## 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 = e^{x} cos x, then find the value of \(\frac { dy }{ dx }\)

Solution:

y = e^{x}.cos x

I II

\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\)(e^{x}.cos x)

= e^{x}.\(\frac { d }{ dx }\)cos x + cos x\(\frac { d }{ dx }\)e^{x}

= e^{x}( – sin x) + cos x\(\frac { d }{ dx }\)e^{x}

= e^{x}[cos x – sin x].

Question 13.

If y = e^{x} cos x, then find the value of \(\frac { dy }{ dx }\)

Solution:

y = e^{x}.sin x

I II

\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\)(e^{x}.sin x)

= e^{x}.\(\frac { d }{ dx }\)sin x + sin x\(\frac { d }{ dx }\)e^{x}

= e^{x}cos x + ^{x}.e^{x}

= e^{x}(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.cosec^{2 }x

\(\frac { dy }{ dx }\) = – cosec x[cot^{2 }x + cosec^{2 }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 sin^{n} x w.r.t. x.

Solution:

Let y = sin^{n} x

\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\)(sin^{n} x )

Put sin x = t,

\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\)t^{n} = \(\frac { d }{ dt }\)t^{n}. \(\frac { dt }{ dx }\) = n sin^{n – 1}\(\frac { d }{ dx }\)(sin x)

⇒ \(\frac { dy }{ dx }\) = n sin^{n – 1}x 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.

(ax^{2} + sin x)(p + q cos x). (NCERT)

Solution:

Let y = (ax^{2} + sin x)(p + q cos x).

∴ \(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\)[(ax^{2} + sin x)(p + q cos x)]

⇒ \(\frac { dy }{ dx }\) = (p + q cos x)\(\frac { d }{ dx }\)(ax^{2} + sin x) + (ax^{2} + sin x)\(\frac { d }{ dx }\) (p + q cos x)

⇒ \(\frac { dy }{ dx }\) = (p + q cos x)(2a + cos x) + (ax^{2} + sin x)(0 – q sin x)

⇒ \(\frac { dy }{ dx }\) = – q sin x(ax^{2} + 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 – sec^{2} x)

= (x – tan x)(1 – sinx) + (x + cos x)(sec^{2} x – 1)

⇒ \(\frac { dy }{ dx }\) = (x – tan x)(1 – sinx) – tan^{2} 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 sec^{2} x + 0

= sec^{2}x. 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,