Given that the function f(x) is continuous on R. Match List-I with List-II.
| List-I | List-II | ||
| (A) | \(f(x)=\begin{cases}kx^2, & \text{if } x \lt 3\\ 3, & \text{if } x \ge 3\end{cases}\) | (I) | \(k=\displaystyle\frac{5}{4}\) |
| (B) | \(f(x)=\begin{cases}kx+1, & \text{if } x \ge 4\\ x+2, & \text{if } x \lt 4\end{cases}\) | (II) | \(k=-1\) |
| (C) | \(f(x)=\begin{cases}\displaystyle\frac{5+x}{2}, & \text{if } x = 0\\ 3k, & \text{if } x \ne 0\end{cases}\) | (III) | \(k=\displaystyle\frac{1}{3}\) |
| (D) | \(f(x)=\begin{cases}x+k, & \text{if } x \gt -2\\ -3, & \text{if } x \le -2\end{cases}\) | (IV) | \(k=\displaystyle\frac{5}{6}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Each piece is a polynomial, so f can fail to be continuous only where the rule changes. There the left-hand limit, right-hand limit and value must be equal.
(A) At x = 3
This matches (III) \(k = \frac{1}{3}\).
\(k(3)^2 = 3\)
\(9k = 3,\quad k = \frac{1}{3}\)
(B) At x = 4
This matches (I) \(k = \frac{5}{4}\).
\(4k + 1 = 4 + 2\)
\(4k = 5,\quad k = \frac{5}{4}\)
(C) At x = 0
This matches (IV) \(k = \frac{5}{6}\). For x ≠ 0 the function is the constant 3k, so the limit at 0 is 3k, while the value is \(f(0) = \frac{5+0}{2} = \frac{5}{2}\):
\(3k = \frac{5}{2},\quad k = \frac{5}{6}\)
(D) At x = −2
This matches (II) \(k = -1\).
\(-2 + k = -3\)
\(k = -1\)