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Question Number 115489 by ruwedkabeh last updated on 26/Sep/20 | ||
$${find}\:{range}\:{for} \\ $$$$\mathrm{1}.\:{f}\left({x}\right)=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}{{x}−\mathrm{1}} \\ $$$$\mathrm{2}.\:{f}\left({x}\right)=\mathrm{ln}\:\left(\sqrt{\mathrm{4}−\mathrm{9}{x}^{\mathrm{2}} }\right) \\ $$ | ||
Answered by PRITHWISH SEN 2 last updated on 26/Sep/20 | ||
$$\mathrm{1}.\:\:\mathrm{Let}\:\mathrm{y}=\:\frac{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{9}}}{\mathrm{x}−\mathrm{1}} \\ $$$$\mathrm{x}^{\mathrm{2}} \left(\mathrm{y}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{2xy}^{\mathrm{2}} +\left(\mathrm{y}^{\mathrm{2}} +\mathrm{9}\right)=\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{real}\:\mathrm{x} \\ $$$$\bigtriangleup\:=\:\left(\mathrm{2y}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{y}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{y}^{\mathrm{2}} +\mathrm{9}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\:\left(\mathrm{y}−\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\left(\mathrm{y}+\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\leqslant\mathrm{0} \\ $$$$\boldsymbol{\mathrm{Range}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\in\:\left[−\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}}\:,\:\:\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\right] \\ $$$$\mathrm{2}.\:\:\mathrm{Again}\:\mathrm{let}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\:=\:\mathrm{ln}\:\left(\sqrt{\mathrm{4}−\mathrm{9x}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\mathrm{e}^{\mathrm{2y}\:} =\:\mathrm{4}−\mathrm{9x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\:=\:\pm\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{4}−\mathrm{e}^{\mathrm{2y}} }\:\: \\ $$$$\:\:\:\:\mathrm{for}\:\mathrm{x}\:\mathrm{to}\:\mathrm{be}\:\mathrm{real} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{y}\:\leqslant\:\mathrm{ln}\:\mathrm{2} \\ $$$$\:\:\boldsymbol{\mathrm{Range}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\in\:\left(−\infty,\:\mathrm{ln}\:\mathrm{2}\right] \\ $$ | ||
Commented by ruwedkabeh last updated on 26/Sep/20 | ||
$${Thank}\:{you}\:{very}\:{much}! \\ $$ | ||