Question

If $ f\left( x \right) $ satisfies $ x + \left| {f\left( x \right)} \right| = 2f\left( x \right) $ then $ {f^{ - 1}}\left( x \right) $ satisfies
A. $ 3x + \left| {{f^{ - 1}}\left( x \right)} \right| = 2{f^{ - 1}}\left( x \right) $
B. $ x + \left| {{f^{ - 1}}\left( x \right)} \right| = 2{f^{ - 1}}\left( x \right) $
C. $ {f^{ - 1}}\left( x \right) - \left| x \right| = 2x $
D. $ 3x - \left| {{f^{ - 1}}\left( x \right)} \right| = 2{f^{ - 1}}\left( x \right) $

Answer 1

Hint: In this problem, first we will consider $ f\left( x \right) = y $ . Then, we will use the definition of the modulus function. Modulus function is given by $ \left| x \right| = \left\{
  x,\quad x \geqslant 0 \\
   - x,\quad x < 0 \\
  \right. $ . Then, we will find an inverse function of $ f\left( x \right) $ . Then, we will check that out of four options, which option is correct.

Complete step-by-step answer:
In this problem, it is given that the function $ f\left( x \right) $ satisfies $ x + \left| {f\left( x \right)} \right| = 2f\left( x \right) \cdots \cdots \left( 1 \right) $ . Let us consider $ f\left( x \right) = y $ . Therefore, from $ \left( 1 \right) $ we can write $ x + \left| y \right| = 2y \cdots \cdots \left( 2 \right) $ . We know that $ \left| x \right| = x $ for $ x \geqslant 0 $ and $ \left| x \right| = - x $ for $ x < 0 $ . We will use this information in equation $ \left( 2 \right) $ .
Case I: Consider $ y \geqslant 0 $
If $ y \geqslant 0 $ then $ \left| y \right| = y $ . Therefore, from equation $ \left( 2 \right) $ we get
 $
  x + y = 2y \\
   \Rightarrow x = 2y - y \\
   \Rightarrow x = y \\
   \Rightarrow {f^{ - 1}}\left( y \right) = y\quad \left[ {\because f\left( x \right) = y \Rightarrow x = {f^{ - 1}}\left( y \right)} \right] \\
  $
Replace $ y $ by $ x $ , we get $ {f^{ - 1}}\left( x \right) = x $ for $ x \geqslant 0 $ .
Case II: Consider $ y < 0 $
If $ y < 0 $ then $ \left| y \right| = - y $ . Therefore, from equation $ \left( 2 \right) $ we get
 $
  x + \left( { - y} \right) = 2y \\
   \Rightarrow x = 2y + y \\
   \Rightarrow x = 3y \\
   \Rightarrow {f^{ - 1}}\left( y \right) = 3y\quad \left[ {\because f\left( x \right) = y \Rightarrow x = {f^{ - 1}}\left( y \right)} \right] \\
  $
Replace $ y $ by $ x $ , we get $ {f^{ - 1}}\left( x \right) = 3x $ for $ x < 0 $ . Therefore, from case I and II, we have $ {f^{ - 1}}\left( x \right) = \left\{
  x,\quad x \geqslant 0 \\
  3x,\quad x < 0 \\
  \right. $ .
Now we will check that out of four options which option is correct.
Let us take $ {f^{ - 1}}\left( x \right) = x $ for $ x \geqslant 0 $ . Note that here $ \left| {{f^{ - 1}}\left( x \right)} \right| = {f^{ - 1}}\left( x \right) $ .
In option A, LHS $ = 3x + \left| {{f^{ - 1}}\left( x \right)} \right| = 3x + {f^{ - 1}}\left( x \right) = 3x + x = 4x $ and RHS $ = 2{f^{ - 1}}\left( x \right) = 2x $ .
Here LHS $ \ne $ RHS. Therefore, option A is wrong.
In option C, LHS $ = {f^{ - 1}}\left( x \right) - \left| x \right| = x - \left| x \right| = x - x = 0 $ and RHS $ = 2x $ .
Here LHS $ \ne $ RHS. Therefore, option C is wrong.
Let us take $ {f^{ - 1}}\left( x \right) = 3x $ for $ x < 0 $ . Note that here $ \left| {{f^{ - 1}}\left( x \right)} \right| = - {f^{ - 1}}\left( x \right) $ .
In option B, LHS $ = x + \left| {{f^{ - 1}}\left( x \right)} \right| = x - {f^{ - 1}}\left( x \right) = x - 3x = - 2x $ and RHS $ = 2{f^{ - 1}}\left( x \right) = 2\left( {3x} \right) = 6x $ .
Here LHS $ \ne $ RHS. Therefore, option B is wrong.
Let us take $ {f^{ - 1}}\left( x \right) = \left\{
  x,\quad x \geqslant 0 \\
  3x,\quad x < 0 \\
  \right. $ . Note that here $ \left| {{f^{ - 1}}\left( x \right)} \right| = {f^{ - 1}}\left( x \right) $ for $ x \geqslant 0 $ and $ \left| {{f^{ - 1}}\left( x \right)} \right| = - {f^{ - 1}}\left( x \right) $ for $ x < 0 $ .
In option D, LHS $ = 3x - \left| {{f^{ - 1}}\left( x \right)} \right| = 3x - {f^{ - 1}}\left( x \right) = 3x - x = 2x $ and RHS $ = 2{f^{ - 1}}\left( x \right) = 2x $ .
Also in option D, LHS $ = 3x - \left| {{f^{ - 1}}\left( x \right)} \right| = 3x - \left( { - {f^{ - 1}}\left( x \right)} \right) = 3x - \left( { - 3x} \right) = 6x $ and RHS $ = 2{f^{ - 1}}\left( x \right) = 2\left( {3x} \right) = 6x $ . Here LHS $ = $ RHS. Therefore, we can say that option D is correct.
So, the correct answer is “Option D”.

Note: A function $ f:A \to B $ is said to be one-one function if for every $ x,y \in A $ , $ x \ne y \Rightarrow f\left( x \right) \ne f\left( y \right) $ or $ f\left( x \right) = f\left( y \right) \Rightarrow x = y $ . A function $ f\left( x \right) $ is one-one if and only if $ {f^{ - 1}}\left( x \right) $ exists.