MHTCET 19 Apr 2026 Shift 2 Paper

Concept:For a finite limit when the denominator tends to zero, the numerator must also contain the same repeated factor.Explanation:The denominator is $x^{2}-2x+1=(x-1)^{2}$.Since the limit is finite, the numerator must have $(x-1)^{2}$ as a factor.As the numerator is a monic cubic polynomial, write $x^{3}+ax^{2}+bx+c=(x-1)^{2}(x+k)$.Then $\lim_{x \to 1} \frac{(x-1)^{2}(x+k)}{(x-1)^{2}} = \lim_{x \to 1}(x+k)=1+k$.Given that the limit equals $2026$, we have $1+k=2026$, so $k=2025$.Now expand: $(x-1)^{2}(x+2025) = x^{3}+2023x^{2}-4049x+2025$.Comparing coefficients with $x^{3}+ax^{2}+bx+c$, we get $a=2023$ and $c=2025$.Therefore, $a-c = 2023 - 2025 = -2$.Answer:Option D, $-2$.