MHTCET 17 Apr 2026 Shift 1 Paper

Concept:For a finite non-zero limit, both numerator and denominator must become zero at $x=1$, and their orders of zeroes must match.Explanation:At $x=1$, the numerator becomes $\sin(0) - 1 + 1 = 0$.Since the given limit is $-2$, a non-zero finite value, the denominator must also become $0$ at $x=1$.So, $2(1)^3 - 7(1)^2 + a(1) + b = 0$, which gives $a + b = 5$.Near $x=1$, we have $3x^2 - 4x + 1 = (x-1)(3x-1) \to 0$.Using $\sin y \sim y$ for small $y$, the numerator behaves like $(3x^2 - 4x + 1) - x^2 + 1 = 2x^2 - 4x + 2 = 2(x-1)^2$.Thus, the numerator has factor $(x-1)^2$.For a non-zero limit, the denominator must also have factor $(x-1)^2$.So, the derivative of the denominator must vanish at $x=1$.The derivative is $6x^2 - 14x + a$.At $x=1$, $6 - 14 + a = 0$, giving $a = 8$.Using $a + b = 5$, we get $b = -3$.The quadratic equation with roots $8$ and $-3$ is $x^2 - (8 + (-3))x + 8(-3) = 0$, i.e. $x^2 - 5x - 24 = 0$.Answer:$x^{2} - 5x - 24 = 0$Hence, the correct option is C.