Let q be the maximum integral value of p [ 0 , 10 ] for which the roots of the equation x 2 - p x + 5 4 p = 0 are rational. Then the area of the region { ( x , y ) : 0 y ( x - q ) 2 , 0 x q } is [2023]

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Solution

Q.
$Let q be the maximum integral value of p in [0,10] for which the roots of the equation x^{2} - p x + \frac{5}{4} p = 0 are rational.$ Then the area of the region
${(x, y) : 0 \leq y \leq {(x - q)}^{2}, 0 \leq x \leq q}$ is [2023]

1 $164$

2 $243$

3 $\frac{125}{3}$

4 $25$

Ans.
(2)

Given equation is $x^{2} - p x + \frac{5}{4} p = 0$

$D = p^{2} - 5 p,$ which is a perfect square when $p = 9$ .

$∴$ $q = 9$

$Required area = \int_{0}^{9} {(x - 9)}^{2} d x$

$= \int_{0}^{9} (x^{2} - 18 x + 81) d x$ $= {[\frac{x^{3}}{3} - 18 \times \frac{x^{2}}{2} + 81 x]}_{0}^{9}$

$= 243 sq. units.$

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