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If-the-range-of-the-function-f-x-x-1-p-x-2-1-does-not-contain-any-values-belonging-to-the-interval-1-1-3-then-true-set-of-values-of-p-is-

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Question Number 33997 by rahul 19 last updated on 29/Apr/18
If the range of the function f(x) = ((x−1)/(p−x^2 +1)) does not contain any values belonging to the interval [−1,((−1)/3)] then true set of values of p is ?
$$\boldsymbol{{I}}\mathrm{f}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\: \\ $$$$\mathrm{f}\left({x}\right)\:=\:\frac{{x}−\mathrm{1}}{\mathrm{p}−{x}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{does}\:\mathrm{not}\:\mathrm{contain}\:\mathrm{any} \\ $$$$\mathrm{values}\:\mathrm{belonging}\:\mathrm{to}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\left[−\mathrm{1},\frac{−\mathrm{1}}{\mathrm{3}}\right]\:{then}\:{true}\:{set}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:\mathrm{p}\:\mathrm{is}\:? \\ $$
Answered by ajfour last updated on 29/Apr/18
for x^2 > p+1 ((x−1)/(p+1−x^2 )) < −1 ⇒ x^2 −p−1 < x−1 x^2 −x−p < 0 ⇒ no value of p. if ((x−1)/(p+1−x^2 )) > −(1/3) x^2 −p−1 > 3x−3 x^2 −3x−p+2 > 0 (for all x) ⇒ 9+4p−8 < 0 ⇒ p < − (1/4) if x^2 < p+1 then x^2 −x+p > 0 ⇒ 1−4p < 0 ⇒ p > (1/4) similarly x^2 −3x−p+2 < 0 no value. hence possible only for p ∈ (−∞, −(1/4)) .
$${for}\:\:{x}^{\mathrm{2}} \:>\:{p}+\mathrm{1} \\ $$$$\:\:\:\frac{{x}−\mathrm{1}}{{p}+\mathrm{1}−{x}^{\mathrm{2}} }\:\:<\:−\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\:{x}^{\mathrm{2}} −{p}−\mathrm{1}\:<\:{x}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −{x}−{p}\:\:<\:\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:{no}\:{value}\:{of}\:{p}. \\ $$$$\:\:\:\:\:{if}\:\:\:\:\:\frac{{x}−\mathrm{1}}{{p}+\mathrm{1}−{x}^{\mathrm{2}} }\:>\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −{p}−\mathrm{1}\:>\:\mathrm{3}{x}−\mathrm{3} \\ $$$$\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{3}{x}−{p}+\mathrm{2}\:>\:\mathrm{0}\:\:\:\left({for}\:{all}\:{x}\right) \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{9}+\mathrm{4}{p}−\mathrm{8}\:<\:\mathrm{0} \\ $$$$\:\:\:\:\Rightarrow\:{p}\:<\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\: \\ $$$${if}\:\:\:{x}^{\mathrm{2}} \:<\:{p}+\mathrm{1} \\ $$$$\:\:\:\:\:\:{then}\:\:{x}^{\mathrm{2}} −{x}+{p}\:\:>\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{1}−\mathrm{4}{p}\:<\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:{p}\:>\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\:{similarly} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{3}{x}−{p}+\mathrm{2}\:<\:\mathrm{0} \\ $$$$\:\:\:\:{no}\:{value}. \\ $$$$\boldsymbol{{hence}}\:\boldsymbol{{possible}}\:\boldsymbol{{only}}\:\boldsymbol{{for}} \\ $$$$\:\:\:\:\:\:\boldsymbol{{p}}\:\in\:\left(−\infty,\:−\frac{\mathrm{1}}{\mathrm{4}}\right)\:. \\ $$
Commented by rahul 19 last updated on 29/Apr/18
Thank you sir!
$$\mathscr{T}\mathrm{hank}\:\mathrm{you}\:\mathrm{sir}! \\ $$

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