Q27 / 75 · JEE Main 2026 (28 January Shift 1) · Mathematics
Let $f$ be a polynomial function such that $f\left(x^{2}+1\right)=x^{4}+5 x^{2}+2$, for all $x \in \mathbb{R}$. Then $\int_{0}^{3} f(x) d x$ is equal to
- A$\frac{5}{3}$
- B$\frac{41}{3}$
- C$\frac{27}{2}$
- Correct answerD$\frac{33}{2}$
Solution
Let $u = x^2 + 1$, then $x^2 = u - 1$.
Substituting into the given condition:
$f(u) = (u-1)^2 + 5(u-1) + 2 =$
$u^2 - 2u + 1 + 5u - 5 + 2$
$= u^2 + 3u - 2$.
Therefore, $f(x) = x^2 + 3x - 2$.
Calculating the integral: $\int_0^3 f(x)dx = \int_0^3 (x^2 + 3x - 2)dx$
$= \left[\frac{x^3}{3} + \frac{3x^2}{2} - 2x\right]_0^3 = 9 + \frac{27}{2} - 6 = \frac{33}{2}$.
Substituting into the given condition:
$f(u) = (u-1)^2 + 5(u-1) + 2 =$
$u^2 - 2u + 1 + 5u - 5 + 2$
$= u^2 + 3u - 2$.
Therefore, $f(x) = x^2 + 3x - 2$.
Calculating the integral: $\int_0^3 f(x)dx = \int_0^3 (x^2 + 3x - 2)dx$
$= \left[\frac{x^3}{3} + \frac{3x^2}{2} - 2x\right]_0^3 = 9 + \frac{27}{2} - 6 = \frac{33}{2}$.
Hence, the correct answer is option: 'D'.