Question

Assertion :The area bounded by the curve $y={\sin }^{-1}x$ & the line x=0 & $\left|y\right|=\frac{\pi }{2}$ is $\sqrt{2}$ square units. Reason: The domain & principal value branch of $y={\sin }^{-1}x$ are [-1,1] & $[\frac{-\pi }{2},\frac{\pi }{2}]$ respectively

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

$y=si{n}^{-1}\left(x\right)$

Hence the required area will be
$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}x.dy$
$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}siny.dy$
$=2{\int }_0^{\frac{\pi }{2}}siny.dy$
$=2[-cosy{]}_0^{\frac{\pi }{2}}$

$=2$.

Hence assertion is false.

Now $f\left(\theta \right)=sin\theta$ has a range of $[-1,1]$

Hence the domain of $y=si{n}^{-1}\left(x\right)$ will be $[-1,1]$.

And $sin\left(\theta \right)$ is one-one function in the interval of $[-\frac{\pi }{2},\frac{\pi }{2}]$.
Hence the range for $y=si{n}^{-1}\left(x\right)$ is $[-\frac{\pi }{2},\frac{\pi }{2}]$.