Question

Suppose f:R\(\rightarrow\)R be given by f(x)={\(e^{(x^{10}-1)}+(x-1)^2sin\frac{1}{1-x},if\,\,\,x\neq1\)}, then

• A. f'(1) does not exist
• B. f'(1) exists and is zero
• C. f'(1) exist and is 9
• D. f'(1) exists and is 10

Answer 1

The Correct Option is D

Let $f(x) = \begin{cases} e^{x^{10} - 1} + (x - 1)^2 \sin\left(\frac{1}{1 - x}\right), & x \ne 1 \end{cases}$ 

Step 1: Check if the function is differentiable at $x = 1$ To find $f'(1)$, we use the definition of derivative: $f'(1) = \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1}$ 

Step 2: Find $f(1)$ To do that, we evaluate the limit: $\lim_{x \to 1} f(x) = \lim_{x \to 1} \left[ e^{x^{10} - 1} + (x - 1)^2 \sin\left(\frac{1}{1 - x}\right) \right]$ As $x \to 1$:

• $x^{10} - 1 \to 0 \Rightarrow e^{x^{10} - 1} \to e^0 = 1$
• $(x - 1)^2 \to 0$
• $\sin\left(\frac{1}{1 - x}\right)$ is bounded in $[-1, 1]$ and oscillates

Hence, $\lim_{x \to 1} (x - 1)^2 \sin\left(\frac{1}{1 - x}\right) = 0$ So, $\lim_{x \to 1} f(x) = 1 + 0 = 1$ 

Therefore, define $f(1) = 1$ 

Step 3: Apply derivative definition $f'(1) = \lim_{x \to 1} \frac{f(x) - 1}{x - 1}$ Substitute $f(x)$: $f'(1) = \lim_{x \to 1} \frac{e^{x^{10} - 1} + (x - 1)^2 \sin\left(\frac{1}{1 - x}\right) - 1}{x - 1}$ Split the limit: $f'(1) = \lim_{x \to 1} \frac{e^{x^{10} - 1} - 1}{x - 1} + \lim_{x \to 1} \frac{(x - 1)^2 \sin\left(\frac{1}{1 - x}\right)}{x - 1}$ 

Step 4: Evaluate the first term Let $u = x^{10} - 1$, then $u \to 0$ as $x \to 1$ $\lim_{x \to 1} \frac{e^{x^{10} - 1} - 1}{x - 1} = \lim_{x \to 1} \left( \frac{e^u - 1}{u} \cdot \frac{x^{10} - 1}{x - 1} \right)$ We know:

• $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$
• $\lim_{x \to 1} \frac{x^{10} - 1}{x - 1} = \frac{d}{dx}(x^{10})\Big|_{x = 1} = 10x^9 = 10$

So, first term = $1 \cdot 10 = 10$ 

Step 5: Evaluate the second term $\lim_{x \to 1} \frac{(x - 1)^2 \sin\left(\frac{1}{1 - x}\right)}{x - 1} = \lim_{x \to 1} (x - 1) \cdot \sin\left(\frac{1}{1 - x}\right)$ Here,

• $(x - 1) \to 0$
• $\sin\left(\frac{1}{1 - x}\right)$ is bounded

So, the product tends to $0$ 

Step 6: Final answer $f'(1) = 10 + 0 = \boxed{10}$