Test Index

Definite Integration

Section: Mathematics

© examsnet.com

Question : 5 of 46

Marks: +1, -0

Let

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

be the function defined as

f(x)=\sqrt{n}

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

n \in \mathbb{N}

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

be a function such that

\int\limits_{x^2}^{x} \sqrt{\; \frac{1-t}{t}} dt < g(x) < 2 \sqrt{x}

x \in (0,1)

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

[JEE Adv 2023 P1]

Does NOT exist
Is equal to 1
Is equal to 2
Is equal to 3

Solution:

We need to solve 1 sided limit here to get some answer, otherwise

\lim\limits_{x \rightarrow 0^{-}}

doesn't exist here (not in domain)

f(x)=\sqrt{ \left(\; \frac{1}{x} \right)-1} \;\; \;\text{ where } \; ( \cdot ) = \; \text{ least integer function } \;

\lim\limits_{x \rightarrow 0^{+}} \int\limits_{x^2}^{x} \sqrt{\; \frac{1-t}{t}} dt \cdot \sqrt{ \left(\; \frac{1}{x} \right)-1 } \leq \lim\limits_{x \rightarrow 0^{+}} f(x) \cdot g(x) \leq \lim\limits_{x \rightarrow 0^{+}} \sqrt{ \left(\; \frac{1}{x} \right)-1 } \times 2 \sqrt{x}

\; \text{ Now } \; \lim\limits_{x \rightarrow 0^{+}} \sqrt{ \left(\; \frac{1}{x} \right)-1 } \times 2 \sqrt{x} = \lim\limits_{x \rightarrow 0^{+}} 2 \sqrt{x} \sqrt{ \left[ \frac{1}{x} \right] } \;\; \left( \frac{1}{x} \notin \mathbb{Z} \right)

= \lim\limits_{x \rightarrow 0^{+}} 2 \sqrt{ x \left( \frac{1}{x} - \left\{ \frac{1}{x} \right\} \right) } = 2

= \lim\limits_{x \rightarrow 0^{+}} 2 \sqrt{ x \left( \frac{1}{x} \right) } = 2 ; \left( \frac{1}{x} \notin \mathbb{Z} \right)

\lim\limits_{x \rightarrow 0^{+}} \int\limits_{x^2}^{x} \sqrt{\; \frac{1-t}{t}} dt \cdot \sqrt{ \; \frac{1}{x} - \left\{ \frac{1}{x} \right\} } = \frac{ \int\limits_{x^2}^{x} \sqrt{ \frac{1-t}{t} } dt \cdot \sqrt{ 1 - x \left\{ \frac{1}{x} \right\} } }{ \sqrt{x} }

\lim\limits_{x \rightarrow 0^{+}} \frac{ \int\limits_{x^2}^{x} \sqrt{ \frac{1-t}{t} } dt }{ \sqrt{x} } = \lim\limits_{x \rightarrow 0^{+}} \frac{ \sqrt{ \frac{1-x}{x} } - 2x \sqrt{ \frac{1-x^2}{x^2} } }{ \frac{1}{2 \sqrt{x}} }

\lim\limits_{x \rightarrow 0^{+}} 2 \sqrt{1-x} - 4 \sqrt{x} \cdot \sqrt{1-x^2} = 2

Similarly for

\frac{1}{x} \in \mathbb{Z}

is equal to 2 .

© examsnet.com

Go to Question:

Prev Question

Next Question