Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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!

$$\mathscr{T}\mathrm{hank}\:\mathrm{you}\:\mathrm{sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com