Correct : a,c

The correct answers are Option A and Option C.
We evaluate the definite integral directly: I(a) = ∫₋₁¹ (3x² − ax + 1) dx, splitting it term by term over the symmetric interval [−1, 1].
∫₋₁¹ 3x² dx = [x³]₋₁¹ = 1 − (−1) = 2. ∫₋₁¹ 1 dx = [x]₋₁¹ = 1 − (−1) = 2. ∫₋₁¹ −ax dx = −a·[x²/2]₋₁¹ = −a·(½ − ½) = 0, because x is an odd function and the interval is symmetric about 0.
Therefore, I(a) = 2 + 2 + 0 = 4 for every real value of a, without exception.
Option A — I(a) is independent of a: True. The term containing a vanishes upon integration, leaving a constant value of 4 regardless of a.
Option B — I(a) can vary with a: False. As shown, I(a) = 4 for all a ∈ (−∞, +∞). There is no variation.
Option C — There exists a such that I(a) is a positive real number: True. Since I(a) = 4 > 0 for every a, it is certainly positive for some (in fact, all) values of a.
Option D — There exists a such that I(a) is a negative real number: False. I(a) = 4 > 0 always; it can never be negative.
The key insight is recognising that −ax is an odd function