JEE Advanced 2023 · Paper 1 · Mathematics Q4

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MathematicsSingle correct+3 / -1Question 4 of 51

Let $f : (0,1) \to \mathbb{R}$ be the function defined as $f(x) = \sqrt n$ if $x \in \left[\dfrac{1}{n+1}, \dfrac1n\right)$ where $n \in \mathbb{N}$ . Let $g : (0,1) \to \mathbb{R}$ be a function such that $\displaystyle\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}\,dt < g(x) < 2\sqrt x$ for all $x \in (0,1)$ . Then $\displaystyle\lim_{x\to0}f(x)g(x)$

does NOT exist

is equal to 1

is equal to 2

is equal to 3

Answer and worked solution

Official answer: C

Size of $f$ . If $x\in\left[\frac1{n+1},\frac1n\right)$ then $\frac1x-1<n\le\frac1x$ . So $f(x)=\sqrt n$ satisfies $\sqrt x\,f(x)\to1$ as $x\to0$ .

Size of $g$ . The upper bound is $2\sqrt x$ . For the lower bound, near $t=0$ we have $\sqrt{\frac{1-t}{t}}=t^{-1/2}(1+O(t))$ , so
$\int_{x^2}^x\sqrt{\frac{1-t}{t}}\,dt=2\sqrt x-2x+O(x^{3/2}).$
By the sandwich theorem, $\dfrac{g(x)}{\sqrt x}\to2$ .

$\lim_{x\to0}f(x)g(x)=\lim\big(\sqrt x f(x)\big)\cdot\frac{g(x)}{\sqrt x}=1\cdot2=2.$

Topics:Limits Fundamental theorem of calculus Types of functions

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JEE Advanced 2023 · Paper 1, Mathematics question 4