Question

For any t Ī R and ƒ a continuous function, let

I₁ =$\int_{\sin^{2}t}^{1 + \cos^{2}t}{xƒ(x(2 - x))dx}$and I₂ = $\int_{\sin^{2}t}^{1 + \cos^{2}t}{ƒ(x(2 - x))dx}$then I₁/I₂ is equal to-

• A. 2
• B. 1
• C. 4
• D. None of these

Answer 1

Answer: (B)

1

Answer 2

I₁ = $\int_{\sin^{2}t}^{1 + \cos^{2}t}{(2 - x)ƒ((2 - x)(2 - (2 - x)))}$ dx

=$\int_{\sin^{2}t}^{1 + \cos^{2}t}{(2 - x)ƒ(x(2 - x))}$dx

= 2$\int_{\sin^{2}t}^{1 + \cos^{2}t}{ƒ(x(2 - x))}$dx – $\int_{\sin^{2}t}^{1 + \cos^{2}t}{xƒ(x(2 - x))}$dx

= 2I₂ – I₁

Therefore, 2I₁ = 2I₂ and so I₁/I₂ = 1.