Match the items of List-I with those of List- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A-(III), B-(V), C-(I), D-(II)) For $A$: Let $y = 	an^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}} = 	an^{-1}\sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}} = 	an

Vedclass · Accepted Answer

Vedclass · Answer

(A-(III), B-(V), C-(I), D-(II)) For $A$: Let $y = 	an^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}} = 	an^{-1}\sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}} = 	an^{-1}(	an(x/2)) = \frac{x}{2}$.Then $\frac{dy}{dx} = \frac{1}{2}$. Thus, $A ightarrow (iii)$.For $B$: Let $y = \frac{3+|x-1|}{3x+4}$. The function $|x-1|$ is not differentiable at $x=1$. Thus, the expression is not differentiable at $x=1$. Thus, $B ightarrow (v)$.For $C$: Let $y = \sinh^{-1} x$. We know that $\sinh^{-1} x = \log(x+\sqrt{1+x^2})$. Thus, $C ightarrow (i)$.For $D$: Let $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight) = 2	an^{-1} x$.Then $\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = 2(1+x^2)^{-1}$.Then $\frac{d^2y}{dx^2} = 2(-1)(1+x^2)^{-2}(2x) = -\frac{4x}{(1+x^2)^2}$. Thus, $D ightarrow (ii)$.The correct matching is $A-(iii), B-(v), C-(i), D-(ii)$.

List-$I$	List-$II$
$A. \frac{d}{dx}\left(\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\right)$	$(i) \log(x+\sqrt{1+x^2})$
$B. \frac{d}{dx}\left(\frac{3+\|x-1\|}{3x+4}\right)$	$(ii) -\frac{4x}{(1+x^2)^2}$
$C. \sinh^{-1} x$	$(iii) \frac{1}{2}$
$D. \frac{d^2}{dx^2}\left(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$	$(iv) \frac{1}{\sqrt{1+x^2}}$
	$(v) \text{not differentiable at } x=1$

Similar Questions

Let $f$ and $g$ be twice differentiable functions on $R$ such that
$f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x$
$f^{\prime}(1)=4, g^{\prime}(1)=3$
$f(2)=12, g(2)=4$
Then which of the following is $NOT$ true?

The number of points, at which the function $f(x) = \max\{6x, 2+3x^2\} + |x-1| |\cos(x^2 - 1/4)|, x \in (-\pi, \pi)$, is not differentiable, is ————

Let $f(x) = x|x|$,$g(x) = \sin x$ and $h(x) = (g \circ f)(x)$. Then

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$ equals

If $f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x$, then $f' \left( \frac{\pi}{4} \right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module