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RD Sharma Solutions Class 12 Mathematics Chapter 11 VSA

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RD Sharma Solutions Class 12 Mathematics Chapter 11 VSA

RD Sharma Solutions Class 12 Mathematics Chapter 11 VSA

Updated on 27 Jan 2022, 02:55 PM IST

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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.

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