Question

If $\alpha$ is only real root of the equation $x^3-\left(\cos 1\right)x^2+\left(\sin 1\right)x+1=0$, then $({\tan }^{-1}\alpha +{\tan }^{-1}\frac{1}{\alpha })$ cannot be equal to _____________.

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

Answer 1

Answer: (A), (B), (D)

The correct options are
A $0$
B $\frac{\pi }{2}$
D $\pi$

Let $f\left(x\right)=x^3+\left(\cos 1\right)x^2+\left(\sin 1\right)x+1$

$\Rightarrow f\left(0\right)=1$

$\Rightarrow f(-\infty )=-\infty$

If $\alpha <0\Rightarrow {\tan }^{-1}\alpha <0$

$\Rightarrow {\tan }^{-1}\left(\frac{1}{\alpha }\right)<0$

$\Rightarrow {\tan }^{-1}\alpha +{\tan }^{-1}\left(\frac{1}{\alpha }\right)<0$ and cannot be equal to for $\alpha =0$

$\Rightarrow {\tan }^{-1}\alpha +{\tan }^{-1}\left(\frac{1}{\alpha }\right)<0$ and cannot be equal to for $\alpha =\frac{\pi }{2}$

$\Rightarrow {\tan }^{-1}\alpha +{\tan }^{-1}\left(\frac{1}{\alpha }\right)<0$ and cannot be equal to for $\alpha =\pi$

Option $\left(a\right),\left(b\right)$ and $\left(d\right)$ are the correct answers.