NIMCET 2025 Paper

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