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Q.
The set of values of $x$ for which $1 + \log x < x$ is
Application of Derivatives
Solution:
Let $f(x) = 1 + log \,x - x$
Clearly $f(x)$ is defined for $x > 0$
$f'\left(x\right)=\frac{1}{x}-1=\frac{1-x}{x}$
$f'\left(x\right) > 0$ for $0 < x < 1$ and $f'\left(x\right) < 0$ for $x > 1$
$\Rightarrow f\left(x\right)$ is increasing for $0 < x < 1$ and is decreasing for $x > 1$ i.e., for $x \in\left(1, \infty\right)$
$\therefore f\left(x\right) < f\left(1\right)$ for all $x \in\left(0, 1\right)$
and $f\left(x\right) < f\left(1\right)$ for, all $x \in\left(1, \infty\right)$
$\Rightarrow f\left(x\right) < f\left(1\right)$ for, all $x \in\left(0, 1\right) \cup\left(1, \infty\right)$
Hence $1+log\,x-x < 1+log\,1-1=0$
i.e., $1+log\,x < x$ for $x \in\left(0, 1\right) \cup\left(1, \infty\right)$