Question

Let ƒ be a positive function and

I₁ = $\int _ { 1 - \mathrm { k } } ^ { \mathrm { k } } \mathrm { x } f ( \mathrm { x } ( 1 - \mathrm { x } ) )$dx, I₂ = dx where 2k – 1 > 0 then I₁/I₂ is -

• A. 2
• B. k
• C. ½
• D. 1

Answer 1

Answer: (C)

½

Answer 2

Since $\int _ { \mathrm { a } } ^ { \mathrm { b } } f ( \mathrm { x } )$ dx = $\int _ { a } ^ { b } f ( a + b - x )$ dx, we have

I₁ = ƒ ((k + 1 – k – x) (1

–(k + 1 – k – x)) dx

= ƒ((1 – x)x) dx

= $\int _ { 1 - \mathrm { k } } ^ { \mathrm { k } } f ( ( 1 - \mathrm { x } ) )$ dx – $\int _ { 1 - \mathrm { k } } ^ { \mathrm { k } } \mathrm { x } f ( ( 1 - \mathrm { x } ) \mathrm { x } )$ dx = I₂ – I₁.

So 2I₁ = I₂ and I₁/I₂ = 1/2.