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Question Number 206833 by BaliramKumar last updated on 27/Apr/24

Answered by MATHEMATICSAM last updated on 27/Apr/24

$$\mathrm{If}0\le \theta \le \frac{\pi }{4}\mathrm{then}\frac{1}{\sqrt{2}}\le \cos \theta \le 1$$$$\mathrm{Or},\frac{1}{2}\le {\cos }^2\theta \le 1$$$$x\cos \theta =x^2+p$$$$\Rightarrow x^2-x\cos \theta +p=0$$$$\mathrm{This}\mathrm{equation}\mathrm{has}\mathrm{real}\mathrm{solution}\mathrm{for}$$$$\mathrm{every}\theta \mathrm{where}0\le \theta \le \frac{\pi }{4}.$$$$\mathrm{So}\mathrm{the}\mathrm{discriminant}\mathrm{of}\mathrm{the}\mathrm{equation}$$$${\cos }^2\theta -4p\ge 0$$$$\Rightarrow 4p\le {\cos }^2\theta \le 1$$$$\Rightarrow 4p\le 1$$$$\Rightarrow p\le \frac{1}{4}$$$$\mathrm{Hence}\mathrm{ans}\mathrm{is}\mathrm{option}\left(\mathrm{d}\right).$$

Commented by A5T last updated on 27/Apr/24

$$\theta =30°\wedge p=\frac{1}{4}contradictsthis.$$$$Wecanalsoalwaysfinda\theta suchthat{cos}^2\theta <4p$$$$whenp>\frac{1}{8}.$$

Commented by MATHEMATICSAM last updated on 27/Apr/24

$$\mathrm{I}\mathrm{saw}\mathrm{it}\mathrm{but}\mathrm{is}\mathrm{there}\mathrm{any}\mathrm{way}\mathrm{to}\mathrm{solve}\mathrm{it}$$$$\mathrm{like}\mathrm{this}\mathrm{one}?$$

Commented by A5T last updated on 27/Apr/24

$$Theproblemwiththisisthatfromthischain$$$$4p\stackrel{?}{⩽}{cos}^2\theta ⩽1;wecannotconcludethat4p⩽1$$$$sincewehavenotactuallyshownwhen$$$$4p⩽{cos}^2\theta anditisjustaconditionforthe$$$$necessarypairs(p,\theta ).$$$$$$$$So,4pmustnotexceedtheminimumboundfor$$$${cos}^2\theta whichis\frac{1}{2}.Otherwise,wecanalways$$$$find\theta suchthat\frac{1}{2}<{cos}^2\theta <4p,whichcontradicts$$$$thecondition4p⩽{cos}^2\theta .$$

Commented by MATHEMATICSAM last updated on 27/Apr/24

$$\mathrm{Hmm}\mathrm{this}\mathrm{is}\mathrm{actually}\mathrm{right}.$$

Answered by A5T last updated on 27/Apr/24

$$x^2-xcos\theta +p=0$$$$Discriminant:{cos}^2\theta -4p⩾0\Rightarrow 4p\stackrel{?}{⩽}{cos}^2\theta ⩽1$$$$\frac{1}{2}⩽{cos}^2\theta ⩽1$$$$Ifp>\frac{1}{8};then4p>\frac{1}{2};whichmeans\mathrm{\exists }\theta such$$$$that\frac{1}{2}<{cos}^2\theta <4pwhichmakesithavenon-real$$$$solutions.So,p⩽\frac{1}{8}\Rightarrow 4p⩽\frac{1}{2}⩽{cos}^2\theta \Rightarrow 4p⩽{cos}^2\theta$$

Commented by BaliramKumar last updated on 27/Apr/24

$$\mathrm{i}\mathrm{think}\mathrm{answer}\mathrm{will}\mathrm{be}\left(\mathrm{b}\right)$$

Commented by A5T last updated on 27/Apr/24

$$How?$$

Commented by BaliramKumar last updated on 27/Apr/24

$$\theta =30°\mathrm{\&}\mathrm{p}=\frac{1}{4}\mathrm{then}\mathrm{roots}\mathrm{are}\mathrm{not}\mathrm{real}$$

Commented by A5T last updated on 27/Apr/24

$$Yea.$$

Answered by Frix last updated on 27/Apr/24

$$x^2-x\cos \theta +p=0$$$$x=\frac{\cos \theta }{2}\pm \frac{\sqrt{{\cos }^2\theta -4p}}{2}$$$$x\in \mathbb{R}\iff {\cos }^2\theta -4p⩾0\Rightarrow p⩽\min \frac{{\cos }^2\theta }{4}$$$$0⩽\theta ⩽\frac{\pi }{4}\Rightarrow \frac{1}{8}⩽\theta ⩽\frac{1}{4}\Rightarrow \min \frac{{\cos }^2\theta }{4}=\frac{1}{8}$$$$\Rightarrow$$$$p⩽\frac{1}{8}$$

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