Let f be a positive function, and let $I_1 = \int\limits_{1-k}^{k}xf[x(1-x)]dx$, $I_2 = \int\limits_{1-k}^{k}f[x(1-x)dx]$, $k >\frac{1}{2}$. If $I_1 = t\, I_2$, then t is equal to

1/2

We have $k >\frac{1}{2}$

⇒ 2k - 1 > 0 ⇒ k > 1 - k.

Hence $I_1 = \int\limits_{1-k}^{k}xf(x(1-x))dx=\int\limits_{1-k}^{k}(k+1-k-x)f(k+1-k-x)(1-k-1+k+x)dx$.

$\int\limits_{1-k}^{k}(1-x)f(x(1-x))dx$

⇒ I₁ = I₂ - I₁

or,  2I₁ = I₂  ⇒ $t=\frac{1}{2}$ 

Hence (C) is the correct answer.