Question

For a real number \(a\), let \[I(a)=\int_{-1}^{1}(3x^2-ax+1)\,dx.\]
Which of the following statements is/are true?

• A. The value of 𝐼(𝑎) is independent of the value of 𝑎
• B. The value of 𝐼(𝑎) can vary with the value of 𝑎
• C. There exists 𝑎∈(−∞, +∞) such that 𝐼(𝑎) is a positive real number
• D. There exists 𝑎∈(−∞, +∞) such that 𝐼(𝑎) is a negative real number

Hint:

The coefficient of the odd term $-ax$ contributes zero when integrated over the symmetric interval $[-1,1]$, so $I(a)$ reduces to a fixed constant that does not depend on $a$ at all.

Answer 1

The Correct Option is A, C

We need to evaluate \(I(a) = \displaystyle\int_{-1}^{1} (3x^2 - ax + 1)\, dx\) and determine which statements about it are true.

Step 1: Split the integral.

\[ I(a) = \int_{-1}^{1} 3x^2\, dx - a\int_{-1}^{1} x\, dx + \int_{-1}^{1} 1\, dx \]

Step 2: Use symmetry. The function \(x\) is odd, and the interval \([-1,1]\) is symmetric about 0, so \(\int_{-1}^{1} x\, dx = 0\) regardless of the coefficient \(a\). This term vanishes entirely.

Step 3: Evaluate the remaining even-function integrals.

\[ \int_{-1}^{1} 3x^2\, dx = \left[x^3\right]_{-1}^{1} = 1 - (-1) = 2 \]\[ \int_{-1}^{1} 1\, dx = \left[x\right]_{-1}^{1} = 1-(-1) = 2 \]

Step 4: Combine.

\[ I(a) = 2 - 0 + 2 = 4 \]

This value does not depend on \(a\) at all - it equals 4 for every real \(a\).

Step 5: Check each option: Statement A ('independent of \(a\)') is true since \(I(a)=4\) always. Statement B ('can vary with \(a\)') is false since it never changes. Statement C ('exists \(a\) making \(I(a)\) positive') is true since \(I(a)=4 > 0\) for every \(a\). Statement D ('exists \(a\) making \(I(a)\) negative') is false since \(I(a)=4\) is never negative.

Final Answer: \(\boxed{\text{Statements A and C are true}}\)