JEE Main 2021 · 18 March · morning · Single correct · +4 / −1
The solutions of the equation | 1 + sin ²x, sin ²x, sin ²x; cos ²x, 1 + cos ²x, cos ²x; 4 sin 2x, 4 sin 2x, 1 + 4 sin 2x | = 0,(0 < x < π ), are
- (a) π/12,π/6
- (b) π/6,5π/6
- (c) 5π/12,7π/12
- (d) 7π/12,11π/12
Answer and solution
Answer (d) 7π/12,11π/12
Solution
Maths sir · 4 steps
Add all three rows: every entry becomes 2 + 4 sin 2x, so pull it out and the rest is easy.
- Add the rows. Apply R₁ → R₁ + R₂ + R₃. In each column the new top entry is 1 + sin² x + cos² x + 4 sin 2x = 2 + 4 sin 2x, the same in all three places.
- Take out the factor. Pull 2 + 4 sin 2x out of R₁, leaving the row (1, 1, 1). Then C₂ → C₂ - C₁ and C₃ → C₃ - C₁ turn the determinant into a lower triangular one with 1s on the diagonal, so it equals 1.
Δ = (2 + 4 sin 2x)|1, 0, 0; cos² x, 1, 0; 4 sin 2x, 0, 1| = 2 + 4 sin 2x
- Solve the equation. 2 + 4 sin 2x = 0 gives sin 2x = -1/2. Since 0 < x < π, we have 0 < 2x < 2π. Sine is negative in the third and fourth quadrants, so 2x = 7π/6 or 11π/6.
- So. x = 7π/12 or 11π/12, which is option (d).
The idea to keep. When every row (or column) adds up to the same expression, add them all into one line first. Take that common factor out, and the determinant that is left is usually trivial.
Answer from the NTA final answer key. Solution by PrepWiser.