Q.16. The area of the region R = {(x,y): , } is equal to :

Q: What is the correct answer for JEE Main Mathematics Q16?

The correct answer is B. Sol. The region R is defined by $xy \le 27$, $1 \le y \le x^2$. This means $y \ge 1$, $y \le x^2$, and $y \le \frac{27}{x}$. We need to find the intersection points of the curves $y=1$, $y=x^2$, and $y=\frac{27}{x}$. 1. $y=1$ and $y=x^2 \Rightarrow x^2=1 \Rightarrow x=1$ (since $x>0$ for $y=27/x$ to be positive). Intersection point (1,1). 2. $y=x^2$ and $y=\frac{27}{x} \Rightarrow x^2 = \frac{2

Q: Q.16. The area of the region R = {(x,y): , } is equal to :

The correct answer is B. Sol. The region R is defined by $xy \le 27$, $1 \le y \le x^2$. This means $y \ge 1$, $y \le x^2$, and $y \le \frac{27}{x}$. We need to find the intersection points of the curves $y=1$, $y=x^2$, and $y=\frac{27}{x}$. 1. $y=1$ and $y=x^2 \Rightarrow x^2=1 \Rightarrow x=1$ (since $x>0$ for $y=27/x$ to be positive). Intersection point (1,1). 2. $y=x^2$ and $y=\frac{27}{x} \Rightarrow x^2 = \frac{2

Q: What are the options for this JEE Main Mathematics question?

A. 78\log_e 3 - \frac{52}{3} · B. 54\log_e 3 - \frac{52}{3} · C. 54\log_e 3 - \frac{26}{3} · D. 54\log_e 3 + \frac{26}{3}

Question

Q.16. The area of the region R = {(x,y): , } is equal to :

FirstInTest · Accepted Answer

Sol. The region R is defined by $xy \le 27$, $1 \le y \le x^2$.
This means $y \ge 1$, $y \le x^2$, and $y \le \frac{27}{x}$.
We need to find the intersection points of the curves $y=1$, $y=x^2$, and $y=\frac{27}{x}$.
1. $y=1$ and $y=x^2 \Rightarrow x^2=1 \Rightarrow x=1$ (since $x>0$ for $y=27/x$ to be positive).
   Intersection point (1,1).
2. $y=x^2$ and $y=\frac{27}{x} \Rightarrow x^2 = \frac{27}{x} \Rightarrow x^3 = 27 \Rightarrow x=3$.
   Intersection point (3,9).
3. $y=1$ and $y=\frac{27}{x} \Rightarrow 1 = \frac{27}{x} \Rightarrow x=27$.
   Intersection point (27,1).
The region is bounded by $y=1$ from below, and by $y=x^2$ and $y=\frac{27}{x}$ from above.
From $x=1$ to $x=3$, $x^2$ is the upper bound. From $x=3$ to $x=27$, $\frac{27}{x}$ is the upper bound.
Area = $\int_1^3 (x^2-1)dx + \int_3^{27} (\frac{27}{x}-1)dx$
$= [\frac{x^3}{3}-x]_1^3 + [27\ln|x|-x]_3^{27}$
$= ((\frac{3^3}{3}-3) - (\frac{1^3}{3}-1)) + ((27\ln(27)-27) - (27\ln(3)-3))$
$= ((9-3) - (\frac{1}{3}-1)) + (27\ln(3^3)-27 - 27\ln(3)+3)$
$= (6 - (-\frac{2}{3})) + (27 \cdot 3\ln(3) - 27 - 27\ln(3) + 3)$
$= (6 + \frac{2}{3}) + (81\ln(3) - 27\ln(3) - 24)$
$= \frac{18+2}{3} + 54\ln(3) - 24$
$= \frac{20}{3} + 54\ln(3) - 24$
$= 54\ln(3) + \frac{20-72}{3} = 54\ln(3) - \frac{52}{3}$.

Answer

78\log_e 3 - \frac{52}{3}

Answer

54\log_e 3 - \frac{52}{3}

Answer

54\log_e 3 - \frac{26}{3}

Answer

54\log_e 3 + \frac{26}{3}

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