JEE Main 2026 (Online) 2nd April Morning ShiftQuestion 1
If $\lim_{x\to2}\frac{\sin\left(x^{3}-5x^{2}+ax+b\right)}{\left(\sqrt{x-1}-1\right)\log_{e}(x-1)}=m$, then $a+b+m$ is equal to :
Limits & Continuity
Solution
Concept. The denominator vanishes like $(x-2)^2$, so the cubic must have a double root at $x=2$. Formulas used. (1) $\sqrt{1+t}-1\sim\frac t2$, $\ln(1+t)\sim t$(2) $\sin u\sim u$ as $u\to0$Step 1 - Denominator. Put $t=x-2$. By (1), $(\sqrt{1+t}-1)\ln(1+t)\sim\frac{t^2}2$. Step 2 - Condition on the cubic. Let $g(x)=x^3-5x^2+ax+b$. For a finite limit, $\sin g(x)$ must vanish like $t^2$; so $g(2)=0$ and $g'(2)=0$. - $g(2)=8-20+2a+b=0\Rightarrow2a+b=12$ - $g'(2)=12-20+a=0\Rightarrow a=8$, so $b=-4$ Step 3 - Leading term. $g(x)=x^3-5x^2+8x-4=(x-2)^2(x-1)$, so $g\sim t^2\cdot1=t^2$ near $x=2$. Step 4 - Limit. By (2), $$m=\lim\frac{t^2}{t^2/2}=2$$ Step 5. $a+b+m=8-4+2=6$. Answer: B