Question

If the area (in sq. units) of the region $\left\{\right(x,y):y^2\le 4x,x+y\le 1,x\ge 0,y\ge 0\}$ is $a\sqrt{2}+b,$ then $a-b$ is equal to :

• A. $\frac{8}{3}$
• B. $6$
• C. $\frac{10}{3}$
• D. $-\frac{2}{3}$

Answer 1

The correct option is B $6$

$C_1:y^2\le 4x$
$C_2:x+y\le 1$
$x,y\ge 0$


$P$ is the point of intersection of these two curve
$y^2=4x,x+y=1$
$y^2-4(1-y)=0\Rightarrow y^2+4y-4=0$
$y=\frac{-4\pm \sqrt{16-4(-4)}}{2}=-2\pm 2\sqrt{2}$
$P:(3-2\sqrt{2},-2+2\sqrt{2}),O:(0,0)R:(1,0)$
Hence, the required area$=\underset{0}{\overset{3-2\sqrt{2}}{\int }}2\sqrt{x}dx+\underset{3-2\sqrt{2}}{\overset{1}{\int }}(1-x)dx$
$=\frac{4}{3}{\left|x^{\frac{3}{2}}\right|}_0^{3-2\sqrt{2}}+{|x-\frac{x^2}{2}|}_{3-2\sqrt{2}}^1$
$=\frac{8\sqrt{2}}{3}-\frac{10}{3}=a\sqrt{2}+b$
$a=\frac{8}{3},b=-\frac{10}{3}$
$\therefore a-b=\frac{8+10}{3}=6$