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2026 PEA Q5

Let f:[0,1]→Rf:[0,1]\to\mathbb{R} be continuous with ∫01f(t) dt=1\displaystyle\int_0^1 f(t)\,dt = 1, and let

P(x)=∑k=1nakxk,∑k=1nak=1. P(x) = \sum_{k=1}^n a_k x^k, \qquad \sum_{k=1}^n a_k = 1.

Then there exists c∈(0,1)c \in (0,1) such that

Reveal answer and solution

Answer

C

Solution

  1. 1

    Define

  2. 2
    Φ(x)  =  ∫0xf(t) dt  −  P(x),x∈[0,1]. \Phi(x) \;=\; \int_0^x f(t)\,dt \;-\; P(x), \qquad x \in [0,1].
  3. 3

    Then Φ\Phi is continuous on [0,1][0,1], differentiable on (0,1)(0,1), and

  4. 4
    Φ(0)=0−P(0)=0,Φ(1)=∫01f(t) dt−P(1)=1−∑k=1nak=1−1=0. \Phi(0) = 0 - P(0) = 0, \qquad \Phi(1) = \int_0^1 f(t)\,dt - P(1) = 1 - \sum_{k=1}^n a_k = 1 - 1 = 0.
  5. 5

    By Rolle's theorem, there exists c∈(0,1)c \in (0,1) with Φ′(c)=0\Phi'(c) = 0, i.e.

  6. 6
    f(c)−P′(c)=0⟺f(c)=P′(c). f(c) - P'(c) = 0 \quad\Longleftrightarrow\quad f(c) = P'(c).

Answer structure / marking notes

Note P(0)=0P(0)=0 because the sum starts at k=1k=1. This is essential for Φ(0)=0\Phi(0)=0.

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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.