Let f(x) be a non-negative differentiable function on [0, ∞) such that f(0)=0 and f(x) ≤ 2 f(x) for all x>0 . Then, on [0, ∞)

Correct answer: (A) f(x) is always a constant function. f^′(x) ≤ 2 f(x) f^′(x) e^-2 x ≤ 2 f(x) e^-2 x d/dx (f(x) e^-2 n ) ≤ 0 ~g(x)=f(x) e^-2 x is non-Increasing function x ≥ 0 ~g(x) ≤ g(0) f(x) e^-2 x ≤ f(0) e^-0 f(x) e^-2 x ≤ 0 f(x) ≤ 0 but given f(x) is not negative ∴ f(x)=0 Constant function

Q54 / 90 · KVPY 2016 (SB/SX) · Mathematics

Let $\mathrm{f}(\mathrm{x})$ be a non-negative differentiable function on $[0, \infty)$ such that $\mathrm{f}(0)=0$ and $\mathrm{f}(\mathrm{x}) \leq 2 \mathrm{f}(\mathrm{x})$ for all $\mathrm{x}>0$. Then, on $[0, \infty)$

Correct answer
A
$\mathrm{f}(\mathrm{x})$ is always a constant function
B
$\mathrm{f}(\mathrm{x})$ is strictly increasing
C
$\mathrm{f}(\mathrm{x})$ is strictly decreasing
D
$\mathrm{f}^{\prime}(\mathrm{x})$ changes sign

Solution

$\mathrm{f}^{\prime}(\mathrm{x}) \leq 2 \mathrm{f}(\mathrm{x})$ $\mathrm{f}^{\prime}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}} \leq 2 \mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}}$ $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{n}}\right) \leq 0$ $\mathrm{~g}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}}$ is non-Increasing function $\mathrm{x} \geq 0$ $\mathrm{~g}(\mathrm{x}) \leq \mathrm{g}(0)$ $\mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}} \leq \mathrm{f}(0) \mathrm{e}^{-0}$ $\mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}} \leq 0$ $\mathrm{f}(\mathrm{x}) \leq 0 \quad$ but given $\mathrm{f}(\mathrm{x})$ is not negative $\therefore \mathrm{f}(\mathrm{x})=0 \quad$ Constant function

Hence, the correct answer is option: 'A'.

Let $\mathrm{f}(\mathrm{x})$ be a non-negative differentiable function on $[0, \infty)$ such that $\mathrm{f}(0)=0$ and $\mathrm{f}(\mathrm{x}) \leq 2 \mathrm{f}(\mathrm{x})$ for all $\mathrm{x}>0$. Then, on $[0, \infty)$

Solution

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