Question

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

Note : consider 8 in radians.

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

Answer 1

The correct option is D $2-\alpha$

$f\left(si{n}^{-1}\left(sin8\right)\right)=f\left(si{n}^{-1}\left(sin(3\pi -8)\right)\right)$
As $-\frac{\pi }{2}<3\pi -8<\frac{\pi }{2}$
$f(3\pi -8)=\alpha \Rightarrow {(3\pi -8)}^{11}+{(3\pi -8)}^9-{(3\pi -8)}^7+{(3\pi -8)}^3+1=\alpha$...(1)
Now,
$f\left(ta{n}^{-1}\left(tan8\right)\right)=f\left(ta{n}^{-1}(tan8-3\pi )\right)$
$=f(8-3\pi )={(8-3\pi )}^{11}+{(8-3\pi )}^9-{(8-3\pi )}^7+{(8-3\pi )}^3+1=2-({(8-3\pi )}^{11}+{(8-3\pi )}^9-{(8-3\pi )}^7+{(8-3\pi )}^3+1)$

From (1):

$f\left(ta{n}^{-1}\left(tan8\right)\right)=2-\alpha$