Q79 / 155 · WBJEE 2021 · Mathematics
For $y=\sin ^{-1}\left\{\frac{5 x+12 \sqrt{1-x^{2}}}{13}\right\} ;|x| \leq 1$, if $a\left(1-x^{2}\right) y_{2}+b x y_{1}=0$ then $(a, b)=$
- A$(2,1)$
- Correct answerB$(1,-1)$
- C$(-1,1)$
- D$(1,2)$
Solution
$y=\sin ^{-1}\left(\frac{5 x+12 \sqrt{1-x^{2}}}{13}\right)$
Let $\sin \theta_{1}=\frac{5}{13}$ and $\cos \theta_{2}=x$
$\begin{array}{l}
y=\sin ^{-1}\left(\sin \theta_{1} \cdot \cos \theta_{2}+\cos \theta_{1} \cdot \sin \theta_{2}\right. \\
y=\theta_{1}+\theta_{2}=\sin ^{-1} \frac{5}{13}+\cos ^{-1} x
\end{array}$
$y_{1}=-\frac{1}{\sqrt{1-x^{2}}}$
$y_{2}=\frac{-x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}$
$\Rightarrow \mathrm{y}_{2}\left(1-\mathrm{x}^{2}\right)=\mathrm{x} \cdot \mathrm{y}_{1}$
$\Rightarrow \mathrm{y}_{2}\left(1-\mathrm{x}^{2}\right)-\mathrm{xy}_{1}=0$
$\therefore \mathrm{a}=1, \mathrm{~b}=-1$
$(\mathrm{a}, \mathrm{b})=(1,-1)$
Hence, the correct answer is option: 'B'.