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Maths Question 17 – JEE-MAIN 2026

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The product of all possible values of α, for which limx→0(1−cos⁡(αx)cos⁡((α+1)x)cos⁡((α+2)x)sin2⁡((α+1)x))=2 is:

Recognize that the limit is of the 00 indeterminate form as x→0.

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

The given limit is of the form 00 as x→0. We use the Taylor series expansions for small y: cos⁡y≈1−y22 and sin⁡y≈y. For the denominator, sin2⁡((α+1)x)≈((α+1)x)2=(α+1)2x2. Note that α+1≠0, otherwise the denominator is 0 and the limit is not 2. For the numerator, 1−cos⁡(αx)cos⁡((α+1)x)cos⁡((α+2)x) can be approximated by expanding each cosine term and keeping terms up to x2:

1−(1−(αx)22)(1−((α+1)x)22)(1−((α+2)x)22)≈1−(1−(αx)22−((α+1)x)22−((α+2)x)22)≈x22[α2+(α+1)2+(α+2)2]
Step 2: Evaluate the Limit and Form a Quadratic Equation○ Expand

Substitute the approximations into the limit expression:

limx→0x22[α2+(α+1)2+(α+2)2](α+1)2x2=α2+(α+1)2+(α+2)22(α+1)2

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

α2+(α+1)2+(α+2)22(α+1)2=2α2+(α2+2α+1)+(α2+4α+4)=4(α2+2α+1)3α2+6α+5=4α2+8α+4

Rearranging the terms gives the quadratic equation:

α2+2α−1=0
Step 3: Calculate the Product of Roots○ Expand

For a quadratic equation ax2+bx+c=0, the product of the roots is given by ca. For the equation α2+2α−1=0, the product of the roots is:

Product of roots=−11=−1

The roots are α=−1±2, neither of which is −1, thus satisfying the condition α+1≠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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