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

Answered by MATHEMATICSAM last updated on 27/Apr/24

If 0 ≤ θ ≤ (π/4) then (1/( (√2))) ≤ cosθ ≤ 1 Or, (1/2) ≤ cos^2 θ ≤ 1 xcosθ = x^2 + p ⇒ x^2 − xcosθ + p = 0 This equation has real solution for every θ where 0 ≤ θ ≤ (π/4) . So the discriminant of the equation cos^2 θ − 4p ≥ 0 ⇒ 4p ≤ cos^2 θ ≤ 1 ⇒ 4p ≤ 1 ⇒ p ≤ (1/4) Hence ans is option (d).

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

Commented by A5T last updated on 27/Apr/24

θ=30° ∧ p=(1/4) contradicts this. We can also always find a θ such that cos^2 θ<4p when p>(1/8).

$$\theta=\mathrm{30}°\:\wedge\:{p}=\frac{\mathrm{1}}{\mathrm{4}}\:{contradicts}\:{this}. \\ $$$${We}\:{can}\:{also}\:{always}\:{find}\:{a}\:\theta\:{such}\:{that}\:{cos}^{\mathrm{2}} \theta<\mathrm{4}{p} \\ $$$${when}\:{p}>\frac{\mathrm{1}}{\mathrm{8}}. \\ $$

Commented by MATHEMATICSAM last updated on 27/Apr/24

I saw it but is there any way to solve it like this one?

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

Commented by A5T last updated on 27/Apr/24

The problem with this is that from this chain 4p≤^? cos^2 θ≤1; we cannot conclude that 4p≤1 since we have not actually shown when 4p≤cos^2 θ and it is just a condition for the necessary pairs (p,θ). So, 4p must not exceed the minimum bound for cos^2 θ which is (1/2). Otherwise,we can always find θ such that (1/2)<cos^2 θ<4p, which contradicts the condition 4p≤cos^2 θ.

$${The}\:{problem}\:{with}\:{this}\:{is}\:{that}\:{from}\:{this}\:{chain} \\ $$$$\mathrm{4}{p}\overset{?} {\leqslant}{cos}^{\mathrm{2}} \theta\leqslant\mathrm{1};\:{we}\:{cannot}\:{conclude}\:{that}\:\mathrm{4}{p}\leqslant\mathrm{1} \\ $$$${since}\:{we}\:{have}\:{not}\:{actually}\:{shown}\:{when}\: \\ $$$$\mathrm{4}{p}\leqslant{cos}^{\mathrm{2}} \theta\:{and}\:{it}\:{is}\:{just}\:{a}\:{condition}\:{for}\:{the} \\ $$$${necessary}\:{pairs}\:\left({p},\theta\right). \\ $$$$ \\ $$$${So},\:\mathrm{4}{p}\:{must}\:{not}\:{exceed}\:{the}\:{minimum}\:{bound}\:{for} \\ $$$${cos}^{\mathrm{2}} \theta\:{which}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}}.\:{Otherwise},{we}\:{can}\:{always} \\ $$$${find}\:\theta\:{such}\:{that}\:\frac{\mathrm{1}}{\mathrm{2}}<{cos}^{\mathrm{2}} \theta<\mathrm{4}{p},\:{which}\:{contradicts} \\ $$$${the}\:{condition}\:\mathrm{4}{p}\leqslant{cos}^{\mathrm{2}} \theta. \\ $$

Commented by MATHEMATICSAM last updated on 27/Apr/24

Hmm this is actually right.

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

Answered by A5T last updated on 27/Apr/24

x^2 −xcosθ+p=0 Discriminant:cos^2 θ−4p≥0⇒4p≤^? cos^2 θ≤1 (1/2)≤cos^2 θ≤1 If p>(1/8);then 4p>(1/2); which means ∃ θ such that (1/2)<cos^2 θ<4p which makes it have non-real solutions. So,p≤(1/8)⇒4p≤(1/2)≤cos^2 θ⇒4p≤cos^2 θ

$${x}^{\mathrm{2}} −{xcos}\theta+{p}=\mathrm{0} \\ $$$${Discriminant}:{cos}^{\mathrm{2}} \theta−\mathrm{4}{p}\geqslant\mathrm{0}\Rightarrow\mathrm{4}{p}\overset{?} {\leqslant}{cos}^{\mathrm{2}} \theta\leqslant\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\leqslant{cos}^{\mathrm{2}} \theta\leqslant\mathrm{1} \\ $$$${If}\:{p}>\frac{\mathrm{1}}{\mathrm{8}};{then}\:\mathrm{4}{p}>\frac{\mathrm{1}}{\mathrm{2}};\:{which}\:{means}\:\exists\:\theta\:{such} \\ $$$${that}\:\frac{\mathrm{1}}{\mathrm{2}}<{cos}^{\mathrm{2}} \theta<\mathrm{4}{p}\:{which}\:{makes}\:{it}\:{have}\:{non}-{real} \\ $$$${solutions}.\:{So},{p}\leqslant\frac{\mathrm{1}}{\mathrm{8}}\Rightarrow\mathrm{4}{p}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\leqslant{cos}^{\mathrm{2}} \theta\Rightarrow\mathrm{4}{p}\leqslant{cos}^{\mathrm{2}} \theta \\ $$

Commented by BaliramKumar last updated on 27/Apr/24

i think answer will be (b)

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

Commented by A5T last updated on 27/Apr/24

How?

$${How}? \\ $$

Commented by BaliramKumar last updated on 27/Apr/24

θ = 30° & p = (1/4) then roots are not real

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

Commented by A5T last updated on 27/Apr/24

Yea.

$${Yea}. \\ $$

Answered by Frix last updated on 27/Apr/24

x^2 −xcos θ +p=0 x=((cos θ)/2)±((√(cos^2 θ −4p))/2) x∈R ⇔ cos^2 θ −4p≥0 ⇒ p≤min ((cos^2 θ)/4) 0≤θ≤(π/4) ⇒ (1/8)≤θ≤(1/4) ⇒ min ((cos^2 θ)/4) =(1/8) ⇒ p≤(1/8)

$${x}^{\mathrm{2}} −{x}\mathrm{cos}\:\theta\:+{p}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{cos}\:\theta}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{cos}^{\mathrm{2}} \:\theta\:−\mathrm{4}{p}}}{\mathrm{2}} \\ $$$${x}\in\mathbb{R}\:\Leftrightarrow\:\mathrm{cos}^{\mathrm{2}} \:\theta\:−\mathrm{4}{p}\geqslant\mathrm{0}\:\Rightarrow\:{p}\leqslant\mathrm{min}\:\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{4}} \\ $$$$\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{4}}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{8}}\leqslant\theta\leqslant\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\:\mathrm{min}\:\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\Rightarrow \\ $$$${p}\leqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$

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