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Question Number 141412 by bramlexs22 last updated on 18/May/21

Find the range of real number of q such that the function f(x) = cos x(q sin^2 x−5) have minimum value is −5 .

$$\:{Find}\:{the}\:{range}\:{of}\:{real}\:{number} \\ $$$${of}\:{q}\:{such}\:{that}\:{the}\:{function}\: \\ $$$$\:{f}\left({x}\right)\:=\:\mathrm{cos}\:{x}\left({q}\:\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{5}\right)\:{have} \\ $$$${minimum}\:{value}\:{is}\:−\mathrm{5}\:. \\ $$

Commented by MJS_new last updated on 18/May/21

I get −(5/2)≤q≤20 but I′ve got no time to type the path right now

$$\mathrm{I}\:\mathrm{get}\:−\frac{\mathrm{5}}{\mathrm{2}}\leqslant{q}\leqslant\mathrm{20}\:\mathrm{but}\:\mathrm{I}'\mathrm{ve}\:\mathrm{got}\:\mathrm{no}\:\mathrm{time}\:\mathrm{to}\:\mathrm{type} \\ $$$$\mathrm{the}\:\mathrm{path}\:\mathrm{right}\:\mathrm{now} \\ $$

Commented by bramlexs22 last updated on 18/May/21

how sir ?

$${how}\:{sir}\:? \\ $$

Answered by MJS_new last updated on 18/May/21

f(x)=(q−5−qcos^2 x)cos x f ′(x)=(5−q+3qcos^2 x)sin x =0 sin x =0 ⇒ x=nπ f(nπ)=±5 5−q−qcos^2 x =0 ⇒ cos^2 x =((q−5)/(3q)) ⇒ ⇒ x=±arccos ((√(q−5))/( (√(3q)))) f(±arccos ((√(q−5))/( (√(3q)))))=±((2(√(3(q−5)^3 )))/(9(√q))) −((2(√(3(q−5)^3 )))/(9(√q)))=−5 ⇒ q=20 +((2(√(3(q−5)^3 )))/(9(√q)))=−5 ⇒ q=−(5/2) this might be wrong but it′s what I did

$${f}\left({x}\right)=\left({q}−\mathrm{5}−{q}\mathrm{cos}^{\mathrm{2}} \:{x}\right)\mathrm{cos}\:{x} \\ $$$${f}\:'\left({x}\right)=\left(\mathrm{5}−{q}+\mathrm{3}{q}\mathrm{cos}^{\mathrm{2}} \:{x}\right)\mathrm{sin}\:{x}\:=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{sin}\:{x}\:=\mathrm{0}\:\Rightarrow\:{x}={n}\pi \\ $$$${f}\left({n}\pi\right)=\pm\mathrm{5} \\ $$$$ \\ $$$$\mathrm{5}−{q}−{q}\mathrm{cos}^{\mathrm{2}} \:{x}\:=\mathrm{0}\:\Rightarrow\:\mathrm{cos}^{\mathrm{2}} \:{x}\:=\frac{{q}−\mathrm{5}}{\mathrm{3}{q}}\:\Rightarrow \\ $$$$\Rightarrow\:{x}=\pm\mathrm{arccos}\:\frac{\sqrt{{q}−\mathrm{5}}}{\:\sqrt{\mathrm{3}{q}}} \\ $$$${f}\left(\pm\mathrm{arccos}\:\frac{\sqrt{{q}−\mathrm{5}}}{\:\sqrt{\mathrm{3}{q}}}\right)=\pm\frac{\mathrm{2}\sqrt{\mathrm{3}\left({q}−\mathrm{5}\right)^{\mathrm{3}} }}{\mathrm{9}\sqrt{{q}}} \\ $$$$−\frac{\mathrm{2}\sqrt{\mathrm{3}\left({q}−\mathrm{5}\right)^{\mathrm{3}} }}{\mathrm{9}\sqrt{{q}}}=−\mathrm{5}\:\Rightarrow\:{q}=\mathrm{20} \\ $$$$+\frac{\mathrm{2}\sqrt{\mathrm{3}\left({q}−\mathrm{5}\right)^{\mathrm{3}} }}{\mathrm{9}\sqrt{{q}}}=−\mathrm{5}\:\Rightarrow\:{q}=−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{this}\:\mathrm{might}\:\mathrm{be}\:\mathrm{wrong}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{what}\:\mathrm{I}\:\mathrm{did} \\ $$

Commented by bramlexs22 last updated on 19/May/21

f ′(x)=−sin x(q−5−q cos^2 x)+ cos x(2q cos x sin x)=0 ⇒sin x [ q−5−qcos^2 x−2q cos^2 x ]=0

$${f}\:'\left({x}\right)=−\mathrm{sin}\:{x}\left({q}−\mathrm{5}−{q}\:\mathrm{cos}\:^{\mathrm{2}} {x}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:{x}\left(\mathrm{2}{q}\:\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:{x}\:\left[\:{q}−\mathrm{5}−{q}\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{2}{q}\:\mathrm{cos}\:^{\mathrm{2}} {x}\:\right]=\mathrm{0} \\ $$$$ \\ $$

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