Question

The true set of real value of $x$ for which the function, $f\left(x\right)=xlnx-x+1$ is positive is

• A. $(1,\infty )$
• B. $(1/e,\infty )$
• C. $(e,\infty )$
• D. $(0,1)\cup (1,\infty )$

Answer 1

The correct option is D $(0,1)\cup (1,\infty )$

$f\left(x\right)=x\ln x-x+1f\left(1\right)=0$

On differentiating w.r.t $x$, we get

$f^{\prime }\left(x\right)=x\times \frac{1}{x}+\ln x-1=\ln x$

Therefore,

$f^{\prime }\left(x\right)>0forallx\in \left(1,\infty \right)f^{\prime }\left(x\right)<0forallx\in \left(0,1\right)$

${lim}_{x\to 0}f\left(x\right)=1f\left(x\right)>0$

$forallx\in \left(0,1\right)\cup \left(1,\infty \right)$

Option $D$ is correct answer.