RD Sharma Solutions Class 12 Mathematics Chapter 11 MCQ

Updated on Jan 27, 2022 02:55 PM IST

RD Sharma books are considered the best source of information for CBSE students. They contain comprehensive material that is helpful for students to get a good insight into the subject. They are widely used all over the country and contain detailed exercises on all concepts. Moreover, many faculties use the RD Sharma book as a medium for setting up question papers. This is why it is beneficial for students.

JEE Main: Study Materials | High Scoring Topics | Preparation Guide

JEE Main: Syllabus | Sample Papers | Mock Tests | PYQs

This Story also Contains

RD Sharma Class 12 Solutions Chapter 11 MCQ Higher Order Derivatives - Other Exercise
Higher Order Derivatives Excercise: MCQ
RD Sharma Chapter-wise Solutions

RD Sharma Class 12th Chapter 11 MCQ contains 28 questions that are easy to solve and based on fundamentals. Students can breeze through this exercise with no problem if they are familiar with the basics of this chapter. RD Sharma Solutions It covers topics like finding second-order derivatives and evaluating differential equations. Students can get good knowledge about evaluating trigonometric and logarithmic equations through this exercise.

RD Sharma Class 12 Solutions Chapter 11 MCQ Higher Order Derivatives - Other Exercise

Higher Order Derivatives Excercise: MCQ

Higher Order Derivatives Exercise Multiple Choice Questions Question 1

Answer:
(b)
Hint:
We must know the derivative of

\cos x

and

\sin x

.
Given:

\begin{aligned} x = a \cos n t - b \sin n t \end{aligned}

Explanation:

\begin{aligned} x = a \cos n t - b \sin n t \end{aligned}

\frac{d x}{d t} = a (- \sin n t) \times n - b (\cos n t) \times n

\frac{d x}{d t} = - n a \sin n t - n b \cos n t

Differentiate on both sides,

\begin{aligned} \frac{d^{2} x}{d t^{2}} = - n a (\cos n t) \times n - n b (- \sin n t) \times n \end{aligned}

\frac{d^{2} x}{d t^{2}} = - n^{2} a (\cos n t) + n^{2} b (\sin n t)

\frac{d^{2} x}{d t^{2}} = n^{2} [- a \cos n t + b \sin n t]

\frac{d^{2} x}{d t^{2}} = - n^{2} x

Higher Order Derivatives Exercise Multiple Choice Questions Question 2

Answer:
(d)
Hint:
We must know about the derivative of

t

.
Given:

x = a t^{2}, y = 2 a t

Explanation:

x = a t^{2}

\Rightarrow \frac{d x}{d t} = 2 a t

y = 2 a t

\Rightarrow \frac{d y}{d t} = 2 a

\begin{aligned} \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 a}{2 a t} \end{aligned}

\Rightarrow \frac{d y}{d x} = \frac{1}{t}

Differentiate on both sides,

\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) \end{aligned}

= \frac{d}{d t} (\frac{d y}{d x}) \times \frac{d t}{d x}

\begin{aligned} = \frac{d}{d t} (\frac{1}{t}) \times \frac{1}{2 a t} \end{aligned}

= \frac{- 1}{t^{2}} \times \frac{1}{2 a t}

= \frac{- 1}{2 a t^{3}}

Higher Order Derivatives Exercise Multiple Choice Questions Question 3

Answer:
(b)
Hint:
We must know the derivative of

x

and

y

.
Given:

\begin{aligned} y = a x^{n + 1} + b x^{- n} \end{aligned}

Explanation:

\begin{aligned} y = a x^{n + 1} + b x^{- n} \end{aligned}

\frac{d y}{d x} = (n + 1) a x^{n + 1 - 1} + (- n) b x^{- n - 1}

\frac{d y}{d x} = (n + 1) a x^{n} + (- n) b x^{- (n + 1)}

\begin{aligned} \frac{d^{2} y}{d x^{2}} = n (n + 1) a x^{n - 1} - n b x^{- (n + 1) - 1} (- (n + 1)) \end{aligned}

\frac{d^{2} y}{d x^{2}} = n (n + 1) a x^{n - 1} + n (n + 1) b x^{- n - 2}

\begin{aligned} x^{2} \frac{d^{2} y}{d x^{2}} = n (n + 1) [a x^{n - 1} + b x^{- (n + 2)}] x^{2} \end{aligned}

x^{2} \frac{d^{2} y}{d x^{2}} = n (n + 1) [a x^{n - 1 + 2} + b x^{- n - 2 + 2}]

x^{2} \frac{d^{2} y}{d x^{2}} = n (n + 1) [a x^{n + 1} + b x^{- n}]

Higher Order Derivatives Exercise Multiple Choice Question question 4

Answer:
(b)
Hint:
We must know the derivative of

\cos x

.
Given:

= \frac{d^{20}}{d x^{20}} (2 \cos x \cos 3 x)

Explanation:
First let us solve the inner function,

= \frac{d}{d x} (2 \cos x \cos 3 x)

Using identity

(2 \cos x \cos 3 x)

\begin{aligned} = \cos (- 2 x) + \cos (4 x) \end{aligned}

= \cos (2 x) + \cos (4 x)

So, now we have to compute

20^{t h}

derivative of

\cos (2 x) + \cos (4 x)

Since, successive derivative of

\cos x

cycle in

4 : - \sin x, - \cos x, \sin x, \cos x \cdot \cdot \cdot

4^{t h}

derivative of

\cos x

20 = {(5.4)}^{t h}

derivative of

\cos x

is also

\cos x

.
Keep chain in mind,

\begin{aligned} 2^{20} \cos (2 x) + 4^{20} \cos (4 x) \end{aligned}

\Rightarrow 2^{20} [\cos (2 x) + 2^{20} \cos (4 x)]

Higher Order Derivatives Exercise Multiple Choice Question question 5

Answer:
(b)
Hint:
We must know the derivative of

x

and

y

Given:

x = t^{2}, y = t^{3}

Explanation:

x = t^{2}, y = t^{3}

\begin{aligned} \frac{d x}{d t} = 2 t, \frac{d y}{d t} = 3 t^{2} \end{aligned}

\frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{3 t^{2}}{2 t}

= \frac{3}{2} t

Differentiate on both sides,

\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{3}{2} t) \end{aligned}

= \frac{d}{d t} (\frac{3}{2} t) \times \frac{d t}{d x}

= \frac{3}{2} t^{2} \times \frac{1}{2 t}

= \frac{3}{4} t

Higher Order Derivatives Exercise Multiple Choice Question question 6

Answer:
(b)
Hint:
We must know the derivative of

x

.
Given:

\begin{aligned} y = a + b x^{2} \end{aligned}

a, b

are arbitrary constant.
Explanation:

\begin{aligned} y = a + b x^{2} \end{aligned}

\frac{d y}{d x} = 2 b x

\frac{1}{x} \cdot \frac{d y}{d x} = 2 b

\begin{aligned} \frac{d^{2} y}{d x^{2}} = 2 b = \frac{1}{x} \cdot \frac{d y}{d x} \end{aligned}

= x^{2} \frac{d^{2} y}{d x^{2}} = \frac{d y}{d x}

Higher Order Derivatives Excercise Multiple Choice Questions Question 7

Answer:
(c)
Hint:
We must know about the derivative of

\cos x

and

\sin x

.
Given:

\begin{aligned} f (x) = (\cos x + i \sin x) (\cos 2 x + i \sin 2 x) (\cos 3 x + i \sin 3 x) \dots . (\cos n x + i \sin n x) \end{aligned}

and

f (1) = 1

, then

f^{11} (1)

is equal to.
Explanation:

\begin{aligned} f (x) = (\cos x + i \sin x) (\cos 2 x + i \sin 2 x) (\cos 3 x + i \sin 3 x) \dots . (\cos n x + i \sin n x) \end{aligned}

\Rightarrow \cos (x + 2 x + 3 x + \dots . + n x) + i \sin (x + 2 x + 3 x + \dots . + n x)

Using Fourier series,

\Rightarrow [\cos \frac{n (n + 1)}{2} x + i \sin \frac{n (n + 1)}{2} x]

Differentiate on both sides,

\begin{aligned} f^{'} (x) = \frac{n (n + 1)}{2} [- \sin \frac{n (n + 1)}{2} x + i \cos \frac{n (n + 1)}{2} x] \end{aligned}

f^{''} (x) = - {(\frac{n (n + 1)}{2})}^{2} [\cos \frac{n (n + 1)}{2} x + i \sin \frac{n (n + 1)}{2} x]

\begin{aligned} f^{''} (x) & = - {(\frac{n (n + 1)}{2})}^{2} f (x) \end{aligned}

∴ f^{''} (x) = - {(\frac{n (n + 1)}{2})}^{2} f (1)

= - {(\frac{n (n + 1)}{2})}^{2}

Higher Order Derivatives Excercise Multiple Choice Questions Question 8

Answer:
(a)
Hint:
We must know about the derivative of

\cos x

and

\sin x

.
Given:

\begin{aligned} y = a \sin m x + b \cos m x \end{aligned}

Explanation:

\begin{aligned} y = a \sin m x + b \cos m x \end{aligned}

\frac{d y}{d x} = a (\cos m x) m + b (- \sin m x) m

\frac{d y}{d x} = m a \cos m x - m b \sin m x

\begin{aligned} \frac{d^{2} y}{d x^{2}} = m a (- \sin m x) \times m - m b (\cos m x) \times m \end{aligned}

\frac{d^{2} y}{d x^{2}} = - a m^{2} (\sin m x) - m^{2} b (\cos m x)

\begin{aligned} \frac{d^{2} y}{d x^{2}} = - m^{2} [a \sin m x + b \cos m x] \end{aligned}

\frac{d^{2} y}{d x^{2}} = - m^{2} y

Higher Order Derivatives Excercise Multiple Choice Questions Question 9

Answer:
(a)
Hint:
We must know about the derivative of

\sin^{- 1} x

.
Given:

f (x) = \frac{\sin^{- 1} x}{\sqrt{1 - x^{2}}}

, then

(1 - x^{2}) f^{'} (x) - x f (x) = ?

Explanation:

f (x) = \frac{\sin^{- 1} x}{\sqrt{1 - x^{2}}}

Differentiate with respect to

x

\begin{aligned} y \sqrt{1 - x^{2}} = \sin^{- 1} x \end{aligned}

y \cdot \frac{1}{2 \sqrt{1 - x^{2}}} (- 2 x) + \sqrt{1 - x^{2}} \frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{2}}}

- x y + (1 - x^{2}) \frac{d y}{d x} = 1

Therefore,

(1 - x^{2}) f^{'} (x) - x f (x) = 1

Higher Order Derivatives Exercise Multiple Choice Question Question 10

Answer:
(c)
Hint:
We must know about the derivative of logarithm and

\tan x

.
Given:

y = \tan^{- 1} {\frac{\log_{e} (\frac{e}{x^{2}})}{\log_{e} (e x^{2})} + \tan^{- 1} (\frac{3 + 2 \log_{e} x}{1 - 6 \log_{e} x})}

Explanation:

y = \tan^{- 1} {\frac{\log_{e} (\frac{e}{x^{2}})}{\log_{e} (e x^{2})} + \tan^{- 1} (\frac{3 + 2 \log_{e} x}{1 - 6 \log_{e} x})}

\begin{aligned} y = \tan^{- 1} (\frac{1 - 2 \log_{e} x}{1 + 2 \log_{e} x}) + \tan^{- 1} (\frac{3 + 2 \log_{e} x}{1 - 6 \log_{e} x}) \end{aligned}

y = \tan^{- 1} (\frac{(\frac{1 - 2 \log_{e} x}{1 + 2 \log_{e} x}) + (\frac{3 + 2 \log_{e} x}{1 - 6 \log_{e} x})}{1 - (\frac{1 - 2 \log_{e} x}{1 + 2 \log_{e} x}) (\frac{3 + 2 \log_{e} x}{1 - 6 \log_{e} x})})

\begin{aligned} y = \tan^{- 1} {\frac{(1 - 2 \log_{e} x) (1 - 6 \log_{e} x) + (3 + 2 \log_{e} x) (1 + 2 \log_{e} x)}{(1 + 2 \log_{e} x) (1 - 6 \log_{e} x) - (1 - 2 \log_{e} x) (3 + 2 \log_{e} x)}} \end{aligned}

y = \tan^{- 1} {\frac{1 - 8 \log_{e} x + 12 {(\log_{e} x)}^{2} + 3 + 8 \log_{e} x + 4 {(\log_{e} x)}^{2}}{1 - 4 \log_{e} x - 12 {(\log_{e} x)}^{2} - 3 + 4 \log_{e} x + 4 {(\log_{e} x)}^{2}}}

\begin{aligned} y = \tan^{- 1} {\frac{4 + 16 {(\log_{e} x)}^{2}}{- 2 - 8 {(\log_{e} x)}^{2}}} \end{aligned}

y = \tan^{- 1} {\frac{4 (1 + 4 {(\log_{e} x)}^{2})}{- 2 (1 + 4 {(\log_{e} x)}^{2})}}

\begin{aligned} y = \tan^{- 1} [- 2] \end{aligned}

\frac{d y}{d x} = 0, \frac{d^{2} y}{d x^{2}} = 0

Higher Order Derivatives Exercise Multiple Choice Question Question 11

Answer:
(a)
Hint:
We know about the derivative of polynomials.
Given:

f (x)

be a polynomial,

f {(e)}^{x}

.
Explanation:

y = f {(e)}^{x}

\begin{matrix} \frac{d}{d x} (f (e^{x})) = f^{'} (e^{x}) \times \frac{d}{d x} (e^{x}) \end{matrix}

= f^{'} (e^{x}) \cdot e^{x}

Similarly,

\begin{aligned} \frac{d^{2}}{d x^{2}} (f (e^{x})) = f " (e^{x}) \frac{d}{d x} (e^{x}) e^{x} + \frac{d}{d x} (e^{x}) f^{'} (e^{x}) \end{aligned}

\Rightarrow f^{''} (e^{x}) e^{2 x} + f^{'} (e^{x}) (e^{x})

Higher Order Derivatives Exercise Multiple Choice Question Question 12

Answer: Option (c)
Hint:
We must know about the derivative of

\cos x

and logarithm.
Given:

y = a \cos (\log_{e} x) + b \sin (\log_{e} x)

Explanation:

y = a \cos (\log_{e} x) + b \sin (\log_{e} x)

Differentiate both side with respect to

x

\frac{d y}{d x} = \frac{a [- \sin (\log x)]}{x} - \frac{b [\cos (\log x)]}{x}

$\frac{d y}{d x}=\frac{-[a \sin (\log x)+b \cos (\log x)]}{x} $

\begin{aligned} x \frac{d y}{d x} = - [a \sin (\log x) + b \cos (\log x)] \end{aligned}

Again differentiating with respect to

x

\begin{aligned} x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = - [\frac{a \cos (\log x)}{x} - \frac{b \sin (\log x)}{x}] \end{aligned}

\begin{aligned} x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = - [\frac{a \cos (\log x)}{x} - \frac{b \sin (\log x)}{x}] \end{aligned}

x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = \frac{- y}{x}

x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + \frac{y}{x} = 0

\begin{aligned} x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0 \end{aligned}

x^{2} y_{2} + x y_{1} + y = 0

Hence option the value of

x^{2} y_{2} + x y_{1}

(- y)

Higher Order Derivatives Exercise Multiple Choice Questions Question 13

Answer:
(a)
Hint:
We must know about the derivative of

x

Given:

x = 2 a t, y = a t^{2}

, where

a

is constant.
Explanation:

x = 2 a t, y = a t^{2}

\frac{d x}{d t} = 2 a, \frac{d y}{d t} = 2 a t

\begin{aligned} \frac{d y}{\frac{d t}{d x}} d t = \frac{2 a t}{2 a} \end{aligned}

= t

\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x})

\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{d t} (\frac{d y}{d x}) \frac{d t}{d x} \end{aligned}

= \frac{d}{d t} (t) \frac{d t}{d x}

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

\begin{aligned} = \frac{1}{2 a} \end{aligned}

Higher Order Derivatives Exercise Multiple Choice Questions Question 14

Answer:
(a)
Hint:
We must know about the derivative.
Given:

x = f (t), y = g (t)

Explanation:

x = f (t), y = g (t)

$\ \frac{d x}{d t}=f^{\prime}(t), \frac{d y}{d t}=g^{\prime}(t) $

\frac{d y}{d t} \cdot \frac{d t}{d x} = \frac{g^{'} (t)}{f^{'} (t)}

\begin{aligned} \frac{d y}{d x} = \frac{g^{'} (t)}{f^{'} (t)} \end{aligned}

Differentiate on both sides,

\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x})

= \frac{d}{d t} (\frac{g^{'} (t)}{f^{'} (t)}) \times \frac{d t}{d x}

= \frac{f^{'} (t) g^{''} (t) - g^{'} (t) f^{''} (t)}{{(f^{'} (t))}^{2}} \times \frac{1}{f^{'} (t)}

= \frac{f^{'} (t) g^{''} (t) - g^{'} (t) f^{''} (t)}{{(f^{'} (t))}^{3}}

Higher Order Derivatives Exercise Multiple Choice Questions Question 15

Answer:
(c)
Hint:
We must have known about the derivative of inverse trigonometric functions like

\sin^{- 1} x

.
Given:

y = \sin (m \sin^{- 1} x)

Explanation:

y = \sin (m \sin^{- 1} x)

Differentiating both sides with respect

x

\frac{d y}{d x} = \cos (m \sin^{- 1} x) \times \frac{m}{\sqrt{1 - x^{2}}}

\begin{aligned} \frac{d y}{d x} = \frac{m \cos (m \sin^{- 1} x)}{\sqrt{1 - x^{2}}} \end{aligned}

Again differentiate with respect to

x

\frac{d^{2} y}{d x^{2}} = m [(\sqrt{1 - x^{2}}) (- \sin (m (\sin^{- 1} x)) \times \frac{m}{\sqrt{1 - x^{2}}}) - \cos (m (\sin^{- 1} x)) \times \frac{1}{2 \sqrt{1 - x^{2}}} \times (0 - 2 x)]

\frac{d^{2} y}{d x^{2}} = m [\frac{(- m \sin (m (\sin^{- 1} x))) + \cos (m (\sin^{- 1} x)) \times \frac{x}{\sqrt{1 - x^{2}}}}{1 - x^{2}}]

\begin{aligned} \frac{d^{2} y}{d x^{2}} = m [\frac{(- m \sin (m (\sin^{- 1} x)))}{1 - x^{2}} + \frac{x \cos (m (\sin^{- 1} x))}{(1 - x^{2}) \sqrt{1 - x^{2}}}] \end{aligned}

\frac{d^{2} y}{d x^{2}} = m [\frac{(- m \sin (m (\sin^{- 1} x)))}{1 - x^{2}} + \frac{x \frac{d y}{d x}}{(1 - x^{2})}]

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} = m x \frac{d y}{d x} - m^{2} \sin (m (\sin^{- 1} x))

\begin{aligned} (1 - x^{2}) \frac{d^{2} y}{d x^{2}} = m x \frac{d y}{d x} - m^{2} y \end{aligned}

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - m x \frac{d y}{d x} + m^{2} y = 0

\begin{aligned} (1 - x^{2}) y^{2} - m x y^{1} = - m^{2} y \end{aligned}

Higher Order Derivatives Exercise Multiple Choice Question question 16

Answer:
(a)
Hint:
We must have known about the derivative of trigonometric function like

\sin^{- 1} x

.
Given:

y = {(\sin^{- 1} x)}^{2}

Explanation:

y = {(\sin^{- 1} x)}^{2}

Differentiate with respect to

x

\frac{d y}{d x} = 2 \sin^{- 1} x \times \frac{1}{\sqrt{1 - x^{2}}}

Again differentiate with respect to

x

\frac{d^{2} y}{d x^{2}} = \frac{2 \frac{1}{\sqrt{1 - x^{2}}} \times \sqrt{1 - x^{2}}}{(1 - x^{2})} - \sin^{- 1} (\frac{x - 2 x}{2 \sqrt{1 - x^{2}}})

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} = 2 [1 + \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}}]

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - \frac{2 x \sin^{- 1} x}{\sqrt{1 - x^{2}}} = 2

\begin{matrix} (1 - x^{2}) y_{2} = 2 + x y_{1} \end{matrix}

Higher Order Derivatives Exercise Multiple Choice Question question 17

Answer:
(c)
Hint:
We must have known about the derivative of

\sin x, \cos x

and

\tan x

.
Given:
$y=e^{\tan x} $
Explanation:
$y=e^{\tan x} $

y_{1} = \frac{d y}{d x} = e^{\tan x} \times \sec^{2} x

\begin{aligned} \frac{d y}{d x} = e^{\tan x} (\sec^{2} x) \end{aligned}

Again differentiate with respect to

x

,
$\frac{d^{2} y}{d x^{2}}=\left[e^{\tan x}\left(\sec ^{2} x\right)\left(\sec ^{2} x\right)+e^{\tan x}\left(2 \sec ^{2} x \tan x\right)\right] $

\begin{aligned} \frac{d^{2} y}{d x^{2}} = e^{\tan x} [\sec^{4} x + 2 \sec^{2} x \tan x] \end{aligned}

\frac{d^{2} y}{d x^{2}} = e^{\tan x} \cdot \sec^{2} x [\sec^{2} x + 2 \tan x]

\frac{d^{2} y}{d x^{2}} = e^{\tan x} \cdot \sec^{2} x [\frac{1}{\cos^{2} x} + 2 \frac{\sin x}{\cos x}]

\begin{aligned} \frac{d^{2} y}{d x^{2}} = e^{lan x} \cdot \sec^{2} x [\frac{1 + 2 \sin x \cos x}{\cos^{2} x}] \end{aligned}

\cos^{2} x \frac{d^{2} y}{d x^{2}} = y_{1} [1 + 2 \sin x \cos x]

\begin{aligned} \cos^{2} x \frac{d^{2} y}{d x^{2}} = y_{1} [1 + \sin 2 x] \end{aligned}

[∵ 2 \sin x \cos x = \sin 2 x]

Higher Order Derivatives Exercise Multiple Choice Question question 18

Answer:
(b)
Hint:
We must know about the derivative of trigonometric function.
Given:

y = \frac{2}{\sqrt{a^{2} - b^{2}}} \tan^{- 1} (\sqrt{\frac{a - b}{a + b}} \tan \frac{x}{2})

Explanation:

y = \frac{2}{\sqrt{a^{2} - b^{2}}} \tan^{- 1} (\sqrt{\frac{a - b}{a + b}} \tan \frac{x}{2})

\begin{aligned} \frac{d y}{d x} = \frac{2}{\sqrt{a^{2} - b^{2}}} \cdot \frac{1}{1 + (\frac{a - b}{a + b}) \tan^{2} (\frac{x}{2})} \times \sqrt{\frac{a - b}{a + b}} \frac{1}{2} \sec^{2} (\frac{x}{2}) \end{aligned}

\frac{d y}{d x} = \frac{2}{\sqrt{a^{2} - b^{2}}} \cdot \frac{(a + b)}{(a + b) + (a - b) \tan^{2} (\frac{x}{2})} \times \sqrt{\frac{a - b}{a + b}} \frac{1}{2} \sec^{2} (\frac{x}{2})

\frac{d y}{d x} = \sqrt{\frac{1}{a^{2} - b^{2}} \times \frac{a - b}{a + b} \times (a + b)^{2}} \times \frac{1 + \tan^{2} (\frac{x}{2})}{a (1 + \tan^{2} (\frac{x}{2}) + b (1 - \tan^{2} (\frac{x}{2})))}

\frac{d y}{d x} = \frac{1}{a + b (\frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})})}

\begin{aligned} \frac{d y}{d x} = & \frac{1}{a + b \cos x} \end{aligned}

Differentiate again,

\frac{d^{2} y}{d x^{2}} = - \frac{- b \sin x}{(a + b \cos x)^{2}}

\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{b \sin x}{(a + b \cos x)^{2}} \end{aligned}

Higher Order Derivatives Exercise Multiple Choice Question question 19

Answer:
(a)
Hint:
We must know about the derivative of

x

Given:
$y=\frac{a x+b}{x^{2}+c} $
Explanation:
$y=\frac{a x+b}{x^{2}+c} $

\begin{aligned} (x^{2} + c) y = a x + b \end{aligned}

Differentiate with respect to

x

2 x y + (x^{2} + c) \frac{d y}{d x} = a

Again differentiate,

2 y + 2 x y_{1} + 2 x y_{1} + (x^{2} + c) y_{2} = 0

\begin{aligned} 2 y + 4 x y_{1} + (x^{2} + c) y_{2} = 0 \end{aligned}

Differentiate again with respect to

x

2 y_{1} + 4 y_{1} + 4 x y_{2} + (x^{2} + c) y_{3} + 2 x y_{2} = 0

\begin{aligned} 6 y_{1} + 6 x y_{2} + (x^{2} + c) y_{3} = 0 \end{aligned}

6 y_{1} + 6 x y_{2} + (\frac{- 2 y - 4 x y_{1}}{y_{2}}) y_{3} = 0

6 y_{1} y_{2} + 6 x {(y_{2})}^{2} - 2 y - 4 x y_{1} y_{3} = 0

3 y_{1} y_{2} + 3 x {(y_{2})}^{2} - y - 2 x y_{1} y_{3} = 0

\begin{aligned} (2 x y_{1} + y) y_{3} = (y_{1} + x y_{2}) 3 y_{2} \end{aligned}

Higher Order Derivatives Exercise Multiple Choice Question question 20

Answer:
(a)
Hint:
We must know about the derivative of logarithm.
Given:

y = \log_{e} {(\frac{x}{a + b x})}^{x}

Explanation:

y = \log_{e} {(\frac{x}{a + b x})}^{x}

$\ y=x \log _{e}\left(\frac{x}{a+b x}\right) $

\begin{aligned} \frac{y}{x} = \log x - \log (a + b x) \end{aligned}

Differentiate with respect to

x

$\frac{x \frac{d y}{d x}-y}{x^{2}}=\frac{1}{x}-\frac{b}{a+b x} $

\begin{aligned} x \frac{d y}{d x} - y = x - \frac{b x^{2}}{a + b x} \end{aligned}

Differentiate with respect to

x

x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} - \frac{d y}{d x} = 1 - (\frac{(a + b x) 2 b x - b x^{2} (b)}{(a + b x)^{2}})

x \frac{d^{2} y}{d x^{2}} = (\frac{(a + b x)^{2} - (a + b x) 2 b x - b^{2} x^{2}}{(a + b x)^{2}})

\begin{aligned} x \frac{d^{2} y}{d x^{2}} = (\frac{(a + b x) [a + b x - 2 b x] + b^{2} x^{2}}{(a + b x)^{2}}) \end{aligned}

x \frac{d^{2} y}{d x^{2}} = (\frac{(a + b x) (a - b x) + b^{2} x^{2}}{(a + b x)^{2}})

\begin{aligned} x \frac{d^{2} y}{d x^{2}} = (\frac{a^{2} - b^{2} x^{2} + b^{2} x^{2}}{(a + b x)^{2}}) \end{aligned}

x \frac{d^{2} y}{d x^{2}} = (\frac{a^{2}}{(a + b x)^{2}}) = {(\frac{a}{a + b x})}^{2}

$\ x \frac{d^{2} y}{d x^{2}}=\left(\frac{a}{a+b x}\right)^{2} $ … (i)
And

\begin{aligned} y = x \log (\frac{x}{a + b x}) \end{aligned}

Differentiate with respect to

x

\frac{d y}{d x} = \frac{x (a + b x - b x)}{(a + b x)^{2}} (\frac{a + b x}{x}) + \log (\frac{x}{a + b x})

\frac{d y}{d x} = \frac{a}{a + b x} + \frac{y}{x}

\begin{aligned} \frac{d y}{d x} - \frac{y}{x} = \frac{a}{a + b x} \end{aligned}

From (i)

x \frac{d^{2} y}{d x^{2}} = {(\frac{d y}{d x} - \frac{y}{x})}^{2}

\begin{aligned} x^{2} \frac{d^{2} y}{d x^{2}} = \frac{1}{x^{2}} {(x \frac{d y}{d x} - y)}^{2} \end{aligned}

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

Higher Order Derivatives Exercise Multiple Choice Question question 21

Answer: (c)
Hint:
We must know about the derivative of trigonometric function.
Given:

x = f (t) \cos t - f^{'} (t) \sin t

y = f (t) \sin t - f^{'} (t) \cos t

Explanation:

x = f (t) \cos t - f^{'} (t) \sin t

y = f (t) \sin t - f^{'} (t) \cos t

\frac{d x}{d t} = f^{'} (t) \cos t - f (t) \sin t - f^{''} (t) \sin t - f^{'} (t) \cos t

\begin{aligned} \frac{d x}{d t} = - f (t) \sin t - f^{''} (t) \sin t \end{aligned}

\frac{d x}{d t} = - \sin t [f (t) + f^{''} (t)]

\frac{d y}{d t} = f^{'} (t) \sin t - f (t) \cos t - f^{''} (t) \cos t - f^{'} (t) \sin t

\begin{aligned} \frac{d y}{d t} = - f (t) \cos t - f^{''} (t) \cos t \end{aligned}

\frac{d y}{d t} = - \cos t [f (t) + f^{''} (t)]

{(\frac{d x}{d t})}^{2} = {[- \sin t [f (t) + f^{''} (t)]]}^{2}

\begin{aligned} {(\frac{d y}{d t})}^{2} = {[- \cos t [f (t) + f^{''} (t)]]}^{2} \end{aligned}

{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} = (\sin t)^{2} {[f (t) + f^{''} (t)]}^{2} + (\cos t)^{2} {[f (t) + f^{''} (t)]}^{2}

\begin{aligned} {(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} = {[f (t) + f^{''} (t)]}^{2} [(\sin t)^{2} + (\cos t)^{2}] \end{aligned}

= {[f (t) + f^{''} (t)]}^{2}

[∵ (\sin t)^{2} + (\cos t)^{2} = 1]

Higher Order Derivatives Exercise Multiple Choice Question question 22

Answer:
(b)
Hint:
We must have known about the derivative of

x

Given:

y^{\frac{1}{n}} + y^{\frac{- 1}{n}} = 2 x

Explanation:

y^{\frac{1}{n}} + y^{\frac{- 1}{n}} = 2 x

Differentiate with respect to

x

[\frac{1}{n} y^{\frac{1}{n} - 1} - \frac{1}{n} y^{\frac{- 1}{n} - 1}] \frac{d y}{d x} = 2

\frac{1}{n} y^{- 1} [y^{\frac{1}{n}} - y^{\frac{- 1}{n}}] \frac{d y}{d x} = 2

\frac{1}{n y} [y^{\frac{1}{n}} - y^{\frac{- 1}{n}}] \frac{d y}{d x} = 2

\begin{aligned} [y^{\frac{1}{n}} - y^{\frac{- 1}{n}}] \frac{d y}{d x} = 2 x y \end{aligned}

Squaring both sides,

[y^{\frac{2}{n}} - y^{\frac{- 2}{n}} - 2] {(\frac{d y}{d x})}^{2} = 4 x^{2} y^{2}

… (i)
Now
Given

y^{\frac{1}{n}} + y^{\frac{- 1}{n}} = 2 x

Squaring both sides,

y^{\frac{2}{n}} + y^{\frac{- 2}{n}} + 2 = 4 x^{2}

\begin{aligned} y^{\frac{2}{n}} + y^{\frac{- 2}{n}} = 4 x^{2} - 2 \end{aligned}

Putting this value in (i)

4 x^{2} y^{2} = [4 x^{2} - 2 - 2] {(\frac{d y}{d x})}^{2}

4 x^{2} y^{2} = [4 x^{2} - 4] {(\frac{d y}{d x})}^{2}

\begin{aligned} x^{2} y^{2} = [x^{2} - 1] {(\frac{d y}{d x})}^{2} \end{aligned}

Differentiate with respect to

x

2 n^{2} y (\frac{d y}{d x}) = 2 (\frac{d y}{d x}) (\frac{d^{2} y}{d x^{2}}) [n^{2} y - 1] + 2 x {(\frac{d y}{d x})}^{2}

n^{2} y = (\frac{d^{2} y}{d x^{2}}) [n^{2} y - 1] + x (\frac{d y}{d x})

[n^{2} y - 1] y_{2} + x y_{1} = n^{2} y

\begin{aligned} [x^{2} - 1] y_{2} + x y_{1} = n^{2} y \end{aligned}

Higher Order Derivatives Exercise Multiple Choice Question question 23

Answer:
(c)
Hint:
We must know about the rules of finding the derivative.
Given:

\frac{d}{d x} {x^{n} - a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots . + (- 1)^{n} a_{n}} e^{x} = x^{n} e^{x}

Explanation:

\frac{d}{d x} {x^{n} - a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots . + (- 1)^{n} a_{n}} e^{x} = x^{n} e^{x}

\frac{d}{d x} [x^{n} - a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots . + (- 1)^{n} a_{n}] e^{x}

\begin{aligned} \frac{d}{d x} [x^{n} - n x^{n - 1} + n (n - 1) x^{n - 2} + \dots . . + (- 1)^{n} a_{n}] e^{x} \end{aligned}

Comparing the coefficients of above two equations,

a_{1} = n, a_{2} = n (n - 1)

Similarly,

a_{r} = n (n - 1) (n - 2) (n - 3) \dots (n - r + 1)

\begin{aligned} a_{r} = \frac{n!}{(n - r)!} \end{aligned}

Higher Order Derivatives Exercise Multiple Choice Question question 24

Answer:
(a)
Hint:
We must know about the derivative of logarithm.
Given:

y = x^{n - 1} \log x

Explanation:

y = x^{n - 1} \log x

\frac{d y}{d x} = y_{1} = (n - 1) x^{n - 2} \log x + \frac{x^{n - 1}}{x}

\begin{aligned} y_{1} = \frac{(n - 1) x^{n - 1} \log x + x^{n - 1}}{x} \end{aligned}

x y_{1} = (n - 1) y + x^{n - 1}

[∵ y = x^{n - 1} \log x]

x y_{2} + y_{1} = (n - 1) y_{1} + (n - 1) x^{n - 2}

\begin{aligned} x y_{2} + y_{1} = (n - 1) y_{1} + \frac{(n - 1) x^{n - 1}}{x} \end{aligned}

x^{2} y_{2} + x y_{1} = x (n - 1) y_{1} + (n - 1) x^{n - 1}

x^{2} y_{2} + x y_{1} = x (n - 1) y_{1} + (n - 1) [x y_{1} - (n - 1) y]

x^{2} y_{2} + x y_{1} = x (n - 1) y_{1} + (n - 1) x y_{1} - (n - 1)^{2} y

\begin{aligned} x^{2} y_{2} + x y_{1} = 2 x (n - 1) y_{1} - (n - 1)^{2} y \end{aligned}

x^{2} y_{2} + x y_{1} - 2 x (n - 1) y_{1} = - (n - 1)^{2} y

x^{2} y_{2} + x y_{1} (1 - 2 n + 2) = - (n - 1)^{2} y

\begin{aligned} x^{2} y_{2} + (3 - 2 n) x y_{1} = - (n - 1)^{2} y \end{aligned}

Higher Order Derivatives Exercise Multiple Choice Questions Question 25

Answer:
(c)
Hint:
We must know about the derivative of logarithm.
Given:

x y - \log_{e} y = 1

satisfy the equation

x (y y_{2} + y_{1}^{2}) - y_{2} + λ y y_{1} = 0

Explanation:

x y - \log_{e} y = 1

\begin{aligned} x y_{1} + y - \frac{y_{1}}{y} = 0 \end{aligned}

\frac{x y y_{1} + y^{2} - y_{1}}{y} = 0

\begin{aligned} x y y_{1} + y^{2} - y_{1} = 0 \end{aligned}

y y_{1} + x y_{1} y_{1} + x y y_{2} + 2 y y_{1} - y_{2} = 0

\begin{aligned} x (y_{1}^{2} + y y_{2}) - y_{2} + 3 y y_{1} = 0 \end{aligned}

Compare with given equation,

x (y y_{2} + y_{1}^{2}) - y_{2} + λ y y_{1} = 0

\begin{aligned} λ = 3 \end{aligned}

Higher Order Derivatives Exercise Multiple Choice Questions Question 26

Answer:
(a)
Hint:
We must know about the derivative values of every function.
Given:

y^{2} = a x^{2} + b x + c

Explanation:

y^{2} = a x^{2} + b x + c

2 y \frac{d y}{d x} = 2 a x + b

\begin{aligned} 2 {(\frac{d y}{d x})}^{2} + 2 y \frac{d^{2} y}{d x^{2}} = 2 a \end{aligned}

y \frac{d^{2} y}{d x^{2}} = a - {(\frac{d y}{d x})}^{2}

\begin{aligned} y \frac{d^{2} y}{d x^{2}} = a - {(\frac{2 a x + b}{2 y})}^{2} \end{aligned}

$y \frac{d^{2} y}{d x^{2}}=\frac{4 a y^{2}-(2 a x+b)^{2}}{4 y^{2}} $

\begin{aligned} 4 y^{3} \frac{d^{2} y}{d x^{2}} = 4 a (a x^{2} + b x + c) - (4 a^{2} x^{2} + 4 a x b + b^{2}) \end{aligned}

$4 y^{3} \frac{d^{2} y}{d x^{2}}=4 a c-b^{2} $

\begin{aligned} y^{3} \frac{d^{2} y}{d x^{2}} = \frac{4 a c - b^{2}}{4} \end{aligned}

= Constant

Higher Order Derivatives Excercise: 1.3

Higher Order Derivatives Exercise Multiple Choice Questions Question 27

Answer:
(a)
Hint:
We must know about the derivative rules of exponential functions.
Given:

y = A e^{5 x} + B e^{- 5 x}

Explanation:

y = A e^{5 x} + B e^{- 5 x}

\frac{d y}{d x} = A e^{5 x} \cdot 5 + B e^{- 5 x} \cdot (- 5)

\begin{aligned} \frac{d y}{d x} = 5 A e^{5 x} - 5 B e^{- 5 x} \end{aligned}

Again differentiate with respect to

x

\frac{d^{2} y}{d x^{2}} = 5 A e^{5 x} (5) - 5 B e^{- 5 x} (- 5)

\begin{aligned} \frac{d^{2} y}{d x^{2}} = 25 A e^{5 x} + 25 B e^{- 5 x} \end{aligned}

$\frac{d^{2} y}{d x^{2}}=25\left[A e^{5 x}+B e^{-5 x}\right] $

\begin{aligned} \frac{d^{2} y}{d x^{2}} = 25 y \end{aligned}

Higher Order Derivatives Exercise Multiple Choice Question Question 28

Answer:
(d)
Hint:
We must know about the derivative rules of logarithm.
Given:

y = \log_{e} (\frac{x^{2}}{e^{2}})

Explanation:

y = \log_{e} (\frac{x^{2}}{e^{2}})

Differentiate with respect to

x

\frac{d y}{d x} = \frac{1}{\frac{x^{2}}{e^{2}}} \cdot \frac{1}{e^{2}} 2 x

\begin{aligned} \frac{d y}{d x} = \frac{2}{x} \end{aligned}

Again differentiate,

\frac{d^{2} y}{d x^{2}} = 2 (\frac{- 1}{x^{2}})

\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{- 2}{x^{2}} \end{aligned}

RD Sharma Class 12th Chapter 11 MCQ material contains solutions that are designed by experts who have years of experience with CBSE question papers. As it complies with the CBSE syllabus and covers, all chapters students can refer to this material for following up and marking their progress in classes. In addition, this material can serve as an excellent guidebook that can help students excel in their exams.

Every answer from RD Sharma Class 12th Chapter 11 MCQ solutions goes through a series of checks to ensure that the result is accurate and according to the standards. Moreover, as maths contains hundreds of sums, solving every one of them is a tiring task. This is why Career360 has offered this material to enable students to study efficiently and save time. In addition, the solutions are updated to the latest version, which means that the answers are from the newest version of the book.

RD Sharma Class 12th Chapter 11 MCQ material makes revision easier as all the solutions are uploaded in one place, which is easier for students to refer to quickly. Another advantage of this material is that students can complete the portion while saving time for revision as it contains solved questions that are easy to understand. Furthermore, every solution has a step-by-step explanation which is helpful for students to understand even if they are weak at maths.

Additionally, the entire material is uploaded on Career360’s website and is accessible for three students to take advantage of this and prepare for their exams. Students can use RD Sharma Class 12th Chapter 11 MCQ solutions to find different ways to solve a question and choose the easiest one. Thousands of students have already started using this material, so everyone must try it to score well in exams.

RD Sharma Chapter-wise Solutions

NEET Highest Scoring Chapters & Topics

This ebook serves as a valuable study guide for NEET exams, specifically designed to assist students in light of recent changes and the removal of certain topics from the NEET exam.

Download E-book

Frequently Asked Questions (FAQs)

1. Does the material contain all solutions?

Yes, RD Sharma Class 12 Chapter 11 MCQ material covers the entire syllabus. This material helps students simplify and streamline their preparation.

2. Where can I find the solutions for the next exercise?

You can find the solutions to all the exercises on Career360’s website, accessible free of cost.

3. Is RD Sharma class 12 chapter 11 MCQ helpful for homework?

Yes, Class 12 RD Sharma Chapter 11 MCQ Solutions help solve homework as the teachers use the same book for assigning and taking references to evaluate their homework. The solution available helps in solving questions efficiently and takes less time.

4. Where can I find this material?

RD Sharma Class 12 Solutions Higher Order Derivatives MCQ is available on the Career360 website free of cost

5. Is the RD Sharma class 12 chapter 11 MCQ of the latest syllabus?

Yes, RD Sharma Class 12 Solutions Chapter 11 MCQ is updated to the latest version. This is why students can rest assured that the answers they are referring to are from the newest edition of the book.

Articles

NCERT Solutions for Exercise 8.2 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Apr 27, 2025

NCERT Solutions for Exercise 7.2 Class 10 Maths Chapter 7 - Coordinate Geometry

Apr 27, 2025

NCERT Solutions for Exercise 8.1 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Apr 27, 2025

NCERT Solutions for Exercise 5.2 Class 10 Maths Chapter 5 - Arithmetic Progressions

Apr 27, 2025

Allen ASAT Preparation Tips 2025 - Check ASAT Sample Papers, Tips & Syllabus

Apr 27, 2025

ANTHE Preparation Tips 2025 - How to Crack Aakash National Scholarship Exam

Apr 27, 2025

Vedantu Scholarship Test Preparation Tips 2025 - Top Essential VSAT Preparation Study Tips

Apr 27, 2025

VMC NAT Preparation Tips 2025- Study Tips for Vidyamandir Scholarship Exam

Apr 27, 2025

Get answers from students and experts

Exams

Top Schools

Products & Resources

NCERT Study Material

Exams

Colleges

Predictors

Resources

Exams

Colleges & Courses

Predictors

Resources

Exams

Colleges

Predictors

Resources

Exams

Colleges

Predictors & E-Books

Resources

Exams

Predictors & Articles

Colleges

Resources

Exams

Colleges

Resources

Exams

Resources

Top Courses & Careers

Colleges

Exams

Colleges

Resources

Quick Links

Exams

Colleges

Resources

Exams

Colleges

Resources

Diploma Colleges

Exams

Resources

Upcoming Events

Other Exams

Exams

Colleges

Top Countries

Student Visas

Exams

Colleges

Upcoming Events

Resources

Engineering Preparation

Medical Preparation

Online Courses

Products

Top Streams

Specializations

Resources

Top Providers

RD Sharma Solutions Class 12 Mathematics Chapter 11 MCQ

RD Sharma Class 12 Solutions Chapter 11 MCQ Higher Order Derivatives - Other Exercise

Higher Order Derivatives Excercise: MCQ

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

Articles

Upcoming School Exams

National Institute of Open Schooling 10th examination

National Institute of Open Schooling 12th Examination

National Means Cum-Merit Scholarship

Goa Board Higher Secondary School Certificate Examination

Telangana Open School Society Intermediate Examination

Also Read

Study Resources, Applications and Opportunities

ALLEN Coaching

NEET/JEE Coaching Scholarship

NEET Most scoring concepts

NEET Previous 10 Year Questions