The product of all possible values of $\alpha$, for which $\lim _{x \rightarrow 0}\left(\frac{1-\cos (\alpha x) \cos ((\alpha+1) x) \cos ((\alpha+2) x)}{\sin ^{2}((\alpha+1) x)}\right)=2$, is :

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Limits

The product of all possible values of

α

, for which

lim_{x \to 0} (\frac{1 - c o s ( α x ) c o s (( α + 1 ) x ) c o s (( α + 2 ) x )}{s i n ^{2} (( α + 1 ) x )}) = 2

, is :

Answer: (C)

x \to 0

cos (α x) cos ((α + 1) x) cos ((α + 2) x) \approx 1 - \frac{x ^{2}}{2} [α^{2} + (α + 1)^{2} + (α + 2)^{2}]

and

sin^{2} ((α + 1) x) \approx (α + 1)^{2} x^{2}

, so the limit equals

\frac{α ^{2} + ( α + 1 ) ^{2} + ( α + 2 ) ^{2}}{2 ( α + 1 ) ^{2}}

. Setting this equal to

2

gives

α^{2} + (α + 1)^{2} + (α + 2)^{2} = 4 (α + 1)^{2}

, which simplifies to

α^{2} + 2 α - 1 = 0

. The product of all possible values of

α

- 1

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option

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