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

Question

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

Smart Achievers · Accepted Answer

(1)

$D = p^{2} - 5 p . If p = 9, then D is perfect square$

${(x, y) : 0 \leq y \leq {(x - q)}^{2}, 0 \leq x \leq q}$

Solution

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

1 $243$

2 $25$

3 $\frac{125}{3}$

4 $164$

Ans.
(1)

$D = p^{2} - 5 p . If p = 9, then D is perfect square$

$Area = \int_{0}^{9} {(x - 9)}^{2} d x = 243$

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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 x2+px+5p4=0 are rational then the area of region {(x,y):0≤y≤(x-q)2, 0≤x≤q} is (in square units)

1 243

2 25

3 1253

4 164

Ans. (1) D=p2-5p. If p=9, then D is perfect square Area=∫09(x-9)2 dx=243