Question

Let $q$ be the maximum integral value of $p$ in $[0,10]$ for which the roots of the equation $x^2-p x+\frac{5}{4} p=0$ are rational. Then the area of the region $\left\{(x, y): 0 \leq y \leq(x-q)^2, 0 \leq x \leq q\right\}$ is

• A. 164
• B. 243
• C. $\frac{125}{3}$
• D. 25

Answer 1

Answer: (B)

$ x^2-p x+\frac{5 p}{4}=0 $
$ D=p^2-5 p=p(p-5) $
$ \therefore q=9 $
$ 0 \leq y \leq(x-9)^2$

Area $=\int\limits_0^9(x-9)^2 dx =243$