Question

If $f:R\to \left[-\frac{\pi }{4},\frac{\pi }{2}\right)$ 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. $\alpha +\frac{1}{\alpha }=2\mathrm{cosec}2$
• 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.