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Question Number 66115 by Rio Michael last updated on 09/Aug/19

$$find\mid z\mid wherez=\frac{{(1+i\sqrt{3})}^3}{{(1-i)}^3}$$$$findthemaximumvalueof12sinx-5cosx$$

Commented by mathmax by abdo last updated on 09/Aug/19

$$\mid z\mid =\frac{\mid 1+i\sqrt{3}{\mid }^3}{\mid 1-i{\mid }^3}and\mid 1+i\sqrt{3}\mid =\sqrt{1+3}=2$$$$\mid 1-i\mid =\sqrt{2}\Rightarrow \mid z\mid =\frac{2^3}{{\left(\sqrt{2}\right)}^3}=\frac{2^3}{2\sqrt{2}}=\frac{4}{\sqrt{2}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}$$

Commented by mathmax by abdo last updated on 09/Aug/19

$$letf\left(x\right)=12sinx-5cosxfis2\pi periodicf\left(x\right)=0\iff x=arctan\left(\frac{5}{12}\right)$$$$f^{{}^{\prime }}\left(x\right)=12cosx+5sinx$$$$f^{{}^{\prime }}\left(x\right)=0\iff 12cosx+5sinx=0\Rightarrow 5sinx=-12cosx\Rightarrow$$$$tanx=-\frac{12}{5}\Rightarrow x=-arctan\left(\frac{12}{5}\right)varationon[-\pi ,\pi ]$$$$x-\pi -arctan\left(\frac{12}{5}\right)0arctan\left(\frac{5}{12}\right)\pi$$$$f\left(x\right)5decr{\alpha }_0?-5incr0inc5$$$$f(-arctan\left(\frac{12}{5}\right))=-12sin\left(arctan\left(\frac{12}{5}\right)\right)-5cos\left(arctan\left(\frac{12}{5}\right)\right)$$$$=-12\frac{\frac{12}{5}}{\sqrt{1+{\left(\frac{12}{5}\right)}^2}}-5\times \frac{1}{\sqrt{1+{\left(\frac{12}{5}\right)}^2}}={\alpha }_0$$$$f\left(arctan\left(\frac{5}{12}\right)\right)=12sin\left(arctan\left(\frac{5}{12}\right)\right)-5cos\left(arctan\left(\frac{5}{12}\right)\right)$$$$=12\times \frac{\frac{5}{12}}{\sqrt{1+{\left(\frac{5}{12}\right)}^2}}-5\times \frac{1}{\sqrt{1+{\left(\frac{5}{12}\right)}^2}}=0$$$$\Rightarrow max{}_{x\in R}f\left(x\right)=5$$

Commented by Rio Michael last updated on 09/Aug/19

$$thankyousir$$

Commented by Prithwish sen last updated on 09/Aug/19

$$\mathrm{f}\left(\mathrm{x}\right)=12\mathrm{sinx}-5\mathrm{cosx}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=12\mathrm{cosx}+5\mathrm{sinx}\mathrm{and}$$$${\mathrm{f}}^{{}^{″}}\left(\mathrm{x}\right)=-12\mathrm{sinx}+5\mathrm{cosx}$$$$\mathrm{for}\max .\mathrm{and}\min .\mathrm{value}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=0\Rightarrow \mathrm{tanx}=-\frac{12}{13}$$$$\mathrm{i}.\mathrm{e}\mathrm{sinx}=\frac{12}{13}\mathrm{and}\mathrm{cosx}=-\frac{5}{13}.....\left(\mathrm{i}\right)$$$$\mathrm{or}\mathrm{sinx}=-\frac{12}{13}\mathrm{and}\mathrm{cosx}=\frac{5}{13}......\left(\mathrm{ii}\right)$$$$\mathrm{for}\left(\mathrm{i}\right){\mathrm{f}}^{″}\left(\mathrm{x}\right)=-13<0\mathrm{i}.\mathrm{e}\max .\mathrm{f}\left(\mathrm{x}\right)$$$$\mathrm{for}\left(\mathrm{ii}\right){\mathrm{f}}^{″}\left(\mathrm{x}\right)=13>0\mathrm{i}.\mathrm{e}\min .\mathrm{f}\left(\mathrm{x}\right)$$$$\therefore \max .\mathrm{f}\left(\mathrm{x}\right)=12.\frac{12}{13}+5.\frac{5}{13}=13$$

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