Q233 / 270 · NDA 2025 (Phase 2) · Mathematics
For the following two (02) items:
Consider the function $f(x) = \begin{cases} 4(5^x), & x < 0 \\ 8k + x, & x \geq 0 \end{cases}$.
If the function is continuous, then what is the value of $k$?
Consider the function $f(x) = \begin{cases} 4(5^x), & x < 0 \\ 8k + x, & x \geq 0 \end{cases}$.
If the function is continuous, then what is the value of $k$?
- Correct answerA$0.5$
- B$1$
- C$1.5$
- D$2$
Solution
For the function to be continuous at $x = 0$, the left-hand limit must be equal to the right-hand limit and the value of the function at $x = 0$.
Left-hand limit at $x = 0$ is given by:
$\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} 4(5^x) = 4(5^0) = 4$
Right-hand limit at $x = 0$ is given by:
$\lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} (8k + x) = 8k + 0 = 8k$
Equating the left-hand limit and the right-hand limit:
$8k = 4$
$k = \dfrac{4}{8} = 0.5$
Answer: $0.5$
Left-hand limit at $x = 0$ is given by:
$\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} 4(5^x) = 4(5^0) = 4$
Right-hand limit at $x = 0$ is given by:
$\lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} (8k + x) = 8k + 0 = 8k$
Equating the left-hand limit and the right-hand limit:
$8k = 4$
$k = \dfrac{4}{8} = 0.5$
Answer: $0.5$
Hence, the correct answer is option: 'A'.