RD Sharma Solutions Class 12 Mathematics Chapter 11 VSA

# 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.

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.

## Higher Order Derivatives Excercise: VSA

Higher Order Derivatives exercise very short answer type question 1

$\lambda =n(n+1)$
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

$\lambda =n^{2}$
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

$\frac{d^{2}y}{dx^{2}}=\frac{3}{4t}$
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

\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

\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

$\frac{d^{2} y}{d x^{2}} \text { in terms of } y \text { is } y$
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

\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

$\frac{d^{2} y}{d x^{2}}=\left\{\begin{array}{l} -2,01, x<0 \end{array}\right.$
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,
$\frac{d^{2} y}{d x^{2}}=\left\{\begin{array}{l} -2,01, x<0 \end{array}\right.$

Higher Order Derivatives exercise very short answer type question 9

$\frac{d^{2} y}{d x^{2}}= \begin{cases}\frac{1}{x^{2}} & \text { if } 01\end{cases}$
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,
$y=\left\{\begin{array}{cc} -\log e^{x} & \text { if } 01 \end{array}\right.$
Diff $y$ w.r.to $x$
$\frac{d y}{d x}=\left\{\begin{array}{lc} -1 / x & \text { if } 01 \end{array}\right.$
Again diff w.r.to $x$
$\frac{d^{2} y}{d x^{2}}= \begin{cases}\frac{1}{x^{2}} & \text { if } 01\end{cases}$
Thus
$\frac{d^{2} y}{d x^{2}}= \begin{cases}\frac{1}{x^{2}} & \text { if } 01\end{cases}$

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