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MCQMediumJEE 2024Trigonometric Equations

JEE Mathematics 2024 Question with Solution

If 2sin⁡3x+sin⁡2xcos⁡x+4sin⁡nx−4=02\sin^3 x + \sin^2 x \cos x + 4\sin^n x - 4 = 0 has exactly 33 solutions in the interval (0,nπ/2)(0, n\pi/2), n∈Nn \in N, then the roots of the equation x2+nx+(n−3)=0x^2 + nx + (n-3) = 0 belong to:

  • A

    (0,∞)(0, \infty)

  • B

    (−∞,0)(-\infty, 0)

  • C

    −17/2,17/2-\sqrt{17}/2, \sqrt{17}/2

  • D

    ZZ

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: the solution concludes that the correct option is B.

Find: The interval to which the roots of x2+nx+(n−3)=0x^2 + nx + (n-3) = 0 belong.

From the solution, the working states that n=3n = 3. Substituting this into the quadratic equation gives

x2+3x+(3−3)=0x^2 + 3x + (3-3) = 0

so

x2+3x=0x^2 + 3x = 0

Factoring,

x(x+3)=0x(x+3) = 0

Hence the roots are

x=0,−3x = 0, -3

The source solution then matches the answer with option B and concludes that the roots belong to (−∞,0)(-\infty, 0).

Therefore, the correct option is B.

Note: The solution contains inconsistencies in the trigonometric working and interval endpoint discussion, but it explicitly declares option B as correct, so that answer is retained.

Common mistakes

  • Using the raw quadratic alone without first determining nn from the trigonometric condition. The parameter nn must be fixed before discussing the roots' interval.

  • Assuming that because one root is 00, the interval must automatically include 00. The source answer marks option B, so students should check the exact option wording instead of inferring a different interval.

  • Confusing sin⁡2xcos⁡x\sin^2 x \cos x with sin⁡2xcos⁡x\sin 2x \cos x. The question text and the solution do not display the same trigonometric expression, so careless substitution can lead to a wrong value of nn.

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