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2026 PEA Q7

Suppose f:[a,b]→Rf:[a,b]\to\mathbb{R} is continuous and

∫axf(t) dt  =  ∫xbf(t) dtfor all x∈[a,b]. \int_a^x f(t)\,dt \;=\; \int_x^b f(t)\,dt \quad \text{for all } x \in [a,b].

Then f(x)f(x) must be

Reveal answer and solution

Answer

C

Solution

  1. 1

    Differentiate both sides with respect to xx using the fundamental theorem of calculus:

  2. 2
    ddx∫axf(t) dt=f(x),ddx∫xbf(t) dt=−f(x). \frac{d}{dx}\int_a^x f(t)\,dt = f(x), \qquad \frac{d}{dx}\int_x^b f(t)\,dt = -f(x).
  3. 3

    Hence f(x)=−f(x)f(x) = -f(x), so 2f(x)=02f(x) = 0, i.e.\ f(x)=0f(x) = 0 for every x∈[a,b]x \in [a,b].

Answer structure / marking notes

An odd function about a+b2\tfrac{a+b}{2} would satisfy the identity only at the single point x=a+b2x = \tfrac{a+b}{2}, not for all xx.

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2026/ISI_MSQE_PEA_2026_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.