Question

Let \( f \) be a polynomial function such that \[ f(x^2+1)=x^4+5x^2+2,\quad \text{for all } x\in\mathbb{R}. \] Then \[ \int_0^3 f(x)\,dx \] is equal to:

• A. \( \dfrac{5}{3} \)
• B. \( \dfrac{27}{2} \)
• C. \( \dfrac{33}{2} \)
• D. \( \dfrac{41}{3} \)

Hint:

When a polynomial identity involves an expression like \( f(x^2+1) \), substitute \( t=x^2+1 \) to convert it into a standard polynomial comparison problem.

Answer 1

The Correct Option is B

Concept: If a polynomial identity holds for all real \( x \), then the corresponding polynomials must be identical. We first determine the explicit form of \( f(x) \) by comparing coefficients, and then evaluate the definite integral.
Step 1: Find the polynomial \( f(x) \) Let: \[ t=x^2+1 \Rightarrow x^2=t-1 \] Then: \[ f(t)=x^4+5x^2+2 \] Substitute \( x^2=t-1 \): \[ x^4=(t-1)^2=t^2-2t+1 \] Hence, \[ f(t)=(t^2-2t+1)+5(t-1)+2 \] \[ f(t)=t^2+3t-2 \] Therefore, \[ f(x)=x^2+3x-2 \]
Step 2: Evaluate the definite integral \[ \int_0^3 f(x)\,dx=\int_0^3 (x^2+3x-2)\,dx \] \[ =\left[\frac{x^3}{3}+\frac{3x^2}{2}-2x\right]_0^3 \] \[ =\left(\frac{27}{3}+\frac{27}{2}-6\right)-0 \] \[ =9+\frac{27}{2}-6=\frac{27}{2} \]