JEE Practice Q & A (Derivatives: Single Choice Type)

Hint

(a)

\(
\begin{aligned}
y & =\tan ^{-1}\left\{\sqrt{\frac{1+\cos x}{1-\cos x}}\right\} \\
& =\tan ^{-1}\left\{\sqrt{\frac{2 \cos ^2 x / 2}{2 \sin ^2 x / 2}}\right\} \\
& =\tan ^{-1}\left|\cot \frac{x}{2}\right|=\tan ^{-1}\left(\cot \frac{x}{2}\right) \\
\Rightarrow y & =\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right\}=\frac{\pi}{2}-\frac{x}{2} \\
\therefore \frac{d y}{d x} & =0-\frac{1}{2}=-\frac{1}{2}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& f(x)=\left|x^2-5 x+6\right|=\left\{\begin{array}{cc}
x^2-5 x+6 & \text { if } x \geq 3 \text { or } x \leq 2 \\
-\left(x^2-5 x+6\right), & \text { if } 2<x<3
\end{array}\right. \\
& \Rightarrow \quad f^{\prime}(x)= \begin{cases}(2 x-5), & \text { if } x>3 \text { or } x<2 \\
-(2 x-5), & \text { if } 2<x<3\end{cases}
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& \text { We have } y=\tan ^{-1}\left(\frac{\log e-\log x^2}{\log e+\log x^2}\right)+\tan ^{-1}\left(\frac{3+2 \log x}{1-6 \log x}\right) \\
& =\tan ^{-1}\left(\frac{1-2 \log x}{1+2 \log x}\right)+\tan ^{-1}\left(\frac{3+2 \log x}{1-6 \log x}\right) \\
& =\tan ^{-1} 1-\tan ^{-1}(2 \log x)+\tan ^{-1} 3+\tan ^{-1}(2 \log x) \\
& =\tan ^{-1} 1+\tan ^{-1} 3 \\
& \Rightarrow \frac{d y}{d x}=0 \Rightarrow \frac{d^2 y}{d x^2}=0
\end{aligned}
\)

Hint

(c)
\(
\begin{aligned}
& y^{\prime}(x)=f^{\prime}(f(f(f(x)))) f^{\prime}(f(f(x))) f^{\prime}(f(x)) f^{\prime}(x) \\
& \Rightarrow y^{\prime}(0)=f^{\prime}(f(f(f(0)))) f^{\prime}(f(f(0))) f^{\prime}(f(0)) f^{\prime}(0) \\
& =f^{\prime}(f(f(0))) f^{\prime}(f(0)) f^{\prime}(0) f^{\prime}(0)
\end{aligned}
\)
\(
\begin{aligned}
& =f^{\prime}(f(0)) f^{\prime}(0) f^{\prime}(0) f^{\prime}(0) \\
& =f^{\prime}(0) f^{\prime}(0) f^{\prime}(0) f^{\prime}(0) \\
& =\left(f^{\prime}(0)\right)^4=2^4=16
\end{aligned}
\)

Hint

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

Hint

(c)

\(
\begin{aligned}
& y=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!} \\
& \Rightarrow \frac{d y}{d x}=0+1+x+\frac{x^2}{2!}+\cdots+\frac{x^{n-1}}{(n-1)!} \\
& \Rightarrow \frac{d y}{d x}+\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!} \\
& \Rightarrow \frac{d y}{d x}=y-\frac{x^n}{n!}
\end{aligned}
\)

Hint

(d) \(y=a \sin x+b \cos x\)
Differentiating with respect to \(x\), we get
\(
\frac{d y}{d x}=a \cos x-b \sin x
\)
Now,
\(
\begin{aligned}
\left(\frac{d y}{d x}\right)^2 & =(a \cos x-b \sin x)^2 \\
& =a^2 \cos ^2 x+b^2 \sin ^2 x-2 a b \sin x \cos x, \text { and } \\
y^2 & =(a \sin x+b \cos x)^2 \\
& =a^2 \sin ^2 x+b^2 \cos ^2 x+2 a b \sin x \cos x
\end{aligned}
\)
\(
\begin{aligned}
& \text { So, }\left(\frac{d y}{d x}\right)^2+y^2=a^2\left(\sin ^2 x+\cos ^2 x\right)+b^2\left(\sin ^2 x+\cos ^2 x\right) \\
& \Rightarrow\left(\frac{d y}{d x}\right)^2+y^2=\left(a^2+b^2\right)=\text { constant. }
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& y=\sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}}=\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{1-\tan x}{1+\tan x}=\tan \left(\frac{\pi}{4}-x\right) \\
& \Rightarrow \frac{d y}{d x}=-\sec ^2\left(\frac{\pi}{4}-x\right)
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left[\left(x+\sqrt{x^2+a^2}\right)^n\right] \\
& =n\left(x+\sqrt{x^2+a^2}\right)^{n-1} \cdot \frac{d}{d x}\left(x+\sqrt{x^2+a^2}\right) \\
& =n\left(x+\sqrt{x^2+a^2}\right)^{n-1}\left(\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}\right)
\end{aligned}
\)

\(
\begin{aligned}
& =\frac{n\left(x+\sqrt{x^2+a^2}\right)^n}{\sqrt{x^2+a^2}} \\
& =\frac{n y}{\sqrt{x^2+a^2}}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) } f(x)=\sqrt{1+\cos ^2\left(x^2\right)} \\
& \Rightarrow f^{\prime}(x)=\frac{1}{2 \sqrt{1+\cos ^2(x)^2}}\left(2 \cos x^2\right)\left(-\sin x^2\right)(2 x) \\
& \Rightarrow f^{\prime}(x)=\frac{-x \sin 2 x^2}{\sqrt{1+\cos ^2\left(x^2\right)}} \\
& \Rightarrow f^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)=\frac{-\frac{\sqrt{\pi}}{2} \sin \frac{2 \pi}{4}}{\sqrt{1+\cos ^2 \frac{\pi}{4}}}=\frac{-\frac{\sqrt{\pi}}{2} 1}{\sqrt{\frac{3}{2}}} \\
& \therefore f^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)=-\sqrt{\frac{\pi}{6}}
\end{aligned}
\)

Hint

(a)

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

Hint

(c)

\(
\begin{aligned}
&\begin{aligned}
& y=\frac{\log \tan x}{\log \sin x} \\
& \frac{d y}{d x}=\frac{(\log \sin x)\left(\frac{\sec ^2 x}{\tan x}\right)-(\log \tan x)(\cot x)}{(\log \sin x)^2}
\end{aligned}\\
&\text { (On simplification) }
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& y=\frac{\sin ^{-1} x}{\sqrt{1-x^2}} \\
& \frac{d y}{d x}=\frac{\sqrt{1-x^2} \frac{1}{\sqrt{1-x^2}}-\left(\sin ^{-1} x\right) \frac{1}{2} \frac{(-2 x)}{\sqrt{1-x^2}}}{1-x^2} \\
& \left(1-x^2\right) \frac{d y}{d x}=1+x\left(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\right)=1+x y
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& y=\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right] \\
& =\cot ^{-1}\left[\frac{2+2 \cos x}{2 \sin x}\right]=\cot ^{-1}\left[\frac{1+\cos x}{\sin x}\right] \\
& =\cot ^{-1}\left[\cot \frac{x}{2}\right]=\frac{x}{2}
\end{aligned}
\)
\(
\therefore \frac{d y}{d x}=\frac{1}{2}
\)

Hint

(c)

\(
\begin{aligned}
& y=x^{\left(x^x\right)} \\
& \log y=x^x \log x \\
& \frac{1}{y} \frac{d y}{d x}=\frac{d z}{d x} \log x+\frac{1}{x} z\left(\text { where } x^x=z\right) \\
& \frac{d y}{d x}=x^{\left(x^x\right)}\left[x^x(\log e x) \log x+x^{x-1}\right] \left(\because \frac{d z}{d x}=x^x \log e x\right)
\end{aligned}
\)

Hint

(b) Let \(y=\sin ^2 \cot ^{-1}\left\{\sqrt{\frac{1-x}{1+x}}\right\}\)
Put \(x=\cos \theta \Rightarrow \theta=\cos ^{-1} x\)
\(
\begin{aligned}
& \Rightarrow y=\sin ^2 \cot ^{-1}\left\{\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\right\}=\sin ^2 \cot ^{-1}\left(\tan \frac{\theta}{2}\right) \\
& \Rightarrow y=\sin ^2\left(\frac{\pi}{2}-\frac{\theta}{2}\right) \\
& =\cos ^2\left(\frac{\theta}{2}\right)=\frac{1+\cos \theta}{2}=\frac{1+x}{2} \\
& \Rightarrow \frac{d y}{d x}=\frac{1}{2}
\end{aligned}
\)

Hint

(c)
\(
\begin{aligned}
y & =a e^{m x}+b e^{-m x} \\
\Rightarrow & \frac{d y}{d x}=a m e^{m x}-m b e^{-m x}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Again } \frac{d^2 y}{d x^2}=a m^2 e^{m x}+m^2 b e^{-m x} \\
& \quad \Rightarrow \frac{d^2 y}{d x^2}=m^2\left(a e^{m x}+b e^{-m x}\right) \Rightarrow \frac{d^2 y}{d x^2}=m^2 y \\
& \quad \Rightarrow \frac{d^2 y}{d x^2}-m^2 y=0
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& \text { Let } y=\log x \\
& \Rightarrow y_1=\frac{1}{x}, y_2=\frac{-1}{x^2}, y_3=\frac{2}{x^3}, \cdots, y_n=\frac{(-1)^{n-1}(n-1)!}{x^n} .
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& y=\sqrt{\log x+y} \\
& \Rightarrow y^2=\log x+y \\
& 2 y \frac{d y}{d x}=\frac{1}{x}+\frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{x(2 y-1)}
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& f\left(\log _e x\right)=\log _e\left(\log _e x\right) \\
& \therefore \frac{d f\left(\log _e x\right)}{d x}=\frac{1}{\log _e x} \times \frac{1}{x}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&y=\sec \left(\tan ^{-1} x\right)=\sec \left(\sec ^{-1} \sqrt{1+x^2}\right) \doteq \sqrt{1+x^2}\\
&\text { Differentiating w.r.t. } x \text {, we have } \frac{d y}{d x}=\frac{x}{\sqrt{i+x^2}}\\
&\Rightarrow\left(\frac{d y}{d x}\right)_{x=1}=\frac{1}{\sqrt{2}}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& y=f\left(x^2\right) \Rightarrow \frac{d y}{d x}=f^{\prime}\left(x^2\right) 2 x=2 x \sqrt{2\left(x^2\right)^2-1} \\
& \text { At } x=1, \frac{d y}{d x}=2 \times 1 \times \sqrt{2-1}=2
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{f^{\prime}\left(x^3\right) 3 x^2}{g^{\prime}\left(x^2\right) 2 x} & =\frac{\cos x^3 3 x^2}{\sin x^2 2 x} \\
& =\frac{3}{2} x \cos x^3 \operatorname{cosec} x^2
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& \frac{d x}{d y}=\frac{\frac{d x}{d t}}{\frac{d y}{d t}}=\frac{\cos t-t \sin t}{1+\cos t} \\
& \therefore \frac{d^2 x}{d y^2}=\frac{\frac{d}{d t}\left(\frac{d x}{d y}\right)}{\frac{d y}{d t}} \\
& =\frac{\frac{(-2 \sin t-t \cos t)(1+\cos t)-(\cos t-t \sin t)(-\sin t)}{(1+\cos t)^2}}{1+\cos t}
\end{aligned}
\)
\(
\text { Now, put } t=\pi / 2, \text { we get } = -\frac{\pi+4}{2}
\)

Hint

(c)
\(
\begin{aligned}
f(x) & =\sqrt{1-\sin 2 x}=\sqrt{(\cos x-\sin x)^2} \\
& =|\cos x-\sin x|
\end{aligned}
\)
\(
\begin{aligned}
& =\left\{\begin{array}{cc}
\cos x-\sin x, & \text { for } 0 \leq x \leq \pi / 4 \\
-(\cos x-\sin x), & \text { for } \pi / 4<x \leq \pi / 2
\end{array}\right. \\
\therefore f^{\prime}(x) & =\left\{\begin{array}{lc}
-(\cos x+\sin x), & \text { for } 0<x<\pi / 4 \\
(\cos x+\sin x), & \text { for } \pi / 4<x<\pi / 2 .
\end{array}\right.
\end{aligned}
\)

Hint

(d) Let \(u=y^2\) and \(v=x^2\)
\(
\begin{aligned}
& \therefore \frac{d u}{d x}=\frac{d^1}{d x} y^2=\left(\frac{d}{d y} y^2\right)\left(\frac{d y}{d x}\right) \\
& =2 y(1-2 x)=2\left(x-x^2\right)(1-2 x)=2 x(1-x)(1-2 x) \dots(1)
\end{aligned}
\)
and \(\frac{d v}{d x}=2 x \dots(2)\)
Hence, \(\frac{d u}{d v}=\frac{\left(\frac{d u}{d x}\right)}{\left(\frac{d v}{d x}\right)}=\frac{2 x(1-x)(1-2 x)}{2 x}\) (from (1) and (2))
\(
=(1-x)(1-2 x)=1-3 x+2 x^2
\)

Hint

(a)
\(
\begin{aligned}
f(x) & =\cos ^{-1}\left[\cos \left(\frac{\pi}{2}-\sqrt{\frac{1+x}{2}}\right)\right]+x^x=\frac{\pi}{2}-\sqrt{\frac{1+x}{2}}+x^x \\
\Rightarrow f^{\prime}(x) & =-\frac{1}{\sqrt{2}} \times \frac{1}{2 \sqrt{1+x}}+x^x(1+\log x) \\
\Rightarrow f^{\prime}(1) & =-\frac{1}{4}+1=\frac{3}{4}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
D & =\left|\begin{array}{ccc}
\sin p x & p \cos p x & -p^2 \sin p x \\
-p^3 \cos p x & p^4 \sin p x & p^5 \cos p x \\
-p^6 \sin p x & -p^7 \cos p x & p^8 \sin p x
\end{array}\right| \\
& =p^9\left|\begin{array}{ccc}
\sin p x & p \cos p x & -p^2 \sin p x \\
-\cos p x & p \sin p x & p^2 \cos p x \\
-\sin p x & -p \cos p x & p^2 \sin p x
\end{array}\right| \\
& =-p^9\left|\begin{array}{ccc}
\sin p x & p \cos p x & -p^2 \sin p x \\
\cos p x & p \sin p x & p^2 \cos p x \\
\sin p x & p \cos p x & -p^2 \sin p x
\end{array}\right|=0
\end{aligned}
\)

Hint

(b) \(2 x f^{\prime}\left(x^2\right)=3 x^2 \Rightarrow 4 f^{\prime}(2)=12 \Rightarrow f^{\prime}(4)=3 .\)

Hint

(c)

\(
\begin{aligned}
& \left(a^2-2 a-15\right) e^{a x}+\left(b^2-2 b-15\right) e^{b x}=0 \\
& \Rightarrow\left(a^2-2 a-15\right)=0 \text { and } b^2-2 b-15=0 \\
& \Rightarrow(a-5)(a+3)=0 \text { and }(b-5)(b+3)=0 \\
& \Rightarrow a=5 \text { or }-3 \text { and } b=5 \text { or }-3 \\
& \therefore a \neq b \text { hence } a=5 \text { and } b=-3 \\
& \text { or } a=-3 \text { and } b=5 \\
& \Rightarrow a b=-15
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& y=\frac{(a-x)^{3 / 2}+(x-b)^{3 / 2}}{\sqrt{a-x}+\sqrt{x-b}} \\
& \frac{(\sqrt{a-x}+\sqrt{x-b})(a-x-\sqrt{a-x} \sqrt{x-b}+x-b)}{\sqrt{a-x}+\sqrt{x-b}} \\
& a-b-\sqrt{a-x} \sqrt{x-b} \\
& \frac{d y}{d x}=\frac{1}{2 \sqrt{a-x}} \sqrt{x-b}-\frac{1}{2 \sqrt{x-b}} \sqrt{a-x} \\
& \frac{2 x-a-b}{2 \sqrt{a-x} \sqrt{x-b}}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& f(g(x))=x \\
& f^{\prime}(g(x)) g^{\prime}(x)=1 \\
& \left(e^{g(x)}+1\right) g^{\prime}(x)=1 \\
& \left(e^{g(f(\log 2))}+1\right) g^{\prime}(f(\log 2))=1 \\
& \left(e^{\log 2}+1\right) g^{\prime}(f(\log 2))=1 \\
& g^{\prime}(f(\log 2))=1 / 3
\end{aligned}
\)

Hint

(c)
\(
\begin{aligned}
& f^{\prime}(x)=\left(k x+e^x\right) h^{\prime}(x)+h(x)\left(k+e^x\right) \\
& f^{\prime}(0)=h^{\prime}(0)+h(0)(k+1) \\
& 18=-2+5(k+1) \Rightarrow k=3
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (d) } y=\tan ^{-1}\left(\frac{2^{x+1}-2^x}{1+2^x \cdot 2^{x+1}}\right)=\tan ^{-1} 2^{(x+1)}-\tan ^{-1} \cdot 2^x \\
& \Rightarrow y^{\prime}=\frac{2^{x+1} \ln 2}{1+\left(2^{x+1}\right)^2}-\frac{2^x \ln 2}{1+\left(2^x\right)^2} \\
& \Rightarrow y^{\prime}(0)=-\frac{1}{10} \ln 2
\end{aligned}
\)

Hint

(b) \(f(x)=1+x^2+x^4+x^6+\cdots \infty\), where \(|x| \leq 1\)
\(\Rightarrow f^n(0)=n!\), where \(n\) is even.

Hint

\(
\begin{aligned}
& \text { (b) } y=2 \cos x \cos 3 x=\cos 4 x+\cos 2 x \\
& \Rightarrow \frac{d^{20} y}{d x^{20}}=4^{20} \cos 4 x+2^{20} \cos 2 x
\end{aligned}
\)

Hint

(c) We have \(y=\sqrt{\frac{1-x}{1+x}}\)
Differentiating w.r.t. \(x\), we get
\(
\begin{aligned}
\frac{d y}{d x} & =\frac{1}{2}\left(\frac{1-x}{1+x}\right)^{1 / 2-1} \frac{d}{d x}\left(\frac{1-x}{1+x}\right) \\
& =\frac{1}{2} \sqrt{\frac{1+x}{1-x}} \times \frac{(1+x)(-1)-(1-x)(1)}{(1+x)^2} \\
& =-\sqrt{\frac{1+x}{1-x}} \frac{1}{(1+x)^2} \\
\Rightarrow & \left(1-x^2\right) \frac{d y}{d x}=-\sqrt{\frac{1+x}{1-x}} \frac{1}{(1+x)^2}\left(1-x^2\right) \\
\Rightarrow & (1-x)^2 \frac{d y}{d x}=-\sqrt{\frac{1-x}{1+x}} \\
\Rightarrow & \left(1-x^2\right) \frac{d y}{d x}=-y \\
\Rightarrow & \left(1-x^2\right) \frac{d y}{d x}+y=0
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& \frac{d y}{d x}=-[(2-x)(3-x) \cdots(n-x)+(1-x)(3-x) \cdots \\
& (n-x)+\cdots(1-x)(2-x) \cdots(n-1-x)] \\
& \text { At } x=1 \\
& \frac{d y}{d x}=-[(n-1)!+0+\cdots+0]=(-1)(n-1)!
\end{aligned}
\)

Hint

(a) Let \(\cos \alpha=\frac{5}{13}\), then \(\sin \alpha=\frac{12}{13}\).
So, \(y=\cos ^{-1}(\cos \alpha \cdot \cos x-\sin \alpha \cdot \sin x)\)
\(\Rightarrow y=\cos ^{-1}\{\cos (x+\alpha)\}=x+\alpha \quad(\because x+\alpha\) is in the first or the second quadrant)
\(
\Rightarrow \frac{d y}{d x}=1
\)

Hint

(a)

\(
\begin{aligned}
&\text { Let } x=\sec \theta\\
&\begin{aligned}
& \text { Then } \begin{aligned}
y & =\tan ^{-1} \sqrt{\frac{\sec \theta+1}{\sec \theta-1}} \\
& =\tan ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\tan ^{-1}\left(\cot \frac{\theta}{2}\right) \\
\Rightarrow y= & \tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right\}=\frac{\pi}{2}-\frac{1}{2} \sec ^{-1} x \\
\Rightarrow \frac{d y}{d x} & =-\frac{1}{2} \times \frac{1}{|x| \sqrt{x^2-1}}
\end{aligned}
\end{aligned}
\end{aligned}
\)

Hint

(d) We have \(\sin ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\log a\).
\(
\begin{aligned}
& \Rightarrow \frac{x^2-y^2}{x^2+y^2}=\sin (\log a) \\
& \Rightarrow \frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}=\sin (\log a), \text { on putting } y=x \tan \theta \\
& \Rightarrow \cos 2 \theta=\sin (\log a) \\
& \Rightarrow 2 \theta=\cos ^{-1}(\sin (\log a)) \\
& \Rightarrow \theta=\frac{1}{2} \cos ^{-1}(\sin (\log a)) \\
& \Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)=\frac{1}{2} \cos ^{-1}(\sin (\log a)) \\
& \Rightarrow \frac{y}{x}=\tan \left(\frac{1}{2} \cos ^{-1}(\sin (\log a))\right)
\end{aligned}
\)
Differentiating w.r.t. \(x\)
\(
\begin{aligned}
& \Rightarrow \frac{x \frac{d y}{d x}-y}{x^2}=0 \\
& \Rightarrow x \frac{d y}{d x}-y=0 \\
& \Rightarrow \frac{d y}{d x}=\frac{y}{x}
\end{aligned}
\)

Hint

(b)
\(
\begin{aligned}
& y=\cos ^{-1}(\cos x)=\cos ^{-1}\{\cos [2 \pi-(2 \pi-x)]\} \\
& =\cos ^{-1}[\cos (2 \pi-x)] \\
& =2 \pi-x \\
& \therefore \frac{d y}{d x}=-1 \quad \text { at } x=\frac{5 \pi}{4}
\end{aligned}
\)

Hint

(a) Let \(t=\cos 2 \theta\)
Then \(e^x=\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}\)
\(
=\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\tan \theta}{1+\tan \theta}=\tan \left(\frac{\pi}{4}-\theta\right)
\)
\(
\tan \frac{y}{2}=\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}=\tan \theta
\)
At \(t=\frac{1}{2}, \cos 2 \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{6}\)
Then \(x=\log \tan \frac{\pi}{12}, y=\frac{\pi}{3}\)
Differentiating w.r.t. \(\theta, e^x \frac{d x}{d \theta}=-\sec ^2\left(\frac{\pi}{4}-\theta\right)\) and
\(
\frac{1}{2} \sec ^2 \frac{y}{2} \frac{d y}{d \theta}=\sec ^2 \theta
\)
\(
\therefore \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{2 \sec ^2 \theta \cos ^2 \frac{y}{2}}{-e^{-x} \sec ^2\left(\frac{\pi}{4}-\theta\right)}
\)
\(
\begin{aligned}
& \text { At } t=\frac{1}{2}, \text { i.e., } \theta=\frac{\pi}{6}, \frac{d y}{d x}=\frac{2 \sec ^2 \frac{\pi}{6} \cos ^2 \frac{\pi}{6}}{-e^{-\log \tan \pi / 12} \sec ^2 \frac{\pi}{12}} \\
& \therefore \frac{d y}{d x}=\frac{2}{-\cot \frac{\pi}{12} \sec ^2 \frac{\pi}{12}} \\
& =-2 \tan \frac{\pi}{12} \cos ^2 \frac{\pi}{12}=-\sin \frac{\pi}{6}=-\frac{1}{2}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\text { We have } x^2+y^2=t-\frac{1}{t} \text { and } x^4+y^4=t^2+\frac{1}{t^2}\\
&\begin{aligned}
& \Rightarrow\left(x^2+y^2\right)^2=t^2+\frac{1}{t^2}-2 \\
& \Rightarrow\left(x^2+y^2\right)^2=x^4+y^4-2 \\
& \Rightarrow 2 x^2 y^2=-2 \\
& \Rightarrow x^2 y^2=-1 \\
& \Rightarrow y^2=-\frac{1}{x^2} \\
& \Rightarrow 2 y \frac{d y}{d x}=\frac{2}{x^3} \\
& \Rightarrow x^3 y \frac{d y}{d x}=1
\end{aligned}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\text { We have }\\
&\begin{aligned}
& y^{1 / m}=\left(x+\sqrt{1+x^2}\right) \\
\Rightarrow & y=\left(x+\sqrt{1+x^2}\right)^m \\
\Rightarrow & \frac{d y}{d x}=m\left(x+\sqrt{1+x^2}\right)^{m-1}\left(1+\frac{x}{\sqrt{x^2+1}}\right) \\
& =m \frac{\left(x+\sqrt{1+x^2}\right)^m}{\sqrt{1+x^2}} \\
\Rightarrow & \frac{d y}{d x}=\frac{m y}{\sqrt{1+x^2}} \\
\Rightarrow & y_1^2\left(1+x^2\right)=m^2 y^2 \\
\Rightarrow & 2 y_1 y_2\left(1+x^2\right)+2 x y_1^2=2 m^2 y y_1 \\
\Rightarrow & y_2\left(1+x^2\right)+x y_1=m^2 y
\end{aligned}
\end{aligned}
\)

Hint

(a)
\(
\begin{aligned}
& y=f(x)-f(2 x) \Rightarrow y^{\prime}=f^{\prime}(x)-2 f^{\prime}(2 x) \\
& \Rightarrow y^{\prime}(1)=f^{\prime}(1)-2 f^{\prime}(2)=5, \text { and } \dots(1) \\
& y^{\prime}(2)=f^{\prime}(2)-2 f^{\prime}(4)=7 \dots(2)
\end{aligned}
\)
Now
\(
\begin{aligned}
& \text { let } y=f(x)-f(4 x) \\
& \Rightarrow y^{\prime}=f^{\prime}(x)-4 f^{\prime}(4 x) \\
& \Rightarrow y^{\prime}(1)=f^{\prime}(1)-4 f^{\prime}(4) \dots(3)
\end{aligned}
\)
Substituting the value of \(f^{\prime}(2)=7+2 f^{\prime}(4)\) in (1), we get
\(
\begin{aligned}
& f^{\prime}(1)-2\left(7+2 f^{\prime}(4)\right)=5 \\
& f^{\prime}(1)-4 f^{\prime}(4)=19
\end{aligned}
\)

Hint

(a)
\(
\begin{aligned}
f(10)=\sin ^{-1} \cos 10 & =\sin ^{-1} \sin \left(\frac{\pi}{2}-10\right) \\
& =-\sin ^{-1} \sin \left(10-\frac{\pi}{2}\right)
\end{aligned}
\)
\(
\begin{aligned}
& =-\sin ^{-1} \sin \left(3 \pi-10+\frac{\pi}{2}\right)=-\left(3 \pi+\frac{\pi}{2}-10\right)=10-\frac{7 \pi}{2} \\
& f^{\prime}(x)=\frac{-\sin x}{\sqrt{1-\cos ^2 x}}=\frac{-\sin x}{|\sin x|} \Rightarrow f^{\prime}(10)=\frac{-\sin 10}{|\sin 10|}=1 . \\
& \text { So, } f(10)+f^{\prime}(10)=11-\frac{7 \pi}{2}
\end{aligned}
\)

Hint

(a)
\(
\begin{aligned}
& (\sin x)(\cos y)=\frac{1}{2} \\
& \Rightarrow(\cos x)(\cos y)-\sin y \sin x \frac{d y}{d x}=0 \\
& \Rightarrow \frac{d y}{d x}=(\cot x)(\cot y) \\
& \Rightarrow \frac{d^2 y}{d x^2}=-\operatorname{cosec}^2 x \cdot \cot y-\operatorname{cosec}^2 y \cot x \cdot \frac{d y}{d x}
\end{aligned}
\)
\(
\begin{aligned}
\text { Now } & \left(\frac{d y}{d x}\right)_{(\pi / 4, \pi / 4)}=1 \\
& \Rightarrow\left(\frac{d^2 y}{d x^2}\right)_{(\pi / 4, \pi / 4)}=-(2)(1)-(2)(1)(1)=-4
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& \text { Given } f=f^{\prime}+f^{\prime \prime}+f^{\prime \prime \prime}+\ldots \infty \\
& \Rightarrow f^{\prime}=f^{\prime \prime}+f^{\prime \prime \prime}+f^{\prime \prime \prime \prime}+\ldots \infty \\
& \Rightarrow f-f^{\prime}=f^{\prime} \\
& \Rightarrow f=2 f^{\prime}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& \text { Putting } x=\sin \theta \text { and } y=\sin \phi \\
& \cos \theta+\cos \phi=a(\sin \theta-\sin \phi) \\
& \Rightarrow 2 \cos \frac{\theta+\phi}{2} \cos \frac{\theta-\phi}{2}=a\left(2 \cos \frac{\theta+\phi}{2} \sin \frac{\theta-\phi}{2}\right) \\
& \Rightarrow \frac{\theta-\phi}{2}=\cot ^{-1} a \\
& \Rightarrow \theta-\phi=2 \cot ^{-1} a \\
& \Rightarrow \sin ^{-1} x-\sin ^{-1} y=2 \cot ^{-1} a \\
& \Rightarrow \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\sqrt{\frac{1-y^2}{1-x^2}}
\end{aligned}
\)

Hint

(a)

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

Hint

(d)
\(
\frac{d}{d x}\left(\tan ^{-1} \frac{(\sqrt{x}(3-x))}{1-3 x}\right)
\)
\(
\text { Put } \begin{aligned}
& \sqrt{x}=\tan \theta \Rightarrow \theta=\tan ^{-1} \sqrt{x} \\
& \frac{d}{d x}\left(\tan ^{-1} \frac{\left(\tan \theta\left(3-\tan ^2 \theta\right)\right)}{1-3 \tan ^2 \theta}\right) \\
& \frac{d}{d x}\left(\tan ^{-1} \frac{\left(3 \tan \theta-\tan ^3 \theta\right)}{1-3 \tan ^2 \theta}\right) \\
& \frac{d}{d x}\left[\tan ^{-1}(\tan 3 \theta)\right]=\frac{d}{d x}(3 \theta) \\
& =\frac{d}{d x}\left[3 \tan ^{-1} \sqrt{x}\right]=\frac{3}{2 \sqrt{x}(1+x)}
\end{aligned}
\)

Hint

(b) Since \(g(x)\) is the inverse of function \(f(x)\), therefore \(g o f(x)=I(x)\) for all \(x\)
Now \(g \circ f(x)=I(x), \forall x\)
\(
\begin{aligned}
& \Rightarrow(g o f)^{\prime}(x)=1, \forall x \\
& \Rightarrow g^{\prime}(f(x)) f^{\prime}(x)=1, \forall x \\
& \Rightarrow g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)}, \forall x \\
& \Rightarrow g^{\prime}(f(c))=\frac{1}{f^{\prime}(c)}(\text { putting } x=c)
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
&\begin{aligned}
& f(x)=x+\tan x \\
& f\left(f^{-1}(y)\right)=f^{-1}(y)+\tan f^{-1}(y) \\
& y=g(y)+\tan g(y) \\
& x=g(x)+\tan g(x)
\end{aligned}\\
&\text { Differentiating, we get } 1=g^{\prime}(x)+\sec ^2 g(x) g^{\prime}(x)\\
&\Rightarrow g^{\prime}(x)=\frac{1}{1+\sec ^2 g(x)}=\frac{1}{2+[g(x)-x]^2}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&y \sqrt{x^2+1}=\log \left\{\sqrt{x^2+1}-x\right\}\\
&\text { Differentiating both sides w.r.t. } x \text {., we get }\\
&\begin{aligned}
& \frac{d y}{d x} \sqrt{x^2+1}+y \frac{1}{2 \sqrt{x^2+1}} 2 x=\frac{1}{\sqrt{x^2+1}-x} \times\left\{\frac{1}{2} \frac{2 x}{\sqrt{x^2+1}}-1\right\} \\
& \Rightarrow\left(x^2+1\right) \frac{d y}{d x}+x y=\sqrt{x^2+1} \frac{-1}{\sqrt{x^2+1}} \\
& \Rightarrow\left(x^2+1\right) \frac{d y}{d x}+x y+1=0
\end{aligned}
\end{aligned}
\)

Hint

(a)
\(
y=\frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}
\)
\(
\begin{aligned}
\Rightarrow \quad y & =\frac{(\sqrt{a+x}-\sqrt{a-x})^2}{(a+x)-(a-x)} \\
\Rightarrow \quad y & =\frac{(a+x)+(a-x)-2\left(\sqrt{a^2-x^2}\right)}{2 x} \\
& =\frac{2 a-2 \sqrt{a^2-x^2}}{2 x}=\frac{a-\sqrt{a^2-x^2}}{x} \dots(1)
\end{aligned}
\)
\(
\begin{aligned}
&\text { Differentiating w.r.t. } x \text {, we get }\\
&\begin{aligned}
\frac{d y}{d x} & =\frac{x\left[-\frac{1}{2 \sqrt{a^2-x^2}}(-2 x)\right]-\left(a-\sqrt{a^2-x^2}\right)}{x^2} \\
& =\frac{x^2-a \sqrt{a^2-x^2}+a^2-x^2}{x^2 \sqrt{a^2-x^2}}=\frac{a\left(a-\sqrt{a^2-x^2}\right)}{x^2 \sqrt{a^2-x^2}}
\end{aligned}
\end{aligned}
\)
\(
=\frac{a}{x \sqrt{a^2-x^2}}\left[\frac{a-\sqrt{a^2-x^2}}{x}\right]=\frac{a y}{x \sqrt{a^2-x^2}} \quad[b y(1)]
\)

Hint

(a) As \(f(x)=x^4 \tan \left(x^3\right)-x \ln \left(1+x^2\right)\) is odd, \(\Rightarrow \frac{d^3 f(x)}{d x^3}\) is even \(\Rightarrow \frac{d^4 f(x)}{d x^4}=0\) at \(x=0\).

Hint

(a)

\(
\begin{aligned}
& \text { Given that } g^{-1}(x)=f(x) \Rightarrow x=g(f(x)) \text { or } g^{\prime}(f(x)) f^{\prime}(x)=1 \\
& \Rightarrow g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} \\
& \Rightarrow g^{\prime \prime}(f(x)) f^{\prime}(x)=\frac{-f^{\prime \prime}(x)}{\left[f^{\prime}(x)\right]^2} \Rightarrow g^{\prime \prime}(f(x))=\frac{-f^{\prime \prime}(x)}{\left[f^{\prime}(x)\right]^3}
\end{aligned}
\)

Hint

(b) For \(x>1\), we have \(f(x)=|\log | x| |=\log x\)
\(
\Rightarrow f^{\prime}(x)=\frac{1}{x}
\)
For \(x<-1\), we have \(f(x)=|\log | x| |=\log (-x)\)
\(
\Rightarrow f^{\prime}(x)=\frac{1}{x}
\)
For \(0<x<1\), we have \(f(x)=|\log | x| |=-\log x\)
\(
\Rightarrow f^{\prime}(x)=\frac{-1}{x}
\)
For \(-1<x<0\), we have \(f(x)=-\log (-x)\)
\(
\Rightarrow f^{\prime}(x)=\frac{-1}{x}
\)
\(
\text { Hence, } f^{\prime}(x)= \begin{cases}\frac{1}{x}, & |x|>1 \\ -\frac{1}{x}, & |x|<1\end{cases}
\)

Hint

(c)

\(
\begin{aligned}
&\text { In neighbourhood of } x=\frac{2 \pi}{3},|\cos x|=-\cos x \text { and }|\sin x|\\
&\begin{aligned}
& =\sin x \\
& \Rightarrow y=-\cos x+\sin x \\
& \Rightarrow \frac{d y}{d x}=\sin x+\cos x \\
& \Rightarrow \text { At } x=\frac{2 \pi}{3}, \frac{d y}{d x}=\sin \frac{2 \pi}{3}+\cos \frac{2 \pi}{3}=\frac{\sqrt{3}-1}{2} .
\end{aligned}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\text { Since } \boldsymbol{g} \text { is the inverse function of } f \text {, we have } f\{g(x)\}=x\\
&\begin{aligned}
& \Rightarrow \frac{d}{d x}(f\{g(x)\})=1 \\
& \Rightarrow f^{\prime}\{g(x)\} \cdot g^{\prime}(x)=1 \\
& \Rightarrow \sin \{g(x)\} g^{\prime}(x)=1 \\
& \Rightarrow g^{\prime}(x)=\frac{1}{\sin \{g(x)\}}
\end{aligned}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\text { We have } x=\phi(t), y=\psi(t) \text {. Therefore, }\\
&\begin{aligned}
& \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\psi^{\prime}}{\phi^{\prime}} \\
& \Rightarrow \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right)=\frac{d}{d t}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right) \frac{d t}{d x} \\
& =\frac{\phi^{\prime} \psi^{\prime \prime}-\psi^{\prime} \phi^{\prime \prime}}{\phi^{\prime 2}} \frac{1}{\phi^{\prime}}=\frac{\phi^{\prime} \psi^{\prime \prime}-\psi^{\prime} \phi^{\prime \prime}}{\phi^{\prime 3}}
\end{aligned}
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& f(x)=e^x-e^{-x}-2 \sin x-\frac{2}{3} x^3 \\
& f^{\mathrm{I}}(x)=e^x+e^{-x}-2 \cos x-2 x^2 \\
& f^{\mathrm{II}}(x)=e^x-e^{-x}+2 \sin x-4 x \\
& f^{\mathrm{II}}(x)=e^x+e^{-x}+2 \cos x-4 \\
& f^{\mathrm{VI}}(x)=e^x-e^{-x}-2 \sin x \\
& f^{\mathrm{V}}(x)=e^x+e^{-x}-2 \cos x \\
& f^{\mathrm{VI}}(x)=e^x-e^{-x}+2 \sin x \\
& f^{\mathrm{VII}}(x)=e^x+e^{-x}+2 \cos x \\
& \text { Clearly, } f^{\mathrm{VII}}(0) \text { is non-zero. }
\end{aligned}
\)

Hint

(b) Given \(f\left(\frac{5 x-3 y}{2}\right)=\frac{5 f(x)-3 f(y)}{2}\)
\(\Rightarrow f\left(\frac{5 x-3 y}{5-3}\right)=\frac{5 f(x)-3 f(y)}{5-3}\), which satisfies section
formula for abscissa on L.H.S. and ordinate on R.H.S.
Hence, \(f(x)\) must be the linear function (as only straight
line satisfies such section formula).
Hence, \(f(x)=a x+b\)
But \(f(0)=3 \Rightarrow b=3, f^{\prime}(0)=2 \Rightarrow a=2\).
Thus, \(f(x)=2 x+3 \Rightarrow\) Period of \(\sin (f(x))=\sin (2 x+3)\) is \(\pi\).

Hint

(c)

\(
\begin{aligned}
& D^*(x)=\lim _{h \rightarrow 0} \frac{f^2(x+h)-f^2(x)}{h} \\
&=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}(f(x+h)+f(x)) \\
&=2 f(x) \lim _{h \rightarrow 0} \cdot \frac{f(x+h)-f(x)}{h} \\
&=2 f(x) \times f^{\prime}(x) \\
& \Rightarrow D^*(x \log x)=2 x \log x(1+\log x) \\
&\left.\Rightarrow D^* f(x)\right|_{x=e}=4 e
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& \sqrt{x}=\cos \theta \\
& x \in\left(0, \frac{1}{2}\right) \Rightarrow \sqrt{x}=\cos \theta \in\left(0, \frac{1}{\sqrt{2}}\right) \\
& \Rightarrow \theta \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \\
& \Rightarrow 2 \theta \in\left(\frac{\pi}{2}, \pi\right) \\
& \Rightarrow f(x)=2 \sin ^{-1} \sqrt{1-\cos ^2 \theta}+\sin ^{-1}\left(2 \sqrt{\cos ^2 \theta \sin ^2 \theta}\right) \\
& \quad \quad=2 \sin ^{-1}(\sin \theta)+\sin ^{-1}(2 \sin \theta \cos \theta) \\
& \quad=2 \theta+\sin ^{-1}(\sin 2 \theta) \\
& \quad=2 \theta+\pi-2 \theta \\
& \quad=\pi \\
& \Rightarrow f^{\prime}(x)=0
\end{aligned}
\)

Hint

(a)
\(
\begin{aligned}
& F^{\prime}(x)=\left[f\left(\frac{x}{2}\right) \cdot f^{\prime}\left(\frac{x}{2}\right)+g\left(\frac{x}{2}\right) g^{\prime}\left(\frac{x}{2}\right)\right] \\
& \text { Here } g(x)=f^{\prime}(x)
\end{aligned}
\)
\(
\begin{aligned}
&\begin{aligned}
& \text { and } g^{\prime}(x)=f^{\prime \prime}(x)=-f(x) \\
& \text { so } F^{\prime}(x)=f\left(\frac{x}{2}\right) g\left(\frac{x}{2}\right)-f\left(\frac{x}{2}\right) g\left(\frac{x}{2}\right)=0
\end{aligned}\\
&\Rightarrow F(x) \text { is a constant function. }\\
&\Rightarrow F(10)=5
\end{aligned}
\)

Hint

(b) Let \(y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\), and \(z=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right)\)
Putting \(x=\tan \theta\) in \(y\), we get
\(
\begin{aligned}
& y=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{1}{2} \tan ^{-1} x \\
& \Rightarrow \frac{d y}{d x}=\frac{1}{2\left(1+x^2\right)}
\end{aligned}
\)
Putting \(x=\sin \theta\) in \(z\), we get
\(
\begin{aligned}
z & =\tan ^{-1}\left(\frac{2 \sin \theta \cos \theta}{\cos 2 \theta}\right)=\tan ^{-1}(\tan 2 \theta)=2 \theta=2 \sin ^{-1} x \\
& \Rightarrow \frac{d z}{d x}=\frac{2}{\sqrt{1-x^2}}
\end{aligned}
\)
Thus, \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}=\frac{1}{4\left(1+x^2\right)} \sqrt{1-x^2} \Rightarrow\left(\frac{d y}{d z}\right)_{x=0}=\frac{1}{4}\)

Hint

(c)
\(
\begin{aligned}
& f(x)=x e^x \\
& f^{\prime}(x)=e^x+x e^x \\
& f^{\prime \prime}(x)=e^x+e^x+x e^x \\
& f^{\prime \prime \prime}(x)=2 e^x+e^x+x e^x=3 e^x+x e^x \\
& \cdots \\
& \cdots \\
& f^{\prime \prime}(x)=n e^x+x e^x
\end{aligned}
\)
\(
\begin{aligned}
& \text { Now, } f^n(x)=0 \\
& \Rightarrow n e^x+x e^x=0 \Rightarrow x=-n
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& y^2=a x^2+b x+c \\
& \Rightarrow 2 y \frac{d y}{d x}=2 a x+b \\
& \Rightarrow 2\left(\frac{d y}{d x}\right)^2+2 y \frac{d^2 y}{d x^2}=2 a \\
& \Rightarrow y \frac{d^2 y}{d x^2}=a-\left(\frac{d y}{d x}\right)^2 \\
& \Rightarrow y \frac{d^2 y}{d x^2}=a-\left(\frac{2 a x+b}{2 y}\right)^2 \\
& \Rightarrow y \frac{d^2 y}{d x^2}=\frac{4 a y^2-(2 a x+b)^2}{4 y^2} \\
& \Rightarrow 4 y^3 \frac{d^2 y}{d x^2}=4 a\left(a x^2+b x+c\right)-\left(4 a^2 x^2+4 a b x+b^2\right) \\
& \Rightarrow 4 y^3 \frac{d^2 y}{d x^2}=4 a c-b^2=\text { constant. }
\end{aligned}
\)

Hint

(c)
\(
y=\sin x+e^x \Rightarrow \frac{d y}{d x}=\cos x+e^x
\)
\(
\Rightarrow \frac{d x}{d y}=\left(\cos x+e^x\right)^{-1} \dots(1)
\)
\(
\begin{aligned}
&\text { Substituting the value of } \frac{d y}{d x} \text { from (1) }\\
&\frac{d^2 x}{d y^2}=\frac{\left(\sin x-e^x\right)}{\left(\cos x+e^x\right)^2}\left(\cos x+e^x\right)^{-1}=\frac{\sin x-e^x}{\left(\cos x+e^x\right)^3}
\end{aligned}
\)

Hint

(d) \(u=x^2+y^2, x=s+3 t, y=2 s-t\)
Now, \(\frac{d x}{d s}=1, \frac{d y}{d s}=2 \dots(1)\)
\(
\frac{d^2 x}{d s^2}=0, \frac{d^2 y}{d s^2}=0 \dots(2)
\)
Now \(u=x^2+y^2, \frac{d u}{d s}=2 x \frac{d x}{d s}+2 y \frac{d y}{d s}\)
\(
\frac{d^2 u}{d s^2}=2\left(\frac{d x}{d s}\right)^2+2 x \frac{d^2 x}{d s^2}+2\left(\frac{d y}{d s}\right)^2+2 y\left(\frac{d^2 y}{d s^2}\right)
\)
From (1) and (2), \(\frac{d^2 u}{d s^2}=2 \times 1+0+2 \times 4+0=10\).

Hint

(c) Here, \(y=t^{10}+1\) and \(x=t^8+1\)
\(
t^8=x-1 \Rightarrow t^2=(x-1)^{1 / 4}
\)
So, \(y=(x-1)^{5 / 4}+1\)
Differentiate both sides w.r.t. \(x\), we get \(\frac{d y}{d x}=\frac{5}{4}(x-1)^{1 / 4}\)
Again, differentiate both sides w.r.t. \(x\), we get
\(
\begin{aligned}
\frac{d^2 y}{d x^2} & =\frac{5}{16}(x-1)^{-3 / 4} \\
\Rightarrow \frac{d^2 y}{d x^2} & =\frac{5}{16(x-1)^{3 / 4}}=\frac{5}{16\left(t^2\right)^3}=\frac{5}{16 t^6}
\end{aligned}
\)

Hint

(b) From the given relation \(\frac{y}{x}=\log x-\log (a+b x)\)
Differentiating w.r.t. \(x\), we get \(\frac{\left(x \frac{d y}{d x}\right)-y}{x^2}=\)
\(
\frac{1}{x}-\frac{b}{a+b x}=\frac{a}{x(a+b x)}
\)
\(
\begin{aligned}
&\therefore x \frac{d y}{d x}-y=\frac{a x}{a+b x} \dots(1) \\
&\text { Differentiating again w.r.t. } x \text {, we get }\\
&\begin{aligned}
& x \frac{d^2 y}{d x^2}+\frac{d y}{d x}-\frac{d y}{d x}=\frac{(a+b x) a-a x . b}{(a+b x)^2} \\
\Rightarrow & x \frac{d^2 y}{d x^2}=\frac{a^2}{(a+b x)^2}
\end{aligned}
\end{aligned}
\)
\(
\Rightarrow x^3 \frac{d^2 y}{d x^2}=\frac{a^2 x^2}{(a+b x)^2}=\left(x \frac{d y}{d x}-y\right)^2[\text { by (1) }]
\)

Hint

(a)

\(
\begin{aligned}
& u(x)=7 v(x) \Rightarrow u^{\prime}(x)=7 v^{\prime}(x) \Rightarrow p=7 \text { (given) } \\
& \text { Again } \frac{u(x)}{v(x)}=7 \Rightarrow\left(\frac{u(x)}{v(x)}\right)^{\prime}=0 \Rightarrow q=0 \\
& \text { Now } \frac{p+q}{p-q}=\frac{7+0}{7-0}=1
\end{aligned}
\)

Hint

(a) \(a x^2+2 h x y+b y^2=1\)
Differentiating both sides w.r.t. \(x\), we get
\(
\begin{aligned}
& 2 a x+2 h x \frac{d y}{d x}+2 h y+2 b y \frac{d y}{d x}=0 \\
& \Rightarrow \frac{d y}{d x}=-\frac{a x+h y}{h x+b y}
\end{aligned}
\)
Again differentiating w.r.t. \(x\), we get
\(
\begin{aligned}
& \Rightarrow \frac{d^2 y}{d x^2} \\
& =-\left[\frac{(h x+b y)\left(a+h \frac{d y}{d x}\right)-(a x+h y)\left(h+b \frac{d y}{d x}\right)}{(h x+b y)^2}\right] \\
& =-\frac{\left[y\left(a b-h^2\right)+\frac{d y}{d x}\left(h^2 x-a b x\right)\right]}{(h x+b y)^2} \\
& =\frac{\left(h^2-a b\right)\left(y-x \frac{d y}{d x}\right)}{(h x+b y)^2} \\
& =\frac{\left(h^2-a b\right)}{(h x+b y)^2}\left[y+x \frac{a x+h y}{h x+b y}\right] \\
& =\frac{h^2-a b}{(h x+b y)^2}
\end{aligned}
\)

Hint

(b)

\(
\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{3 t^2}{2 t}=\frac{3}{2} \sqrt{x} \Rightarrow \frac{d^2 y}{d x^2}=\frac{3}{4 \sqrt{x}}=\frac{3}{4 t}
\)

Hint

(b) \(y=x+e^x \Rightarrow \frac{d y}{d x}=1+e^x \Rightarrow \frac{d x}{d y}=\frac{1}{1+e^x}\)
\(
\begin{aligned}
& \Rightarrow \frac{d}{d y}\left(\frac{d x}{d y}\right)=\frac{d}{d y}\left(\frac{1}{1+e^x}\right) \Rightarrow \frac{d^2 x}{d y^2}=\frac{d}{d x}\left(\frac{1}{1+e^x}\right) \frac{d x}{d y} \\
& \Rightarrow \frac{d^2 x}{d y^2}=\frac{-e^x}{\left(1+e^x\right)^2} \frac{1}{\left(1+e^x\right)}=-\frac{e^x}{\left(1+e^x\right)^3}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) In the neighbourhood of } x=7 \pi / 6 \text {, we have } f(x)=\mid \sin x+\\
&\begin{aligned}
& \cos x \mid=-\sin x-\cos x \\
\Rightarrow & f^{\prime}(x)=-\cos x+\sin x \Rightarrow f^{\prime}(7 \pi / 6)=-\cos (7 \pi / 6)+\sin (7 \pi / 6) \\
= & \frac{\sqrt{3}-1}{2}
\end{aligned}
\end{aligned}
\)

Hint

(a) \(y=f(x)\) is an even function and \(y=g(x)\) is an odd function.
\(\Rightarrow h(x)=f(x) g(x)\) is an odd function.
\(\Rightarrow h(x)=-h(-x)\)
\(\Rightarrow h^{\prime}(x)=h^{\prime}(-x)\)
\(\Rightarrow h^{\prime \prime}(x)=-h^{\prime \prime}(-x)\)
\(\Rightarrow h^{\prime \prime \prime}(x)=h^{\prime \prime \prime}(-x)\)
Now, we cannot determine the value of \(h^{\prime \prime \prime}(0)\).

Hint

(c)

\(
\begin{aligned}
& \frac{d y}{d x}=\frac{-\frac{1}{p^2}}{\frac{\mathrm{i}}{p}}=-\frac{1}{p}=-y \Rightarrow \frac{d^2 y}{d x^2}=-\frac{d y}{d x} \\
& \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& y=2 \ln (1+\cos x) \\
& \frac{d y}{d x}=\frac{-2 \sin x}{1+\cos x} \\
& \frac{d^2 y}{d x^2}=-2\left[\frac{(1+\cos x) \cos x-\sin x(-\sin x)}{(1+\cos x)^2}\right] \\
& =-2\left[\frac{\cos x+1}{(1+\cos x)^2}\right]=\frac{-2}{(1+\cos x)} \\
& \text { Now } 2 e^{-y / 2}=2 \cdot e^{-\frac{\ln (1+\cos x)^2}{2}}=\frac{2}{(1+\cos x)} \\
& \frac{d^2 y}{d x^2}+\frac{2}{e^{y / 2}}=0
\end{aligned}
\)

Hint

(a) Let \(g(x)=(\sin x)^{\ln x}=e^{\ln x \cdot \ln (\sin x)}\)
\(f(x)=g^{\prime}(x)=(\sin x)^{\ln x}\left[\cot x(\ln x)+\frac{\ln (\sin x)}{x}\right]\)
Hence, \(f\left(\frac{\pi}{2}\right)=g^{\prime}\left(\frac{\pi}{2}\right)=1(0+0)=0\).