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

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(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^

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$1$

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(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

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