## MP Board Class 11th Maths Important Questions Chapter 13 Limits and Derivatives

### Limits and Derivatives Important Questions

Limits and Derivatives Short Answer Type Questions

Evaluate the following limits :

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
If = 405, then find the value of n.
Solution:

Question 10.
Find the value of :

Solution:

Question 11.

Solution:

Question 12.
If y = ex cos x, then find the value of $$\frac { dy }{ dx }$$
Solution:
y = ex.cos x
I II
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$(ex.cos x)
= ex.$$\frac { d }{ dx }$$cos x + cos x$$\frac { d }{ dx }$$ex
= ex( – sin x) + cos x$$\frac { d }{ dx }$$ex
= ex[cos x – sin x].

Question 13.
If y = ex cos x, then find the value of $$\frac { dy }{ dx }$$
Solution:
y = ex.sin x
I II
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$(ex.sin x)
= ex.$$\frac { d }{ dx }$$sin x + sin x$$\frac { d }{ dx }$$ex
= excos x + x.ex
= ex(cos x + sin x).

Question 14.
Differentiate sin(x + a) w.r.t. x.
Solution:
Let y = sin(x + a)
y = sin x cos a + cos x sin a
∴$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$(sin x cos a + cos x sin a)
= cos a$$\frac { d }{ dx }$$ sin x + sin$$\frac { d }{ dx }$$cos x
= cos a cos x – sin a sin x
⇒ $$\frac { dy }{ dx }$$ = cos(x + a).

Question 15.
Differentiate cosecx.cotx w.r.t. x.
Solution:
Let y = cosec x. cot x
⇒ $$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$[cosec x. cot x]
⇒ $$\frac { dy }{ dx }$$ = cot x$$\frac { d }{ dx }$$ cosec x + cosec x$$\frac { d }{ dx }$$ cot x
= – cot xcosec x cot x – cosec x.cosec2 x
$$\frac { dy }{ dx }$$ = – cosec x[cot2 x + cosec2 x].

Question 16.
Differentiate $$\frac { cos x }{ 1 + sin x }$$ w.r.t. x.
Solution:
Let y = $$\frac { cos x }{ 1 + sin x }$$

Question 17.
Differentiate $$\frac { sec c – 1 }{ sec x + 1 }$$ w.r.t. x.
Solution:
Let y = $$\frac { sec c – 1 }{ sec x + 1 }$$

Question 18.
Differentiate sinn x w.r.t. x.
Solution:
Let y = sinn x
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$(sinn x )
Put sin x = t,
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$tn = $$\frac { d }{ dt }$$tn. $$\frac { dt }{ dx }$$ = n sinn – 1$$\frac { d }{ dx }$$(sin x)
⇒ $$\frac { dy }{ dx }$$ = n sinn – 1x cos x.

Limits and Derivatives Long Answer Type Questions

Question 1.
Let f(x) =
, then find the value of a and b.
Solution:

Question 2.
Find the value of , where

Solution:

Question 3.
f(x) is defined such that
and is exist x = 2, then find the value of k.
Solution:

Question 4.
If the function f(x) satisfies= π, then find the value of
Solution:

Question 5.
Find the differential coefficient of the following functions by using first principle method :
(i) sin(x + 1), (ii) cos(x – $$\frac { π }{ 8 }$$,
Solution:
(i) Let f(x) = sin(x + 1)
∴ f(x + h) = sin[x + h + 1]
By definition of first principle,

By definition of first principle,

Find the differential coefficient of the following functions

Question 6.
(ax2 + sin x)(p + q cos x). (NCERT)
Solution:
Let y = (ax2 + sin x)(p + q cos x).
∴ $$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$[(ax2 + sin x)(p + q cos x)]
⇒ $$\frac { dy }{ dx }$$ = (p + q cos x)$$\frac { d }{ dx }$$(ax2 + sin x) + (ax2 + sin x)$$\frac { d }{ dx }$$ (p + q cos x)
⇒ $$\frac { dy }{ dx }$$ = (p + q cos x)(2a + cos x) + (ax2 + sin x)(0 – q sin x)
⇒ $$\frac { dy }{ dx }$$ = – q sin x(ax2 + sin x) + (p + q cos x)(2a + cos x)

Question 7.
(x + cos x)(x – tan x) (NCERT)
Solution:
Let y = (x + cos x)(x – tan x)
∴ $$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$[(x + cos x)(x – tan x)]
⇒ $$\frac { dy }{ dx }$$ = (x – tan x)$$\frac { d }{ dx }$$(x + cos x) + (x + cos x)$$\frac { d }{ dx }$$(x – tan x)
⇒ $$\frac { dy }{ dx }$$ = (x – tan x)(1 – sinx) + (x + cos x)(1 – sec2 x)
= (x – tan x)(1 – sinx) + (x + cos x)(sec2 x – 1)
⇒ $$\frac { dy }{ dx }$$ = (x – tan x)(1 – sinx) – tan2 x(x + cos x)

Question 8.
$$\frac { 4x + 5sin x }{ 3x + 7 cos x}$$
Solution:
Let y = $$\frac { 4x + 5sin x }{ 3x + 7 cos x}$$

Question 9.
$$\frac{x^{2} \cos \frac{\pi}{4}}{\sin x}$$
Solution:
Let y = $$\frac{x^{2} \cos \frac{\pi}{4}}{\sin x}$$

Question 10.
$$\frac { x }{ 1 + tan x}$$
Solution:
Let y = $$\frac { x }{ 1 + tan x}$$

Question 11.
Find the differential coefficient of $$\frac { sec x – 1 }{ sec x + 1}$$
Solution:
Let y = $$\frac { sec x – 1 }{ sec x + 1}$$

Question 12.
If y = $$\frac { x }{ x + 4 }$$, then prove that:
x$$\frac { dy }{ dx}$$ = y(1 – y)
Solution:
Given : y = $$\frac { x }{ x + 4 }$$

Question 13.
If y = $$\sqrt {x}$$ + $$\frac{1}{\sqrt{x}}$$, then prove that : 2x$$\frac { dy }{ dx}$$ + y – 2 $$\sqrt {x}$$ = 0
Solution:
Given : y = $$\sqrt {x}$$ + $$\frac{1}{\sqrt{x}}$$

Question 14.
Find the differential coefficient of $$\frac { sin(x + a) }{ cos x }$$
Solution:
Let y = $$\frac { sin(x + a) }{ cos x }$$
⇒ y = $$\frac { sin x + cos a + cos x sin a }{ cos x }$$
⇒ y = $$\frac { sin x + cos a }{ cos x }$$ + $$\frac { cos x + sin a }{ cos x }$$
⇒ y = cos a tan x + sin a
∴ $$\frac { dy }{ dx}$$ = $$\frac { d }{ dx}$$(cos a tan x + sin a)
= cos a $$\frac { d }{ dx}$$ tan x + $$\frac { d }{ dx}$$ sin a
= cos a x sec2 x + 0
= sec2x. cos a.

Question 15.
If f(x) = $$\frac { { x }^{ 100 } }{ 100 } +\frac { { x }^{ 99 } }{ 100 }$$+……..$$\frac { { x }^{ 2 } }{ 2 }$$ + x +1, then prove that:
f'(1) = 100 f'(0). (NCERT)
Solution:
Put x = 1, we get
f'(1) = 1 + 1 + ………. 1 + 1 (100 times)
f’(1) = 100 …. (1)
Put x = 0, we get
f'(0) = 0 + 0 + ……… 0 + 1
f’(0) = 1 …. (2)
From equation (1) and (2),
f'(1) = 100 f'(0).

Question 16.
Find the differential coefficient of cos x by first principle method. (NCERT)
Solution:
Let f(x) = cosx
∴ f(x + h) = cos(x + h)
By definition of first principle

Question 17.
Find the differential coefficient of f(x) = $$\frac { x + 1 }{ x – 1 }$$ by first principle method.
Solution:
Given : f(x) = $$\frac { x + 1 }{ x – 1 }$$
f(x + h) = $$\frac { x + h + 1}{ x + h – 1 }$$
By definition of first principle,