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Area Under the Curves question

2023 · 30 Jan · Shift 2 · Q29

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Area Under the Curves
2023 · 30 Jan · Shift 2 · Q29

Area Under the Curves question

2023 · 30 Jan · Shift 2 · Q29

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1

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
$\frac{125}{3}$
B
243
C
164
D
25

View written solutionFree

Correct answer: B

Condition for rational roots

Given $x^2-px+\frac{5}{4}p=0$ for the roots to be rational, the discriminant must be a perfect square rational number.

The discriminant is $\Delta=p^2-4\cdot 1\cdot \frac{5}{4}p=p^2-5p=p(p-5).$

So we need $p(p-5)$ to be a perfect square.

Also, since $p$ is an integral value in $[0,10]$ , test integers $p=0,1,2,\dots,10$ .

Check integer values of $p$

We compute $p(p-5)$ :

$p=0$ : $0( -5)=0$ , perfect square.
$p=1$ : $1(-4)=-4$ , not possible.
$p=2$ : $2(-3)=-6$ , not possible.
$p=3$ : $3(-2)=-6$ , not possible.
$p=4$ : $4(-1)=-4$ , not possible.
$p=5$ : $5(0)=0$ , perfect square.
$p=6$ : $6(1)=6$ , not a square.
$p=7$ : $7(2)=14$ , not a square.
$p=8$ : $8(3)=24$ , not a square.
$p=9$ : $9(4)=36$ , perfect square.
$p=10$ : $10(5)=50$ , not a square.

Thus the maximum integral value is $q=9.$

Set up the area integral

We need the area of $\{(x,y):0\le y\le (x-q)^2,\ 0\le x\le q\}.$

Since $q=9$ , this becomes $0\le y\le (x-9)^2,\quad 0\le x\le 9.$

Hence area is $A=\int_0^9 (x-9)^2\,dx.$

Evaluate the integral

Let $u=x-9$ . Then when $x=0$ , $u=-9$ , and when $x=9$ , $u=0$ . So

=\left[\frac{u^3}{3}\right]_{-9}^{0} =0-\left(-\frac{729}{3}\right)=243.$$ Alternatively, $$\int_0^9 (9-x)^2dx=\frac{9^3}{3}=243.$$ 5. **Final answer** The required area is $$\boxed{243}.$$ So the correct option is **B**.

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