Question

There is a polynomial function given in which if $f\left( x \right)={{x}^{11}}+{{x}^{9}}-{{x}^{7}}+{{x}^{3}}+1$ and $f\left( {{\sin }^{-1}}\left( \sin 8 \right) \right)=\alpha$ , $\alpha$ is a constant, the value of $f\left( {{\tan }^{-1}}\left( \tan 8 \right) \right)$ is equal to:
Note: consider 8 in radians.
(a) $\alpha$
(b) $\alpha -2$
(c) $\alpha +2$
(d) $2-\alpha$

Answer 1

Hint: First of all convert 8 which is in radians to degrees then the angle in degrees will be 458.36. The angle that we have got in degree lie in second quadrant so we can write $\sin 8$ as $\sin \left( 3\pi -8 \right)$ and we can write $\tan 8$ as $-\tan \left( 3\pi -8 \right)$ . The value of ${{\sin }^{-1}}\left( \sin \left( 3\pi -8 \right) \right)$ will be $3\pi -8$ and the value of ${{\tan }^{-1}}\left( -\tan \left( 3\pi -8 \right) \right)$ will be $-\left( 3\pi -8 \right)$ . It is given that $f\left( {{\sin }^{-1}}\left( \sin 8 \right) \right)=\alpha$ then we can write it as $f\left( 3\pi -8 \right)=\alpha$ and we can write $f\left( {{\tan }^{-1}}\left( \tan 8 \right) \right)$ as $f\left( -\left( 3\pi -8 \right) \right)$ . Let us assume $3\pi -8$ as t then substitute t in $f\left( x \right)$ and –t in $f\left( x \right)$ and add them. After adding $f\left( t \right)\And f\left( -t \right)$ you will get 2 then substitute $f\left( t \right)$ as $\alpha$ then you will get the value of $f\left( -t \right)$ or $f\left( {{\tan }^{-1}}\left( \tan 8 \right) \right)$ .

Complete step-by-step answer:
A polynomial function given in the question is:
$f\left( x \right)={{x}^{11}}+{{x}^{9}}-{{x}^{7}}+{{x}^{3}}+1$
It is also given that $f\left( {{\sin }^{-1}}\left( \sin 8 \right) \right)=\alpha$ and we are asked to find the value of $f\left( {{\tan }^{-1}}\left( \tan 8 \right) \right)$ .
As the angle of sine is given in radians which is 8 radians so converting the radians into degrees we get,
$8\left( \dfrac{{{180}^{0}}}{\pi } \right)$
$\begin{aligned} & =8\left( \dfrac{{{180}^{0}}}{3.14} \right) \\\ & ={{458.36}^{0}} \\\ \end{aligned}$
The angle that we have calculated above lies in second quadrant so we can write sine as:
$\sin 8=\sin \left( 3\pi -8 \right)$
Taking inverse of sine on both the sides we get,
$\begin{aligned} & {{\sin }^{-1}}\left( \sin 8 \right)={{\sin }^{-1}}\left( \sin \left( 3\pi -8 \right) \right) \\\ & \Rightarrow {{\sin }^{-1}}\left( \sin 8 \right)=3\pi -8 \\\ \end{aligned}$
Similarly, we can write $\tan 8$ as:
$\tan 8=-\tan \left( 3\pi -8 \right)$
Taking inverse of tan on both the sides we get,
$\begin{aligned} & {{\tan }^{-1}}\tan 8={{\tan }^{-1}}\left( -\tan \left( 3\pi -8 \right) \right) \\\ & \Rightarrow {{\tan }^{-1}}\tan 8=-\left( 3\pi -8 \right) \\\ \end{aligned}$
It is given that:
$f\left( {{\sin }^{-1}}\left( \sin 8 \right) \right)=\alpha$
Substituting ${{\sin }^{-1}}\left( \sin 8 \right)$ as $3\pi -8$ in the above equation we get,
$f\left( 3\pi -8 \right)=\alpha$ ……… eq. (1)
Substituting ${{\tan }^{-1}}\tan 8=-\left( 3\pi -8 \right)$ in $f\left( {{\tan }^{-1}}\left( \tan 8 \right) \right)$ we get,
$f\left( -\left( 3\pi -8 \right) \right)$
If we assume $f\left( 3\pi -8 \right)$ as $f\left( t \right)$ then $f\left( -\left( 3\pi -8 \right) \right)$ will be $f\left( -t \right)$ . Now, substituting the value of “t” in place of x in $f\left( x \right)$ we get,
$f\left( t \right)={{t}^{11}}+{{t}^{9}}-{{t}^{7}}+{{t}^{3}}+1$ …………. Eq. (2)
Substituting the value of –t in place of x in $f\left( x \right)$ we get,
$f\left( -t \right)=-{{t}^{11}}-{{t}^{9}}+{{t}^{7}}-{{t}^{3}}+1$ ………. Eq. (3)
Adding eq. (1) and eq. (2) we get,
$f\left( t \right)+f\left( -t \right)=2$
Now, replacing t with $3\pi -8$ in the above equation we get,
$f\left( 3\pi -8 \right)+f\left( -\left( 3\pi -t \right) \right)=2$
Using the relation in eq. (2) and substituting in the above equation we get,
$\begin{aligned} & \alpha +f\left( -\left( 3\pi -8 \right) \right)=2 \\\ & \Rightarrow f\left( -\left( 3\pi -8 \right) \right)=2-\alpha \\\ \end{aligned}$
Hence, the correct option (d).

Note: You might think of solving this problem as follows:
In the function $f\left( {{\sin }^{-1}}\left( \sin 8 \right) \right)=\alpha$ , the value of ${{\sin }^{-1}}\left( \sin 8 \right)$ is 8 then $f\left( 8 \right)=\alpha$ We have to find the value of the function $f\left( {{\tan }^{-1}}\left( \tan 8 \right) \right)$ , in this function the value of ${{\tan }^{-1}}\left( \tan 8 \right)$ is 8 so the value of this function $f\left( {{\tan }^{-1}}\left( \tan 8 \right) \right)$ is $f\left( 8 \right)$ and which is equal to $\alpha$ .
In the above solution, the wrong step is that we cannot write ${{\tan }^{-1}}\left( \tan 8 \right)$ as 8 because the angle given us in radians and as we have shown above that this angle is of ${{458.36}^{0}}$ which lies in the second quadrant so basically we should write $\tan 8$ as $-\tan \left( 3\pi -8 \right)$ and then proceed.