JEE MainMathematics

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Area under a Parabola Determined by Rational Quadratic Roots

Q: Can p take fractional values?

The question asks for the 'maximum integral value of p ', so p must be an integer.

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

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A243B164C125/3D25

Step-by-step solutionView

Correct answer

The maximum integer $p \in [0, 10]$ yielding rational roots is $q = 9$ , which gives an area under the curve of $\int_{0}^{9} (x - 9)^{2} d x = 243$ .

Option analysis

Why each option works or fails

A · 243

This is the correct value. The discriminant $p^{2} - 5 p$ is a perfect square for $p = 9$ , and evaluating $\int_{0}^{9} (x - 9)^{2} d x = \frac{9^{3}}{3} = 243$ .

B · 164

The student correctly finds $q = 9$ and attempts $\frac{9^{3}}{3} = 243$ , but makes an arithmetic error during division or cubing (e.g., confusing $9^{3} / 3$ with $2 \cdot 81 + 2$ or similar arithmetic blunders to get 164). Carefully compute $\frac{9^{3}}{3} = 9^{2} \times 3 = 81 \times 3 = 243$ .

C · 125/3

The student finds $p = 5$ as a valid integer giving a perfect square discriminant ( $5^{2} - 5 (5) = 0$ ), but fails to find the maximum possible value $q = 9$ , leading to $\int_{0}^{5} (x - 5)^{2} d x = \frac{125}{3}$ . Check all integers up to 10. For $p = 9$ , the discriminant is $81 - 45 = 36 = 6^{2}$ , which is larger than $p = 5$ .

D · 25

The student selects $p = 5$ and further forgets to divide by 3 when integrating, computing $(5)^{2} = 25$ , or evaluates the integrand at an endpoint without integrating. Maximize $p$ across $[0, 10]$ to get $q = 9$ , and integrate correctly using $\int_{0}^{q} (x - q)^{2} d x = \frac{q^{3}}{3}$ .

Reviewed route

Solution

StepWorking

01Given

Given the quadratic equation $x^{2} - p x + \frac{5}{4} p = 0$ with $p \in [0, 10]$ and an integer. $q$ is the maximum integral value of $p$ such that the roots of the equation are rational. The target region is defined by $0 \leq y \leq (x - q)^{2}$ and $0 \leq x \leq q$ .

02Approach

For a quadratic with rational coefficients to have rational roots, its discriminant $D$ must be the square of a rational number. Since $p$ is an integer, $D = p^{2} - 5 p$ must be a perfect square of an integer. We test integral values of $p \in [0, 10]$ from 10 downwards to find $q$ , then compute the area via definite integration: $Area = \int_{0}^{q} (x - q)^{2} d x$ .

03Execute

Calculate the discriminant: $D = (- p)^{2} - 4 (1) (\frac{5}{4} p) = p^{2} - 5 p$ . We require $p (p - 5) = k^{2}$ for some integer $k \geq 0$ . Checking integers $p \in [0, 10]$ in descending order: - For $p = 10$ : $D = 100 - 50 = 50$ (not a square) - For $p = 9$ : $D = 81 - 45 = 36 = 6^{2}$ (a perfect square). Hence, the maximum integral value is $q = 9$ .

04Execute

Compute the area of the region $0 \leq y \leq (x - 9)^{2}$ for $0 \leq x \leq 9$ : $Area = \int_{0}^{9} (x - 9)^{2} d x = {[\frac{(x - 9)^{3}}{3}]}_{0}^{9} = 0 - (- \frac{9^{3}}{3}) = \frac{729}{3} = 243.$

✓Verify

By substitution $u = 9 - x$ , $\int_{0}^{9} (x - 9)^{2} d x = \int_{0}^{9} u^{2} d u = \frac{9^{3}}{3} = 243$ . Both integration bounds and antiderivative signs are confirmed.

Hints that build this answer step by step

For the quadratic equation $x^{2} - p x + \frac{5}{4} p = 0$ with rational coefficients to have rational roots, what condition must its discriminant $D$ satisfy?
$D = p^{2} - 5 p$ must be the square of a rational number (an integer for integer $p$ ).
What is the maximum integer $p \in [0, 10]$ such that $p^{2} - 5 p$ is a perfect square?
$q = 9$
With $q = 9$ , what is the value of the integral representing the area, $\int_{0}^{9} (x - 9)^{2} d x$ ?
$243$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Can

p

take fractional values?

The question asks for the 'maximum integral value of $p$ ', so $p$ must be an integer.

Answer

The maximum integer $p \in [0, 10]$ yielding rational roots is $q = 9$ , which gives an area under the curve of $\int_{0}^{9} (x - 9)^{2} d x = 243$ .

Why each option works or fails

A · correct: 243 - This is the correct value. The discriminant $p^{2} - 5 p$ is a perfect square for $p = 9$ , and evaluating $\int_{0}^{9} (x - 9)^{2} d x = \frac{9^{3}}{3} = 243$ .
B: 164 - The student correctly finds $q = 9$ and attempts $\frac{9^{3}}{3} = 243$ , but makes an arithmetic error during division or cubing (e.g., confusing $9^{3} / 3$ with $2 \cdot 81 + 2$ or similar arithmetic blunders to get 164). Carefully compute $\frac{9^{3}}{3} = 9^{2} \times 3 = 81 \times 3 = 243$ .
C: 125/3 - The student finds $p = 5$ as a valid integer giving a perfect square discriminant ( $5^{2} - 5 (5) = 0$ ), but fails to find the maximum possible value $q = 9$ , leading to $\int_{0}^{5} (x - 5)^{2} d x = \frac{125}{3}$ . Check all integers up to 10. For $p = 9$ , the discriminant is $81 - 45 = 36 = 6^{2}$ , which is larger than $p = 5$ .
D: 25 - The student selects $p = 5$ and further forgets to divide by 3 when integrating, computing $(5)^{2} = 25$ , or evaluates the integrand at an endpoint without integrating. Maximize $p$ across $[0, 10]$ to get $q = 9$ , and integrate correctly using $\int_{0}^{q} (x - q)^{2} d x = \frac{q^{3}}{3}$ .

Step-by-step solution

Given: Given the quadratic equation $x^{2} - p x + \frac{5}{4} p = 0$ with $p \in [0, 10]$ and an integer. $q$ is the maximum integral value of $p$ such that the roots of the equation are rational. The target region is defined by $0 \leq y \leq (x - q)^{2}$ and $0 \leq x \leq q$ .
Approach: For a quadratic with rational coefficients to have rational roots, its discriminant $D$ must be the square of a rational number. Since $p$ is an integer, $D = p^{2} - 5 p$ must be a perfect square of an integer. We test integral values of $p \in [0, 10]$ from 10 downwards to find $q$ , then compute the area via definite integration: $Area = \int_{0}^{q} (x - q)^{2} d x$ .
Execute: Calculate the discriminant: $D = (- p)^{2} - 4 (1) (\frac{5}{4} p) = p^{2} - 5 p$ . We require $p (p - 5) = k^{2}$ for some integer $k \geq 0$ . Checking integers $p \in [0, 10]$ in descending order: - For $p = 10$ : $D = 100 - 50 = 50$ (not a square) - For $p = 9$ : $D = 81 - 45 = 36 = 6^{2}$ (a perfect square). Hence, the maximum integral value is $q = 9$ .
Execute: Compute the area of the region $0 \leq y \leq (x - 9)^{2}$ for $0 \leq x \leq 9$ : $Area = \int_{0}^{9} (x - 9)^{2} d x = {[\frac{(x - 9)^{3}}{3}]}_{0}^{9} = 0 - (- \frac{9^{3}}{3}) = \frac{729}{3} = 243.$
Verify: By substitution $u = 9 - x$ , $\int_{0}^{9} (x - 9)^{2} d x = \int_{0}^{9} u^{2} d u = \frac{9^{3}}{3} = 243$ . Both integration bounds and antiderivative signs are confirmed.

Area under a Parabola Determined by Rational Quadratic Roots

What feels right?

The maximum integer $p \in [0, 10]$ yielding rational roots is $q = 9$ , which gives an area under the curve of $\int_{0}^{9} (x - 9)^{2} d x = 243$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

✓Verify

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