JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

Solution

Step 1: Finding $q$

Roots are rational $\implies D = k^2$ (perfect square)

Equation: $x^2 - px + \frac{5}{4}p = 0$

D = (-p)^2 - 4(1)\left(\frac{5}{4}p\right) = p^2 - 5p

Checking integral values of $p$ in $[0, 10]$ for $D = k^2$ :

For $p = 10 \implies D = 10^2 - 50 = 50$ (Not a square)
For $p = 9 \implies D = 9^2 - 45 = 36 = 6^2$ (Square!)

\therefore q = 9

Step 2: Calculating Area

Region: $0 \le y \le (x - 9)^2$ for $x \in [0, 9]$

\text{Area} = \int_{0}^{9} (x - 9)^2 \,dx

\text{Area} = \left[ \frac{(x - 9)^3}{3} \right]_{0}^{9}

\text{Area} = \left( \frac{(9 - 9)^3}{3} \right) - \left( \frac{(0 - 9)^3}{3} \right)

\text{Area} = 0 - \left( \frac{-729}{3} \right)

\text{Area} = 243

Correct Option: (1)