KVPY SX SB Exam 2018 Question Paper

Since,
$\mathit{y}={\mathit{x}}^2+\mathit{x}+10$

$\Rightarrow {\mathit{x}}^2+\mathit{x}+\frac{1}{4}=\mathit{y}-10+\frac{1}{4}$
$\Rightarrow {\left(\mathit{x}+\frac{1}{2}\right)}^2=\left(\mathit{y}-\frac{39}{4}\right)$
$\Rightarrow$ Latusrectum of the above parabola is 1

$\therefore$ length of chord is also 1 Area is maximum when chord is latusrectum.

Area of shaded region $=$
Area of rectangle $\mathit{O}\mathit{A}\mathit{B}\mathit{C}-2\underset{-1∕2}{\overset{0}{\int }}$ parabola

$=10-2{\underset{\frac{-1}{2}}{\int }}^0\left({\mathit{x}}^2+\mathit{x}+10\right)\mathit{d}\mathit{x}$
$=10-2{\left[\frac{{\mathit{x}}^3}{3}+\frac{{\mathit{x}}^2}{2}+10\right]}_{\frac{-1}{2}}^0$
$=10-2\left[\frac{-1}{24}+\frac{1}{8}-\frac{10}{2}\right]=\frac{1}{6}$