Question

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

• A. ${\cos }^{-1}\left(\frac{1-{\alpha }^2}{1+{\alpha }^2}\right)=2$
• B. $\alpha +\frac{1}{\alpha }=2\text{cosec}2$
• 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)

${\sin }^{-1}\left(\frac{2\alpha }{{\alpha }^2+1}\right)=\pi -2$

As $f\left(x\right)$ is onto,
$\Rightarrow min(x^4-x^2-\frac{7}{4}+{\tan }^{-1}\alpha )=-1$
$\Rightarrow min({(x^2-\frac{1}{2})}^2-2+{\tan }^{-1}\alpha )=-1$
$\Rightarrow {\tan }^{-1}\alpha =1$
$\Rightarrow \alpha =\tan 1$

${\cos }^{-1}\left(\frac{1-{\tan }^{2}1}{1+{\tan }^{2}1}\right)$
$\Rightarrow {\cos }^{-1}\left(\cos 2\right)=2(\because 2\in (0,\pi \left)\right)$

$\alpha +\frac{1}{\alpha }=\tan 1+\frac{1}{\tan 1}$
$=\frac{{\sin }^{2}1+{\cos }^{2}1}{\sin 1\cos 1}=\frac{2}{\sin 2}=2\text{cosec 2}$

${\sin }^{-1}\left(\frac{2\alpha }{{\alpha }^2+1}\right)={\sin }^{-1}\left(\frac{2\tan 1}{1+{\tan }^{2}1}\right)={\sin }^{-1}\left(\sin 2\right)=\pi -2$

${\tan }^{-1}\left(\frac{2\alpha }{{\alpha }^2-1}\right)={\tan }^{-1}\left(\frac{2\tan 1}{{\tan }^{2}1-1}\right)={\tan }^{-1}(-\tan 2)=\pi -2$