if-f-x-x-2-b-1-x-b-x-2-a-1-x-a-a-b-amp-a-b-R-1-can-take-all-values-except-two-values-amp-such-that-0-then-a-3-b-3-8-ab-

Question Number 198367 by universe last updated on 18/Oct/23

if f (x) = \frac{x^{2} - (b + 1) x + b}{x^{2} - (a + 1) x + a} (a \neq b & a, b \in R \sim {1})

can take all values except two values α & β

such that α + β = 0 then ∣ \frac{a^{3} + b^{3} - 8}{ab} ∣ = ? ?

Commented by Frix last updated on 18/Oct/23

y = \frac{x^{2} - (b + 1) x + b}{x^{2} - (a + 1) x + a} = \frac{(x - 1) (x - b)}{(x - 1) (x - a)} = \frac{x - b}{x - a}

x \neq 1 \land x \neq a

\Leftrightarrow

x = \frac{a y - b}{y - 1} \Rightarrow y \neq 1

{There}^{'} s only 1 value f (x) cannot take .

If you mean values x cannot take, these

are 1, a

Commented by universe last updated on 18/Oct/23

Commented by universe last updated on 18/Oct/23

a n s w e r i s 6

Commented by Frix last updated on 18/Oct/23

This question makes no sense .

Commented by Tinku Tara last updated on 19/Oct/23

for a = b, y = 1 . So there is a case when y = 1

Commented by Frix last updated on 19/Oct/23

But it says a \neq b

Commented by Tinku Tara last updated on 19/Oct/23

Correct, commented without

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