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 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

Option: 1

243


Option: 2

164


Option: 3

\frac{125}{3}


Option: 4

 25


Answers (1)

best_answer

\mathrm{x}^{2}-\mathrm{px}+\frac{5}{4} \mathrm{p}=0
Roots are rational

\mathrm{D}= A perfect square
\mathrm{p}^{2}-4(1) \frac{5}{4} \mathrm{p}
p^{2}-5 p=A perfect square
for p=0, p=5, p=9 the D is a perfect square

\therefore maximum integral of \mathrm{p} is 9.
\mathrm{q}=9
\left\{(\mathrm{x}, \mathrm{y}) ; 0 \leq \mathrm{y} \leq(\mathrm{x}-9)^{2}, 0 \leq \mathrm{x} \leq 9\right\}


\text{Area }=\int_{0}^{9}(x-9)^{2} \mathrm{dx}
\left.\Rightarrow \frac{(\mathrm{x}-9)^{3}}{3}\right]_{0}^{3}
\Rightarrow 0-\frac{(0-9)^{3}}{3}
\Rightarrow \frac{9 \times 9 \times 9}{3}
=243

Posted by

Divya Prakash Singh

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