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JEE Main 2019 · 12 April · Shift II

JEE Main 2019 Mathematics question on Matrices and Determinants

A value of θ∈(0,π/3)\theta \in (0, \pi/3), for which
∣1+cos⁡2θsin⁡2θ4cos⁡6θcos⁡2θ1+sin⁡2θ4cos⁡6θcos⁡2θsin⁡2θ1+4cos⁡6θ∣=0\begin{vmatrix} 1 + \cos^2\theta & \sin^2\theta & 4\cos 6\theta \\ \cos^2\theta & 1 + \sin^2\theta & 4\cos 6\theta \\ \cos^2\theta & \sin^2\theta & 1 + 4\cos 6\theta \end{vmatrix} = 0
is
  1. A.
    π9\dfrac{\pi}{9}
  2. B.
    π18\dfrac{\pi}{18}
  3. C.
    7π24\dfrac{7\pi}{24}
  4. D.
    7π36\dfrac{7\pi}{36}
Show answer and solution

Correct option: A

Solution

Apply R1→R1-R3 and R2→R2-R3 to simplify:

R1 becomes (1, 0, -1), R2 becomes (0, 1, -1), R3 stays (cos²θ, sin²θ, 1+4cos6θ).

Expanding the determinant along the first two rows:
det = 1·[(1+4cos6θ) - (-1)·sin²θ] - 0 + (-1)·[0·sin²θ - cos²θ·1]...

More directly, expanding gives:
det = (1+4cos6θ) + cos²θ + sin²θ = (1+4cos6θ) + 1 + ...

Let me redo carefully. With R1=(1,0,-1), R2=(0,1,-1), R3=(cos²θ, sin²θ, 1+4cos6θ):

det = 1·[1·(1+4cos6θ) - (-1)·sin²θ] - 0 + (-1)·[0·sin²θ - 1·cos²θ]
= [(1+4cos6θ) + sin²θ] + cos²θ
= 1 + 4cos6θ + sin²θ + cos²θ
= 2 + 4cos6θ

Setting this to zero: cos6θ = -1/2, so 6θ = 2π/3 (taking the principal value in the valid range).

Thus θ = π/9, which lies in (0, π/3).

[JEE Main 2019 · 12 April · Shift II]

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