TMUA Logic Drill · Necessary Conditions

Question 9 · Hard · Paper 2 Style

Let f be a function such that f(x) ≤ 0 for all x ≥ 0.

Which one of the following is necessary for ∫₋₁¹ f(x) dx > 0?

A. f(x) = −f(−x) for all x B. f(0) = 0 C. f(x) ≥ 0 for all x < 0 D. ∫_−a⁰ f(x) dx > 0 for some a E. f(x) = f(−x) for all x F. f(−1) > 0

Correct answer: D

Solution:

Since f(x) ≤ 0 for all x ≥ 0, the integral on the right-hand side satisfies

∫₀¹ f(x) dx ≤ 0.

Now split the integral:

∫₋₁¹ f(x) dx = ∫₋₁⁰ f(x) dx + ∫₀¹ f(x) dx.

For the total integral to be positive, the negative-side contribution must be positive enough to overcome the non-positive contribution from 0 ≤ x ≤ 1.

Therefore,

∫₋₁⁰ f(x) dx > 0

must hold. This is exactly option D with a = 1.

The other options are too strong or not required. For example, the function does not need to be odd, even, positive for every negative x, or positive at x = −1. It only needs enough positive area somewhere on the negative side.