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Let f((x+y)/2)=(f(x)+f(y))/2fora l lr e ...

Let `f((x+y)/2)=(f(x)+f(y))/2fora l lr e a lxa n dy` If `f^(prime)(0)` exists and equals `-1a n df(0)=1,t h e n f i n d f(2)dot`

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The correct Answer is:
`(-1)`
Given, ` f((x+y)/2) = (f(x)+f(y))/2, AA x, y in R`
On putting y = 0, we get
` f(x/2) = (f(x)+f(0))/2 = 1/2 [1+f(x)]" "[:' f(0) = 1]`
`rArr" " 2f(x/2) = f(x) +1`
` rArr" " f(x) = 2f(x/2)-1 AA x, y in R` ...(i)
Since, ` f'(0) =- 1 `, we get
`f(x/2)=(f(x)+f(0))/2 = 1/2 [1+f(x)]" "[:' f(0) = 1]`
`rArr" " 2f(x/2) = f(x) +1 `
` rArr" " f(x) = 2f(x/2) -1, AA x, y in R` ....(i)
Since, ` f'(0) =- 1, ` we get
`underset (h to 0) lim (f(0+h)-f(0))/ lim =- 1`
`rArr" " underset ( h to 0) lim (f(h)-1)/h =- 1` .....(ii)
Again, ` f'(x) = underset( h to 0) lim (f(x+h)-f(x))/h`
` = underset( h to 0) lim (f((2x+2h)/2)-f(x))/h`
`= underset( h to 0) lim ((f(2x)+f(2h))/2-f(x))/h`
`= underset( h to 0) lim (1/2 [2f((2x)/2) -1+2 f((2h)/2) -1]-f (x))/h `
[from Eq. (i)]
`=underset( h to 0) lim (1/2[2f(x)-1+2f(h)-1]-f(x))/h`
` = underset( h to 0) lim (f(x)+ f(h) -1-f(x))/h `
` = underset( h to 0) lim (f(h)-1)/h =- 1` [from Eq. (ii)]
`:." " f'(x) =- 1, AA x in R`
` rArr" " int f'(x) dx = int -1 dx `
`rArr" " f(x) =- x + k`, where, k is a constant.
But f(0) = 1,
therefore f(0)=- 0 + k
`rArr" " 1 = k`
`rArr" " f(x) = 1- x, AA x in R rArr f(2) =- 1`
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