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Let `f(x)` be continuous functions `f: RvecR` satisfying `f(0)=1a n df(2x)-f(x)=xdot` Then the value of `f(3)` is `2` b. `3` c. `4` d. 5

A

2

B

3

C

4

D

5

Text Solution

Verified by Experts

The correct Answer is:
C
Given `f(2x)-f(x)=x,`
Replacing x by x/2, we have `f(x)-f((x)/(2))=(x)/(2)`
Replacing x by `(x)/(2)` repeatedly and adding, we get
`(f(x)-f((x)/(2)))+(f((x)/(2))-f((x)/(4)))+...+(f((x)/(2^(n-1)))-f((x)/(2^(n))))`
`=(x)/(2)+(x)/(2^(2))+(x)/(2^(3))+...+(x)/(2^(n))`
`rArr" "f(x)-f((x)/(n^(n)))=(x)/(2)+(x)/(2^(2))+(x)/(2^(3))+...+(x)/(2^(n))`
`therefore" "underset(nrarroo)(lim)(f(x)-f((x)/(2^(n))))=((x)/(2))/(1-(1)/(2))`
`rArr" "f(x)-f(0)=x`
`(becausef(x)" is continuous "rArr underset(nrarroo)(lim)f((x)/(2^(n)))=f(0)=1)`
`rArr" "f(x)=x+f(0)=x+1`
`rArr" "f(3)=4`
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