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Question Number 144833 by gsk2684 last updated on 29/Jun/21

$$\mathrm{find}\mathrm{the}\mathrm{least}\mathrm{value}\mathrm{of}\alpha \mathrm{such}\mathrm{that}\frac{4}{\sin \mathrm{x}}+\frac{1}{1-\sin \mathrm{x}}=\alpha$$$$\mathrm{has}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{solution}\mathrm{in}\left(0\frac{\mathrm{\Pi }}{2}\right)$$

Answered by liberty last updated on 29/Jun/21

$$\alpha =\mathrm{f}\left(\mathrm{x}\right)=\frac{4}{\sin \mathrm{x}}+\frac{1}{1-\sin \mathrm{x}}$$$$\mathrm{let}\mathrm{u}=\sin \mathrm{x}\Rightarrow \mathrm{f}\left(\mathrm{u}\right)={4\mathrm{u}}^{-1}+{(1-\mathrm{u})}^{-1}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{u}\right)=-\frac{4}{{\mathrm{u}}^2}+\frac{1}{{(1-\mathrm{u})}^2}=0$$$$\Rightarrow {\mathrm{u}}^2-4({\mathrm{u}}^2-2\mathrm{u}+1)=0$$$$\Rightarrow -{3\mathrm{u}}^2+8\mathrm{u}-4=0$$$$\Rightarrow {3\mathrm{u}}^2-8\mathrm{u}+4=0$$$$\Rightarrow (3\mathrm{u}-2)(\mathrm{u}-2)=0$$$$\Rightarrow \mathrm{u}=\frac{2}{3};\mathrm{minimum}\mathrm{value}$$$$\mathrm{of}\alpha =\frac{4}{\sin \mathrm{x}}+\frac{1}{1-\sin \mathrm{x}}=6+3=9$$

Commented by gsk2684 last updated on 29/Jun/21

$$\mathrm{thank}\mathrm{you}$$$$\mathrm{kindly}\mathrm{explain}\mathrm{why}\mathrm{did}\mathrm{not}$$$$\mathrm{find}\mathrm{second}\mathrm{derivative}\mathrm{for}\mathrm{verifying}$$$$\mathrm{whether}\mathrm{it}\mathrm{is}\mathrm{positive}\mathrm{or}\mathrm{negative}?$$

Commented by liberty last updated on 30/Jun/21

$$\mathrm{i}\mathrm{used}\mathrm{first}\mathrm{test}\mathrm{derivative}$$

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