If lim xto 2 frac sin(x 3 - 5x 2 + ax + b) (√(x - 1) - 1)log e (x - 1) = m , then a + b + m is equal to:

Mathematics · Limits, Continuity and Differentiability

JEE Main 2026 — 2 April, Morning Shift — Question 33

If $\lim_{x\to 2}\frac{\sin(x^3 - 5x^2 + ax + b)}{(\sqrt{x - 1} - 1)\log_{\mathrm{e}}(x - 1)} = \mathbf{m}$ , then $\mathbf{a} + \mathbf{b} + \mathbf{m}$ is equal to:

Option A:
5
Option B:
6
Correct
Option C:
8
Option D:
10

Answer: B

Step-by-step solution

$\because$ Denominator $=0$ at $\mathrm{x}=2$ ⇒ Numerator $=0$ at $\mathrm{x}=2$ $\Rightarrow 2^{3}-5(2)^{2}+\mathrm{a}(2)+\mathrm{b}=0$

\Rightarrow 2 \mathrm{a}+\mathrm{b}=12 \end{gathered}$$ $\therefore \mathrm{m}=\underset{\mathrm{x} \rightarrow 2}{\operatorname{Lim}} \frac{\frac{\sin \left(\mathrm{x}^{3}-5 \mathrm{x}^{2}+\mathrm{ax}+\mathrm{b}\right)}{\left(\mathrm{x}^{3}-5 \mathrm{x}^{2}+\mathrm{ax}+\mathrm{b}\right)} \cdot\left(\mathrm{x}^{3}-5 \mathrm{x}^{2}+\mathrm{ax}+\mathrm{b}\right)}{\left(\frac{\mathrm{x}-1-1}{\sqrt{\mathrm{x}-1}+1}\right) \cdot \frac{\log _{\mathrm{e}}(1+(\mathrm{x}-2))}{(\mathrm{x}-2)} \cdot(\mathrm{x}-2)}$ $\Rightarrow \mathrm{m}=\operatorname{Lim}_{\mathrm{x} \rightarrow 2} 2\left(\frac{\mathrm{x}^{3}-5 \mathrm{x}^{2}+\mathrm{ax}+\mathrm{b}}{(\mathrm{x}-2)^{2}}\right)$ $\Rightarrow \mathrm{m}=\operatorname{Lim}_{\mathrm{x} \rightarrow 2} 2\left(\frac{3 \mathrm{x}^{2}-10 \mathrm{x}+\mathrm{a}}{2(\mathrm{x}-2)}\right)$ ∵ Denominator $=0$ at $\mathrm{x}=2$ ⇒ Numerator $=0$ at $\mathrm{x}=2$ $\Rightarrow 3(2)^{2}-10(2)+\mathrm{a}=0 \Rightarrow \mathrm{a}=8$ put $\mathrm{a}=8$ in (1), we get $\mathrm{b}=-4$ $\therefore \mathrm{m}=\operatorname{Lim}_{\mathrm{x} \rightarrow 2} 2\left(\frac{6 \mathrm{x}-10}{2}\right) \Rightarrow \mathrm{m}=2$ $\therefore \mathrm{a}+\mathrm{b}+\mathrm{m}=6$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Morning Shift
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Introduction to Limit