JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 18)
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 :
3
4
2
5
Explanation
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
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
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