Area Under Curves Question 8
Question 8 - 30 January - Shift 2
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
$\lbrace (x, y): 0 \leq y \leq(x-q)^{2}, 0 \leq x \leq q \rbrace $ is
(1) 243
(2) 25
(3) $\frac{125}{3}$
(4) 164
Show Answer
Answer: (1)
Solution:
Formula: Area between two curves - Area bounded by a curve with $\mathbf{x}$ - axis, Nature of roots
$x^{2}-px+\frac{5 p}{4}=0, \ p \ \in [0,10]$
$D=p^{2}-5 p=p(p-5), \ p \ \in [0,10]$
$D$ is perfect square iff $p=9$
$\therefore q=9$
$0 \leq y \leq(x-9)^{2}$
Area $=\int_0^{9}(x-9)^{2} d x=243$
