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JEE Mathematics 2024 Question with Solution

The sum of the solutions x∈Rx \in \mathbb{R} of the equation 3cos⁡(2x)+cos⁡3(2x)cos⁡6(x)−sin⁡6(x)=x3−x2+6\frac{3 \cos(2x) + \cos^3(2x)}{\cos^6(x) - \sin^6(x)} = x^3 - x^2 + 6 is:

  • A

    00

  • B

    11

  • C

    −1-1

  • D

    33

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

3cos⁡2x+cos⁡32xcos⁡6x−sin⁡6x=x3−x2+6\frac{3\cos 2x + \cos^3 2x}{\cos^6 x - \sin^6 x} = x^3 - x^2 + 6

Find: The sum of all real solutions.

Using

cos⁡6x−sin⁡6x=(cos⁡2x−sin⁡2x)(cos⁡4x+cos⁡2xsin⁡2x+sin⁡4x)\cos^6 x - \sin^6 x = (\cos^2 x - \sin^2 x)(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x)

and

cos⁡2x−sin⁡2x=cos⁡2x\cos^2 x - \sin^2 x = \cos 2x

we get

cos⁡6x−sin⁡6x=cos⁡2x (cos⁡4x+cos⁡2xsin⁡2x+sin⁡4x)\cos^6 x - \sin^6 x = \cos 2x \,(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x)

Algebraic Simplification

Also,

3cos⁡2x+cos⁡32x=cos⁡2x (3+cos⁡22x)3\cos 2x + \cos^3 2x = \cos 2x\,(3 + \cos^2 2x)

So, for cos⁡2x≠0\cos 2x \neq 0,

cos⁡2x (3+cos⁡22x)cos⁡2x (cos⁡4x+cos⁡2xsin⁡2x+sin⁡4x)=x3−x2+6\frac{\cos 2x\,(3 + \cos^2 2x)}{\cos 2x\,(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x)} = x^3 - x^2 + 6

which becomes

3+cos⁡22xcos⁡4x+cos⁡2xsin⁡2x+sin⁡4x=x3−x2+6\frac{3 + \cos^2 2x}{\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x} = x^3 - x^2 + 6

From the extracted solution,

cos⁡4x+cos⁡2xsin⁡2x+sin⁡4x=1−sin⁡2xcos⁡2x\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x = 1 - \sin^2 x \cos^2 x

and hence

cos⁡2x(3+cos⁡22x)cos⁡2x(1−sin⁡2xcos⁡2x)=x3−x2+6\frac{\cos 2x(3 + \cos^2 2x)}{\cos 2x(1 - \sin^2 x \cos^2 x)} = x^3 - x^2 + 6

Using

1−sin⁡2xcos⁡2x=4−sin⁡22x41 - \sin^2 x \cos^2 x = \frac{4 - \sin^2 2x}{4}

and

sin⁡22x=1−cos⁡22x\sin^2 2x = 1 - \cos^2 2x

we get

4(3+cos⁡22x)4−sin⁡22x=x3−x2+6\frac{4(3 + \cos^2 2x)}{4 - \sin^2 2x} = x^3 - x^2 + 6 4(3+cos⁡22x)3+cos⁡22x=x3−x2+6\frac{4(3 + \cos^2 2x)}{3 + \cos^2 2x} = x^3 - x^2 + 6 4=x3−x2+64 = x^3 - x^2 + 6

Therefore,

x3−x2+2=0x^3 - x^2 + 2 = 0

Factorizing,

(x+1)(x2−2x+2)=0(x + 1)(x^2 - 2x + 2) = 0

The only real root is x=−1x = -1. Hence the sum of the real solutions is −1-1.

Therefore, the correct option is C.

Common mistakes

  • Cancelling cos⁡2x\cos 2x without noting the condition cos⁡2x≠0\cos 2x \neq 0 is incorrect. First identify the restriction, then simplify the fraction only under that condition.

  • Using a wrong identity for cos⁡6x−sin⁡6x\cos^6 x - \sin^6 x leads to an incorrect denominator. Treat it as a difference of cubes in cos⁡2x\cos^2 x and sin⁡2x\sin^2 x.

  • After obtaining x3−x2+2=0x^3 - x^2 + 2 = 0, assuming all factors give real roots is wrong. Check the quadratic factor x2−2x+2x^2 - 2x + 2 and note that its discriminant is negative.

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