Question

The true set of real values of $x$ for which the function, $f\left(x\right)=x\mathrm{lnx}-x+1$

• A. $\left(1,\infty \right)$
• B. $\left(1/e,\infty \right)$
• C. $\left[e,\infty \right]$
• D. $\left(0,1\right)and\left(1,\infty \right)$

Answer 1

Answer: (A)

$\left(1,\infty \right)$

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

$f^{\prime }\left(x\right)>0$

$\Rightarrow f^{\prime }\left(x\right)=x\times \frac{1}{x}+\ln x-1>0$

$\Rightarrow 1+\ln x-1>0$

$\Rightarrow \ln x>0$

$\Rightarrow x>e^0=1$

$\therefore x>1$

Hence $x\in (1,\infty )$