Question

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. $(2,1)$
• B. $(1,-1)$
• C. $(-1,1)$
• D. $(1,2)$

Answer 1

Answer: (B)

$y=\sin ^{-1}\left(\frac{5 x+12 \sqrt{1-x^{2}}}{13}\right)$
Let $\sin \theta_{1}=\frac{5}{13}$ and $\cos \theta_{2}=x$
$y=\sin ^{-1}\left(\sin \theta_{1} \cdot \cos \theta_{2}+\cos \theta_{1} \cdot \sin \theta_{2}\right)$
$y=\theta_{1}+\theta_{2}=\sin ^{-1} \frac{5}{13}+\cos ^{-1} x$
$y_{1}=-\frac{1}{\sqrt{1-x^{2}}}$
$y_{2}=\frac{-x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}$
$\Rightarrow y_{2}\left(1-x^{2}\right)=x \cdot y_{1}$
$\Rightarrow y_{2}\left(1-x^{2}\right)-x y_{1}=0$
$\therefore a=1, b=-1$
$(a, b)=(1,-1)$