f-x-ax-2-bx-c-is-given-a-b-c-a-b-c-R-a-0-and-f-ax-b-f-bx-c-find-1-2-f-b-f-a-

Question Number 174096 by mnjuly1970 last updated on 24/Jul/22

f (x) = {a x}^{2} + b x + c i s g i v e n

a \neq b \neq c, a, b, c \in R

a \neq 0 a n d :

f (a x + b) = f (b x + c)

f i n d : \frac{1}{2} (f (b) - f (a)) = ?

Answered by mahdipoor last updated on 24/Jul/22

i f f (x) = a x x + b x + c a n d

M a x / m i n f = f (\frac{- b}{2 a})

f (m) = f (n) \Leftrightarrow \frac{m + n}{2} = \frac{- b}{2 a}

\Rightarrow\Rightarrow \frac{(a x + b) + (b x + c)}{2} = \frac{- b}{2 a} \Rightarrow

a + b = 0 a n d b + c = \frac{- b}{a} \Rightarrow

a = k b = - k c = 1 + k \Rightarrow

f (x) = k x x - k x + 1 + k

\frac{1}{2} ([k {(- k)}^{2} - k (- k) + 1 + k] -

[k (k^{2}) - k (k) + 1 + k]) = k^{2} = a^{2}

Commented by mnjuly1970 last updated on 24/Jul/22

t h a n k y o u m a s t e r . . f a g h a t = a^{2}

t y p o \dots

Commented by mahdipoor last updated on 24/Jul/22

t n x, i e d i t e d

Answered by cortano1 last updated on 25/Jul/22

f (a x + b) = a {(a x + b)}^{2} + b (a x + b) + c

f (b x + c) = a {(b x + c)}^{2} + b (b x + c) + c

{\begin{cases} f (a x + b) = a^{3} x^{2} + 2 a^{2} b x + {a b}^{2} + a b x + b^{2} + c \\ f (b x + c) = {a b}^{2} x^{2} + 2 a b c x + {a c}^{2} + b^{2} x + b c + c \end{cases}

{\begin{cases} a^{2} = b^{2} \Rightarrow {\begin{cases} a = b (r e j e c t e d) \\ a = - b \end{cases} \\ 2 a^{2} b + a b = b^{2} + 2 a b c \\ {a b}^{2} + b^{2} + c = {a c}^{2} + b c + c \end{cases}

f o r a = - b \Rightarrow {\begin{cases} - 2 a^{3} - a^{2} = a^{2} - 2 a^{2} c \\ a^{3} + a^{2} + c = a c - a c + c \end{cases}

\Rightarrow a^{2} (a + 1) = 0 \Rightarrow a = - 1, b = 1,

\Rightarrow 2 - 1 = 1 - 2 c; c = 0

∴ f (x) = - x^{2} + x

{\begin{cases} f (b) = f (1) = - 1 + 1 = 0 \\ f (a) = f (- 1) = - 1 - 1 = - 2 \end{cases}

\Rightarrow \frac{1}{2} [f (b) - f (a)] = \frac{1}{2} (0 - (- 2)) = 1

Facebook Tweet Pin