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COMEDK20269 May 2026Evening ShiftMathematicsApplication of DerivativesActual

The function f(x) = e^ ax + e^ -ax , x R and a < 0 , is strictly decreasing for all values of ' x ', where

Options

  1. Ax > 0
  2. Bx > 1
  3. Cx < 1
  4. Dx < 0

Correct answer

D. x < 0

Step-by-step solution

Given f(x) = e^ ax + e^ -ax Differentiating with respect to x , we get: f'(x) = a e^ ax - a e^ -ax = a(e^ ax - e^ -ax ) For f(x) to be strictly decreasing, f'(x) a(e^ ax - e^ -ax ) Since a e^ ax - e^ -ax > 0 e^ ax > e^ -ax Since the exponential function is strictly increasing, we can compare the exponents: ax > -ax 2ax > 0 Since a x Therefore, the function is strictly decreasing for x Answer: x < 0

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