Question

For x Ī R and a continuous function ƒ, let

I₁ = $\int_{\sin^{2}t}^{1 + \cos^{2}t}{}$x ƒ {(2 – x)} dx and I₂ = $\int_{\sin^{2}t}^{1 + \cos^{2}t}{}$ƒ{x(2 – x)} dx. Then I₁/I₂ is -

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

1

Answer 2

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

= I₁ = $\int_{\sin^{2}t}^{1 + \cos^{2}t}{}$ (2 – x) ƒ(x(2 – x)) dx = 2 . I₂ – I₁

Ž 2I₁ = 2I₂ Ž $\frac{I_{1}}{I_{2}}$ = 1.