PEA 2026 PEA Q3 | ISI MSQE

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2026 PEA Q3

Let $f:\mathbb{R}\to\mathbb{R}$ satisfy $f(x^3+1) = x^6 + 4x^3 + 8$ for all $x \in \mathbb{R}$ . Then $\displaystyle \int_1^2 f(x)\,dx$ equals

Reveal answer and solution

Answer

Solution

1
Set $u = x^3 + 1$ , so $x^3 = u - 1$ and
2
$f(u) = (u-1)^2 + 4(u-1) + 8 = u^2 - 2u + 1 + 4u - 4 + 8 = u^2 + 2u + 5.$
3
Therefore
4
$\int_1^2 f(x)\,dx = \int_1^2 (x^2 + 2x + 5)\,dx = \left[\frac{x^3}{3} + x^2 + 5x\right]_1^2 = \left(\tfrac{8}{3} + 4 + 10\right) - \left(\tfrac{1}{3} + 1 + 5\right) = \tfrac{7}{3} + 8 = \tfrac{31}{3}.$

Answer structure / marking notes

Setting $u = x^3 + 1$ is bijective on $\mathbb{R}$ , so $f$ is uniquely determined.

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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.