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The number of real solutions of $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$ is
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The correct answer is:
Two
$\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$
$\tan ^{-1} \sqrt{x(x+1)}$ is defined when
$x(x+1) \geq 0$
$\sin ^{-1} \sqrt{x^2+x+1}$ is defined when
$0 \leq x(x+1)+1 \leq 1 \text { or } 0 \leq x(x+1) \leq 0$
From (i) and (ii), $x(x+1)=0$
or $x=0$ and -1 .
Hence number of solution is 2 .
$\tan ^{-1} \sqrt{x(x+1)}$ is defined when
$x(x+1) \geq 0$
$\sin ^{-1} \sqrt{x^2+x+1}$ is defined when
$0 \leq x(x+1)+1 \leq 1 \text { or } 0 \leq x(x+1) \leq 0$
From (i) and (ii), $x(x+1)=0$
or $x=0$ and -1 .
Hence number of solution is 2 .
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