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Question Number 115489 by ruwedkabeh last updated on 26/Sep/20

$$findrangefor$$$$1.f\left(x\right)=\frac{\sqrt{x^2-9}}{x-1}$$$$2.f\left(x\right)=\ln \left(\sqrt{4-9x^2}\right)$$

Answered by PRITHWISH SEN 2 last updated on 26/Sep/20

$$1.\mathrm{Let}\mathrm{y}=\frac{\sqrt{{\mathrm{x}}^2-9}}{\mathrm{x}-1}$$$${\mathrm{x}}^2({\mathrm{y}}^2-1)-{2\mathrm{xy}}^2+({\mathrm{y}}^2+9)=0$$$$\mathrm{for}\mathrm{real}\mathrm{x}$$$$△={\left({2\mathrm{y}}^2\right)}^2-4({\mathrm{y}}^2-1)({\mathrm{y}}^2+9)⩾0$$$$\Rightarrow (\mathrm{y}-\frac{3}{2\sqrt{2}})(\mathrm{y}+\frac{3}{2\sqrt{2}})⩽0$$$$\mathbf{Range}\mathbf{f}\left(\mathbf{x}\right)\in [-\frac{3}{2\sqrt{2}},\frac{3}{2\sqrt{2}}]$$$$2.\mathrm{Again}\mathrm{let}$$$$\mathrm{y}=\ln \left(\sqrt{4-{9\mathrm{x}}^2}\right)$$$$\Rightarrow {\mathrm{e}}^{2\mathrm{y}}=4-{9\mathrm{x}}^2$$$$\mathrm{x}=\pm \frac{1}{3}\sqrt{4-{\mathrm{e}}^{2\mathrm{y}}}$$$$\mathrm{for}\mathrm{x}\mathrm{to}\mathrm{be}\mathrm{real}$$$$\mathrm{y}⩽\ln 2$$$$\mathbf{Range}\mathbf{of}\mathbf{f}\left(\mathbf{x}\right)\in (-\mathrm{\infty },\ln 2]$$

Commented by ruwedkabeh last updated on 26/Sep/20

$$Thankyouverymuch!$$

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