Q.19. The area of the region {(x, y) : , , } is :

Q: What is the correct answer for JEE Main Mathematics Q19?

The correct answer is B. Sol. The region is bounded by $y = 6-x$ (or $x = 6-y$), $y^2 = 4x-3$ (or $x = \frac{y^2+3}{4}$), and $x \ge 0$. We need to find the intersection points of $x = 6-y$ and $x = \frac{y^2+3}{4}$. $6-y = \frac{y^2+3}{4} \Rightarrow 24-4y = y^2+3 \Rightarrow y^2+4y-21 = 0 \Rightarrow (y+7)(y-3) = 0$. Since $y \ge 0$ (from $0 \le y \le 6-x$ and $x \ge 0$), we take $y=3$. When $y=3$, $x = 6-3 = 3$. So, th

Q: Q.19. The area of the region {(x, y) : , , } is :

The correct answer is B. Sol. The region is bounded by $y = 6-x$ (or $x = 6-y$), $y^2 = 4x-3$ (or $x = \frac{y^2+3}{4}$), and $x \ge 0$. We need to find the intersection points of $x = 6-y$ and $x = \frac{y^2+3}{4}$. $6-y = \frac{y^2+3}{4} \Rightarrow 24-4y = y^2+3 \Rightarrow y^2+4y-21 = 0 \Rightarrow (y+7)(y-3) = 0$. Since $y \ge 0$ (from $0 \le y \le 6-x$ and $x \ge 0$), we take $y=3$. When $y=3$, $x = 6-3 = 3$. So, th

Question

Q.19. The area of the region {(x, y) : , , } is :

FirstInTest · Accepted Answer

Sol. The region is bounded by $y = 6-x$ (or $x = 6-y$), $y^2 = 4x-3$ (or $x = \frac{y^2+3}{4}$), and $x \ge 0$. We need to find the intersection points of $x = 6-y$ and $x = \frac{y^2+3}{4}$. $6-y = \frac{y^2+3}{4} \Rightarrow 24-4y = y^2+3 \Rightarrow y^2+4y-21 = 0 \Rightarrow (y+7)(y-3) = 0$. Since $y \ge 0$ (from $0 \le y \le 6-x$ and $x \ge 0$), we take $y=3$. When $y=3$, $x = 6-3 = 3$. So, the intersection point is (3,3). The area can be calculated by integrating with respect to $y$. The limits for $y$ are from $0$ to $3$. The area $A = \int_0^3 [(6-y) - (\frac{y^2+3}{4})] dy$. $A = \int_0^3 (6-y - \frac{y^2}{4} - \frac{3}{4}) dy = \int_0^3 (\frac{21}{4} - y - \frac{y^2}{4}) dy$. $A = [\frac{21}{4}y - \frac{y^2}{2} - \frac{y^3}{12}]_0^3 = \frac{21}{4}(3) - \frac{3^2}{2} - \frac{3^3}{12} = \frac{63}{4} - \frac{9}{2} - \frac{27}{12} = \frac{189 - 54 - 27}{12} = \frac{108}{12} = 9$.

Answer

8

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9

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12

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15

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