Sol. The curves y = [a] x 2 and y = [a] x 2 represent parabolas which are symmetric about
y- axis. The equation y 2 –3y + 2 = 0 gives a pair of straight lines y = 1, y = 2 which are parallel to x- axis.

The shaded region in fig. determines the area bounded by the two parabolas and two lines. Let us slice this region into horizontal strips. For the approximating rectangle shown in fig., we have

Length = x 2 –x 1 , Width = Δ y, Area= (x 2 –x 1 ) dy

As it can move vertically between y = 1 and y = 2. So,

Required area

= 2 dy

= 2 dy

= 2 dy

=

= ×

= (2 3/2 –1)

=

Clearly, A will be greatest, if [a] is least. It is given that a ≥ 1. Therefore, the least value of [a] is 1.

Thus, A is greatest when [a] = 1 ⇒ a ∈ [1, 2)