Let Z denote the set of integers. Then match | vedclass

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

Question

(C) We have $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$.Since $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)=\cos ^{-1}\left(\f

Vedclass · Accepted Answer

$A-V, B-I, C-IV, D-II$

Vedclass · Answer

(C) We have $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}ight)+\sin ^{-1} \frac{1}{3}$.Since $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}ight)=\cos ^{-1}\left(\frac{1}{3}ight)$, we have $\cos ^{-1}\left(\frac{1}{3}ight)+\sin ^{-1}\left(\frac{1}{3}ight)=\frac{\pi}{2}$. Thus, $A-V$.$(B)$ Let $	heta=\sin ^{-1}\left(\frac{(-1)^n}{2}ight), n \in Z$. Then $\sin 	heta=\frac{(-1)^n}{2}=\sin\left((-1)^n \frac{\pi}{6}ight)$. The general solution is $	heta=k \pi+(-1)^k\left((-1)^n \frac{\pi}{6}ight)$, which simplifies to $k \pi \pm \frac{\pi}{6}, k \in Z$. This matches $I$.$(C)$ Let $\alpha=	an ^{-1}\left(\sec \frac{\pi}{4}+	an \frac{\pi}{4}ight) = 	an ^{-1}(\sqrt{2}+1)$. Since $	an \frac{3\pi}{8} = \sqrt{2}+1$, we have $\alpha = \frac{3\pi}{8}$. Thus, $C-IV$.$(D)$ Let $t=\sin ^{-1}|\sin x|$. Then $t=\sqrt{t} \Rightarrow t^2-t=0 \Rightarrow t(t-1)=0$. So $t=0$ or $t=1$. $\sin ^{-1}|\sin x|=0 \Rightarrow |\sin x|=0 \Rightarrow x=k\pi$. $\sin ^{-1}|\sin x|=1 \Rightarrow |\sin x|=\sin 1 \Rightarrow x=k\pi \pm 1$. Combining these, the general solution is $x=k\pi \pm 1, k \in Z$. Thus, $D-II$.

List-$I$	List-$II$
$A$. $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$	$I$. $k \pi \pm(-1)^k \frac{\pi}{6}, k \in Z$
$B$. $\sin ^{-1}\left(\frac{(-1)^n}{2}\right), n \in Z$	$II$. $k \pi \pm 1, k \in Z$
$C$. $\tan ^{-1}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)$	$III$. $\frac{3}{2}$
$D$. $\sin ^{-1}\|\sin x\|=\sqrt{\sin ^{-1}\|\sin x\|} \Rightarrow x \in$	$IV$. $\frac{3 \pi}{8}$
	$V$. $\frac{\pi}{2}$

Similar Questions

The number of real roots of the equation $\sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0$ that are greater than or equal to $1$ are

The possible values of $x$, which satisfy the trigonometric equation $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$ are

The imaginary part of $\tan^{-1}\left(\frac{5i}{3}\right)$ is

Let $\cos ^{-1}(x) + \cos ^{-1} (2x) + \cos ^{-1}(3x) = \pi.$ If $x$ satisfies the cubic equation $ax^3 + bx^2 + cx - 1 = 0,$ then the value of $(a + b + c)$ is -

If $f(x) = \cot^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$ and $g(x) = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$, then $\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)}$ for $0 < a < \frac{1}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module