The product of all possible values of alpha, for which lim x->0 ( (1 - cos(alpha x)cos((alpha+1)x)cos((alpha+2)x)) / sin^2((alpha+1)x) ) = 2 is:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q17 – JEE-MAIN 2026

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Home Maths Limits and DerivativesQuestion #262

The product of all possible values of

α

, for which

lim_{x \to 0} (\frac{1 - \cos (α x) \cos ((α + 1) x) \cos ((α + 2) x)}{\sin^{2} ((α + 1) x)}) = 2

is:

Clue 1 of 3: Indeterminate Form

Recognize that the limit is of the $\frac{0}{0}$ indeterminate form as $x \to 0$ .

🧠 Full Solution Path

Step 1: Apply Taylor Series Expansion for the Limit✦ Active

The given limit is of the form $\frac{0}{0}$ as $x \to 0$ . We use the Taylor series expansions for small $y$ : $\cos y \approx 1 - \frac{y^{2}}{2}$ and $\sin y \approx y$ . For the denominator, $\sin^{2} ((α + 1) x) \approx ((α + 1) x)^{2} = (α + 1)^{2} x^{2}$ . Note that $α + 1 \neq 0$ , otherwise the denominator is $0$ and the limit is not $2$ . For the numerator, $1 - \cos (α x) \cos ((α + 1) x) \cos ((α + 2) x)$ can be approximated by expanding each cosine term and keeping terms up to $x^{2}$ :

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

Step 2: Evaluate the Limit and Form a Quadratic Equation○ Expand

Substitute the approximations into the limit expression:

lim_{x \to 0} \frac{\frac{x^{2}}{2} [α^{2} + (α + 1)^{2} + (α + 2)^{2}]}{(α + 1)^{2} x^{2}} = \frac{α^{2} + (α + 1)^{2} + (α + 2)^{2}}{2 (α + 1)^{2}}

Given that the limit equals 2, we set up the equation:

\frac{α^{2} + (α + 1)^{2} + (α + 2)^{2}}{2 (α + 1)^{2}} = 2 α^{2} + (α^{2} + 2 α + 1) + (α^{2} + 4 α + 4) = 4 (α^{2} + 2 α + 1) 3 α^{2} + 6 α + 5 = 4 α^{2} + 8 α + 4

Rearranging the terms gives the quadratic equation:

α^{2} + 2 α - 1 = 0

Step 3: Calculate the Product of Roots○ Expand

For a quadratic equation $a x^{2} + b x + c = 0$ , the product of the roots is given by $\frac{c}{a}$ . For the equation $α^{2} + 2 α - 1 = 0$ , the product of the roots is:

Product of roots = \frac{- 1}{1} = - 1

The roots are $α = - 1 \pm \sqrt{2}$ , neither of which is $- 1$ , thus satisfying the condition $α + 1 \neq 0$ . The product of all possible values of $α$ is $- 1$ .

💡 Teacher's Secret Hint

Remember to check for any conditions on $α$ that might invalidate the initial approximations or the problem statement.

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Maths Question 17 – JEE-MAIN 2026

Clue 1 of 3: Indeterminate Form

🧠 Full Solution Path