Match the max/min value of function in Col - I with corresponding values in Col-II

Question

Professor PC Thomas Classes · Answer

(A) $f(x)=\frac{x}{4+x+x^{2}}$ on $[0,\infty)$
$f^{\prime}(x)=\frac{(4+x+x^{2})-x(2x+1)}{(4+x+x^{2})^{2}}=0 \Rightarrow 4-x^{2}=0 \Rightarrow x=\pm2$
if $x=2\Rightarrow\frac{2}{4+2+4}=\frac{2}{10}=1/5$
$x=-2\Rightarrow\frac{-2}{6}=-1/3$
(B) $f^{\prime}(x)=\frac{1-ln~x}{x^{2}}>0 \Rightarrow lnx<1 \Rightarrow x<e$
$	herefore$ Maximum value of $\frac{ln~x}{x}=\frac{1}{e}$
(C) $y=1/x\Rightarrow f(x) = \frac{3}{e^{3}}x+\frac{27e}{x} \Rightarrow f^{\prime}(x)=\frac{3}{e^{3}}-\frac{27e}{x^{2}}=0,x
e0 \Rightarrow x=3e^{2}$
$f^{\prime\prime}(x)=+\frac{54e}{x^{3}}\Rightarrow f(3e^{2}) = \frac{3}{e^{3}}	imes3e^{2}+27	imes e	imes\frac{1}{3e^{2}} =\frac{9}{e}+\frac{9}{e}=\frac{18}{e}$
(D) $R	heta+2R=4e \Rightarrow 	heta=\frac{4e-2R}{R}$
Area $=\frac{1}{2}R^{2}	heta = \frac{1}{2}R^{2}(\frac{4e-2R}{R}) = 2eR-R^{2}=f(R)$
$f^{\prime}(R)=2e-2R=0 \Rightarrow R=e$