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If-a-b-are-real-numbers-amp-4cos-2-4a-2-9b-2-5-a-3b-then-the-value-of-a-b-will-be-a-7-6-b-5-4-c-11-6-d-17-12-

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Question Number 194552 by BaliramKumar last updated on 09/Jul/23
If a, b are real numbers & 4cos^2 θ = ((4a^2 +9b^2 +5)/(a+3b)), then the value of (a+b) will be: (a) (7/6) (b) (5/4) (c) ((11)/6) (d) ((17)/(12))
$$\mathrm{If}\:{a},\:{b}\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}\:\&\:\mathrm{4cos}^{\mathrm{2}} \theta\:=\:\frac{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} +\mathrm{5}}{{a}+\mathrm{3}{b}},\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left({a}+{b}\right)\:\mathrm{will}\:\mathrm{be}: \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{7}}{\mathrm{6}}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{5}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\frac{\mathrm{11}}{\mathrm{6}}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\frac{\mathrm{17}}{\mathrm{12}} \\ $$
Commented by Frix last updated on 09/Jul/23
Let′s just solve the equation for b b=((2cos^2 θ)/3)±((√(−4a^2 +4acos^2 θ +4cos^4 θ −5))/3) b∈R ⇒ −4a^2 +4acos^2 θ +4cos^4 θ −5≥0 a^2 −acos^2 θ −cos^4 θ +(5/4)≤0 a^2 −acos^2 θ +((cos^4 θ)/4)≤(5/4)(cos^4 θ −1) (a−((cos^2 θ)/2))^2 ≤(5/4)(cos^4 θ −1) But 0≤lhs ∧ −(5/4)≤rhs≤0 ⇒ Only solution when lhs=rhs=0 lhs=0 ⇒ a=((cos^2 θ)/2) rhs=0 ⇒ cos θ =±1 ⇒ a=(1/2)∧b=(2/3) ⇒ a+b=(7/6)
$$\mathrm{Let}'\mathrm{s}\:\mathrm{just}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{for}\:{b} \\ $$$${b}=\frac{\mathrm{2cos}^{\mathrm{2}} \:\theta}{\mathrm{3}}\pm\frac{\sqrt{−\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{a}\mathrm{cos}^{\mathrm{2}} \:\theta\:+\mathrm{4cos}^{\mathrm{4}} \:\theta\:−\mathrm{5}}}{\mathrm{3}} \\ $$$${b}\in\mathbb{R}\:\Rightarrow\:−\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{a}\mathrm{cos}^{\mathrm{2}} \:\theta\:+\mathrm{4cos}^{\mathrm{4}} \:\theta\:−\mathrm{5}\geqslant\mathrm{0} \\ $$$${a}^{\mathrm{2}} −{a}\mathrm{cos}^{\mathrm{2}} \:\theta\:−\mathrm{cos}^{\mathrm{4}} \:\theta\:+\frac{\mathrm{5}}{\mathrm{4}}\leqslant\mathrm{0} \\ $$$${a}^{\mathrm{2}} −{a}\mathrm{cos}^{\mathrm{2}} \:\theta\:+\frac{\mathrm{cos}^{\mathrm{4}} \:\theta}{\mathrm{4}}\leqslant\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{cos}^{\mathrm{4}} \:\theta\:−\mathrm{1}\right) \\ $$$$\left({a}−\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{cos}^{\mathrm{4}} \:\theta\:−\mathrm{1}\right) \\ $$$$\mathrm{But}\:\mathrm{0}\leqslant\mathrm{lhs}\:\wedge\:−\frac{\mathrm{5}}{\mathrm{4}}\leqslant\mathrm{rhs}\leqslant\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{Only}\:\mathrm{solution}\:\mathrm{when}\:\mathrm{lhs}=\mathrm{rhs}=\mathrm{0} \\ $$$$\mathrm{lhs}=\mathrm{0}\:\Rightarrow\:{a}=\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{2}} \\ $$$$\mathrm{rhs}=\mathrm{0}\:\Rightarrow\:\mathrm{cos}\:\theta\:=\pm\mathrm{1} \\ $$$$\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{2}}\wedge{b}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\:{a}+{b}=\frac{\mathrm{7}}{\mathrm{6}} \\ $$
Commented by BaliramKumar last updated on 09/Jul/23
Thanks Sir
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$
Commented by Frix last updated on 09/Jul/23
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Answered by MM42 last updated on 09/Jul/23
0≤((4a^2 +9b^2 +5)/(a+3b))≤4 ⇒4a^2 −4a+9b^2 −12b+5≤0 ⇒4(a−(1/2))^2 +9(b−(2/3))^2 ≤0 ⇒a=(1/2) & b=(2/3) ⇒a+b=(7/6)
$$\mathrm{0}\leqslant\frac{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} +\mathrm{5}}{{a}+\mathrm{3}{b}}\leqslant\mathrm{4} \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{9}{b}^{\mathrm{2}} −\mathrm{12}{b}+\mathrm{5}\leqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\left({a}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{9}\left({b}−\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\&\:\:{b}=\frac{\mathrm{2}}{\mathrm{3}}\:\:\Rightarrow{a}+{b}=\frac{\mathrm{7}}{\mathrm{6}}\: \\ $$
Commented by BaliramKumar last updated on 10/Jul/23
thanks
$$\mathrm{thanks} \\ $$
Answered by witcher3 last updated on 10/Jul/23
5=4+1 1+4a^2 ≥4a 9b^2 +4≥12b ⇒4cos^2 (θ)≥4⇒cos^2 (θ)=1 ⇒a=(1/2) b=(2/3)
$$\mathrm{5}=\mathrm{4}+\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{4a}^{\mathrm{2}} \geqslant\mathrm{4a} \\ $$$$\mathrm{9b}^{\mathrm{2}} +\mathrm{4}\geqslant\mathrm{12b} \\ $$$$\Rightarrow\mathrm{4cos}^{\mathrm{2}} \left(\theta\right)\geqslant\mathrm{4}\Rightarrow\mathrm{cos}^{\mathrm{2}} \left(\theta\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{a}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{b}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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