Maths MCQ

[Hint: Put x = a cos θ]

$
\begin{aligned}
& \text { (a) : Let } y=\sqrt{x^2+16} \text { and } z=\frac{x}{x-1} \\
& \frac{d y}{d x}=\frac{1}{2 \sqrt{x^2+16}} \times 2 x, \frac{d z}{d x}=\frac{(x-1)-x}{(x-1)^2} \\
& \Rightarrow \frac{d y}{d x}=\frac{x}{\sqrt{x^2+16}} \text { and } \frac{d z}{d x}=\frac{-1}{(x-1)^2}
\end{aligned}
$
Now, $\frac{d y}{d z}=\frac{-x}{\sqrt{x^2+16}} \times(x-1)^2$
$
\therefore\left[\frac{d y}{d z}\right]_{x=5}=\frac{-5 \times(4)^2}{\sqrt{41}}=\frac{-80}{\sqrt{41}}
$

$\begin{aligned} & \text   (b): \text { Let } P(x)=a(x-2)^2+b
\\ & \Rightarrow \quad P^{\prime}(x)=2 a(x-2) \text { and } P^{\prime \prime}(x)=2 a
\\ & \text { Now, } P(2)=-1 \Rightarrow b=-1 \\ & \quad P^{\prime \prime}(2)=2 \Rightarrow 2 a=2 \Rightarrow a=1
\\ & \therefore \quad P(x)=(x-2)^2-1
\\ & \text { Now, } P(1.001)=(1.001-2)^2-1=0.998001-1
\\ & \quad=-0.001999 \approx-0.002\end{aligned}$

(c) : Given that $y=f\left(\frac{2 x+3}{3-2 x}\right)$
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=f^{\prime}\left(\frac{2 x+3}{3-2 x}\right) \frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right) \\
& \quad=\sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]\left[\frac{(3-2 x) 2-(2 x+3)(-2)}{(3-2 x)^2}\right] \\
& \quad=\frac{12}{(3-2 x)^2} \sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right] \\
& \Rightarrow\left(\frac{d y}{d x}\right)_{x=1}=\frac{12}{(3-2)^2} \sin (\log 5)=12 \sin (\log 5)
\end{aligned}
$

(b) : $y=\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2$ ...(i)
On differentiating w.r.t. ' $x$ ', we get
$
y_1=\frac{2 \sin ^{-1} x}{\sqrt{1-x^2}}-\frac{2 \cos ^{-1} x}{\sqrt{1-x^2}} ...(ii)
$
Again, differentiating w.r.t. ' $x$ ' we get
$
\begin{aligned}
& \frac{\frac{2}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2}-\frac{1}{2 \sqrt{1-x^2}}(-2 x) 2 \sin ^{-1} x}{\left(1-x^2\right)} \\
& \frac{-\left[\frac{-2}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2}+\frac{1}{2 \sqrt{1-x^2}}(2 x) 2 \cos ^{-1} x\right]}{1-x^2} \\
\Rightarrow & y^2=\frac{2+\frac{2 x \sin ^{-1} x}{\sqrt{1-x^2}}+2-\frac{2 x \cos ^{-1} x}{\sqrt{1-x^2}}}{1-x^2} \\
\Rightarrow & \left(1-x^2\right) y_2=4+x\left[\frac{2 \sin ^{-1} x}{\sqrt{1-x^2}}-\frac{2 \cos ^{-1} x}{\sqrt{1-x^2}}\right] \\
\Rightarrow & \left(1-x^2\right) y_2=4+x y_1 [Using (ii)] \\
\Rightarrow & \left(1-x^2\right) y_2-x y_1=4
\end{aligned}
$

(a) : Given, $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$
$
\begin{aligned}
& \text { Now, } \frac{d x}{d t}=\frac{1}{2 \cdot \sqrt{a^{\sin ^{-1} t}}} \times a^{\sin ^{-1} t}(\log a) \cdot \frac{1}{\sqrt{1-t^2}} \\
& =\frac{1}{2 x} x^2(\log a) \cdot \frac{1}{\sqrt{1-t^2}}=\frac{x}{2} \log a \cdot \frac{1}{\sqrt{1-t^2}} \\
& \text { and } \frac{d y}{d t}=\frac{1}{2 \cdot \sqrt{a^{\cos ^{-1} t}}} \times a^{\cos ^{-1} t} \log a \cdot \frac{-1}{\sqrt{1-t^2}} \\
& =\frac{1}{2 y} \times y^2 \log a \cdot\left(-\frac{1}{\sqrt{1-t^2}}\right)=-\frac{y}{2} \log a \times \frac{1}{\sqrt{1-t^2}} \\
&
\end{aligned}
$
So, $\frac{d y}{d x}=-\frac{y}{x}$

(a): We know that, $\tan ^{-1} x+\cot ^{-1} x=\pi / 2$ for $x \in R$ So, $x y=\pi / 2$
Now differentiate $x y=\pi / 2$ with respect to $x$, we get
$
x \cdot \frac{d y}{d x}+y=0 \Rightarrow \frac{d y}{d x}=\frac{-y}{x} \Rightarrow\left(\frac{d y}{d x}\right)_{(4,2)}=\frac{-2}{4}=-\frac{1}{2}
$

(b) : We have, $x=e^{x / y}$
By taking $\log$ on both sides, we get $\log x=\log e^{x / y}$
$\log x=\frac{x}{y}$ or $y \log x=x$
Differentiating w.r.t $x$ ',
$
\frac{d y}{d x} \log x+\frac{y}{x}=1 \Rightarrow \frac{d y}{d x}=\frac{1-\frac{y}{x}}{\log x}=\frac{x-y}{x \log x}
$

(d) : Let $y=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ Differentiate w.r.t. $x$, we get
$
\begin{aligned}
\frac{d y}{d x} & =\frac{1}{1+\left(\frac{x}{\sqrt{1-x^2}}\right)^2} \times \frac{d}{d x}\left(\frac{x}{\sqrt{1-x^2}}\right) \\
& =\frac{\left(1-x^2\right)}{1-x^2+x^2} \times \frac{\sqrt{1-x^2} \cdot 1+x \cdot \frac{1}{2 \sqrt{1-x^2}} \times 2 x}{\left(1-x^2\right)} \\
\Rightarrow & \frac{d y}{d x}=\frac{\left(1-x^2\right)+x^2}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}...(i)
\end{aligned}
$
Let $u=\sin ^{-1}\left(3 x-4 x^3\right)$
Put $x=\sin \theta$,
$
\begin{aligned}
\therefore \quad u & =\sin ^{-1}\left(3 \sin \theta-4 \sin ^3 \theta\right) \\
& =\sin ^{-1}(\sin 3 \theta)=3 \theta=3 \sin ^{-1} x
\end{aligned}
$
Differentiate w.r.t. $x$, we get
$
\frac{d u}{d x}=3 \cdot \frac{1}{\sqrt{1-x^2}}...(ii)
$
On dividing (i) by (ii), we get
$
\frac{d y}{d u}=\frac{1}{\sqrt{1-x^2}} \times \frac{\sqrt{1-x^2}}{3}=\frac{1}{3}
$

(c) : Let $y=\log _{e^2}(\log x)=\frac{\log (\log x)}{\log e^2}=\frac{\log (\log x)}{2}$
Now, $\frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}[\log (\log x)]=\frac{1}{2} \times \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)$
$
=\frac{1}{2 \log x} \times \frac{1}{x}=\frac{1}{x \log x^2}
$