GATE CS 2020 Q1 — Maxima and Minima, Mean Value Theorem, Integration

A function $f(x)$ is increasing on an interval $[a, b]$ if its derivative satisfies $f'(x) \ge 0$ for all $x \in [a, b]$.

Let us analyze each function on the interval $[0, 1]$:

1. Function I: $f_1(x) = e^{-x}$

   Taking the derivative:

   $$f_1'(x) = -e^{-x}$$

   Since $e^{-x} > 0$ for all real $x$, we have $f_1'(x) < 0$ for all $x \in [0, 1]$. Thus, $f_1(x)$ is strictly decreasing on $[0, 1]$.

2. Function II: $f_2(x) = x^2 - \sin x$

   Taking the derivative:

   $$f_2'(x) = 2x - \cos x$$

   Let us evaluate the derivative at $x = 0$:

   $$f_2'(0) = 2(0) - \cos(0) = -1 < 0$$

   Since the derivative is negative at the beginning of the interval, $f_2(x)$ is not increasing everywhere on $[0, 1]$.

3. Function III: $f_3(x) = \sqrt{x^3 + 1}$

   Taking the derivative using the chain rule:

   $$f_3'(x) = \frac{1}{2\sqrt{x^3 + 1}} \cdot (3x^2) = \frac{3x^2}{2\sqrt{x^3 + 1}}$$

   For any $x \in [0, 1]$:

   • The numerator $3x^2 \ge 0$ (with equality only at $x = 0$).
   • The denominator $2\sqrt{x^3 + 1} > 0$.

   Therefore, $f_3'(x) \ge 0$ for all $x \in [0, 1]$, which means $f_3(x)$ is increasing everywhere on $[0, 1]$.

Hence, only function III is increasing everywhere in $[0, 1]$.