RD Sharma Solutions Class 12 Mathematics Chapter 11 VSA

Edited By Satyajeet Kumar | Updated on Jan 27, 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
$\frac{d^{2} y}{d x^{2}}= \begin{cases}\frac{1}{x^{2}} & \text { if } 0<x<1 \\ \frac{-1}{x^{2}} & \text { if } x>1\end{cases}$

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