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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 : 
  1. 2
  2. 5
  3. 3
  4. 4


Solution

The correct option is C 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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