JEE Main 2023 (30 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: 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 ${(x, y) : 0 \leq y \leq (x - q)^{2}, 0 \leq x \leq q}$ is

Q: Why did we test $p$ starting from 10? Is there a faster algebraic way?

You could set $p^2 - 5p = k^2$ and solve it as a Diophantine equation, but in the JEE context, when the range is small (like $[0, 10]$), testing values is often the most efficient and error-proof strategy. Don't overcomplicate when a simple check saves time and prevents algebraic slips!

Q: I got a negative area when I integrated. Did I do something wrong?

Ah, the classic sign trap! When you integrate $\int_{0}^{9} (x-9)^2 \, dx$, you are calculating the signed area. Since the parabola is entirely above the x-axis in this interval, the area must be positive. If you got a negative result, check your limits of integration or the order of subtraction. Remember, Area = $\int_{lower}^{upper} f(x) dx$. If you swap the limits, you flip the sign!

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. When you see 'rational roots' for a quadratic equation with rational coefficients, your mind should immediately jump to the Discriminant (

D

). What does the quadratic formula tell us about the nature of roots? If the term under the square root is not a perfect square, can the roots ever be rational?

Hint 2

Hint 2: The Arsenal. You have the equation

x^{2} - p x + \frac{5}{4} p = 0

. Calculate the discriminant

D = b^{2} - 4 a c

. Once you simplify this to a function of

p

, remember that for the roots to be rational,

D

must be a perfect square (let's call it

k^{2}

). Test the integer values of

p

in the range

[0, 10]

starting from the maximum to find the largest

p

that satisfies this condition.

Hint 3

Hint 3: The Execution. Once you identify

q = 9

, the problem transforms into a geometric one. You are looking for the area under the curve

y = (x - 9)^{2}

from

x = 0

x = 9

. Set up your definite integral as

A = \int_{0}^{9} (x - 9)^{2} d x

. Remember to handle the limits carefully!

Visualized Solution

The Quadratic Equation

Given equation: $x^{2} - p x + \frac{5}{4} p = 0$
Constraint: $p \in [0, 10]$ and $p$ is an integer.
We need to find the maximum value of $p$ for which roots are rational.

Condition for Rational Roots

For a quadratic equation with rational coefficients, roots are rational if and only if the Discriminant ( $D$ ) is a perfect square.
$D = b^{2} - 4 a c$

Calculate the Discriminant $D$

Substitute $a = 1$ , $b = - p$ , $c = \frac{5}{4} p$ into $D = b^{2} - 4 a c$ .
$D = (- p)^{2} - 4 (1) (\frac{5}{4} p)$

Simplify the Discriminant

$D = p^{2} - 5 p$
We need $p^{2} - 5 p$ to be a perfect square, say $k^{2}$ .

Finding the Maximum Integral $p$

Test integers $p \in [0, 10]$ starting from the maximum.
If $p = 10$ , $D = 100 - 50 = 50$ (Not a perfect square).
If $p = 9$ , $D = 81 - 45 = 36 = 6^{2}$ (Perfect square!).
Therefore, the maximum integral value is $q = 9$ .

Define the Bounded Region

Substitute $q = 9$ into the given region.
Region: ${(x, y) : 0 \leq y \leq (x - 9)^{2}, 0 \leq x \leq 9}$

Visualize the Parabola

The curve $y = (x - 9)^{2}$ is an upward-opening parabola.
Vertex is at $(9, 0)$ .

The Y-Intercept

At $x = 0$ , $y = (0 - 9)^{2} = 81$ .
This gives us the upper bound on the $y$ -axis for our region.

Identify the Area to Calculate

We need the area under $y = (x - 9)^{2}$ .
Bounded by the $x$ -axis ( $y \geq 0$ ).
From $x = 0$ to $x = 9$ .

Set up the Definite Integral

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

Integrate the Function

Use the power rule: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1}$
$A = [\frac{( x - 9 ) ^{3}}{3}]_{0}^{9}$

Apply the Limits

Substitute upper limit $x = 9$ : $\frac{( 9 - 9 ) ^{3}}{3} = 0$
Substitute lower limit $x = 0$ : $\frac{( 0 - 9 ) ^{3}}{3} = \frac{- 729}{3}$

Final Calculation

$A = 0 - (\frac{- 729}{3})$
$A = 0 - (- 243)$
$A = 243$

Final Conclusion

The maximum integral value $q = 9$ .
The area of the bounded region is $243$ square units.
Key Takeaway: Rational roots imply the discriminant is a perfect square.

00:00 / 00:00

The Sigma Insight: Area Bounded by Curves

Solution Diagram

Analyzing the Setup

We are investigating the quadratic equation

x^{2} - p x + \frac{5}{4} p = 0

, where

p

is an integer in the interval

[0, 10]

. Our objective is to determine the maximum value of

p

that ensures the roots of this equation are rational.

The Discriminant's Secret

For a quadratic equation with rational coefficients to have rational roots, its discriminant

D = b^{2} - 4 a c

must be a perfect square. Given

a = 1

b = - p

, and

c = \frac{5}{4} p

, we calculate the discriminant as follows:

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

Simplifying this expression, the factors of

4

cancel out, yielding:

D = p^{2} - 5 p

We require

D = k^{2}

for some non-negative integer

k

The Hunt for $p$

We test integer values of

p

within the interval

[0, 10]

, starting from the maximum value to find the largest

p

that satisfies the condition.

p = 10

, then

D = 1 0^{2} - 5 (10) = 100 - 50 = 50

. Since

50

is not a perfect square,

p = 10

is not the solution.

p = 9

, then

D = 9^{2} - 5 (9) = 81 - 45 = 36

. Because

36 = 6^{2}

, the condition is satisfied. Thus, the maximum integer value is $p = 9$ .

Visualizing the Geometry

With

p = 9

, the problem shifts to finding the area of the region defined by

0 \leq y \leq (x - 9)^{2}

and

0 \leq x \leq 9

This region is bounded by the parabola

y = (x - 9)^{2}

and the x-axis. The curve has its vertex at

(9, 0)

and intersects the y-axis at

(0, 81)

The Calculus of Victory

To calculate the area

A

under the curve, we evaluate the definite integral:

A = \int_{0}^{9} (x - 9)^{2} d x

Applying the power rule for integration, we obtain:

A = [\frac{( x - 9 ) ^{3}}{3}]_{0}^{9}

Evaluating at the boundaries using the Fundamental Theorem of Calculus:

A = (\frac{( 9 - 9 ) ^{3}}{3}) - (\frac{( 0 - 9 ) ^{3}}{3})

A = 0 - (\frac{- 729}{3}) = 0 - (- 243) = 243

The final area of the region is $243$ square units.

Common Traps & FAQs

Wait, why does the discriminant have to be a perfect square? Can't it just be any positive number?

That is a fantastic question! Remember the quadratic formula:

x = \frac{- b \pm D}{2 a}

. If

D

is positive but *not* a perfect square, then

D

remains an irrational number. Adding or subtracting an irrational number from a rational number (

- b /2 a

) will always result in an irrational root. To ensure the roots are rational, the square root must 'disappear,' which only happens if

D

is a perfect square (like

0, 1, 4, 9, 16, \dots

Why did we test

p

starting from 10? Is there a faster algebraic way?

You could set

p^{2} - 5 p = k^{2}

and solve it as a Diophantine equation, but in the JEE context, when the range is small (like

[0, 10]

), testing values is often the most efficient and error-proof strategy. Don't overcomplicate when a simple check saves time and prevents algebraic slips!

I got a negative area when I integrated. Did I do something wrong?

Ah, the classic sign trap! When you integrate

\int_{0}^{9} (x - 9)^{2} d x

, you are calculating the signed area. Since the parabola is entirely above the x-axis in this interval, the area must be positive. If you got a negative result, check your limits of integration or the order of subtraction. Remember, Area =

\int_{l o w er}^{u pp er} f (x) d x

. If you swap the limits, you flip the sign!

Animated Solution for Mathematics - Definite Integration: 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 ${(x, y) : 0 \leq y \leq (x - q)^{2}, 0 \leq x \leq q}$ is

Select Answer:

Visualized Solution

The Quadratic Equation

Condition for Rational Roots

Calculate the Discriminant $D$

Simplify the Discriminant

Finding the Maximum Integral $p$

Define the Bounded Region

Visualize the Parabola

The Y-Intercept

Identify the Area to Calculate

Set up the Definite Integral

Integrate the Function

Apply the Limits

Final Calculation

Final Conclusion

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Discriminant's Secret

The Hunt for $p$

Visualizing the Geometry

The Calculus of Victory

Similar Questions

The area of the region ${(x, y) : ∣ x - y ∣ \leq y \leq 4 x}$ is

The area (in sq. units) of the region ${x \in R : x \geq 0, y \geq 0, y \geq x - 2 and y \leq x}$ , is

Find the area bounded by the curves, $x^{2} + y^{2} = 25, 4 y = 4 - x^{2}$ and $x = 0$ above the $x$ -axis.

Let $g (x) = cos^{2} x, f (x) = x$ , and $α, β (α < β)$ be the roots of the quadratic equation $18 x^{2} - 9 π x + π^{2} = 0$ . Then the area (in sq. units) bounded by the curve $y = (g \circ f) (x)$ and the lines $x = α, x = β$ and $y = 0$ , is :

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Sketch the region bounded by the curves $y = x^{2}$ and $y = \frac{2}{1 + x ^{2}}$ . Find the area.

Select Answer:

Visualized Solution

The Quadratic Equation

Condition for Rational Roots

Calculate the Discriminant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Simplify the Discriminant

Finding the Maximum Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Define the Bounded Region

Visualize the Parabola

The Y-Intercept

Identify the Area to Calculate

Set up the Definite Integral

Integrate the Function

Apply the Limits

Final Calculation

Final Conclusion

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Discriminant's Secret

The Hunt for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Visualizing the Geometry

The Calculus of Victory

Similar Questions

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Calculate the Discriminant $D$

Finding the Maximum Integral $p$

The Hunt for $p$

The area of the region ${(x, y) : ∣ x - y ∣ \leq y \leq 4 x}$ is

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Find the area bounded by the curves, $x^{2} + y^{2} = 25, 4 y = 4 - x^{2}$ and $x = 0$ above the $x$ -axis.

The area (in sq. units) of the region $A = {(x, y) : x^{2} \leq y \leq x + 2}$ is

The area of the region ${(x, y) : x^{2} + 4 x + 2 \leq y \leq ∣ x + 2∣}$ is equal to

Let the area of the region ${(x, y) : 0 \leq x \leq 3, 0 \leq y \leq min {x^{2} + 2, 2 x + 2}}$ be $A$ . Then $12 A$ is equal to

The area of the region $R = {(x, y) : x y \leq 8, 1 \leq y \leq x^{2}, x \geq 0}$ is

Sketch the region bounded by the curves $y = x^{2}$ and $y = \frac{2}{1 + x ^{2}}$ . Find the area.