The number of solutions of |sin^{-1}x| = |x| | vedclass

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

Question

(B) To find the number of solutions for $|\sin^{-1}x| = |x|$,we analyze the graphs of $f(x) = |\sin^{-1}x|$ and $g(x) = |x|$.$1$. The domain of $\sin^

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) To find the number of solutions for $|\sin^{-1}x| = |x|$,we analyze the graphs of $f(x) = |\sin^{-1}x|$ and $g(x) = |x|$.$1$. The domain of $\sin^{-1}x$ is $[-1, 1]$.$2$. At $x = 0$,both $|\sin^{-1}0| = 0$ and $|0| = 0$. Thus,$x = 0$ is a solution.$3$. For $x \in (0, 1]$,we know that $\sin^{-1}x > x$. Since both sides are positive,$|\sin^{-1}x| = \sin^{-1}x$ and $|x| = x$. Thus,$\sin^{-1}x = x$ has no solution for $x > 0$ because $\sin^{-1}x$ is strictly greater than $x$ for all $x \in (0, 1]$.$4$. For $x \in [-1, 0)$,we have $|\sin^{-1}x| = |-\sin^{-1}(-x)| = |\sin^{-1}(-x)|$. Since $-x > 0$,$\sin^{-1}(-x) > -x$,so $|\sin^{-1}x| > |x|$.$5$. Therefore,the only point of intersection is at the origin $(0, 0)$.$6$. The number of solutions is $1$.

The number of solutions of $|\sin^{-1}x| = |x|$ is

Similar Questions

$\coth^{-1}(2) + \operatorname{cosech}^{-1}(-2\sqrt{2}) = $

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

If $\frac{(x + 1)^2}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$,then $\sin^{-1}\left(\frac{A}{C}\right) = $

If $\tan \theta = - \frac{1}{\sqrt{3}}$,$\sin \theta = \frac{1}{2}$,and $\cos \theta = - \frac{\sqrt{3}}{2}$,then the principal value of $\theta$ is:

$\tan \left( 2{{\cos }^{ - 1}}\frac{3}{5} \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module