Let $\mathbb{R}$ denote the set of all real numbers. Define the function $f:\mathrm{ℝ}\to \mathrm{ℝ}$ by $f\left(x\right)=\left{\begin{matrix}2-2{x}^{2}-{x}^{2}\sin \frac{1}{x}, & \text{ if }x\neq 0 \ 2, & \text{ if }x=0\end{matrix}\right.$ Then which one of the following statements is TRUE? [JEE Advanced 2025]

Question

YESPYQ · Accepted Answer

Correct answer: c) For any positive real number $\delta$, the function $f$ is NOT an increasing function on the interval $(-\delta ,0)$. (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 . ∴ Differentiable at $\mathrm{x}=0$ (D) $\mathrm{f}(0)=2,\ \mathrm{f}(0+\mathrm{h}) therefore \(x=0$ is a point of local maxima (B) Now, ${f}^{'}(x)=-\left(4x+2x\sin \frac{1}{x}\right)+\cos \frac{1}{x}$ for any $\delta>0$, choose $x=\frac{1}{2 \pi n} \in(0, \delta)$ with $n$ large. Then $\sin \frac{1}{x}=0, \cos \frac{1}{x}=1$, so $f^{\prime}(x)=1-4 x>0$ A decreasing function cannot have $f^{\prime}(x)>0$ at a point where derivative exists. (C) for any $\delta>0$, choose $x=-\frac{1}{(2 n+1) \pi} \in(-\delta, 0)$ with $n$ large. Then $\cos \frac{1}{x}=-1$ and $f^{\prime}(x)=-4 x-2 x \sin \frac{1}{x}-1 \leq 6|x|-1<0$ for sufficiently small $|x|$ Hence $f$ cannot be increasing on ( $-\delta, 0$ )

Let \(\mathbb{R}\) denote the set of all real numbers. Define the function \(f:\mathrm{ℝ}\to \mathrm{ℝ}\) by \(f\left(x\…

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