Question

Given \[f:\left[ {0,\infty } \right) \to R\] be a strictly increasing function such that the functions \[g\left( x \right) = f\left( x \right) - 3x \] and \[h\left( x \right) = f\left( x \right) - {x^3}\] are both strictly increasing function. Then the function \[F\left( x \right) = f\left( x \right) - {x^2} – x\] is
A. increasing in (0,1) and decreasing in \[\left( {1,\infty } \right)\]
B. decreasing in (0,1) and increasing in \[\left( {1,\infty } \right)\]
C. increasing throughout \[\left( {0,\infty } \right)\]
D. decreasing throughout \[\left( {0,\infty } \right)\]

Answer 1

- Hint: In this problem, we need to use the definition of the strictly increasing function to find whether the given function is increasing or decreasing in the given interval. A function is said to be strictly increasing if the first derivative of the function is always greater than 0.

Complete step-by-step solution -
The derivative of the function \[g\left( x \right) = f\left( x \right) - 3x\] can be calculated as shown below.
\[
  \,\,\,\,\,\,g'\left( x \right) = f'\left( x \right) - 3 \\
   \Rightarrow f'\left( x \right) = g'\left( x \right) + 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right) \\
\]
The derivative of the function \[h\left( x \right) = f\left( x \right) - {x^3}\] can be calculated as shown below.
\[
  \,\,\,\,\,\,h'\left( x \right) = f'\left( x \right) - 3{x^2} \\
   \Rightarrow f'\left( x \right) = h'\left( x \right) + 3{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 2 \right) \\
\]
Similarly, the derivative of the function \[F\left( x \right) = f\left( x \right) - {x^2} – x\] is calculated as shown below.
\[
  \,\,\,\,\,\,\,F'\left( x \right) = f'\left( x \right) - 2x - 1 \\
   \Rightarrow F'\left( x \right) = \dfrac{1}{2}\left[ {2f'\left( x \right)} \right] - 2x - 1 \\
   \Rightarrow F'\left( x \right) = \dfrac{1}{2}\left[ {f'\left( x \right) + f'\left( x \right)} \right] - 2x - 1 \\
\]
Now, from equation (1) and (2),
\[
  \,\,\,\,\,\,\,F'\left( x \right) = \dfrac{1}{2}\left[ {g'\left( x \right) + 3 + h'\left( x \right) + 3{x^2}} \right] - 2x - 1 \\
   \Rightarrow F'\left( x \right) = \dfrac{1}{2}\left[ {g'\left( x \right) + h'\left( x \right) + 3{x^2} - 4x + 1} \right] \\
   \Rightarrow F'\left( x \right) = \dfrac{1}{2}\left[ {g'\left( x \right) + h'\left( x \right) + 3{{\left( {x - \dfrac{2}{3}} \right)}^2} + \dfrac{1}{6}} \right] \\
\]
Now, since \[g'\left( x \right),h'\left( x \right) > 0\], therefore,
\[F'\left( x \right) > 0\]

Thus, \[F\left( x \right)\] is increasing in interval \[\left( {0,\infty } \right)\], hence, option (C) is the correct answer.

Note: The function \[f\left( x \right)\] is said to be strictly increasing function, if \[f'\left( x \right) > 0\] for all the value of\[x\], in other hand, the function \[f\left( x \right)\] is said to be strictly decreasing function, if \[f'\left( x \right) < 0\].