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

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Limits XII

Let

f : (0, 1) \to R

be the functions defined as

f (x) = n

x \in [\frac{1}{n + 1}, \frac{1}{n})

where

n \in N

. Let

g : (0, 1) \to R

be a function such that

\int_{x^{2}}^{x} \frac{1 - t}{t} dt < g (x) < 2 x

for all

x \in (0, 1)

. Then

lim_{x \to 0} f (x) g (x)

Answer: (C)

We have

\int_{x^{2}}^{x} \frac{1 - t}{t} d t < g (x) < 2 x

.

Evaluating the integral:

\int_{x^{2}}^{x} \frac{1 - t}{t} d t = sin^{- 1} x + x 1 - x - sin^{- 1} x - x 1 - x^{2}

Multiplying the inequality by

f (x) = n

where

x \in [\frac{1}{n + 1}, \frac{1}{n})

n \int_{x^{2}}^{x} \frac{1 - t}{t} d t < f (x) g (x) < 2 n x

x \to 0

, we have

n \to \infty

and

n \cdot x \to 0

, so

n x \to 0

.

Dividing by

x

\frac{n}{x} \int_{x^{2}}^{x} \frac{1 - t}{t} d t < \frac{f ( x ) g ( x )}{x} < 2 n

Using L'Hôpital's rule or Taylor expansion on the integral, the left side approaches 2 as

x \to 0

.

By squeeze theorem:

lim_{x \to 0} f (x) g (x) = 2

.

Answer: (C)

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Single correct option