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

$$Findtherangeofrealnumber$$$$ofqsuchthatthefunction$$$$f\left(x\right)=\cos x(q\sin {}^{2}x-5)have$$$$minimumvalueis-5.$$

Commented by MJS_new last updated on 18/May/21

$$\mathrm{I}\mathrm{get}-\frac{5}{2}⩽q⩽20\mathrm{but}{\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{got}\mathrm{no}\mathrm{time}\mathrm{to}\mathrm{type}$$$$\mathrm{the}\mathrm{path}\mathrm{right}\mathrm{now}$$

Commented by bramlexs22 last updated on 18/May/21

$$howsir?$$

Answered by MJS_new last updated on 18/May/21

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

Commented by bramlexs22 last updated on 19/May/21

$$f{}^{\prime }\left(x\right)=-\sin x(q-5-q\cos {}^{2}x)+$$$$\cos x\left(2q\cos x\sin x\right)=0$$$$\Rightarrow \sin x[q-5-q\cos {}^{2}x-2q\cos {}^{2}x]=0$$$$$$

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