JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Limits

As $x\to 2$ the denominator $(\sqrt{x-1}-1)\ln (x-1)\to 0$, so the numerator must vanish too: $\sin (8-20+2a+b)=0$ needs $2a+b=12$. Put $u=x-2$; then $\sqrt{x-1}-1\approx \frac{u}{2}$ and $\ln (x-1)\approx u$, so the denominator behaves like $\frac{u^2}{2}$. For a finite limit the cubic $g(x)=x^3-5x^2+ax+b$ must also satisfy $g^{\prime }(2)=0$: $12-20+a=0$ gives $a=8$, hence $b=12-2a=-4$. With $g(x)=x^3-5x^2+8x-4$, near $x=2$ one has $g(x)\approx \frac{g^{\prime \prime }(2)}{2}{u}^2=u^2$ and $\sin g\approx u^2$, so $m=\lim \frac{u^2}{u^2/2}=2$. Therefore $a+b+m=8-4+2=6$.