Question

Let $f\left(x\right)=$max$\{x^2,{(1-x)}^2,2x(1-x)\}$, where $0\le x\le 1$ the area of the region bounded by the curve $y=f\left(x\right)$ between $x\in [0,\frac{1}{3}],[\frac{1}{3},\frac{2}{3}],[\frac{2}{3},1]$ and $x-$axis represented by $S_1,S_2,S_3$ then

• A. $S_1,S_2,S_3$ are in G.P
• B. $S_1,S_2,S_3$ are in H.P
• C. $S_1=S_3$
• D. $S_1+S_2+S_3=\frac{17}{27}$

Answer 1

Answer: (C), (D)

The correct options are
C $S_1=S_3$
D $S_1+S_2+S_3=\frac{17}{27}$

$S_1=\left|{\int }_0^{\frac{1}{3}}{(1-x)}^{2}dx\right|$
$=\left|{[-\frac{1}{3}{(1-x)}^3]}_0^{\frac{1}{3}}\right|$
$=|-\frac{1}{3}\{\frac{8}{27}-1\}|$
$=\frac{19}{81}$
$S_2=\left|{\int }_{\frac{1}{3}}^{\frac{2}{3}}2x(1-x)dx\right|$
$={|x^2-\frac{2x^3}{3}|}_{\frac{1}{3}}^{\frac{2}{3}}$
$=(\frac{4}{9}-\frac{16}{81})-(\frac{1}{9}-\frac{2}{81})$
$=|\frac{3}{9}-\frac{14}{81}|=\frac{13}{81}$
and $S_3=\left|{\int }_{\frac{2}{3}}^1{x}^{2}dx\right|$
$=\frac{1}{3}(1-\frac{8}{27})=\frac{19}{81}$
Hence, $S_1=S_3=\frac{19}{81}$
and $S_2=\frac{13}{81}$
$\therefore S_1+S_2+S_3=\frac{19}{81}+\frac{13}{81}+\frac{19}{81}=\frac{51}{81}=\frac{17}{27}$