JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $f(x) = \frac{x + |x|}{2}$ and $g(x) = \begin{cases} x, & x < 0 \\ x^2, & x \geq 0 \end{cases}$ . Then area bounded by the curve $y = (f \cdot g)(x)$ and the lines $y = 0, 2y - x = 15$ is equal to:

Choose the correct answer:

Explanation

1. Function ko Simplify karna

f(x) = \frac{x+|x|}{2} \text{ and } g(x) = \begin{cases} x, & x < 0 \\ x^2, & x \geq 0 \end{cases}

fog(x) = f[g(x)] = \begin{cases} g(x), & g(x) \geq 0 \\ 0, & g(x) < 0 \end{cases}

fog(x) = \begin{cases} x^2, & x \geq 0 \\ 0, & x < 0 \end{cases}

2. Given Lines aur Intersection

Diye gaye samikaran hain:

2y - x = 15 \text{ and } y = 0

Image ke anusar, line $y = \frac{x+15}{2}$ aur curve $y = x^2$ ka intersection point $(3, 9)$ hai kyunki:

2(9) - 3 = 18 - 3 = 15

3. Area ki Calculation

Area ko do bhagon mein banta gaya hai: $x = -15$ se $0$ tak ka triangle aur $x = 0$ se $3$ tak ka curve ke beech ka hissa.

Area = \int_{0}^{3} \left( \frac{x+15}{2} - x^2 \right) dx + \frac{1}{2} \times \frac{15}{2} \times 15

= \left[ \frac{x^2}{4} + \frac{15x}{2} - \frac{x^3}{3} \right]_{0}^{3} + \frac{225}{4}

= \left( \frac{9}{4} + \frac{45}{2} - \frac{27}{3} \right) + \frac{225}{4}

= \frac{9}{4} + \frac{45}{2} - 9 + \frac{225}{4}

Sabhi terms ko solve karne par:

= \frac{9 + 90 - 36 + 225}{4} = \frac{288}{4} = 72

\text{Area} = 72

Choose the correct answer:

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