Question

The value(s) of the parameter $\alpha (\alpha \ge 2)$ for which the area of region bounded by pair of straight lines $y^2-3y+2=0$ and the curves $y=\left[\alpha \right]x^2$, $y=\frac{1}{2}\left[\alpha \right]x^2$ is greatest, where $[.]$ denotes the greatest integer function, is

• A. $\alpha \in [2,3)$
• B. $\alpha \in \{2,3,4,\dots \}$
• C. $\alpha =2$
• D. $\alpha \in (2,3)$

Answer 1

The correct option is A $\alpha \in [2,3)$

$y^2-3y+2=0\Rightarrow (y-1)(y-2)=0$
$y=1$ and $y=2$

Area, $A=2\underset{1}{\overset{2}{\int }}(\frac{(2y{)}^{\frac{1}{2}}}{[\alpha {]}^{\frac{1}{2}}}-\frac{y^{\frac{1}{2}}}{[\alpha {]}^{\frac{1}{2}}})dy$

$=\frac{2}{[\alpha {]}^{\frac{1}{2}}}\underset{1}{\overset{2}{\int }}(\sqrt{2}-1)y^{\frac{1}{2}}dy$
$A$ is greatest when $\left[\alpha \right]$ is minimum.
Given $\alpha \ge 2$
So, $A$ is greatest when $\left[\alpha \right]=2;\alpha \in [2,3)$