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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, dydx=f(x) with y(0)=1,then y(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 B 3
f(xy)=f(x)f(y)
Put x=0,y=0
f(0)=(f(0))2
f(0)=1 as f(0)0

Now, put y=0
f(0)=f(x)f(0)
f(x)=1

As given, dydx=f(x)
y=f(x)dx=1 dx
y=x+c
At x=0, y(0)=1 c=1
y(x)=x+1

y(14)+y(34)=14+1+34+1=3

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