Question Number 68991 by pranay02 last updated on 17/Sep/19
![if the range of f(x, y) = sec^(−1) (x+(1/x))+sec^(−1) (y+(1/y)), xy<0 is (a, b) and (a+b) equals ((λπ)/(10)), then λ is eqal to](https://www.tinkutara.com/question/Q68991.png)
$${if}\:{the}\:{range}\:{of}\:{f}\left({x},\:{y}\right)\:=\:{sec}^{−\mathrm{1}} \left({x}+\frac{\mathrm{1}}{{x}}\right)+{sec}^{−\mathrm{1}} \left({y}+\frac{\mathrm{1}}{{y}}\right),\:{xy}<\mathrm{0}\:{is}\:\left({a},\:{b}\right)\:{and}\:\left({a}+{b}\right)\:{equals}\:\frac{\lambda\pi}{\mathrm{10}},\:{then}\:\lambda\:{is}\:{eqal}\:{to}\: \\ $$
Answered by MJS last updated on 18/Sep/19
![g(t)=sec^(−1) (t+(1/t)) =cos^(−1) ((t/(t^2 +1))) (π/3)≤g(t)≤((2π)/3) the minimum is at ((1),((π/3)) ) the maximum is at (((−1)),(((2π)/3)) ) t<0 ⇒ (π/2)<g(t)≤((3π)/2) t>0 ⇒ (π/3)≤g(t)<(π/2) ⇒ ((5π)/6)<f(x, y)<2π for xy<0 ((5π)/6)+2π=((17π)/6)=((λπ)/(10)) ⇒ λ=((85)/3)](https://www.tinkutara.com/question/Q69026.png)
$${g}\left({t}\right)=\mathrm{sec}^{−\mathrm{1}} \:\left({t}+\frac{\mathrm{1}}{{t}}\right)\:=\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\frac{\pi}{\mathrm{3}}\leqslant{g}\left({t}\right)\leqslant\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{at}\:\begin{pmatrix}{\mathrm{1}}\\{\frac{\pi}{\mathrm{3}}}\end{pmatrix} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{is}\:\mathrm{at}\:\begin{pmatrix}{−\mathrm{1}}\\{\frac{\mathrm{2}\pi}{\mathrm{3}}}\end{pmatrix} \\ $$$${t}<\mathrm{0}\:\Rightarrow\:\frac{\pi}{\mathrm{2}}<{g}\left({t}\right)\leqslant\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${t}>\mathrm{0}\:\Rightarrow\:\frac{\pi}{\mathrm{3}}\leqslant{g}\left({t}\right)<\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{5}\pi}{\mathrm{6}}<{f}\left({x},\:{y}\right)<\mathrm{2}\pi\:\mathrm{for}\:{xy}<\mathrm{0} \\ $$$$\frac{\mathrm{5}\pi}{\mathrm{6}}+\mathrm{2}\pi=\frac{\mathrm{17}\pi}{\mathrm{6}}=\frac{\lambda\pi}{\mathrm{10}}\:\Rightarrow\:\lambda=\frac{\mathrm{85}}{\mathrm{3}} \\ $$