JEE Main 2023 · 1 February · morning · Single correct · +4 / −1
Let f(x) = | 1 + sin ²x, cos ²x, sin 2x; sin ²x, 1 + cos ²x, sin 2x; sin ²x, cos ²x, 1 + sin 2x |, x ∈ [ π/6,π/3 ]. If α and β respectively are the maximum and the minimum values of f, then
- (a) α ² - β ² = 4√3
- (b) β ² - 2√α = 19/4
- (c) β ² + 2√α = 19/4
- (d) α ² + β ² = 9/2
Answer and solution
Answer (b) β ² - 2√α = 19/4
Solution
Maths sir · 4 steps
Add all three columns into the first one: every row sums to 2 + sin 2x, and the rest collapses to 1.
- Add the columns. Apply C₁ → C₁ + C₂ + C₃. Each row of the new first column becomes 1 + sin² x + cos² x + sin 2x = 2 + sin 2x, so take that factor out. The first column is now all 1s.
- Clear the rows. Now apply R₁ → R₁ - R₃ and R₂ → R₂ - R₃. The rows become (0, 0, -1), (0, 1, -1) and (1, cos² x, 1 + sin 2x). Expanding along the first column gives 1 × (0 × (-1) - (-1) × 1) = 1.
f(x) = 2 + sin 2x
- Find the extremes. For x ∈ [π/6, π/3] we have 2x ∈ [π/3, 2π/3], so sin 2x runs from √3/2 up to 1. That gives α = 3 and β = 2 + √3/2.
- So. β² = 4 + 2√3 + 3/4 = 19/4 + 2√3 and 2√α = 2√3. Their difference is β² - 2√α = 19/4, which is option (b).
The idea to keep. When every row is a common row plus 1 on the diagonal, add all columns into one. The common sum factors out and what remains is usually 1.
Answer from the NTA final answer key. Solution by PrepWiser.