LimitsmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Limit with sin of a Cubic: a + b + m = 6 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
If lim⁡x→2sin⁡(x3−5x2+ax+b)(x−1−1)log⁡e(x−1)=m\displaystyle\lim_{x\to2}\dfrac{\sin\left(x^3-5x^2+ax+b\right)}{\left(\sqrt{x-1}-1\right)\log_e(x-1)}=m, then a+b+ma+b+m is equal to
A55
B66correct
C88
D1010
Solution
Step 1: As x→2x\to2, the denominator →(1−1)log⁡e1=0\to(\sqrt1-1)\log_e1=0. For the limit to exist, the numerator's argument must vanish at x=2x=2:
23−5(2)2+2a+b=0 ⇒ 8−20+2a+b=0 ⇒ 2a+b=12.(1)2^3-5(2)^2+2a+b=0\ \Rightarrow\ 8-20+2a+b=0\ \Rightarrow\ 2a+b=12.\quad(1)
 Step 2: Near x=2x=2, sin⁡(⋅)≈(⋅)\sin(\cdot)\approx(\cdot), and the denominator behaves like (x−2)2(x-2)^2 (since x−1−1∼12(x−2)\sqrt{x-1}-1\sim\tfrac12(x-2) and log⁡e(x−1)∼(x−2)\log_e(x-1)\sim(x-2)). So
m=lim⁡x→2x3−5x2+ax+b(x−2)2.m=\lim_{x\to2}\frac{x^3-5x^2+ax+b}{(x-2)^2}.
 This is finite only if the cubic has a double root at x=2x=2, i.e. its derivative also vanishes there:
3(2)2−10(2)+a=0 ⇒ 12−20+a=0 ⇒ a=8.3(2)^2-10(2)+a=0\ \Rightarrow\ 12-20+a=0\ \Rightarrow\ a=8.
 Then from (1), b=12−2(8)=−4b=12-2(8)=-4. Step 3: Evaluate mm by applying L'Hôpital (or factoring): differentiating numerator and (x−2)2(x-2)^2,
m=lim⁡x→23x2−10x+a2(x−2)=lim⁡x→26x−102=12−102=2.m=\lim_{x\to2}\frac{3x^2-10x+a}{2(x-2)}=\lim_{x\to2}\frac{6x-10}{2}=\frac{12-10}{2}=2.
 Step 4:
a+b+m=8+(−4)+2=6.a+b+m=8+(-4)+2=6.
 Correct answer: (2)
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