Question

If $si{n}^{-1}\left(cos\left\{\frac{x^2+3\left|x\right|+4}{x^2+3\left|x\right|+5}\right\}\right)=sin\left(si{n}^{-1}\left\{\frac{1-9\left|x\right|}{9\left|x\right|}\right\}\right)+\frac{\pi }{2},$ then the product of the possible value of x is

• A. 9
• B. 25
• C. 36
• D. 4

Answer 1

The correct option is B 25

Simplifying, we get

$\frac{\pi }{2}-\frac{x^2+3|x|+4}{x^2+3|x|+5}=\left(\frac{1-9|x|}{9|x|}\right)+\frac{\pi }{2}$

$-\frac{x^2+3|x|+4}{x^2+3|x|+5}=\frac{1-9|x|}{9|x|}$

$\frac{x^2+3|x|+5-1}{x^2+3|x|+5}=\frac{9|x|-1}{9|x|}$

$1-\frac{1}{x^2+3|x|+5}=1-\frac{1}{9|x|}$

$\frac{-1}{x^2+3|x|+5}=-\frac{1}{9|x|}$

$x^2+3|x|+5=9|x|$

$x^2-6|x|+5=0$

$(|x|-1)(|x|-5)=0$

$x=\pm 1$ and $x=\pm 5$

Hence product is 25.