JEE Main 2025 · 3 April · morning · Single correct · +4 / −1
If y(x)=|sin x, cos x, sin x+cos x+1; 27, 28, 27; 1, 1, 1|, x ∈ ℝ , then (d² y)/(d x²)+y is equal to
- (a) 28
- (b) 27
- (c) -1
- (d) 1
Answer and solution
Answer (c) -1
Solution
Maths sir · 4 steps
Clear the third column first with C₃ → C₃ - C₁ - C₂, and y becomes a one-line function of x.
- Simplify a column. Apply C₃ → C₃ - C₁ - C₂. This does not change the determinant. The third column becomes 1, 27 - 27 - 28 = -28 and 1 - 1 - 1 = -1.
y = |sin x, cos x, 1; 27, 28, -28; 1, 1, -1|
- Expand along row 1. y = sin x (-28 + 28) - cos x (-27 + 28) + 1 (27 - 28). This gives y = -cos x - 1.
- Differentiate twice. dy/dx = sin x and d²y/dx² = cos x. So d²y/dx² + y = cos x - cos x - 1.
- So. d²y/dx² + y = -1, which is option (c).
The idea to keep. When one column is a sum of the other columns plus constants, subtract those columns out before you expand. The determinant stays the same and the algebra becomes much shorter.
Answer from the NTA final answer key. Solution by PrepWiser.