NIMCET 2022 — Mathematics PYQ

Solution

To determine if the function is continuous or differentiable at $x=0$, we must analyze the limit as $x\to 0$.

1. Checking Continuity

For $f(x)$ to be continuous at $x=0$, we need ${\lim }_{x\to 0}f(x)=f(0)$.

The value of $f(0)$ is given as $e^2$.

Now, calculate the limit:

$$\underset{x\to 0}{\lim }(1+2x{)}^{\frac{1}{x}}$$

This is in the indeterminate form $1^{\infty }$. We use the standard limit formula ${\lim }_{u\to 0}(1+u{)}^{\frac{1}{u}}=e$.

Let $u=2x$. As $x\to 0$, $u\to 0$. Also, $\frac{1}{x}=\frac{2}{u}$.

$$\underset{u\to 0}{\lim }(1+u{)}^{\frac{2}{u}}={\left(\underset{u\to 0}{\lim }(1+u{)}^{\frac{1}{u}}\right)}^2=e^2$$

Since ${\lim }_{x\to 0}f(x)=f(0)=e^2$, the function is continuous at $x=0$.

2. Checking Differentiability

To check differentiability, we calculate the derivative at $x=0$ using the definition:

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

$$f^{\prime }(0)=\underset{h\to 0}{\lim }\frac{(1+2h{)}^{\frac{1}{h}}-e^2}{h}$$

Using Taylor series expansion for $(1+2h{)}^{\frac{1}{h}}=e^{\frac{1}{h}\ln (1+2h)}$:

$$\ln (1+2h)=2h-\frac{(2h{)}^2}{2}+\frac{(2h{)}^3}{3}-\cdots =2h-2h^2+\frac{8}{3}{h}^3-\dots$$

$$\frac{1}{h}\ln (1+2h)=2-2h+\frac{8}{3}{h}^2-\dots$$

$$f(h)=e^{2-2h+\frac{8}{3}{h}^2}=e^2\cdot e^{-2h+\frac{8}{3}{h}^2}\approx e^2(1-2h+\frac{8}{3}{h}^2+\frac{(-2h{)}^2}{2})=e^2(1-2h+\frac{14}{3}{h}^2)$$

Substituting this into the derivative limit:

$$f^{\prime }(0)=\underset{h\to 0}{\lim }\frac{e^2(1-2h+\dots )-e^2}{h}=\underset{h\to 0}{\lim }\frac{-2h{e}^2}{h}=-2e^2$$

Since the derivative exists, the function is differentiable at $x=0$.

Correct Option: B) continuous at $x=0$