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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\epsilon \left [ -1,1 \right ], and \left [ x \right ] denotes the greatest integer less than or equal to x, is :
 
Option: 1 0
Option: 2 2
Option: 3 infinite
Option: 4 4

Answers (1)

best_answer

Given equation is 

\\\begin{aligned} &\sin ^{-1}\left[\mathrm{x}^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[\mathrm{x}^{2}-\frac{2}{3}\right]=\mathrm{x}^{2}\\ &\text { Now, } \sin ^{-1}\left[x^{2}+\frac{1}{3}\right] \text { is defined if } \end{aligned}\\\begin{aligned} &-1 \leq x^{2}+\frac{1}{3}<2 \Rightarrow \frac{-4}{3} \leq x^{2}<\frac{5}{3}\\ &\Rightarrow 0 \leq x^{2}<\frac{5}{3} \end{aligned}

\begin{aligned} &\text { and } \cos ^{-1}\left[\mathrm{x}^{2}-\frac{2}{3}\right] \text { is defined if }\\ &-1 \leq x^{2}-\frac{2}{3}<2 \Rightarrow \frac{-1}{3} \leq x^{2}<\frac{8}{3}\\ &\Rightarrow 0 \leq x^{2}<\frac{8}{3} \end{aligned}

from above, we get 

0 \leq x^{2}<\frac{5}{3}

\begin{aligned} &\mathbf{case\; 1} \text { if } 0 \leq x^{2}<\frac{2}{3}\\ &\sin ^{-1}(0)+\cos ^{-1}(-1)=x^{2}\\ &\Rightarrow x+\pi=x^{2}\\ &\Rightarrow x^{2}=\pi\\ &\text { but } \pi \notin\left[0, \frac{2}{3}\right)\\ &\Rightarrow \text { No value of 'x' } \end{aligned}

\begin{aligned} &\mathbf{case \;2}\text { if } \frac{2}{3} \leq \mathrm{x}^{2}<\frac{5}{3}\\ &\sin ^{-1}(1)+\cos ^{-1}(0)=x^{2}\\ &\Rightarrow \frac{\pi}{2}+\frac{\pi}{2}=\mathrm{x}^{2}\\ &\Rightarrow x^{2}=\pi\\ &\text { but } \pi \notin\left[\frac{2}{3}, \frac{5}{3}\right)\\ &\Rightarrow \text { No value of 'x' } \end{aligned}

So, number of solution i s0 

Posted by

himanshu.meshram

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