Question

If $f\left(x\right)=x^{11}+x^9-x^7+x^3+1$ and $f\left({\sin }^{-1}\right(\sin 8\left)\right)=\alpha ,$ where $\alpha$ is constant, then $f\left({\tan }^{-1}\right(\tan 8\left)\right)$ is equal to

• A. $\alpha$
• B. $\alpha -2$
• C. $\alpha +2$
• D. $2-\alpha$

Answer 1

The correct option is D $2-\alpha$

We observe $8\notin [-\frac{\pi }{2},\frac{\pi }{2}]$
But $3\pi -8\in [-\frac{\pi }{2},\frac{\pi }{2}]$
$\therefore f\left({\sin }^{-1}\right(\sin 8\left)\right)=f\left({\sin }^{-1}\right(\sin (3\pi -8)\left)\right)=f(3\pi -8)$
$f(3\pi -8)=\alpha$
$\Rightarrow (3\pi -8{)}^{11}+(3\pi -8{)}^9-(3\pi -8{)}^7+(3\pi -8{)}^3+1=\alpha \cdots \left(1\right)$

Now, $f\left({\tan }^{-1}\right(\tan 8\left)\right)$
$=f\left({\tan }^{-1}\right(\tan (8-3\pi )\left)\right)$
$=f(8-3\pi )$
$=(8-3\pi {)}^{11}+(8-3\pi {)}^9-(8-3\pi {)}^7+(8-3\pi {)}^3+1$
$=2-\left(\right(3\pi -8{)}^{11}+(3\pi -8{)}^9-(3\pi -8{)}^7+(3\pi -8{)}^3+1)$

From $\left(1\right),$ $f\left({\tan }^{-1}\right(\tan 8\left)\right)=2-\alpha$