Question

The product of the real roots of the equation ${\tan }^{-1}\left(\cot \left(\frac{3x^2+9|x|+21}{x^2+3|x|+8}\right)\right)=\frac{\pi }{2}+\sec \left({\sec }^{-1}\left(\frac{3-27|x|}{9|x|}\right)\right)$ is

• A. 64
• B. 64.00
• C. 64.0

Answer 1

Answer: (A), (B), (C)

Given: ${\tan }^{-1}\left(\cot \left(\frac{3x^2+9|x|+21}{x^2+3|x|+8}\right)\right)=\frac{\pi }{2}+\sec \left({\sec }^{-1}\left(\frac{3-27|x|}{9|x|}\right)\right)$
$\Rightarrow \frac{\pi }{2}-{\cot }^{-1}\left(\cot \left(\frac{3x^2+9|x|+21}{x^2+3|x|+8}\right)\right)=\frac{\pi }{2}+\sec \left({\sec }^{-1}\left(\frac{3-27|x|}{9|x|}\right)\right)$
$\Rightarrow -{\cot }^{-1}\left(\cot (3-\frac{3}{x^2+3|x|+8})\right)=\sec \left({\sec }^{-1}(\frac{1}{3|x|}-3)\right)$
$\Rightarrow \frac{3}{x^2+3|x|+8}=\frac{1}{3|x|}$
$\Rightarrow x^2-6|x|+8=0$
$\Rightarrow |x|=2,4$
$\Rightarrow |x|=\pm 2,\pm 4$
Hence, the product of roots is $64.$