Let f:[0,1]→R be such that f(xy)=f(x)f(y)∀x,y∈[0,1] and f(0)≠0 . If y=y(x) satisfies the differential equation dydx=f(x) with y_(0)=1 then y(1/4)+y(3/4) is
f(x)=1⇒dydx=f(x)=1 given ∫dy=∫dxy=x+1⇒y(0)=1→c=1y=x+1⇒y(1/4)+y(3/4)=14+34+2=3