MHTCET 11 Apr 2026 Shift 1 Paper

Concept:Use definite integration followed by the quadratic sum-of-roots formula.Explanation:We are given:$$\int_{1}^{a}(2x+1)\,dx=5$$First, find the antiderivative of $2x+1$:$$\int (2x+1)\,dx = x^2 + x$$Now apply the limits 1 to $a$:$$\int_{1}^{a}(2x+1)\,dx = \left[x^2 + x\right]_{1}^{a}$$$$= (a^2 + a) - (1^2 + 1) = a^2 + a - 2$$Set this equal to 5:$$a^2 + a - 2 = 5$$$$a^2 + a - 7 = 0$$The sum of the values of $a$ is the sum of the roots of this quadratic.For $Ax^2 + Bx + C = 0$, the sum of roots is $-\frac{B}{A}$.Here, $A = 1$ and $B = 1$.So, the sum of values of $a$ is:$$-\frac{1}{1} = -1$$Answer:The sum of the values of $a$ is $-1$.Correct option: B. $-1$