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Question:

Let f:[0,1]→R be such that f(xy)=f(x).f(y) for all x,y∈[0,1], and f(0)≠0. If y=y(x) satisfies the differential equation dy/dx=f(x) with y(0)=1, then y(1/4)+y(3/4) is equal to

3

2

5

4

Solution:

f(xy)=f(x).f(y)
f(0)=1 as f(0)≠0 ⇒f(x)=1
dy/dx=f(x)=1 ⇒y=x+c
At x=0, y=1 ⇒c=1
y=x+1 ⇒y(1/4)+y(3/4)=1/4+1+3/4+1=3