Question

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

• A. 1
• B. $\tan 1$
• C. $1+\tan 1$
• D. $1-\tan 1$

Answer 1

Answer: (B)

$\Theta f(x)$ is surjective
$\therefore \text{ Range of }f(x)=\left[\frac{-\pi }{4},\frac{\pi }{2}\right)$
$f(x)={\tan }^{-1}\left({\left(x^2-\frac{1}{2}\right)}^2-2+{\tan }^{-1}\alpha \right)$
$\therefore -2+{\tan }^{-1}\alpha =-1$
$\Rightarrow {\tan }^{-1}\alpha =1\Rightarrow \alpha =\tan 1$