The number of solutions of the equation sin ^ | vedclass

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(B) Given equation: $\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]=x^{2}$.For the domain of $\sin^{-1}(u)$,we requ

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(B) Given equation: $\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]=x^{2}$.For the domain of $\sin^{-1}(u)$,we require $-1 \leq u \leq 1$. Since $u$ is an integer (due to the greatest integer function),$u \in \{-1, 0, 1\}$.Case $1$: Let $x^2 + \frac{1}{3} = k_1$ and $x^2 - \frac{2}{3} = k_2$,where $k_1, k_2 \in \mathbb{Z}$.Given $x \in [-1, 1]$,$x^2 \in [0, 1]$.Then $x^2 + \frac{1}{3} \in [1/3, 4/3]$,so $[x^2 + 1/3] \in \{0, 1\}$.And $x^2 - 2/3 \in [-2/3, 1/3]$,so $[x^2 - 2/3] \in \{-1, 0\}$.Case $I$: $[x^2 + 1/3] = 0$ and $[x^2 - 2/3] = -1$.This implies $0 \leq x^2 + 1/3 Substituting into the equation: $\sin^{-1}(0) + \cos^{-1}(-1) = x^2 \Rightarrow 0 + \pi = x^2 \Rightarrow x^2 = \pi$.Since $\pi \approx 3.14$,it is not in $[0, 2/3)$. No solution.Case $II$: $[x^2 + 1/3] = 1$ and $[x^2 - 2/3] = 0$.This implies $1 \leq x^2 + 1/3 Substituting into the equation: $\sin^{-1}(1) + \cos^{-1}(0) = x^2 \Rightarrow \pi/2 + \pi/2 = x^2 \Rightarrow x^2 = \pi$.Since $\pi \approx 3.14$,it is not in $[2/3, 1]$ (as $x \in [-1, 1]$ implies $x^2 \leq 1$). No solution.Thus,the number of solutions is $0$.

The number of solutions of the equation $\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]=x^{2}$ for $x \in[-1,1],$ where $[x]$ denotes the greatest integer function,is:

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