Let ℝ denote the set of all real numbers. Define the function f: R arrow R by f(x)=\ array cc 2-2 x^2-x^2 (1)/(x) & if x ≠ 0 2 & if x=0 array . Then which one of the following statements is TRUE ?

Correct answer: C. For any positive real number , the function f is NOT an increasing function on the interval (- , 0) (A) RHD at x=0: _ h arrow 0 (2-2 h^2-h^2 (1)/(h))-2 h =0 Similarly LHD at x=0 is also equal to 0 . Differentiable at x=0 (B) f^ (x)=-4 x-2 x (1)/(x)-x^2( (1)/(x))((-1)/(x^2)) aligned & f^ (x)=-(4 x+2 x (1)/(x))+ (1)/(x) & f^ (x)=-(2 x(4- (1)/(x)))+ (1)/(x) aligned for x ∈(0, ) We can't say f(x) is decreasing on (0, ) as (1)/(x) oscillates. (C) for x ∈(- , 0) , for any >0 f(x) is not increasing on (- , 0) as (1)/(x) oscillates from -1 to 1. (D) f(0)=2 aligned & f(0+h) <2 & f(0-h) <2 aligned x=0 is local maxima

JEE Advanced 2025Paper 1

Exam: JEE Advanced
Year: 2025
Paper: Paper 1
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2025 · Paper 1 · Mathematics · Application of Derivatives

Let ℝ denote the set of all real numbers. Define the function f: R...

Let

\mathbb{R}

denote the set of all real numbers. Define the function

f: \mathbb{R \rightarrow \mathbb { R }}

f(\mathrm{x})=\left\{

\begin{array}{cc}2-2 x^2-x^2 \sin\frac{1}{x} & \text{ if } x \neq 0 \\ 2 & \text{ if } x=0\end{array}

\right.

Then which one of the following statements is TRUE ?

Options

A $\text{ The function } f \text{ is NOT differentiable at } x=0$
BThere is a positive real number $\delta$ , such that $f$ is a decreasing function on the interval $(0, \delta)$
CFor any positive real number $\delta$ , the function $f$ is NOT an increasing function on the interval $(-\delta, 0)$
D $x=0$ is a point of local minima of $f$

Correct answer

C. For any positive real number $\delta$ , the function $f$ is NOT an increasing function on the interval $(-\delta, 0)$

Step-by-step solution

(A) RHD at

x=0: \lim _{h \rightarrow 0} \frac{\left(2-2 h^2-h^2 \sin \frac{1}{h}\right)-2}

{h}=0

Similarly LHD at

\mathrm{x}=0

is also equal to 0 .

\therefore

Differentiable at

\mathrm{x}=0

(B)

f^{\prime}(x)=-4 x-2 x \sin \frac{1}{x}-x^2\left(\cos \frac{1}{x}\right)\left(\frac{-1}{x^2}\right)

\begin{aligned} & f^{\prime}(x)=-\left(4 x+2 x \sin \frac{1}{x}\right)+\cos \frac{1}{x} \\ & f^{\prime}(x)=-\left(2 x\left(4-\sin \frac{1}{x}\right)\right)+\cos \frac{1}{x} \end{aligned}

for

\mathrm{x} \in(0, \delta) \Rightarrow

We can't say

f(x)

is decreasing on

(0, \delta)

as

\cos \frac{1}{x}

oscillates. (C) for

x \in(-\delta, 0)

, for any

\delta>0 \Rightarrow f(x)

is not increasing on

(-\delta, 0)

as

\cos \frac{1}{x}

oscillates from -1 to 1. (D)

f(0)=2

\begin{aligned} & f(0+h) <2 \\ & f(0-h) <2 \end{aligned}

\therefore \mathrm{x}=0 is local maxima

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