f-x-p-f-x-p-6x-4-f-20-29p-f-p-2p-

Question Number 95662 by john santu last updated on 26/May/20

f (x + p) + f (x - p) = 6 x - 4

f (20) = 29 p

\frac{f (p)}{2 p} = ?

Commented by i jagooll last updated on 26/May/20

replace x + p = x

(1) f (x) + f (x - 2 p) = 6 x - 6 p - 4

replace x - p = x

(2) f (x + 2 p) + f (x) = 6 x + 6 p - 4

\Rightarrow (1) - (2)

f (x - 2 p) - f (x + 2 p) = - 12 p

let x - 2 p = 20 \Rightarrow x = 20 + 2 p

f (20) - f (20 + 4 p) = - 12 p

29 p + 12 p = f (20 + 4 p)

f (20 + 4 p) = 41 p

let 20 + 4 p = t \Rightarrow p = \frac{t - 20}{4}

f (t) = 41 (\frac{t - 20}{4}) = \frac{41 t - 820}{4}

\frac{f (t)}{2 t} = \frac{41 t - 820}{8 t}

Commented by PRITHWISH SEN 2 last updated on 26/May/20

put p = 0

2 f (x) = 6 x - 4 \Rightarrow f (x) = 3 x - 2 \Rightarrow f (20) = 58

\Rightarrow p = 2

∴ \frac{f (p)}{2 p} = \frac{3 \times 2 - 2}{2 \times 2} = 1 please check

Commented by i jagooll last updated on 27/May/20

how can p = 0 and p = 2 ?

it meant 0 = 2 ?

Commented by PRITHWISH SEN 2 last updated on 27/May/20

for the 1 st equation p is the point which can

take any value .

But for the second equation it is a constant .

now fixed the value of p at 2 then you get

for f (x) = 3 x - 2

then f (x + 2) + f (x - 2) = 3 (x + 2) - 2 + 3 (x - 2) - 2

= 6 x - 4

and f (20) = 3 \times 20 - 2 = 58 = 29 p where p = 2

Commented by i jagooll last updated on 27/May/20

alright sir . thank you

Answered by john santu last updated on 28/May/20

let f (x) = ax + b

f (x + p) = ax + ap + b (i)

f (x - p) = ax - ap + b (ii)

(i) + (ii) \Rightarrow 2 ax + 2 b = 6 x - 4

a = 3, b = - 2

f (x) = 3 x - 2

f (20) = 58 = 29 p; p = 2

\frac{f (p)}{2 p} = \frac{f (2)}{4} = \frac{6 - 2}{4} = 1