Let (a, b) be the point of intersection of the curve and the straight line y – 2x – 6 = 0 in the second quadrant. Then the integral is equal to : [2025]

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Solution

Q.
Let (a, b) be the point of intersection of the curve $x^{2} = 2 y$ and the straight line y – 2x – 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 : [2025]

1 21

2 18

3 24

4 27

Ans.
(3)

We have, $x^{2} = 2 y$ and y – 2x – 6 = 0

Substitute y = 2x + 6 in $x^{2} = 2 y$ , we get

$x^{2} = (4 x + 12)$

$\Rightarrow x^{2} - 4 x - 12 = 0 \Rightarrow (x - 6) (x + 2) = 0$

$\Rightarrow x = 6 or - 2 \Rightarrow y = 18 or 2$

$∴$ Intersection points are (6, 18) and (–2, 2)

Since, (a, b) is a point in second quadrant

$∴$ (a, b) = (–2, 2)

Now, $I = \int_{- 2}^{2} \frac{9 x^{2}}{1 + 5^{x}} d x$ ... (i)

$∴ I = \int_{- 2}^{2} \frac{9 x^{2}}{1 + 5^{- x}} d x = \int_{- 2}^{2} \frac{9 x^{2} 5^{x}}{5^{x} + 1} d x$ ... (ii)

On adding (i) and (ii), we get

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

$= {[\frac{9 x^{3}}{3}]}_{- 2}^{2} = 3 [8 - (- 8)] = 48$

$\Rightarrow I = \frac{48}{2} = 24$ .

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