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

The Correct Option is B

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

Area $=\underset{0}{\overset{9}{\int }}(x-9{)}^{2}dx=243$
So , the correct option is (B) : 243

Answer 2

△ = p² - 5p is perfect square (∵ roots are rational)
∴ q = 9, (∵0 ≤ p ≤ 10)
Therefore, Area of the region :
\(\int\limits_{0}^9\left(x-9\right)^2dx=\frac{(x-9)^3}{3}^9_0=243\)
So, the correct answer is (B) : 243.