Exhaustive value of $x$ such that $\cos^{-1}\left(\dfrac{8x}{1+16x^2}\right)=-\dfrac{\pi}{2}+2\tan^{-1}(4x)$
$\left(\dfrac{1}{8},\infty\right)$
$\left(\dfrac{1}{16},\dfrac{1}{8}\right)$
$\left[\dfrac{1}{4},\infty\right)$
$\left(\dfrac{1}{8},\dfrac{1}{4}\right)$
Step-by-Step Solution
Key Concept: Use $\cos^{-1}(t)=-\pi/2+2\tan^{-1}$ form; domain conditions
Identity holds when $4x\ge 1$: $x\ge 1/4$. Exhaustive set: $[1/4,\infty)$.
Correct Answer: 3