Question

For $y={\sin }^{-1}\left\{\frac{5x+12\sqrt{1-x^2}}{13}\right\};|x|\le 1$, if $a\left(1-x^2\right)y_2+bx{y}_1=0$ then $(a,b)=$

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

Answer 1

Answer: (B)

$y={\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)$
Let $\sin {\theta }_1=\frac{5}{13}$ and $\cos {\theta }_2=x$
$y={\sin }^{-1}\left(\sin {\theta }_1\cdot \cos {\theta }_2+\cos {\theta }_1\cdot \sin {\theta }_2\right)$
$y={\theta }_1+{\theta }_2={\sin }^{-1}\frac{5}{13}+{\cos }^{-1}x$
$y_1=-\frac{1}{\sqrt{1-x^2}}$
$y_2=\frac{-x}{\left(1-x^2\right)\sqrt{1-x^2}}$
$\Rightarrow y_2\left(1-x^2\right)=x\cdot y_1$
$\Rightarrow y_2\left(1-x^2\right)-x{y}_1=0$
$\therefore a=1,b=-1$
$(a,b)=(1,-1)$