Let f : [0,1] [ ] R be such that f(xy) = f(x).f(y), for all x, y [ ] [0, 1], and f(0) [ ] …

Question

Let f : [0,1] $$ \to $$ R be such that f(xy) = f(x).f(y),  for all x, y $$ \in $$ [0, 1], and f(0) $$ \ne $$ 0. If y = y(x) satiesfies the differential equation, $${{dy} \over {dx}}$$ = f(x) with y(0) = 1, then y$$\left( {{1 \over 4}} \right)$$ + y$$\left( {{3 \over 4}} \right)$$ is equal to :

Accepted Answer

If f(xy) = f(x) f(y) $$\forall $$ x, y $$ \in $$ R and f(0) $$ \ne $$ 0

put x = y = 0

$$ \Rightarrow $$  f(0) = [f(0)]2

$$ \Rightarrow $$  f(0) = 1

put y = 0 $$ \Rightarrow $$ f(0) = f(x) f(0)

$$ \Rightarrow $$ f(x) = 1

given that $${{dy} \over {dx}}$$ = f(x)

$$ \therefore $$  $${{dy} \over {dx}}$$ = 1 $$ \Rightarrow $$ y = x + k

given that y(0) = 1

$$ \therefore $$  k = 1

hence y = x + 1

y$$\left( {{1 \over 4}} \right)$$ + y$$\left( {{3 \over 4}} \right)$$ = $$\left( {{1 \over 4} + 1} \right)$$ + $$\left( {{3 \over 4} + 1} \right)$$ = 3

Solutions · 1