LimitsmediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Limit Condition Product of α = -1 | JEE Main 2026

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

Question
The product of all possible values of α\alpha, for which lim⁡x→0(1−cos⁡(αx)cos⁡((α+1)x)cos⁡((α+2)x)sin⁡2((α+1)x))=2\displaystyle\lim_{x\to0}\left(\dfrac{1-\cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}\right)=2, is
A−2-2
B11
C−1-1correct
D54\dfrac54
Solution
Step 1: As x→0x\to0, cos⁡(kx)=1−k2x22+O(x4)\cos(kx)=1-\dfrac{k^2x^2}{2}+O(x^4), so 1−cos⁡(k1x)cos⁡(k2x)cos⁡(k3x)=(k12+k22+k32)x22+O(x4)1-\cos(k_1x)\cos(k_2x)\cos(k_3x)=\dfrac{(k_1^2+k_2^2+k_3^2)x^2}{2}+O(x^4), with k1=α, k2=α+1, k3=α+2k_1=\alpha,\ k_2=\alpha+1,\ k_3=\alpha+2; and sin⁡2((α+1)x)=(α+1)2x2+O(x4)\sin^2((\alpha+1)x)=(\alpha+1)^2x^2+O(x^4). Step 2: x2x^2 cancels:
lim⁡x→012[α2+(α+1)2+(α+2)2]x2(α+1)2x2=α2+(α+1)2+(α+2)22(α+1)2=2.\lim_{x\to0}\frac{\tfrac12\big[\alpha^2+(\alpha+1)^2+(\alpha+2)^2\big]x^2}{(\alpha+1)^2x^2}=\frac{\alpha^2+(\alpha+1)^2+(\alpha+2)^2}{2(\alpha+1)^2}=2.
 Step 3: × 2(α+1)2\times\,2(\alpha+1)^2: α2+(α+1)2+(α+2)2=4(α+1)2\alpha^2+(\alpha+1)^2+(\alpha+2)^2=4(\alpha+1)^2.
⇒α2+(α+2)2=3(α+1)2.\Rightarrow\alpha^2+(\alpha+2)^2=3(\alpha+1)^2.
 Step 4: α2+(α2+4α+4)=3(α2+2α+1)⇒2α2+4α+4=3α2+6α+3\alpha^2+(\alpha^2+4\alpha+4)=3(\alpha^2+2\alpha+1)\Rightarrow 2\alpha^2+4\alpha+4=3\alpha^2+6\alpha+3.
⇒α2+2α−1=0.\Rightarrow\alpha^2+2\alpha-1=0.
 Step 5: Product of roots =ca=−11=−1=\dfrac{c}{a}=\dfrac{-1}{1}=-1. Correct answer: (3)
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