Question

If $f : R \rightarrow\left[-\frac{\pi}{4}, \frac{\pi}{2}\right)$ is a function defined by $f ( x )=\tan ^{-1}\left( x ^4- x ^2-\frac{7}{4}+\tan ^{-1} \alpha\right)$ and $f$ is surjective then

• A. $\cos ^{-1}\left(\frac{1-\alpha^2}{1+\alpha^2}\right)=2$
• B. $\alpha+\frac{1}{\alpha}=2 \operatorname{cosec} 2$
• C. $\sin ^{-1}\left(\frac{2 \alpha}{\alpha^2+1}\right)=\pi-2$
• D. $\tan ^{-1}\left(\frac{2 \alpha}{\alpha^2-1}\right)=2-\pi$

Answer 1

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

$f(x)=\tan ^{-1}\left(x^4-x^2+\frac{1}{4}-2+\tan ^{-1} \alpha\right)=\tan ^{-1}\left(\left(x^2-\frac{1}{2}\right)^2+\tan ^{-1} \alpha-2\right)$
For ' $f$ ' to be surjective, $\tan ^{-1} \alpha-2=-1 \Rightarrow \alpha=\tan 1$ Now, verify the options.