QJEEAdv23Ma6bd1: JEE Advanced 2023 Mathematics - Class 12 JEEAdv23Ma6bd1 Solution: Let be the function defined as if where . Let be a function…

Question

Verified Solution • 2025–26 Session moderate

Let

f:(0,1) \rightarrow \mathbb{R}

be the function defined as

f(x)=\sqrt{n}

if

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

where

n \in \mathbb{N}

. Let

g:(0,1) \rightarrow \mathbb{R}

be a function such that

\int\limits_{x^2}^x \sqrt{\frac{1-t}{t}} d t &lt; g(x) &lt; 2 \sqrt{x}

for all

x \in(0,1)

. Then

\lim\limits_{x \rightarrow 0} f(x) g(x)

Accepted Answer

\begin{aligned} &amp; f:(0,1) \rightarrow R, \\\\ &amp; f(x)=\sqrt{n}, x \in\left[\frac{1}{n+1}, \frac{1}{n}\right), n \in N\end{aligned}

g:(0,1) \rightarrow R

where

\int_{x^2}^x \sqrt{\frac{1-t}{t}} d t &lt; g(x) &lt; 2 \sqrt{x}, x \in(0,1)

Now (According to the question)

\begin{aligned}&amp; \lim _{x \rightarrow \infty} f(x) \cdot g(x) \\\\&amp; \Rightarrow \text { Put } x=\frac{1}{n} \\\\&amp; \lim _{n \rightarrow \infty} f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right)\end{aligned}

\begin{aligned} \lim _{n \rightarrow \infty} \sqrt{n-1} \int_{\frac{1}{n^2}}^{\frac{1}{n}} \sqrt{\frac{1-t}{t} d t} \leq \lim _{n \rightarrow \infty} &amp; f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right) \leq \lim _{n \rightarrow \infty} \sqrt{n}-1 \frac{2}{\sqrt{n}}\end{aligned}

\Rightarrow \lim\limits_{n \rightarrow \infty} \frac{\int_{\frac{1}{n^2}}^{\frac{1}{n}} \sqrt{\frac{1-t}{t}} d t}{\frac{1}{\sqrt{n-1}}} \leq \lim\limits_{n \rightarrow \infty} f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right) \leq 2

\begin{gathered}\Rightarrow \frac{\lim\limits_{n \rightarrow \infty} \frac{-1}{n^2} \sqrt{n-1}+\frac{2}{n^3} \sqrt{n^2-1}}{\frac{1}{2(n-1)^{\frac{3}{2}}}} \leq \lim _{n \rightarrow \infty} f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right) \leq 2\end{gathered}

\begin{aligned} &amp; \therefore \lim _{n \rightarrow \infty} \frac{2(n-1)^2}{n^2}-\frac{4(n-1)^{\frac{3}{2}} \sqrt{n^2-1}}{n^3}=2 \\\\ &amp; \therefore \lim _{n \rightarrow \infty} f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right)=2 \text { (Using Sandwich Theorem) }\end{aligned}

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