MHTCET 13 Apr 2026 Shift 2 Paper

Concept:Evaluate the definite integral in terms of $a$, then solve the resulting cubic equation.Explanation:Integrate the polynomial term by term.$\int \left(x^{2} - 4x + 1\right) dx = \frac{x^{3}}{3} - 2x^{2} + x$.Apply the limits from $0$ to $a$.$\left[\frac{x^{3}}{3} - 2x^{2} + x\right]_{0}^{a} = \frac{a^{3}}{3} - 2a^{2} + a$.Set this equal to $6$.$\frac{a^{3}}{3} - 2a^{2} + a = 6$.Multiply the entire equation by $3$.$a^{3} - 6a^{2} + 3a = 18$.Rearrange to form a cubic equation.$a^{3} - 6a^{2} + 3a - 18 = 0$.Factor by grouping.$a^{2}(a - 6) + 3(a - 6) = 0$.$(a - 6)(a^{2} + 3) = 0$.So, $a - 6 = 0$ or $a^{2} + 3 = 0$.Since $a^{2} + 3 = 0$ gives no real value of $a$, we get $a = 6$.Answer:The real value of $a$ is $6$, which is Option B.