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JEE Main 202628 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2026 · 28 January · Morning Shift · Mathematics · Definite Integration

Let f be a polynomial function such that f(x^ 2 +1)=x^ 4 +5 x^ 2 +2...

Let ff be a polynomial function such that f(x2+1)=x4+5x2+2f\left(x^{2}+1\right)=x^{4}+5 x^{2}+2, for all x∈Rx \in \mathbb{R}. Then ∫03f(x)dx\int_{0}^{3} f(x) d x is equal to

Options

  1. A53\frac{5}{3}
  2. B413\frac{41}{3}
  3. C272\frac{27}{2}
  4. D332\frac{33}{2}

Correct answer

D. 332\frac{33}{2}

Step-by-step solution

Let u=x2+1u = x^2 + 1, then x2=u−1x^2 = u - 1.
Substituting into the given condition:
f(u)=(u−1)2+5(u−1)+2=f(u) = (u-1)^2 + 5(u-1) + 2 =
u2−2u+1+5u−5+2u^2 - 2u + 1 + 5u - 5 + 2
=u2+3u−2= u^2 + 3u - 2.
Therefore, f(x)=x2+3x−2f(x) = x^2 + 3x - 2.
Calculating the integral: ∫03f(x)dx=∫03(x2+3x−2)dx\int_0^3 f(x)dx = \int_0^3 (x^2 + 3x - 2)dx
=[x33+3x22−2x]03=9+272−6=332= \left[\frac{x^3}{3} + \frac{3x^2}{2} - 2x\right]_0^3 = 9 + \frac{27}{2} - 6 = \frac{33}{2}.
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