Question

Let $f(x)=\frac{x^2}{4}(2 \ln x-1)-e x+k, k \in R$. 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

$f ( x )=\frac{ x ^2}{4}(2 \ln x -1)- cx + k$
$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 $
$f ^{\prime}( x )>0 \text { in }( e , \infty) \Rightarrow f ( x ) \text { is } \uparrow$
$\left. f ( x )\right|_{\min .}= f ( e )=\frac{ e ^2}{4}- e ^2+ k = k -\frac{3}{4} e ^2$
For $\sqrt{ f ( x )}$ to be defined for all $x \in(0, \infty)$
$f ( x ) \geq 0 \Rightarrow k \geq \frac{3}{4} e ^2 $
$\left. k \right|_{\text {min. }}=\frac{3 e ^2}{4}=\frac{ ae ^2}{ b }$
$\therefore a + b =3+4=7 $