Question

Differece between the greatest and the least values of the function $f\left(x\right)=x(lnx-2)$ on $(1,e^2)$ is :

• A. 2
• B. e
• C. $e^2$
• D. 1

Answer 1

Answer: (B)

e

$f\left(x\right)=x(lnx-2)$

$f^{\prime }\left(x\right)=1(lnx-2)+x\left(\frac{1}{x}\right)$

$f^{\prime }\left(x\right)=lnx-2+1$

$f^{\prime }\left(x\right)=lnx-1$

Let $f^{\prime }\left(x\right)=0$ for critical points.

$\Rightarrow$ $lnx-1=0$

$\Rightarrow$ $lnx=1$

$\Rightarrow$ $x=e$ [ Since $lne=1$ ]

Now, we are going to find greatest and least values of functions at points $x=e,1,e^2$

Now,

$f\left(x\right)=x(lnx-2)$

$\Rightarrow$ $f\left(e\right)=e(lne-2)$

$\Rightarrow$ $f\left(e\right)=e(1-2)$

$\Rightarrow$ $f\left(e\right)=-e$ ---- ( 1 )

$f\left(x\right)=x(lnx-2)$

$\Rightarrow$ $f\left(1\right)=1(ln1-2)$

$\Rightarrow$ $f\left(1\right)=1(0-2)$

$\Rightarrow$ $f\left(1\right)=-2$ ---- ( 2 )

$f\left(x\right)=x(lnx-2)$

$\Rightarrow$ $f\left(e^2\right)=e^2(ln{e}^2-2)$

$\Rightarrow$ $f\left(e^2\right)=e^2(2lne-2)$

$\Rightarrow$ $f\left(e^2\right)=0$ ---- ( 3 )

From ( 1 ), ( 2 ) and ( 3 )

Greatest value $=0$

Least value $=-e$

Required difference $=0-(-e)=e$