NIMCET 2025 — Mathematics PYQ

To find the correct statement, let us analyze the behavior of the function $f(x)$ at $x=0$ regarding both continuity and differentiability.

1. Analysis of Continuity at $x=0$

For $f(x)$ to be continuous at $x=0$, the limiting value must equal the functional value:

$$\underset{x\to 0}{\lim }f(x)=f(0)=0$$

Let's evaluate the limit:

$$\underset{x\to 0}{\lim }{x}^{\alpha }\sin \left(\frac{1}{x^{\beta }}\right)$$

We know that the sine function is bounded between $-1$ and $1$ for any real input:

$$-1\le \sin \left(\frac{1}{x^{\beta }}\right)\le 1$$

• Case 1: If \alpha > 0

  As $x\to 0$, $x^{\alpha }\to 0$. Multiplying a term that approaches $0$ by a bounded oscillating term yields $0$.

  $$\underset{x\to 0}{\lim }{x}^{\alpha }\sin \left(\frac{1}{x^{\beta }}\right)=0\times (\text{bounded quantity})=0$$

  Since the limit equals $f(0)$, the function is continuous at $x=0$ whenever \alpha > 0, regardless of the value of $\beta$.

• Case 2: If $\alpha \le 0$

  The limit will either not exist or blow up to infinity because $x^{\alpha }$ will either be $1$ (for $\alpha =0$) or move to the denominator (for \alpha < 0), causing infinitely large oscillations near $x=0$.

Conclusion on Continuity: $f(x)$ is continuous at $x=0$ if and only if \alpha > 0. This directly validates Option (D) and invalidates Option (A).

2. Analysis of Differentiability at $x=0$

Using the first principle of derivatives, the derivative at $x=0$ is given by:

$$f^{\prime }(0)=\underset{x\to 0}{\lim }\frac{f(x)-f(0)}{x-0}$$

Substitute the values:

$$f^{\prime }(0)=\underset{x\to 0}{\lim }\frac{x^{\alpha }\sin (1/x^{\beta })-0}{x}$$

$$f^{\prime }(0)=\underset{x\to 0}{\lim }{x}^{\alpha -1}\sin \left(\frac{1}{x^{\beta }}\right)$$

For this limit to exist and equal a finite value (specifically $0$), the exponent of $x$ must be strictly greater than zero:

\alpha - 1 > 0 \implies \alpha > 1

If $\alpha \le 1$, the limit does not exist due to oscillation or divergence. Therefore, the function is differentiable at $x=0$ only when \alpha > 1. This invalidates Option (B) because it claims differentiability holds for all \alpha > 0.

3. Analysis of Option (C)

At $x=10$, the term $\sin (1/x^{\beta })=\sin (1/10^{\beta })$ is a perfectly well-defined, non-zero constant for most values of $\beta$. The function behaves like a simple power function $x^{\alpha }$ multiplied by a constant in the neighborhood of $x=10$. It is continuous for many other ranges of $\alpha$ and $\beta$ (e.g., any $\alpha \in \mathbb{R}$ when \beta > 0), making the condition "only \alpha > 0 \text{ \& } \beta > 0" false.

Correct Answer

Correct Option: (D)