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, dydx=f(x) with y(0)=1,theny(14)+y(34) is equal to :
A
2
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B
3
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C
4
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D
5
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Solution
The correct option is B3 f(xy)=f(x)⋅f(y) Put x=0,y=0 f(0)=(f(0))2 ⇒f(0)=1asf(0)≠0
Now, put y=0 f(0)=f(x)⋅f(0) ⇒f(x)=1
As given, dydx=f(x) ⇒y=∫f(x)dx=∫1dx ⇒y=x+c At x=0,y(0)=1⇒c=1 ∴y(x)=x+1