Let Then which of the following is true

is not continuous at Explanation: 1. Checking Continuity of at A function is continuous at if . Since , the term is a bounded quantity oscillating between and . Multiplying it by (which approaches as ): Given that , we have: Hence, is continuous at . (Option A is incorrect) 2. Checking Differentiability of at Using the first principle definition of the derivative at : Substitute the values of and : Using the same squeeze theorem logic, is bounded between and , and : Since the limit exists and is finite, is differentiable at and . (Option B is incorrect) 3. Checking Continuity of at First, let's find the general derivative expression for using the product rule and chain rule: Now, to check if is continuous at , we find : For the first term: . For the second term: does not exist because the function oscillates infinitely bet

How do I solve NIMCET 2024 Mathematics questions?

Practice NIMCET 2024 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

What is the difficulty level of NIMCET Mathematics questions?

NIMCET Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Tip:A–D to answerE for explanationV for videoS to reveal answer

NIMCET 2024 — Mathematics PYQ

NIMCET | Mathematics | 2024

Let $f (x) = {x^{2} sin \frac{1}{x}, 0, am p; x \neq = 0 am p; x = 0$

Then which of the following is true

Choose the correct answer:

Explanation

1. Checking Continuity of $f (x)$ at $x = 0$

A function is continuous at $x = 0$ if $lim_{x \to 0} f (x) = f (0)$ .

$x \to 0 lim f (x) = x \to 0 lim (x^{2} sin \frac{1}{x})$

Since $- 1 \leq sin \frac{1}{x} \leq 1$ , the term $sin \frac{1}{x}$ is a bounded quantity oscillating between $- 1$ and $1$ .

Multiplying it by $x^{2}$ (which approaches $0$ as $x \to 0$ ):

$x \to 0 lim f (x) = 0 \times (a bounded value between - 1 and 1) = 0$

Given that $f (0) = 0$ , we have:

$x \to 0 lim f (x) = f (0)$

Hence, $f (x)$ is continuous at $x = 0$ . (Option A is incorrect)

2. Checking Differentiability of $f (x)$ at $x = 0$

Using the first principle definition of the derivative at $x = 0$ :

$f^{'} (0) = h \to 0 lim \frac{f ( 0 + h ) - f ( 0 )}{h}$

Substitute the values of $f (h) = h^{2} sin \frac{1}{h}$ and $f (0) = 0$ :

$f^{'} (0) = h \to 0 lim \frac{h ^{2} sin \frac{1}{h} - 0}{h}$

$f^{'} (0) = h \to 0 lim (h sin \frac{1}{h})$

Using the same squeeze theorem logic, $sin \frac{1}{h}$ is bounded between $- 1$ and $1$ , and $lim_{h \to 0} h = 0$ :

$f^{'} (0) = 0 \times (bounded value) = 0$

Since the limit exists and is finite, $f (x)$ is differentiable at $x = 0$ and $f^{'} (0) = 0$ . (Option B is incorrect)

3. Checking Continuity of $f^{'} (x)$ at $x = 0$

First, let's find the general derivative expression $f^{'} (x)$ for $x \neq = 0$ using the product rule and chain rule:

$f^{'} (x) = \frac{d}{d x} (x^{2} sin \frac{1}{x})$

$f^{'} (x) = 2 x \cdot sin \frac{1}{x} + x^{2} \cdot cos \frac{1}{x} \cdot (- \frac{1}{x ^{2}})$

$f^{'} (x) = 2 x sin \frac{1}{x} - cos \frac{1}{x} (for x \neq = 0)$

Now, to check if $f^{'} (x)$ is continuous at $x = 0$ , we find $lim_{x \to 0} f^{'} (x)$ :

$x \to 0 lim f^{'} (x) = x \to 0 lim (2 x sin \frac{1}{x} - cos \frac{1}{x})$

For the first term: $lim_{x \to 0} 2 x sin \frac{1}{x} = 0$ .
For the second term: $lim_{x \to 0} cos \frac{1}{x}$ does not exist because the function oscillates infinitely between $- 1$ and $1$ as $x$ approaches $0$ .

Since $lim_{x \to 0} f^{'} (x)$ does not exist, it cannot equal $f^{'} (0) = 0$ .

Hence, $f^{'} (x)$ is not continuous at $x = 0$ .

Correct Answer: (C) $f^{'} (x)$ is not continuous at $x = 0$