Question

$A:\frac{d}{dx}{\sin }^2\left[{\cot }^{-1}\sqrt{\frac{1+x}{1-x}}\right]$

$B:\frac{d}{dx}\left(\frac{1+\tan x}{1-\tan x}\right)atx=0$

$C$ : For the curve $\sqrt{x}+\sqrt{y}=1,\frac{dy}{dx}$ at $(\frac{1}{4},\frac{1}{4})$ is

$D:\frac{d}{dx}({\tan }^{-1}\frac{3x}{1+x^2}+{\cot }^{-1}\frac{3x}{1+x^2})$
Arrange the above values in the increasing order of the magnitude

• A. A, B, C, D
• B. C, A, D, B
• C. C, D, B, A
• D. A, D, B, C

Answer 1

Answer: (B)

C, A, D, B

$A)\frac{d}{dx}\left(si{n}^2\right(si{n}^{-1}\left)\left(\sqrt{\frac{1-x}{2}}\right)\right)$

$\Rightarrow \frac{d}{dx}\left(\frac{1-x}{2}\right)=\frac{-1}{2}$

$B)\frac{d}{dx}tan(\frac{\pi }{4}+x)=se{c}^2(\frac{\pi }{4}+x)$

Put $x=0$

$=2$

$C)y=(1-\sqrt{x}{)}^2$

$\Rightarrow \frac{dy}{dx}=(1-\sqrt{x})\left(\frac{-1}{\sqrt{x}}\right)$

Put $x=\frac{1}{4}$

$=\frac{1}{2}(-2)=-1$

$D.\frac{d}{dx}\left(\frac{\pi }{2}\right)$

$=0$