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Question Number 16179 by Tinkutara last updated on 24/Jun/17

If a > 0, b > 0 and the minimum value of a sin^2 θ + b cosec^2 θ is equal to maximum value of a sin^2 θ + b cos^2 θ, then (a/b) is equal to [Answer: 4]

$$\mathrm{If}\:{a}\:>\:\mathrm{0},\:{b}\:>\:\mathrm{0}\:\mathrm{and}\:\mathrm{the}\:\mathrm{minimum} \\ $$ $$\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:{b}\:\mathrm{cosec}^{\mathrm{2}} \:\theta\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ $$\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:{b}\:\mathrm{cos}^{\mathrm{2}} \:\theta, \\ $$ $$\mathrm{then}\:\frac{{a}}{{b}}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\left[\boldsymbol{\mathrm{Answer}}:\:\mathrm{4}\right] \\ $$

Answered by ajfour last updated on 24/Jun/17

let sin^2 θ=t f(t)=at+(b/t) is minimum when f ′(t)=a−(b/t^2 )=0 ⇒ t_0 =(√(b/a)) f(t_0 )=2a((√(b/a)) )=2(√(ab)) g(t)=at+b(1−t) = (a−b)t+b here i need to know if a>b or not if a>b maximum of g(t) is = (a−b)(1)+b =a since min. of f(t)=max. of g(t) 2a((√(b/a)) ) = a or (a/b) = 4 (a>b) but if a<b then max. of g(t) = (a−b)(0)+b =b (remember t=sin^2 θ) then from given condition 2a((√(b/a)) )=b or (√(b/a)) =2 ⇒ (a/b) = (1/4) (a<b) .

$$\mathrm{let}\:\mathrm{sin}\:^{\mathrm{2}} \theta=\mathrm{t} \\ $$ $$\:\mathrm{f}\left(\mathrm{t}\right)=\mathrm{at}+\frac{\mathrm{b}}{\mathrm{t}}\:\:\:\mathrm{is}\:\mathrm{minimum}\:\mathrm{when} \\ $$ $$\:\mathrm{f}\:'\left(\mathrm{t}\right)=\mathrm{a}−\frac{\mathrm{b}}{\mathrm{t}^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:\mathrm{t}_{\mathrm{0}} =\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\: \\ $$ $$\mathrm{f}\left(\mathrm{t}_{\mathrm{0}} \right)=\mathrm{2a}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\:\right)=\mathrm{2}\sqrt{\mathrm{ab}}\: \\ $$ $$\:\mathrm{g}\left(\mathrm{t}\right)=\mathrm{at}+\mathrm{b}\left(\mathrm{1}−\mathrm{t}\right)\:=\:\left(\mathrm{a}−\mathrm{b}\right)\mathrm{t}+\mathrm{b} \\ $$ $$\mathrm{here}\:\mathrm{i}\:\mathrm{need}\:\mathrm{to}\:\mathrm{know}\:\mathrm{if}\:\mathrm{a}>\mathrm{b}\:\mathrm{or}\:\mathrm{not} \\ $$ $$\mathrm{if}\:\mathrm{a}>\mathrm{b}\:\:\:\mathrm{maximum}\:\mathrm{of}\:\mathrm{g}\left(\mathrm{t}\right)\:\mathrm{is} \\ $$ $$\:\:\:=\:\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{1}\right)+\mathrm{b}\:=\mathrm{a} \\ $$ $$\mathrm{since}\:\mathrm{min}.\:\mathrm{of}\:\mathrm{f}\left(\mathrm{t}\right)=\mathrm{max}.\:\mathrm{of}\:\mathrm{g}\left(\mathrm{t}\right) \\ $$ $$\:\:\:\mathrm{2a}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\:\right)\:=\:\mathrm{a} \\ $$ $$\mathrm{or}\:\:\:\frac{\mathrm{a}}{\mathrm{b}}\:=\:\mathrm{4}\:\:\:\:\:\:\:\:\:\left(\mathrm{a}>\mathrm{b}\right) \\ $$ $$\mathrm{but}\:\mathrm{if}\:\mathrm{a}<\mathrm{b}\:\mathrm{then} \\ $$ $$\:\mathrm{max}.\:\mathrm{of}\:\mathrm{g}\left(\mathrm{t}\right)\:=\:\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{0}\right)+\mathrm{b}\:=\mathrm{b} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{remember}\:\mathrm{t}=\mathrm{sin}\:^{\mathrm{2}} \theta\right) \\ $$ $$\mathrm{then}\:\:\mathrm{from}\:\mathrm{given}\:\mathrm{condition} \\ $$ $$\mathrm{2a}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\:\right)=\mathrm{b} \\ $$ $$\mathrm{or}\:\:\:\:\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\:=\mathrm{2}\:\:\Rightarrow\:\:\:\frac{\mathrm{a}}{\mathrm{b}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\left(\mathrm{a}<\mathrm{b}\right)\:. \\ $$

Commented byTinkutara last updated on 24/Jun/17

Thanks Sir! I also got confused with 4 and (1/4) .

$$\mathrm{Thanks}\:\mathrm{Sir}!\:\mathrm{I}\:\mathrm{also}\:\mathrm{got}\:\mathrm{confused}\:\mathrm{with} \\ $$ $$\mathrm{4}\:\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$

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