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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-

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Question Number 198367 by universe last updated on 18/Oct/23
if f(x) = ((x^2 −(b+1)x+b)/(x^2 −(a+1)x+a)) (a≠b & a,b ∈ R ∼ {1}) can take all values except two values α & β such that α+β = 0 then ∣((a^3 +b^3 −8)/(ab))∣ = ??
$$\:\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{{x}^{\mathrm{2}} −\left({b}+\mathrm{1}\right){x}+{b}}{{x}^{\mathrm{2}} −\left({a}+\mathrm{1}\right){x}+{a}}\:\:\left({a}\neq{b}\:\&\:{a},{b}\:\in\:\mathbb{R}\:\sim\:\left\{\mathrm{1}\right\}\right) \\ $$$$\:\:\mathrm{can}\:\mathrm{take}\:\mathrm{all}\:\mathrm{values}\:\mathrm{except}\:\mathrm{two}\:\mathrm{values}\:\alpha\:\&\:\beta \\ $$$$\:\:\mathrm{such}\:\mathrm{that}\:\alpha+\beta\:=\:\mathrm{0}\:\:\mathrm{then}\:\mid\frac{\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} −\mathrm{8}}{\mathrm{ab}}\mid\:\:=\:\:?? \\ $$
Commented by Frix last updated on 18/Oct/23
y=((x^2 −(b+1)x+b)/(x^2 −(a+1)x+a))=(((x−1)(x−b))/((x−1)(x−a)))=((x−b)/(x−a)) x≠1∧x≠a ⇔ x=((ay−b)/(y−1)) ⇒ y≠1 There′s only 1 value f(x) cannot take. If you mean values x cannot take, these are 1, a
$${y}=\frac{{x}^{\mathrm{2}} −\left({b}+\mathrm{1}\right){x}+{b}}{{x}^{\mathrm{2}} −\left({a}+\mathrm{1}\right){x}+{a}}=\frac{\left({x}−\mathrm{1}\right)\left({x}−{b}\right)}{\left({x}−\mathrm{1}\right)\left({x}−{a}\right)}=\frac{{x}−{b}}{{x}−{a}} \\ $$$${x}\neq\mathrm{1}\wedge{x}\neq{a} \\ $$$$\Leftrightarrow \\ $$$${x}=\frac{{ay}−{b}}{{y}−\mathrm{1}}\:\Rightarrow\:{y}\neq\mathrm{1} \\ $$$$\mathrm{There}'\mathrm{s}\:\mathrm{only}\:\mathrm{1}\:\mathrm{value}\:{f}\left({x}\right)\:\mathrm{cannot}\:\mathrm{take}. \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{values}\:{x}\:\mathrm{cannot}\:\mathrm{take},\:\mathrm{these} \\ $$$$\mathrm{are}\:\mathrm{1},\:{a} \\ $$
Commented by universe last updated on 18/Oct/23
Commented by universe last updated on 18/Oct/23
answer is 6
$${answer}\:{is}\:\mathrm{6} \\ $$
Commented by Frix last updated on 18/Oct/23
This question makes no sense.
$$\mathrm{This}\:\mathrm{question}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{sense}. \\ $$
Commented by Tinku Tara last updated on 19/Oct/23
for a=b, y=1. So there is a case when y=1
$$\mathrm{for}\:\mathrm{a}=\mathrm{b},\:{y}=\mathrm{1}.\:\mathrm{So}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{case}\:\mathrm{when}\:\mathrm{y}=\mathrm{1} \\ $$
Commented by Frix last updated on 19/Oct/23
But it says a≠b
$$\mathrm{But}\:\mathrm{it}\:\mathrm{says}\:{a}\neq{b} \\ $$
Commented by Tinku Tara last updated on 19/Oct/23
Correct, commented without readimg question
$$\mathrm{Correct},\:\mathrm{commented}\:\mathrm{without} \\ $$$$\mathrm{readimg}\:\mathrm{question} \\ $$

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