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Question Number 68991 by pranay02 last updated on 17/Sep/19

$$iftherangeoff(x,y)={sec}^{-1}(x+\frac{1}{x})+{sec}^{-1}(y+\frac{1}{y}),xy<0is(a,b)and(a+b)equals\frac{\lambda \pi }{10},then\lambda iseqalto$$

Answered by MJS last updated on 18/Sep/19

$$g\left(t\right)={\sec }^{-1}(t+\frac{1}{t})={\cos }^{-1}\left(\frac{t}{t^2+1}\right)$$$$\frac{\pi }{3}⩽g\left(t\right)⩽\frac{2\pi }{3}$$$$\mathrm{the}\mathrm{minimum}\mathrm{is}\mathrm{at}\left(\begin{array}{c}1\\ \frac{\pi }{3}\end{array}\right)$$$$\mathrm{the}\mathrm{maximum}\mathrm{is}\mathrm{at}\left(\begin{array}{c}-1\\ \frac{2\pi }{3}\end{array}\right)$$$$t<0\Rightarrow \frac{\pi }{2}<g\left(t\right)⩽\frac{3\pi }{2}$$$$t>0\Rightarrow \frac{\pi }{3}⩽g\left(t\right)<\frac{\pi }{2}$$$$\Rightarrow \frac{5\pi }{6}<f(x,y)<2\pi \mathrm{for}xy<0$$$$\frac{5\pi }{6}+2\pi =\frac{17\pi }{6}=\frac{\lambda \pi }{10}\Rightarrow \lambda =\frac{85}{3}$$

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