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

\alpha

, for which

\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

-2

1

-1

correct

\dfrac54

Solution

Step 1: As

x\to0

\cos(kx)=1-\dfrac{k^2x^2}{2}+O(x^4)

, so

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

k_1=\alpha,\ k_2=\alpha+1,\ k_3=\alpha+2

; and

\sin^2((\alpha+1)x)=(\alpha+1)^2x^2+O(x^4)

. Step 2:

x^2

cancels:

\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:

\times\,2(\alpha+1)^2

\alpha^2+(\alpha+1)^2+(\alpha+2)^2=4(\alpha+1)^2

\Rightarrow\alpha^2+(\alpha+2)^2=3(\alpha+1)^2.

Step 4:

\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

\Rightarrow\alpha^2+2\alpha-1=0.

Step 5: Product of roots

=\dfrac{c}{a}=\dfrac{-1}{1}=-1

. Correct answer: (3)

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