JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 16

Denominator. As $x\to2$: $\sqrt{x-1}-1\sim\frac{x-2}2$ and $\ln(x-1)\sim x-2$, so the denominator $\sim\frac{(x-2)^2}2$.

Numerator must vanish to second order. With $g=x^3-5x^2+ax+b$: $g(2)=0$ and $g'(2)=0$ give $a=8$, $b=-4$. Then $g\approx\frac{g''(2)}2(x-2)^2=(x-2)^2$.

Limit. $m=\dfrac{(x-2)^2}{(x-2)^2/2}=2$, so $a+b+m=8-4+2=6$.

(B)