Each function f in List-I is defined for \(x \ne a\) as shown, and f(a) = k. Match each function with the value of k that makes f continuous at the given point a.
| List-I | List-II | ||
|---|---|---|---|
| Function and point | Value of k | ||
| (A) | \(f(x) = \displaystyle\frac{\sin 4x}{x},\ a = 0\) | (I) | \(\displaystyle\frac{1}{2}\) |
| (B) | \(f(x) = \displaystyle\frac{x^{2} - 1}{x - 1},\ a = 1\) | (II) | 2 |
| (C) | \(f(x) = \displaystyle\frac{e^{3x} - 1}{x},\ a = 0\) | (III) | 3 |
| (D) | \(f(x) = \displaystyle\frac{\tan x}{2x},\ a = 0\) | (IV) | 4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
For continuity at a, k must equal \(\lim_{x \to a} f(x)\).
(A) \(f(x) = \frac{\sin 4x}{x},\ a = 0\)
This matches (IV) 4. Using \(\lim_{t \to 0}\frac{\sin t}{t} = 1\) with t = 4x:
\( \displaystyle \frac{\sin 4x}{x} = 4 \cdot \frac{\sin 4x}{4x} \to 4 \)
(B) \(f(x) = \frac{x^{2} - 1}{x - 1},\ a = 1\)
This matches (II) 2:
\( \displaystyle \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \to 2 \)
(C) \(f(x) = \frac{e^{3x} - 1}{x},\ a = 0\)
This matches (III) 3. Using \(\lim_{t \to 0}\frac{e^{t} - 1}{t} = 1\) with t = 3x:
\( \displaystyle \frac{e^{3x} - 1}{x} = 3 \cdot \frac{e^{3x} - 1}{3x} \to 3 \)
(D) \(f(x) = \frac{\tan x}{2x},\ a = 0\)
This matches (I) \(\frac{1}{2}\), since \(\lim_{x \to 0}\frac{\tan x}{x} = 1\):
\( \displaystyle \frac{\tan x}{2x} = \frac{1}{2} \cdot \frac{\tan x}{x} \to \frac{1}{2} \)