Question

Let $f(x)=\frac{x^2}{4}(2\ln x-1)-ex+k,k\in R$. If least value of $k$ for which $\sqrt{f(x)}$ is defined for all $x\in (0,\infty )$ is $\frac{a}{b}{e}^2$, where $a,b\in N$, then find least value of $(a+b)$.

Answer 1

Answer: 7

$f(x)=\frac{x^2}{4}(2\ln x-1)-cx+k$
$f^{\prime }(x)=\frac{x^2}{4}\left(\frac{2}{x}\right)+(2\ln x-1)\cdot \frac{x}{2}-e$
$=\frac{x}{2}+x\ln x-\frac{x}{2}-e$
$=x\ln x-e$

$f^{\prime }(x)=0\Rightarrow x\ln x-e=0\Rightarrow \ln x-\frac{e}{x}=0$
$f^{\prime }(x)<0\text{ in }(0,e)\Rightarrow f(x)\text{ is }\downarrow$
$f^{\prime }(x)>0\text{ in }(e,\infty )\Rightarrow f(x)\text{ is }\uparrow$
${f(x)|}_{\min .}=f(e)=\frac{e^2}{4}-e^2+k=k-\frac{3}{4}{e}^2$
For $\sqrt{f(x)}$ to be defined for all $x\in (0,\infty )$
$f(x)\ge 0\Rightarrow k\ge \frac{3}{4}{e}^2$
${k|}_{\text{min. }}=\frac{3e^2}{4}=\frac{a{e}^2}{b}$
$\therefore a+b=3+4=7$