JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Definite Integrals

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:

Answer: (A)

Solution

Solving

x^{2} = 2 y

with

y = 2 x + 6

gives

x^{2} = 4 x + 12

, so

x^{2} - 4 x - 12 = 0

and

x = 6

x = - 2

. The second-quadrant point has

x = - 2, y = 2

, so the limits are

a = - 2

b = 2

. Writing

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

and replacing

x

- x

gives

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

. Adding the two forms,

2 I = \int_{- 2}^{2} 9 x^{2} d x = 18 \int_{0}^{2} x^{2} d x = 18 \cdot \frac{8}{3} = 48

, so

I = 24

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option

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