JEE Main 2022 · 25 June · evening · Single correct · +4 / −1
If bₙ = ∫₀^(π/2) (cos ²nx)/(sin x)dx, n ∈ N, then
- (a) b₃ - b₂, b₄ - b₃, b₅ - b₄ are in A.P. with common difference -2
- (b) 1/(b₃ - b₂),1/(b₄ - b₃),1/(b₅ - b₄) are in an A.P. with common difference 2
- (c) b₃ - b₂, b₄ - b₃, b₅ - b₄ are in a G.P.
- (d) 1/(b₃ - b₂),1/(b₄ - b₃),1/(b₅ - b₄) are in an A.P. with common difference -2
Answer and solution
Answer (d) 1/(b₃ - b₂),1/(b₄ - b₃),1/(b₅ - b₄) are in an A.P. with common difference -2
Solution
Maths sir · 4 steps
Do not find each bₙ. Subtract two neighbours, and the sin x in the denominator cancels out.
- Difference of squares. Use cos² A - cos² B = -sin(A+B)sin(A-B) with A = (n+1)x and B = nx. The numerator of bₙ₊₁ - bₙ becomes -sin((2n+1)x)sin x, and the sin x cancels with the denominator.
bₙ₊₁ - bₙ = -∫₀^(π/2) sin((2n+1)x)dx = -(1 - cos(2n+1)π/2)/(2n+1) = -1/(2n+1)
- Put in n. cos(2n+1)π/2 = 0 for every natural n. So b₃ - b₂ = -1/5, b₄ - b₃ = -1/7 and b₅ - b₄ = -1/9.
- Take reciprocals. The reciprocals are -5, -7, -9. Each term is 2 less than the one before it.
- So. The reciprocals form an A.P. with common difference -2, which is option (d).
The idea to keep. When an integral is defined for every n, look at Iₙ₊₁ - Iₙ first. The awkward denominator often cancels, and the difference is easy to integrate.
Answer from the NTA final answer key. Solution by PrepWiser.