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UPSC NDA II 2025 Mathematics Paper

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Question Numbers: 83-84

For the following two (02) items:

Consider the function

f(x)=\begin{cases} 4(5^{x}), & x<0 \\ 8k+x, & x\ge 0 \end{cases}

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Question : 84 of 120

Marks: +1, -0

What is f'(-1) equal to?

$\frac{2}{5}ln5$
$\frac{3}{5}ln5$
$\frac{4}{5}ln5$
$20ln5$

Solution:

Concept:

For

x<0

, the derivative of

f(x)

is

4\cdot 5^{x}\cdot\ln5

.

Explanation:

The function is defined piecewise.

At

x=-1

, which is less than

0

, use

f'(x)=4\cdot5^{x}\ln5

.

Substitute

x=-1

:

f'(-1)=4\cdot5^{-1}\ln5=\frac{4}{5}\ln5

.

Answer:

\frac{4}{5}\ln5

→ Option C.

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