Q234 / 270 · NDA 2025 (Phase 2) · Mathematics
For the following two (02) items:
Consider the function $f(x) = \begin{cases} 4(5^x), & x < 0 \\ 8k + x, & x \geq 0 \end{cases}$.
What is $f'(-1)$ equal to?
Consider the function $f(x) = \begin{cases} 4(5^x), & x < 0 \\ 8k + x, & x \geq 0 \end{cases}$.
What is $f'(-1)$ equal to?
- A$\dfrac{2}{5}\ln 5$
- B$\dfrac{3}{5}\ln 5$
- Correct answerC$\dfrac{4}{5}\ln 5$
- D$20\ln 5$
Solution
For $x < 0$, the function is given by $f(x) = 4(5^x)$.
Differentiating with respect to $x$, we get:
$f'(x) = 4(5^x \ln 5)$
To find $f'(-1)$, substitute $x = -1$ into the derivative:
$f'(-1) = 4(5^{-1} \ln 5)$
$f'(-1) = \dfrac{4}{5}\ln 5$
Answer: $\dfrac{4}{5}\ln 5$
Differentiating with respect to $x$, we get:
$f'(x) = 4(5^x \ln 5)$
To find $f'(-1)$, substitute $x = -1$ into the derivative:
$f'(-1) = 4(5^{-1} \ln 5)$
$f'(-1) = \dfrac{4}{5}\ln 5$
Answer: $\dfrac{4}{5}\ln 5$
Hence, the correct answer is option: 'C'.