Question

The set of values of $x$ for which $1+\log x<x$ is

• A. $(1,\infty )$
• B. (0, 1)
• C. $(0,1)\cup (1,\infty )$
• D. none of these.

Answer 1

Answer: (C)

Let $f(x)=1+logx-x$
Clearly $f(x)$ is defined for $x>0$
$f^{\prime }\left(x\right)=\frac{1}{x}-1=\frac{1-x}{x}$
$f^{\prime }\left(x\right)>0$ for $0<x<1$ and $f^{\prime }\left(x\right)<0$ 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 $1+logx-x<1+log1-1=0$
i.e., $1+logx<x$ for $x\in \left(0,1\right)\cup \left(1,\infty \right)$