๐’‡(๐Ÿ๐’™โˆ’๐Ÿ‘)=๐Ÿ’๐’™โˆ’๐Ÿ, ๐’‡(๐’™)=? | Olympiad Question

preview_player
ะŸะพะบะฐะทะฐั‚ัŒ ะพะฟะธัะฐะฝะธะต
We derive f(x) from the given condition using two different methods.

ะ ะตะบะพะผะตะฝะดะฐั†ะธะธ ะฟะพ ั‚ะตะผะต
ะšะพะผะผะตะฝั‚ะฐั€ะธะธ
ะะฒั‚ะพั€

Or let f (x)=ax+b then f(2x--3)=a(2x--3)+b=4x--2 2ax--3a+b=4x--2 then 2a=4 and 3a+b=2 a=2 & b=4 f ((x)=2x+4...

ู…ุญู…ุฏุนู„ูŠ-ูู‚ุท
ะะฒั‚ะพั€

What kind of olympiad question is this easy

KookyPiranha
ะะฒั‚ะพั€

f(2x-3)=4x-2
let 2x-3=t
2x=t+3
4x=2t+6
4x-2=2t+4
f(t)=2t+4
f(x)=2x+4

์—์Šคํ”ผ-ht
ะะฒั‚ะพั€

Let's change the variable name from x to t to avoid naming collisions.
f(2t-3)=4t-2
y=f(x)

x=2t-3

y=4t-2

2x=4t-6
4t=2x+6
y=2x+6-2=2x+4

y=f(x) =2x+4

boguslawszostak
ะะฒั‚ะพั€

Thanks sir but it easy, .... in my mind

fouziah
ะะฒั‚ะพั€

Thank you for your solution

This is my solution:

Let f(๐‘ฅ) = a๐‘ฅ + b

Hence,
f(2๐‘ฅ - 3)
= a(2๐‘ฅ - 3) + b

Hence,
a(2๐‘ฅ - 3) + b = 4๐‘ฅ - 2

2a๐‘ฅ - 3a + b = 4๐‘ฅ - 2

2a๐‘ฅ - (3a - b) = 4๐‘ฅ - 2

Comparing coefficients,

๐‘ฅ term:
2a = 4
a = 4/2
= 2

Constant term:
3a - b = 2
3(2) - b = 2
6 - b = 2
b = 4

Hence,
f(๐‘ฅ) = 2๐‘ฅ + 4 (Ans)

absolutezero