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NDA 2 2025 Maths, question 84

Topic: Differential calculus

Passage · Q83 to Q84

Consider the function f(x)={4(5x),x<08k+x,x≥0f(x)=\left\{ \begin{matrix} 4(5^{x}), & x<0 \\ 8k+x, & x\ge 0 \end{matrix} \right.

Q84

What is f′(−1)f'(-1) equal to?

  1. (a) 25ln⁡5\frac{2}{5}\ln 5
  2. (b) 35ln⁡5\frac{3}{5}\ln 5
  3. (c) 45ln⁡5\frac{4}{5}\ln 5Correct answer
  4. (d) 20ln⁡520\ln 5

Solution.

Answer: (c) 45ln⁡5\frac{4}{5}\ln 5

  1. For x<0, f′(x)=4×5ˣ ln5.
  2. Substitution of x=−1 gives (4/5)ln5.
  3. The other branch is irrelevant at this point.