Q120 / 200 · Manipal (MET) 2013 · Mathematics
$\begin{aligned} & y=\int_{1 / 8}^{\sin ^2 x} \sin ^{-1} \sqrt{t} d t+\int_{1 / 8}^{\cos ^2 x} \cos ^{-1} \sqrt{t} d t, \\ & 0 \leq x \leq x / 2 .\end{aligned}$
- Correct answerAIs the equation of a straight line parallel to the $x$-axis
- BIs the euqation of a straight line which is the bisector of first quadrant
- CIs the equation of a straight line which is the bisector of second quadrant
- DNone of the above
Solution
Here, we have to prove that,
$y=$ constant or derivative of $y$
with respect to $x$ is zero.
$y=\int_{1 / 8}^{\sin ^2 x} \sin ^{-1} \sqrt{t} d t+\int_{1 / 8}^{\cos ^2 x} \cos ^{-1} \sqrt{t} d t \quad \ldots(i)$
$\begin{aligned} & \frac{d y}{d x}=\sin ^{-1} \sqrt{\sin ^2 x} \cdot 2 \sin x \cdot \cos x+\cos ^{-1} \sqrt{\cos ^2 x} \\ & (-2 \sin x \cdot \cos x)=2 x \sin x \cdot \cos x-2 x \cdot \sin x \cdot \cos x \\ & =0 \text { for all } x\end{aligned}$
Hence, the correct answer is option: 'A'.