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Question

# Let f(x) be a real-valued differentiable function not identically zero such that f(x+y2n+1)=f(x)+{f(y)}2n+1,nϵN and x, y are any real numbers and f′(0)≥0. Find the value of f(5)

A
0
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B
1
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C
2
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D
5
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Solution

## The correct option is D 5Here, f(x+y2n+1)=f(x)+{f(y)}2n+1 .....(i)Putting x=0,y=0, we getf(0)=f(0)+{f(0)}2n+1⇒f(0)=0f′(0)≥0 (given)⇒limh→0f(x)−f(0)x−0≥0⇒limh→0f(x)x≥0Now, if x>0⇒f(x)≥0 .....(ii)Putting x=0,y=1 in Eq. (i),f(1)=f(0)+{f(1)}2n+1 or f(1)[1−{f(1)}2n]=0∴f(1)=0 or 1 [using Eq. (ii)]Putting y=1 in Eq. (i), for all real x,f(x+1)=f(x)+{f(1)}2n+1 .....(iii)Now, two cases arise either f(1)=0 or 1Case I If f(1)=0⇒f(x+1)=f(x) [using Eq. (iii)]⇒f(1)=f(2)=f(3)=....=0∴f(x) is identically zero. (which is not possible)Case II If f(1)=1⇒f(x+1)=f(x)+1∴f(2)=f1)+1=1+1=2f(3)=f(2)+1=2+1=3f(4)=f(3)+1=3+1=4f(5)=f(4)+1=4+1=5Proceeding in same way, we getf(x)=x and f′(x)=1⇒f′(10)=1Hence, f(5)=5 and f′(10)=1

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