NIMCET 2025 Paper

To check continuity at 0 we need $\lim\limits_{x\to 0} f(x) = f(0) = 0$.We'll use $|\sin(\cdot)| \leq 1$ so $|f(x)| \leq \lvert x^{\alpha} \rvert$ and see when that goes to 0 .To check differentiability at 0 , we use$f'(0) = \lim\limits_{x\to 0} \frac{f(x)-f(0)}{x} = \lim\limits_{x\to 0} x^{\alpha-1} \sin\left(\frac{1}{x^{\beta}}\right)$and again compare $\lvert x^{\alpha-1} \rvert$ to 0 .Short Solution1. Continuity at $x=0$For $x \neq 0$,$f(x) = x^{\alpha} \sin\left(\frac{1}{x^{\beta}}\right)$Use $|\sin(\theta)| \leq 1$ :$\lvert f(x) \rvert \leq \lvert x^{\alpha} \rvert$As $x \to 0$,- If $\alpha > 0$, then $\lvert x^{\alpha} \rvert \to 0 \Rightarrow f(x) \to 0$, so $f$ is continuous at 0 .- If $\alpha \leq 0$, then $\lvert x^{\alpha} \rvert$ does not go to 0 (it either stays 1 or blows up), so the limit is not 0 .Thus, continuity at 0 holds iff $\alpha > 0$, and this doesn't depend on $\beta$.So:- Statement (2) is false (cannot hold for all $\alpha \in \mathbb{R}$ ).- Statement (3) is too restrictive (it unnecessarily forces $\beta > 0$ ).- Statement (4) matches: continuous at 0 for all $c > 0$ and any real $\beta$. 2. Differentiability at $x=0$For completeness, check (1):$f'(0) = \lim\limits_{x\to 0} \frac{f(x)-0}{x} = \lim\limits_{x\to 0} x^{\alpha-1} \sin\left(\frac{1}{x^{\beta}}\right).$Using $|\sin| \leq 1$ :$\left| x^{\alpha-1} \sin\left(\frac{1}{x^{\beta}}\right) \right| \leq \left| x^{\alpha-1} \right|.$This goes to 0 only if $\alpha - 1 > 0 \Rightarrow \alpha > 1$, so differentiability at 0 is not true for all $\alpha > 0$.So (1) is false.