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{align}
  & =8\left( \dfrac{{{180}^{0}}}{3.14} \right) \\
 & ={{458.36}^{0}} \\
\end{align} $
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{align}
  & {{\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{align} $
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{align}
  & {{\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{align} $
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{align}
  & \alpha +f\left( -\left( 3\pi -8 \right) \right)=2 \\
 & \Rightarrow f\left( -\left( 3\pi -8 \right) \right)=2-\alpha \\
\end{align} $
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.