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Question: Given \[f:\left[ {0,\infty } \right) \to R\] be a strictly increasing function such that the functio...

Given f:[0,∞)→Rf:\left[ {0,\infty } \right) \to R be a strictly increasing function such that the functions g(x)=f(x)−3xg\left( x \right) = f\left( x \right) - 3x and h(x)=f(x)−x3h\left( x \right) = f\left( x \right) - {x^3} are both strictly increasing function. Then the function F(x)=f(x)−x2–xF\left( x \right) = f\left( x \right) - {x^2} – x is
A. increasing in (0,1) and decreasing in (1,∞)\left( {1,\infty } \right)
B. decreasing in (0,1) and increasing in (1,∞)\left( {1,\infty } \right)
C. increasing throughout (0,∞)\left( {0,\infty } \right)
D. decreasing throughout (0,∞)\left( {0,\infty } \right)

Explanation

Solution

- 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(x)=f(x)−3xg\left( x \right) = f\left( x \right) - 3x can be calculated as shown below.

      g′(x)=f′(x)−3 ⇒f′(x)=g′(x)+3                                                      ......(1)  \,\,\,\,\,\,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(x)=f(x)−x3h\left( x \right) = f\left( x \right) - {x^3} can be calculated as shown below.

      h′(x)=f′(x)−3x2 ⇒f′(x)=h′(x)+3x2                                                      ......(2)  \,\,\,\,\,\,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(x)=f(x)−x2–xF\left( x \right) = f\left( x \right) - {x^2} – x is calculated as shown below.

       F′(x)=f′(x)−2x−1 ⇒F′(x)=12[2f′(x)]−2x−1 ⇒F′(x)=12[f′(x)+f′(x)]−2x−1  \,\,\,\,\,\,\,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′(x)=12[g′(x)+3+h′(x)+3x2]−2x−1 ⇒F′(x)=12[g′(x)+h′(x)+3x2−4x+1] ⇒F′(x)=12[g′(x)+h′(x)+3(x−23)2+16]  \,\,\,\,\,\,\,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′(x),h′(x)>0g'\left( x \right),h'\left( x \right) > 0, therefore,
F′(x)>0F'\left( x \right) > 0

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

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