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Question Number 101243 by Dwaipayan Shikari last updated on 01/Jul/20

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{solution}\: \\$$$$\:\:\mathrm{xa}^{\frac{\mathrm{1}}{\mathrm{x}}} +\frac{\mathrm{1}}{\mathrm{x}}\mathrm{a}^{\mathrm{x}} =\mathrm{2a}\:\:\:\mathrm{a}\in\left\{−\mathrm{1},\mathrm{0},\mathrm{1}\right\}\:\:\:{and}\:{also}\:{find}\:{when}\:{a}\:\:{is}\:{not}\:{given} \\$$

Commented by mr W last updated on 01/Jul/20

$${for}\:{a}>\mathrm{0}: \\$$$${x}=\mathrm{1}\:{is}\:{the}\:{only}\:{solution}. \\$$

Commented by mr W last updated on 01/Jul/20

$${a}=−\mathrm{1}: \\$$$${x}\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{{x}}} +\frac{\left(−\mathrm{1}\right)^{{x}} }{{x}}=−\mathrm{2} \\$$$$\Rightarrow{x}=\mathrm{1} \\$$$$\\$$$${a}=\mathrm{0}: \\$$$$\Rightarrow{x}\in{R}−\left\{\mathrm{0}\right\} \\$$$$\\$$$${a}=\mathrm{1}: \\$$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{2} \\$$$$\Rightarrow{x}=\mathrm{1} \\$$

Commented by Dwaipayan Shikari last updated on 01/Jul/20

$${You}\:{are}\:{right}\:{sir}.\:{Can}\:{you}\:{find}\:{when}\:{a}\:{is}\:{not}\:{given} \\$$