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Let f:[0,1] rarr R be such that f(xy)=f(...

Let `f:[0,1] rarr 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)` satisfies the differential equation, `dy/dx=f(x)` with `y(0)=1,` then `y(1/4)+y(3/4)` is equal to

A

(a) 5

B

(b) 3

C

(c) 2

D

(d) 4

Text Solution

Verified by Experts

The correct Answer is:
(b)
Given , `f(xy)=f(x)cdot f(y),AAx,yin[0,1] …(i)`
Putting `x = y = 0` in Eq. (i), we get
`f(0) = f(0)cdot f(0)`
`rArr f(0)[f(0)-1]=0`
`rArr f(0)=1as f(0)ne 0`
Now, put y=0 in Eq. (i) , we get
`f(0)=f(x)cdot f(0)`
`rArr f(x)=1`
so, `dy/dx=f(x)rArr dy/dx=1`
`rArr int dy=intdx`
`rArr y=x+ C`
`therefore y(0)=1`
`therefore 1=0+C`
`rArr C=1`
`therefore y=x+1`
Now, `y(1/4)=1/4+1 =5/4 and y(3/4)=3/4+1=7/4`
`rArr y(1/4)+y(3/4)=5/4+7/4=3`
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