.

Write the answer in one word/sentence:

- Write the identity relation on the set A = {a, b, c}
- What is the range of the function f(x) = \(\frac{|x – 1|}{x – 1}\)?
- Write the domain of real valued function f defined by f(x) = \(\sqrt { 25-x^{ 2 } } \)
- Let * be a binay operation defined by a*b = 3a + 4b – 2, then find 4*5
- Let fg : R → R, defined f(x) = 2x + 1 and g(x) = x
^{2}– 2, ∀x ∈ R respectively. Then find fog.

Answer:

- {(a,a), (b,b), (c,c)}
- {-1,1}
- [-5,5]
- 30
- 4x
^{2}+ 4x – 1

**Relations and Functions Long Answer Type Questions – I**

Question 1.

Show that the relation R in set Z of integers given by R = {(a, b) : 2 divides a – b} is an equivalence relation? (NCERT)

Solution:

R is reflexive as 2 divides (a – a) for all Z.

Let (a,b) ∈ R ⇒ 2, divides (a – b)

⇒ 2,divides – (b – a)

(a,b) e R ⇒ (b,a) ∈ R

∴R is symmetric.

Let (a,b) ∈ R ⇒ 2, divides (a -b)

(b,c) ∈ R ⇒ 2, divides (b – c)

(a – b) and (b – c), are divisible by 2

(a – b) + (b – c), are divisible by 2

Hence (a – c) is also divisible by 2

∴ (a,b) ∈ R,(b,c) ∈ R ⇒ (a,c) ∈ R

∴ R is transitive.

∴ R is reflexive, symmetric and transitive. Therefore R is an equivalence relation in Z. Proved.

Question 2.

Let R be the relation defined in the set A = {1,2,3,4, 5,6,7} by R = {(a, b): both a and b are either odd or even}

Show that R is an equivalence relation. Further, show that all the elements of the subset {1,3,5,7} are related to each other and all the elements of the subset {2,4,6} are related to each other but no element of the subset {1,3,5, 7} is related to any element of the subset {2,4, 6}? (NCERT)

Solution:

Given any element a in A, both a and a must be either odd or even. So that (a, a) ∈ R .

Further (a,b) ∈ R ⇒ both a and b must be either odd or even (a,b) ∈ R

Similarly (a,b) ∈ R (b,c) ∈ R ⇒ all the elements of the a, b, c must be either even or odd simultaneously ⇒ (a,c) ∈ R .

Hence R is an equivalence relation.

All elements of {1,3,5,7} are related to each other, as all the elements of this subset are odd. Similarly, all the elements of the subset {2, 4, 6} are related to each other(MPBoardSolutions.com) as all of them are even. Also no elements of the subset{ 1, 3, 5, 7} can be related to any elements of {2,4, 6} as elements of {1, 3, 5, 7} are odd, while elements of {2, 4, 6} are even.

Question 3.

Let N is set of natural numbers. If R is a relation defined in set N × N such that (a, b) R (c, d). If ad (b + c) = bc (a + d). Prove that R is an equivalence? (CBSE 2015)

Solution:

Reflexive: For every (a,b) ∈ N × N

ab (b + a) = ba (a + b)

⇒ (a, b) R {a, b)

∴ R is reflexive.

Symmetric: Let (a,b) (c,d) ∈ N × N

(a,b) R (c,d) ⇒ ab(b + c) = bc (a + d)

⇒ bc (a + d) = ab(b + c)

(a,b) R (c,d) ⇒ (c,d) R (a,b)

∴ R is symmetric.

Transitive:

Let (a, b) R(c,d) and (c, d) R (e,f)

⇒ ad (b + c) = bc (a + d)

and cf (d + e) = de (c+f)

⇒ \(\frac{ad (b+c)}{abcd}\) = \(\frac{bc (a+d)}{abcd}\)

and \(\frac{cf(d+e)}{cdef}\) = \(\frac{de(c+f)}{cdef}\)

⇒ \(\frac{1}{c}\) + \(\frac{1}{b}\) = \(\frac{1}{d}\) + \(\frac{1}{a}\)

and \(\frac{1}{e}\) + \(\frac{1}{d}\) = \(\frac{1}{f}\) + \(\frac{1}{c}\)

⇒ \(\frac{1}{c}\) + \(\frac{1}{b}\) + \(\frac{1}{e}\) + \(\frac{1}{d}\) = \(\frac{1}{d}\) + \(\frac{1}{a}\) + \(\frac{1}{f}\) + \(\frac{1}{c}\), (by adding)

⇒ \(\frac{1}{b}\) + \(\frac{1}{e}\) = \(\frac{1}{a}\) + \(\frac{1}{f}\)

⇒ \(\frac{b+e}{be}\) = \(\frac{a+f}{af}\)

⇒ af (b+e) = be (a+f)

(a,b) R (c,d) R (e,f) ⇒ (a,b) R (e,f)

∴ R is transitive

∴ R is reflecxive, symmetric and transitive, hence, equivalence.

Question 4.

Let A = {1,2,3,4,5} and R = {(a, b) : |a – b| is divided by 2}. Prove that R is an equivalence relation. Also form equivalence class?

Solution:

A = { 1, 2, 3,4, 5 } and R = {(a, b ) : |(a – b) is divisible by 2}

R = {(1, 1), (2,2), (3,3), (4,4), (5,5), (1,3), (1,5), (2,4), (3,5), (3,1) (5,1), (4,2),(5,3)}

∀a eA(a,a) E R R is reflexive.

Since (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) ∈ R

(a,b) ∈ R ⇒ (b,a) ∈ R

(1, 3), (1, 5), (2,4), (3, 5), (3, 1), (5, 1), (4, 2), (5, 3) ∈ R

∴ R is symmetric.

∀(a,b) ∈ R, (b,c) ∈ R ⇒ (a,c) ∈ R

Since (1,3), (3,1) ∈ R ⇒ (1,1) ∈ R

R is transitive.

R is reflexive, symmetric and transitive, hence R is equivalence. Proved.

Equivalence class:

[1] = { a : a and 2 is divided by |a – 1|}

[1] = {a : a ∈ A and a – 1 = 2 k}

[1] = {1, 3, 5}

[2] = { a : a and 2, is divided by |a – 2 |}

[2] = {a : a and a – 2 = 2k}

[2] = {2,4}. Proved.

Question 5.

Let for all n ∈ N

defines a function f : N → N.

Tell, is the function f one – one onto? Justify your answer also? (NCERT)

Solution:

In f : N → N

f (1) = \(\frac{1+1}{2}\) = 1

f (2) = \(\frac{2}{2}\) = 1

Here, f (1) = f (2) ⇒ 1 ≠ 2.

In co – domain there is only one image 1 of the two elements 1 and 2 of domain. Hence, f is not an one – one function.

Case I.

When n is odd

n = 2r + l, r ∈ N

Then, 4r + 1 ∈ N exists such that,

f (4r + 1) = \(\frac{4r+1+1}{2}\) = \(\frac{4r+2}{2}\) = 2r + 1.

Clearly, there is a pre – image in domain of the each element of co – domain. Hence, f is onto function.

Case II.

When n = 2r (even number)

Then, 4r ∈ N exists such that,

f (4r) = \(\frac{4r}{2}\) = 2r

Clearly, there is a pre – image in domain of the each element of co – domain.

Hence, f is onto function.

Thus, f is one – one onto function

Question 6.

Let A = R – {3} and B = R – {1}. Discuss the function f : A → B defined by f (x) = \(\frac{x-2}{x-3}\), is the function is one – one and onto? Justify your answer also? (NCERT)

A = R – {3} and B = R – {1}.

Solution:

f: A → B, f(x) = \(\frac{x-2}{x-3}\) , A = R – {3} and B = R – {1}.

Let x,y ∈ A is such that,

f (x) = f (y)

⇒ \(\frac{x-2}{x-3}\) = \(\frac{y-2}{y-3}\)

⇒ (x – 2) (y – 3) = (y – 2) (x – 3)

⇒ xy – 3x – 2y + 6 = xy – 3y – 2x + 6

⇒ – 3x – 2y = – 3y – 2x

⇒ x = y

Here, f(x) = f(y) ⇒ x = y.

∴ f is one – one function.

Let y ∈ B = R – {1}

f is onto if, x ∈ A is such that,

f (x) = y

\(\frac{x-2}{x-3}\) = y

⇒ x – 2 = y (x-3)

⇒ x – 2 = xy – 3y

⇒ xy – x = xy – 3y – 2

⇒ x (y – 1) = 3y – 2

⇒ x = \(\frac{3y-2}{y-1}\) ∈ A

For each y ∈ B, x ∈ A, then

f (x) = y

∴ f(x) is onto function,

∴ f is one – one function.

Question 7.

Prove that function defined below f : N → N is one – one and onto: (NCERT)

Solution:

Let f(x_{1}) = f(x_{2})

If x_{1} is odd and x_{2} is even, then

x_{1} + 1 = x_{2} – 1

x_{1} – x_{2} = – 2

Which is impossible.

There is no chance that x_{1} is even and x_{2} is odd.

Hence either x_{1} and x_{2} both are even or odd.

Let x_{1}, x_{2} both are odd.

f(x_{1}) = f(x_{2})

⇒ x_{1} + 1 = x_{2} + 1

⇒ x_{1} = x_{2}

Let x_{1} and x_{2} both are even.

f(x_{1}) = f(x_{2})

⇒ x_{1} – 1 = x_{2} – 1

⇒ x_{1} = x_{2}

∴ f is one – one

Odd number 2r + 1 of co – domain N is image of number 2r + 2 of N and any even number of co – domain N. i.e; 2r is image of number 2r – 1 of N.

∴ f is onto.

Question 8.

Discuss on function given by f(x) = 4x + 3, f : R →R. Prove that f is invertible. Find also the inverse of f? (NCERT)

Solution:

f : R → R, f(x) = 4x + 3

Domain and co – domain of the function is R.

Let, x,y ∈R are such that,

f(x) = f(y)

4x + 3 = 4y + 3

⇒ 4x = 4y

∴ x = y

Thus, f(x) = f(y) ⇒ x = y

f is one – one.

Let y is any element of co – domain R.

∴ y = f(x)

y = 4x + 3

⇒ 4x = y – 3

⇒ x = \(\frac{y – 3}{4}\)

y is real. Then, x = \(\frac{y – 3}{4}\) is also real. i.e., there is pre – image in domain of each of the element in co – domain.

i.e., range = co – domain

f is onto.

Hence, f is one – one and onto.

Let f – image of x is y. Then,

y = f(x)

y = 4x + 3, y ∈R

4x = y – 3, [f(x) = y ∴x \(f^{ -1 }\) (y)]

⇒ \(f^{ -1 }\) (y) = \(\frac{y – 3}{4}\)

∴ \(f^{ -1 }\) (x) = \(\frac{x – 3}{4}\)

Question 9.

Let y = {n^{2}: n ∈ N] ⊂ N consider f : N → y as f(n) = n^{2} Show that/is invertible. Find the inverse of f? (NCERT)

Solution:

y = f(n) = n^{2}

n = \(\sqrt { y } \)

From g(y) = \(\sqrt { y } \) is defined g : y → N.

gof (n) = g[f(n)]

= g[n^{2}]

= \(\sqrt { n^{ 2 } } \). [ ∵g (n) = \(\sqrt { n } \) ⇒ g(n^{2}) = \(\sqrt { n^{ 2 } } \) = n]

(gof) n = n

and (fog) y = f [g(y)]

= f \(\sqrt { y } \), [ ∵f (n) = n^{2} ⇒ f ( \(\sqrt { y } \)) = ( \(\sqrt { y } \))^{2} = y]

= y

Clearly, gof = I_{n} and fog = I_{y}

Hence, f is invertibel and f^{-1} = g. Proved.

Question 10.

If f : R → R and g : R → R are defined as f(x) = cosx and g(x) = 3x^{2}. Find gof and fog. Prove that gof ≠ fog? (NCERT)

Solution:

Given:

f(x) = cosx

g(x) = 3x^{2}

(gof) x = g[ f(x)]

(gof) x = g [cosx]

Given: f(x) = cos x …………….. (1)

g (x) = 3x^{2}

g(cosx) = 3 cos^{2}x

From eqns. (1) and (2)

(gof) x = 3 cos^{2}x ………………. (2)

(fog) = f [g(x)]

= f [3x^{2}] …………………….. (3)

Given: g(x) = 3x^{2}

f (x) = cosx

f [3x^{2}] = cos3x^{2} ………………………… (4)

From eqns. (3) and (4),

(fog)x = cos 3x^{2}

For x = 0

3 cos^{2}x ≠ cos 3x^{2}

Hence, gof ≠ fog.

Question 11.

Let f : {1,2,3} → {a,b,c} given by f(1) = a, f(2) = b, f(3) = c. Find f^{-1} and show that (f^{-1})^{-1} = f?

Solution:

Given:

f : {1,2,3} → {a,b,c}

f(1) = a, f(2) = b, f(3) = c

Let g: {a,b,c} → {1,2,3}

g(a) = 1, g(b) = 2, g(c) = 3

(fog)a = f[g(a)]

= f[1] = a

(fog)b = f[g(b)]

= f(2) = b

(fog)c = f[g(b)]

= f(3) = c

and (gof) (1) = g[f(1)]

= g(a) = 1

(gof) (2) = g[f(2)]

= g(b) = 2

(gof) (3) = g[f(3)]

= g(c) = 3

Hence gof = I_{x} and fog = I_{y}

Where x = {1,2,3} and y = {a,b,c}

Inverse of f exists

and f^{-1} = g

∴ f^{-1} → {a,b,c} {1,2,3}

f^{-1}(a) = 1, f^{-1}(b) = 2, f^{-1}(c) = 3

Now, find inverse of f^{-1} i.e.,g.

Let h : {1,2,3} → {a,b,c}

h(1) = a, h(2) = b, h(3) = c

(goh) 1 = g [h(1)] = g(a) = 1

(goh) 2 = g [h(2)] = g(b) = 2

(goh) 3 = g [h(3)] = g(c) = 3

and (hog) a = h[g(a)] = h(1) = a

(hog) b = h [g(b)] = h(2) = b

(hog) c = h [g(c)] = h(3) = c

∴ goh = I_{x} and hog = I_{y}

Where, x = {1,2,3} and y = {a,b,c}

Inverse of g exists and g^{-1} = h = (f^{-1})^{-1} = f

∴ h = f

∴ (f^{-1})^{-1} = f. Proved.

Question 12.

Show that the relation defined in the set A of all polygons as R = {(P_{1},P_{2}): P_{1} and P_{2} have same number of sides } is an equivalence relation. (MPBoardSolutions.com) What is the set of all elements in A related of the right angle triangle T with sides 3,4 and 5? (NCERT)

Solution:

Given, A = set of all polygons

R = {(P_{1}, P_{2}): P_{1} and P_{2} have same number of sides}

In each polygon P the number of sides of polygon P are equal.

(P,P) ∈ R, ∀P ∈ A

Let (P_{1},P_{2}) ∈ R

⇒ The number of sides in polygon P_{1} and Polygon P_{2} are same.

⇒ The sides of polygon P_{2} and polygon P_{1} are same.

(P_{1},P_{2}) ∈ R ⇒ P_{2}P_{1} ∈ R

∴ R is a symmetric relation.

Let (P_{1},P_{2}) ∈ R and P_{2},P_{3}) ∈ R.

The number of sides of polygon P_{1} and P_{2}) are same.

∴ (P_{1},P_{2}) ∈ R , (P_{2},P_{3}) ⇒ (P_{1},P_{3}) ∈ R

∴ Relation R is transitive.

The number of polygon realated with right angle traingle with sides 3, 4, 5 be three. Hence polygon realted with right angles traingle with sides 3, 4, 5 is a traingle. Proved.

Question 13.

If f(x) = \(\frac{4x+3}{6x-4}\) , x ≠ \(\frac{2}{3}\) then, prove that for all x ≠ \(\frac{2}{3}\), fof(x) = x. What is inverse function of f?

Solution:

Let, inverse function of f^{-1} (x) = y .

Then f(y) = x

∴ \(\frac{4y+3}{6y-4}\) = x

⇒ 4y +3 = 6xy – 4x

⇒ 6xy – 4y = 3 + 4x

⇒ y (6x – 4) = 3 + 4x

⇒ y = \(\frac{3+4x}{6x-4}\)

⇒ f^{-1} (x) = \(\frac{3+4x}{6x-4}\) = f(x)

∴ f^{-1} = f.

Question 14.

Prove that the function given by f : [-1,1] → R, f(x) = \(\frac{x}{x+2}\) is one – one. Find the inverse function of function f : [-1,1] → (Range of f).

Solution:

f(x) = \(\frac{x}{x+2}\)

f : [-1,1] → R

Here f(x) = f(y)

⇒ \(\frac{x}{x+2}\) = \(\frac{y}{y+2}\)

⇒ xy + 2x = xy + 2y

⇒ 2x = 2y

⇒ x = y

∴ f is one – one

Let f^{-1}(x) = y

∴ f(y) = x

⇒ \(\frac{y}{y+2}\) = x

⇒ y = xy + 2x

⇒ y (1 – x) = 2x

∴y = \(\frac{2x}{1-x}\)

⇒ f^{-1}(x) = \(\frac{2x}{1-x}\).

Question 15.

If f(x) = \(\frac{x}{1+|x|}\), ∀x ∈ R and g(x) = \(\frac{x}{1-|x|}\) , ∀x ∈ R where -1 < x < 1, then find gof and fog? Show that fog = gof?

Solution:

Given: f(x) = \(\frac{x}{1+|x|}\)

∴ gof (x) = x

∴ From eqns. (1) and (2),

fog = gof.

Question 16.

Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and h(z) = sin z ∀ x, y and z ∈ N. Show that ho(gof) = (hog)of? (NCERT)

Solution:

ho(gof) x = h[gof(x)]

= h[g[f(x)]

= h(g(2x))

= h[3(2x) + 4]

= h[6x + 4]

= sin (6x + 4) ……………………. (1)

Similarly, ((hog)of)x = (hog) f(x)

= (hog) 2x

= h(g(2x))

= h[3(2x) + 4]

= h [6x + 4]

= sin (6x + 4) ……………………… (2)

From eqns. (1) and (2),

ho(gof) = (hog)of. Proved.