Question

Let$f\left(x\right)=cos\left({\tan }^{-1}2x\right)-sin\left\{{\tan }^{-1}\left(\frac{1}{2x+1}\right)\right\}$ and

$a=cos\left({\tan }^{-1}\right(\sin \left({\cot }^{-1}2x\right)\left)\right)$ and $b=\cos (\frac{\pi }{2}+co{s}^{-1}2x)$

The value of x for which$f\left(x\right)=0$is

• A. $-\frac{1}{4}$
• B. $\frac{1}{4}$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $-\frac{1}{4}$

Given

$f\left(x\right)=\cos \left({\tan }^{-1}2x\right)-\sin \left({\tan }^{-1}\left(\frac{1}{2x+1}\right)\right)$

$⟹f\left(x\right)=\cos \left({\cos }^{-1}\left(\frac{1}{\sqrt{1+4x^2}}\right)\right)-\sin \left({\sin }^{-1}\left(\frac{1}{\sqrt{4x^2+4x+2}}\right)\right)$

$⟹\cos \left({\cos }^{-1}\left(\frac{1}{\sqrt{1+4x^2}}\right)\right)-\sin \left({\sin }^{-1}\left(\frac{1}{\sqrt{4x^2+4x+2}}\right)\right)=0$

$⟹\frac{1}{\sqrt{1+4x^2}}-\frac{1}{\sqrt{4x^2+4x+2}}=0$

$⟹\frac{1}{\sqrt{1+4x^2}}=\frac{1}{\sqrt{4x^2+4x+2}}$

$⟹1+4x^2=4x^2+4x+2$

$⟹4x+1=0$

$⟹x=\frac{-1}{4}$