Let y = ϕ(x) be the extremizing function for the functional \(\rm I(y)=\int^1_0y^2\left(\frac{dy}{dx}\right)^2dx, \) subject to y(0) = 0, y(1) = 1. Then ϕ(1/4) is equal to 

1
1/2
2
1/4
3
1/8
4
1/12
5
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