$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ y'=3x^{2} -8x\\ b.\ f'( x) =\frac{2}{( x+1)^{2}}\\ PT\ tiếp\ tuyến\ của\ dths\ y=f( x) \ tại\ x=a >0:\\ d:y=\frac{2( x-a)}{( a+1)^{2}} +\frac{2a}{a+1} =\frac{2x}{( a+1)^{2}} +\frac{-2a+2a^{2} +2a}{( a+1)^{2}}\\ =\frac{2x}{( a+1)^{2}} +\frac{2a^{2}}{( a+1)^{2}}\\ d\cap Ox=A=\left( -a^{2} ;0\right)\\ d\cap Oy=B=\left( 0;\frac{2a^{2}}{( a+1)^{2}}\right)\\ S_{OAB} =\frac{1}{2} .a^{2} .\frac{2a^{2}}{( a+1)^{2}} \ ( do\ a >0)\\ \Rightarrow \frac{1}{2} .a^{2} .\frac{2a^{2}}{( a+1)^{2}} =\frac{1}{4}\\ \Leftrightarrow \frac{a^{4}}{a^{2} +2a+1} =\frac{1}{4}\\ \Leftrightarrow 4a^{4} -a^{2} -2a+1=0\\ \Leftrightarrow ( a-1)( 2a+1)\left( 2a^{2} +a+1\right) =0\\ \Leftrightarrow a=1\ ( TM) \ or\ a=-\frac{1}{2} \ ( loại)\\ Vậy\ d:y=\frac{x}{2} +\frac{1}{2} \end{array}$
