Question

Given p $\le$4 is a positive real. Let A be the area of the bounded region enclosed by the curves y = 1 - |1-x| and y = |2x-p|. Then which among the following best describes A?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Option (c)

Case 1: 0$<$p$\le$1 The area formed is of a triangle with vertices $(\frac{p}{3},\frac{p}{3}).(\frac{p}{2},0)$ and (p, p). Thus, the area is $\frac{p^2}{6}$ sq. units.

Case 2: 1 $\le$ p $\le$ 3The figures formed is a quadrilateral $(\frac{p}{3},\frac{p}{3}),\left(\frac{p}{2}.0\right),(\frac{p+2}{3},\frac{4-p}{3})$ and (1, 1).Thus, the area formed is $\frac{1}{3}-\frac{1}{6}\times (p-2{)}^2$

Case 3: 3$\le$ p $\le$ 4The area formed is the image of case 1.

Area = $\frac{(4-p{)}^2}{6}$

Thus, min (A) = 0 at p = 4 and max (A) = $\frac{1}{3}$ at p = 2. Hence, option (c) is the right answer.