Question

For a given positive integer $n$ let $f:(0,\infty )\to R$ be defined as
$f\left(x\right)=x^{n}lnx$.
Then

• A. $f\left(x\right)$ is an increasing function on $(0,\infty )$
• B. $f$ has only one zero in $(0,\infty )$
• C. $f\left(x\right)$ has a minimum at $x=1/\sqrt[n]{ne}$
• D. The line $y=x-1$ is a tangent to the graph of $f\left(x\right)$ at the point $(1,0)$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $f$ has only one zero in $(0,\infty )$
C The line $y=x-1$ is a tangent to the graph of $f\left(x\right)$ at the point $(1,0)$
D $f\left(x\right)$ has a minimum at $x=1/\sqrt[n]{ne}$

Given $f\left(x\right)=x^n\ln x$,

Differentiating with respect to $x$ and equating it to $0$ gives,

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

$x=e^{-\frac{1}{n}}$,

As the function is increasing that is the maximum of it,

The slope of the tangent at the point $(1,0)$ is $1$,

$\therefore$ the equation is $y=x-1$,

$f\left(1\right)=0$ is the only zero,

after this point the function reaches its maximum and doesn't reach zero again.