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 Question:
The function $g(x)$ is continuous at $x = 3$. This means the limit of $g(x)$ as $x$ approaches 3 must equal its defined value at $x = 3$.
Therefore, $K = \lim_{x \to 3} g(x)$.

Step 2: Key Formula or Approach:
Notice that $g(x)$ for $x \neq 3$ contains an integral whose upper limit depends on $x$.
$$g(x) = \frac{\int_3^{f(x)} 3t^2 dt}{x-3}$$
As $x \to 3$, the denominator goes to 0. The numerator goes to $\int_3^{f(3)} 3t^2 dt = \int_3^3 3t^2 dt = 0$.
This is a $0/0$ indeterminate form. We must use L'H\^{o}pital's Rule and the Newton-Leibniz formula for differentiating an integral.
Leibniz Rule: $\frac{d}{dx} \int_{a(x)}^{b(x)} h(t) dt = h(b(x))b'(x) - h(a(x))a'(x)$.

Step 3: Detailed Explanation:
Apply L'H\^{o}pital's Rule to $\lim_{x \to 3} g(x)$:
$$K = \lim_{x \to 3} \frac{\frac{d}{dx} \int_3^{f(x)} 3t^2 dt}{\frac{d}{dx} (x-3)}$$
Differentiating the numerator using Leibniz Rule:
$$Numerator' = 3(f(x))^2 \cdot f'(x) - 3(3)^2 \cdot 0 = 3(f(x))^2 f'(x)$$
Differentiating the denominator:
$$Denominator' = 1$$
Therefore, the limit becomes:
$$K = \lim_{x \to 3} \left[ 3(f(x))^2 f'(x) \right]$$
Substitute $x = 3$ directly into the expression:
$$K = 3(f(3))^2 \cdot f'(3)$$
We are given $f(3) = 3$ and $f'(3) = 1/27$. Substitute these values:
$$K = 3(3)^2 \cdot \left(\frac{1}{27}\right)$$
$$K = 3(9) \cdot \left(\frac{1}{27}\right)$$
$$K = 27 \cdot \frac{1}{27} = 1$$

Step 4: Final Answer:
The value of $K$ is 1, matching option (a).