Maths Q19 – JEE-MAIN 2025

Let

(a, b)

be the point of intersection of the curve

x^{2} = 2 y

and the straight line

y - 2 x - 6 = 0

in the second quadrant. Then the integral

I = \int_{a}^{b} \frac{9 x^{2}}{1 + 5^{x}} d x

is equal to :

Clue 1 of 3: Intersection Point

First, determine the coordinates $(a, b)$ of the intersection point of the given curve and line in the specified quadrant.

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Step 1: Find the intersection point

(a, b)

✦ Active

Substitute $y = \frac{x^{2}}{2}$ from the curve equation into the line equation $y - 2 x - 6 = 0$ . This gives $\frac{x^{2}}{2} - 2 x - 6 = 0$ , which simplifies to $x^{2} - 4 x - 12 = 0$ . Factoring the quadratic equation yields $(x - 6) (x + 2) = 0$ , so $x = 6$ or $x = - 2$ . For $x = 6$ , $y = 2 (6) + 6 = 18$ , giving point $(6, 18)$ . For $x = - 2$ , $y = 2 (- 2) + 6 = 2$ , giving point $(- 2, 2)$ . Since the point $(a, b)$ is in the second quadrant (where $x < 0$ and $y > 0$ ), we have $a = - 2$ and $b = 2$ .

Step 2: Apply King's Property of Definite Integrals○ Expand

The integral is $I = \int_{- 2}^{2} \frac{9 x^{2}}{1 + 5^{x}} d x$ . Using the property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ , with $a = - 2$ and $b = 2$ , we have $a + b - x = - 2 + 2 - x = - x$ . Applying this to the integral:

I = \int_{- 2}^{2} \frac{9 (- x)^{2}}{1 + 5^{- x}} d x = \int_{- 2}^{2} \frac{9 x^{2}}{1 + \frac{1}{5^{x}}} d x = \int_{- 2}^{2} \frac{9 x^{2}}{\frac{5^{x} + 1}{5^{x}}} d x = \int_{- 2}^{2} \frac{9 x^{2} \cdot 5^{x}}{1 + 5^{x}} d x

💡 Teacher's Secret Hint

Remember that $5^{- x} = \frac{1}{5^{x}}$ and simplify the denominator.

Step 3: Combine and Evaluate the Integral○ Expand

Add the original integral and the transformed integral:

2 I = \int_{- 2}^{2} (\frac{9 x^{2}}{1 + 5^{x}} + \frac{9 x^{2} \cdot 5^{x}}{1 + 5^{x}}) d x = \int_{- 2}^{2} \frac{9 x^{2} (1 + 5^{x})}{1 + 5^{x}} d x = \int_{- 2}^{2} 9 x^{2} d x

Since $9 x^{2}$ is an even function, $\int_{- c}^{c} f (x) d x = 2 \int_{0}^{c} f (x) d x$ . Thus, $2 I = 2 \int_{0}^{2} 9 x^{2} d x$ . Dividing by 2, we get:

I = \int_{0}^{2} 9 x^{2} d x = {[9 \frac{x^{3}}{3}]}_{0}^{2} = {[3 x^{3}]}_{0}^{2} = 3 (2^{3}) - 3 (0^{3}) = 3 (8) - 0 = 24

💡 Teacher's Secret Hint

Recognize that $9 x^{2}$ is an even function to simplify the limits of integration.

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Maths Question 19 – JEE-MAIN 2025

Clue 1 of 3: Intersection Point

🧠 Full Solution Path