Question

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ is differentiable function having $f(3) = 3, f'(3) = 1/27$ and $g(x) = \begin{cases} \int_3^{f(x)} \frac{3t^2}{x-3} dt, & x \neq 3 \\ K, & x = 3 \end{cases}$ is continuous at $x = 3$, then $K = \dots$

• A. 1
• B. 3
• C. 1/3
• D. 9

Hint:

Limits involving integrals bounded by functions of $x$ almost universally require the Newton-Leibniz differentiation rule applied in conjunction with L'H\^{o}pital's rule.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For $g(x)$ to be continuous at $x=3$, $K = \lim_{x \to 3} g(x)$. This is a $0/0$ form, so we use L'Hôpital's Rule and the Leibniz Rule for differentiation under the integral sign.

Step 2: Formula Application:
$K = \lim_{x \to 3} \frac{\int_3^{f(x)} 3t^2 dt}{x-3}$. Differentiating numerator: $\frac{d}{dx}[\int_3^{f(x)} 3t^2 dt] = 3(f(x))^2 \cdot f'(x)$. Differentiating denominator: $\frac{d}{dx}[x-3] = 1$.

Step 3: Explanation:
$K = \lim_{x \to 3} \frac{3(f(x))^2 f'(x)}{1} = 3(f(3))^2 f'(3)$. Substituting values: $K = 3(3^2)(1/27) = 3(9)(1/27) = 27/27 = 1$.

Step 4: Final Answer:
The value of K is 1.