RD Sharma Solutions Class 12 Mathematics Chapter 11 VSA

Edited By Satyajeet Kumar | Updated on Jan 27, 2022 02:55 PM IST

RD Sharma's books are well known throughout the country for their detailed maths textbooks. They are more preferred even than NCERT materials as they are comprehensive and contain almost every concept. A majority of CBSE recommend the students to have a copy of the RD Sharma solution books. Students can learn a great deal about the subject through these books.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

This Story also Contains

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

RD Sharma Class 12th Chapter 11 VSA covers the chapter' Higher-order derivatives. It contains 9 VSAQs that are simple and easy to understand. Students can learn to solve basic differential equations through this exercise. RD Sharma solution Once students go learn the concepts in a proper way, they can start solving sums effortlessly.

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

Higher Order Derivatives Excercise: VSA

Higher Order Derivatives exercise very short answer type question 1

Answer:
$\lambda =n(n+1)$
Hint:
Double differentiate $y$ with respect to $x$ to get;
$\frac{d^{2}y}{dx^{2}}\: then \; substitute\; the\; val\! ue\; in\; x\frac{d^{2}y}{dx^{2}}=\lambda y \; to \; get\; \lambda .$
Given:
$y=ax^{n+1}+bx^{-n}$
Explanation:
It is given that
$y=ax^{n+1}+bx^{-n}\; \; \; \; \; \; \; ......(1)$
$\begin{aligned} &\frac{d y}{d x}=a(n+1) x^{n+1-1}+b(-n) x^{-n-1} \quad\left[\frac{d}{d p} p^{9}=a p^{a-1}\right]\\ &\frac{d y}{d x}=a(n+1) x^{n}-b n x^{-(n+1)}\\ &\text { Again diff. above w.r.to } x\\ &\frac{d^{2} y}{d x^{2}}=n \cdot a(n+1) x^{n-1}-b n[-(n+1)] x^{-(n+1)-1}\\ &\frac{d^{2} y}{d x^{2}}=a n(n+1) x^{n-1}-b n(n+1) x^{-(n+2)} \ldots . .(2) \end{aligned}$
We have,
$x^{2}\frac{d^{2}y}{dx^{2}}=\lambda y \; \; \; \; \; \; .....(3)$
Using (1), (2) & (3) becomes
$\begin{aligned} &\begin{aligned} &x^{2}\left[\operatorname{an}(n+1) x^{n-1}+b n(n+1) x^{-(n+2)}\right] \\ &=\lambda\left[a x^{n+1}+b x^{-n}\right] \\ &n(n+1)\left[a x^{n+1}+b x^{-n}\right]=\lambda\left[a x^{n+1}+b x^{-n}\right] \\ &\lambda=n(n+1) \end{aligned}\\ &\text { Hence, the value of } \lambda \text { is } n(n+1) \text { . } \end{aligned}$

Higher Order Derivatives exercise very short answer type question 2

Answer:
$\lambda =n^{2}$
Hint:
$Double\; di\! f\! \! f\! erentiate\; x\; with\; respect\; to\; t\; to\; get \; \frac{d^{2}x}{dt^{2}},\; then\; substitute\; the\; value\; in\; \frac{d^{2}x}{dt^{2}}=\lambda \; to\; get\; the\; val\! ue\; o\! f\; \lambda .$
Given:
$x=a\: cos\: nt-b\: sin\: nt$
Explanation:
It is given that
$x=a\: cos\: nt-b\: sin\: nt\; \; \; \; \; \; \; \; \; ....(1)$
Diff w.r.t $t$
$\frac{d x}{d t}=a(-\sin n t) n-b n \cos n t \\\\Again\; di\! f\! \! f\; w.r.t \\\\ \frac{d^{2} x}{d t^{2}}=a n(\cos n t)-b n(-\sin n t) \\\\ \frac{d^{2} x}{d t^{2}}=a n^{2} \cos n t+b n^{2} \sin n t \ldots(2) \\\\W\! e \; have, \\\\ \frac{d^{2} x}{d t^{2}}=\lambda x \ldots(3)$
Using (1) & (2), (3) becomes
$\begin{aligned} &-a n^{2} \cos n t+b n^{2} \sin n t=\lambda(a \cos n t-b \sin n t) \\ &-n^{2}(a \cos n t+b \sin n t)=\lambda(a \cos n t-b \sin n t) \\ &\therefore \lambda=-n^{2} \end{aligned}$
Hence, the value of $\lambda$ is $-n^{2}$ .

Higher Order Derivatives exercise very short answer type question 3

Answer:
$\frac{d^{2}y}{dx^{2}}=\frac{3}{4t}$
Hint:
First diff $x$ w.r.t $t$ , then
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \& \text { hence find } \frac{d^{2} y}{d x^{2}}$
Given:
$x=t^{2}\; \; and\; \; y=t^{3}$
Explanation:
It is given that
$x=t^{2}\; \; and\; \; y=t^{3}$
Diff $x$ w.r.t
$\frac{dy}{dt}=3t^{2}$
We know,
$\begin{aligned} \frac{d y}{d x}=& \frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ =& \frac{3 t^{2}}{2 t} \end{aligned}$
$\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\frac{3}{2} \frac{d(t)}{d x} \\ &=\frac{3}{2} \frac{d t}{d x} \\ &=\frac{3}{2}\left(\frac{1}{2 t}\right) \\ &=\frac{3}{4 t} \end{aligned}$
$\text { Hence, } \frac{d^{2} y}{d x^{2}}=\frac{3}{4 t}$

Higher Order Derivatives exercise very short answer type question 4

Answer:
$\begin{aligned} &\text { The value of }\frac{d^{2} y}{d x^{2}} \text { at } x=\frac{1}{2} \end{aligned}$
Hint:
First diff. $x$ &w.r.t $t$ .
$\begin{aligned} \frac{d y}{d x}=& \frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ \end{aligned}$
$\begin{aligned} Di\! f\! f\; it\; w.r.t \; x\; to\; get \frac{d^{2} y}{d x^{2}} \end{aligned}$
Given:
$x=2at \; and \; y=at^{2}$
Explanation:
It is given that
$x=2at \; and \; y=at^{2}$
Diff. $x$ w.r.t $t$
$\\\frac{d x}{d y}=2 a \\\\Di\! f\! f.\; w.r.t\; \; t \\\\\frac{d x}{d y}=2 a t \\\\Now,$
$\begin{aligned} \frac{d y}{d x}=& \frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ =& \frac{2at}{2a} \\ =&t \end{aligned}$
$\begin{aligned} &\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(t) \\ &\frac{d^{2} y}{d x^{2}}=\frac{d t}{d x} \\ &\therefore \frac{d^{2} y}{d x^{2}}=\frac{1}{2 a} \\ &{\left[\because \frac{d x}{d t}=2 a\right]} \end{aligned}$
So,
$\begin{aligned} &\left(\frac{d^{2} y}{d x^{2}}\right)_{x=\frac{1}{2}}=\frac{1}{2 a} \\ &\text { Hence, } \frac{d^{2} y}{d x^{2}} \text { at } x=\frac{1}{2} \text { is } \frac{1}{2 a} . \end{aligned}$

Higher Order Derivatives exercise very short answer type question 5

Answer:
$\begin{aligned} &\text { The value of }\frac{d^{2} y}{d x^{2}}=\frac{f^{\prime} g^{\prime \prime}-g^{\prime} f^{\prime \prime}}{f^{13}} \end{aligned}$
Hint:
Diff $x$ and $y$ w.r.t $t$
$\text { Use } \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \text { again differentiate it to get } \frac{d^{2} y}{d x^{2}}$
Given:
$x=f(t)\: and\: y=g(t)$
Explanation:
It is given that
$x=f(t)\: and\: y=g(t)$
$\frac{dx}{dt}=f'(t)\: and\: \frac{dy}{dt}=g'(t)$
So,
$\begin{aligned} &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ &\frac{d y}{d x}=\frac{g^{\prime}(t)}{f^{\prime}(t)} \end{aligned}$
Diff. w.r.t $x$
$\begin{aligned} \frac{d}{d x}\left(\frac{d y}{d x}\right) &=\frac{d}{d x}\left(\frac{g^{\prime}(t)}{f^{\prime}(t)}\right) \\ &=\frac{d}{d t}\left(\frac{g^{\prime}(t)}{f^{\prime}(t)}\right) \cdot \frac{d t}{d x} \\ &=\frac{f^{\prime} \cdot g^{\prime \prime}-g^{\prime} f^{\prime \prime}}{\left(f^{\prime}\right)^{2}} \cdot\left(\frac{1}{f^{\prime}(t)}\right) \\ \frac{d^{2} y}{d x^{2}}=\frac{f^{\prime} g^{\prime \prime}-g^{\prime} f^{\prime \prime}}{f^{13}} \end{aligned}$
$\begin{aligned} &\text { Thus the value of }\frac{d^{2} y}{d x^{2}}=\frac{f^{\prime} g^{\prime \prime}-g^{\prime} f^{\prime \prime}}{f^{13}} \end{aligned}$

Higher Order Derivatives exercise very short answer type question 6

Answer:
$\frac{d^{2} y}{d x^{2}} \text { in terms of } y \text { is } y$
Hint:
differentiate the equation of $y$ two times to get
$\frac{d^{2}y}{dx}$
& compare its value with $y$
Given:
$y=1-x+\frac{x^{2}}{2 !}+\frac{-x^{3}}{3 !}+\frac{x^{4}}{4 !} \ldots . .\: to\: \infty$
Explanation:
it is given that
$y=1-x+\frac{x^{2}}{2 !}+\frac{-x^{3}}{3 !}+\frac{x^{4}}{4 !} \ldots . \: to\: \infty \; \; \; \; \; \;\; \; \; \ldots\ldots(1)$
Diff $y$ w.r.t to $x$
$\begin{aligned} &\frac{d y}{d x}=-1+\frac{2 x}{2 !}-\frac{-3 x^{2}}{3 !}+\frac{4 x^{3}}{4 !}-\ldots \ldots \text { to } \infty \\&\text { Again diff it w.r.t } x\\ &\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{2}{2 !}-\frac{-6 x^{2}}{3 !}+\frac{4 \cdot 3 \cdot x^{2}}{4 !}-\ldots \ldots \text { To } \infty \\ &\quad=\quad 1-\frac{6 x}{6}+\frac{12 x^{2}}{4 \times 3 \times 2 !}-\ldots . . \text { to } \infty \\&\quad=\quad 1-x+\frac{x^{2}}{2!}-\ldots \text { to } \infty \\ &\quad=\quad \therefore \frac{d^{2} y}{d x}=y \ldots .\; \; \; \; \; \; \; \; (F\! rom\; 1) \end{aligned} \end{aligned}$
$Hence,\; \; \frac{d^{2} y}{d x^{2}} = y$

Higher Order Derivatives exercise very short answer type question 7

Answer:
$\begin{aligned} &\text { The value of } \frac{d^{2} x}{d y^{2}} \text { is } \frac{-e^{x}}{\left(1+e^{x}\right)^{3}} \end{aligned}$
Hint:
First differentiate given equation w.r.t to $x$
$\begin{aligned} &\text { Use } \frac{d y}{d x} =\frac{1}{\frac{dy}{dx}} \end{aligned}$
Given:
$y=x+e^{x}$
Explanation:
It is given that
$y=x+e^{x}$
Diff w. r .t be $x$
$\begin{aligned} &\frac{d y}{d x}=1+e^{x} \\ &\therefore \frac{d y}{d x}=\frac{1}{\frac{d y}{d x}} \\ &\therefore \frac{d x}{d y}=\frac{1}{1+e^{x}} \end{aligned}$
Diff w.r. to $y$
$\begin{aligned} &\frac{d}{d y}\left(\frac{d x}{d y}\right)=\frac{d}{d y}\left(\frac{1}{1+e^{x}}\right) \\\\ &\frac{d^{2} x}{d y^{2}}=\frac{d}{d x}\left(\frac{1}{1+e^{x}}\right) \frac{d x}{d y} \\ &=\frac{-1}{\left(1+e^{x}\right)^{2}} e^{x}\left(\frac{1}{1+e^{x}}\right) \\ &=-\left(\frac{1}{1+e^{x}}\right)^{3} e^{x} \\ &\quad \\ &=\frac{-e^{x}}{\left(1+e^{x}\right)^{3}} \end{aligned}$
$\begin{aligned} Thus,\; \; \frac{d^{2}x}{dy^{2}}=\frac{-e^{x}}{\left(1+e^{x}\right)^{3}} \end{aligned}$

Higher Order Derivatives exercise very short answer type question 8

Answer:
$\frac{d^{2} y}{d x^{2}}=\left\{\begin{array}{l} -2,0<x<1 \\ 2, x>1, x<0 \end{array}\right.$
Hint:
$\text { if } y=|p| \: {\text {then }} y=\left\{\begin{array}{c} p, \text { if } 0 \leq p \leq 1 \\ -p, \text { if } p<0 \end{array}\right.$
Given:
$y=\left | x-x^{2} \right |$
Explanation:
It is given that
$y=\left | x-x^{2} \right |$
This can be written as
$y=\left\{\begin{array}{c} x-x^{2} \text { if } 0 \leq x \leq 1 \\ -\left(x-x^{2}\right) \text { if } x<0 \text { or } x>1 \end{array}\right.$
Diff $y$ w.r to $x$
$\frac{d y}{d x}=\left\{\begin{array}{c} 1-2 x \text { if } 0<\mathrm{x} \leq 1 \\ -(1-2 x) \text { if } x>0 \text { or } x>1 \end{array}\right.$
$Di\! f\! f\; w.r.t\; to\; x$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\left\{\begin{array}{c} -2 \text { if } 0 \leq x \leq 1 \\ -(-2) \text { if } x<0 \text { or } x>1 \end{array}\right. \end{aligned}$
Thus,
$\frac{d^{2} y}{d x^{2}}=\left\{\begin{array}{l} -2,0<x<1 \\ 2, x>1, x<0 \end{array}\right.$

Higher Order Derivatives exercise very short answer type question 9

Answer:
$\frac{d^{2} y}{d x^{2}}= \begin{cases}\frac{1}{x^{2}} & \text { if } 0<x<1 \\ \frac{-1}{x^{2}} & \text { if } x>1\end{cases}$
Hint:
$\text { if } y=|p| \text { then } y=\left\{\begin{array}{c} p \text { if } 0<\mathrm{p}<1 \\ -p\; p<0 \text { or } p>1 \end{array}\right.$
Given:
$y=|log\: ex|$
Explanation:
It is given that
$y=|log\: ex|$
This can be written as,
$y=\left\{\begin{array}{cc} -\log e^{x} & \text { if } 0<x<1 \\ \log e^{x} & \text { if } x>1 \end{array}\right.$
Diff $y$ w.r.to $x$
$\frac{d y}{d x}=\left\{\begin{array}{lc} -1 / x & \text { if } 0<x<1 \\ 1 / x & \text { if } x>1 \end{array}\right.$
Again diff w.r.to $x$
$\frac{d^{2} y}{d x^{2}}= \begin{cases}\frac{1}{x^{2}} & \text { if } 0<x<1 \\ \frac{-1}{x^{2}} & \text { if } x>1\end{cases}$
Thus
$\frac{d^{2} y}{d x^{2}}= \begin{cases}\frac{1}{x^{2}} & \text { if } 0<x<1 \\ \frac{-1}{x^{2}} & \text { if } x>1\end{cases}$

RD Sharma Class 12th Chapter 11 VSA solutions available at the Career360 website covers all the solutions for the questions given in the textbook. As teachers can't cover all topics in their lectures, Career360 has provided this material for easy reference.

The following are the advantages of these solutions:

1. Expert-created solutions

RD Sharma Class 12th Chapter 11 VSA solutions provided by Career360 are expert-created, which means that they contain accurate answers. Moreover, as it is based on the CBSE board and covers the entire syllabus, students can refer to it to stay on par with their lectures.

2. Easy reference

The students can rely on the RD Sharma Class 12th Chapter 11 VSA material as it contains all the solved sums of the particular chapter. No other additional reference material is required.

3. Convenient to use

Being available in PDF form, the students need not carry a big solution book wherever they go. The material can be stored in their mobile phones that are easily portable.

4. Easy to track the progress

With the help of RD Sharma Class 12th Chapter 11 VSA material. The students can easily track their progress. This can be done by analyzing how much their scores in mock tests have improved.

5. Free of cost

The biggest advantage of using the RD Sharma Class 12 Chapter 11 VSA solutions book is that, it is easily available for the students to download from the career 360 website.

RD Sharma Chapter-wise Solutions

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book

Frequently Asked Questions (FAQs)

1. Who can use this material?

Students who follow CBSE syllabus can use Class 12 RD Sharma Chapter 11 VSA Solutions to ease their preparation and score well in exams

2. Can the class 12 students rely on the RD Sharma solution books for reference?

The RD Sharma reference materials are the most used set of books by the 12th graders. Many students have admitted that this material was of great help to them, this makes the book more reliable.

3. At what price is the RD Sharma class 12 Chapter 11 solution available online?

RD Sharma Class 12 Solutions Higher Order Derivatives VSA solutions are free of cost and available on the official website of Career360.

4. Can I finish my homework using these solutions?

As RD Sharma Class 12 Solutions Chapter 11 VSA material covers the entire syllabus, students can use it as a guide to complete their homework.

5. Can I trust these solutions?

Yes, the RD Sharma Solution materials can be trusted as all the answers are provided by the mathematical experts. The students need not worry regarding the accuracy of the solved sums in this book.