QJEEM24Maf801: JEE Main 2024 Mathematics - Class 12 JEEM24Maf801 Solution: The area enclosed by the curves and is equal to :

To find the enclosed area between the two curves $x y+4 y=16$ and $x+y=6$ , we need to determine the region of intersection and integrate the difference of the functions over the interval where they intersect.

First, let's solve the equations simultaneously to find the points of intersection.

The second curve $x+y=6$ can be rewritten as $y=6-x$ .

Substitute $y=6-x$ into the first equation:

$x(6-x) + 4(6-x) = 16$

$6x - x^2 + 24 - 4x = 16$

$-x^2 + 2x + 8 = 0$

This can be simplified to:

$x^2 - 2x - 8 = 0$

Using the quadratic formula, we can find the roots of this equation:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}$

$x = \frac{2 \pm \sqrt{4 + 32}}{2}$

$x = \frac{2 \pm \sqrt{36}}{2}$

$x = \frac{2 \pm 6}{2}$

So we have two solutions:

$x_1 = \frac{2 + 6}{2} = 4$

$x_2 = \frac{2 - 6}{2} = -2$

For $x_1 = 4$ , substitute this back into the line equation $x+y=6$ :

$4+y=6$

$y=2$

For $x_2 = -2$ , do the same:

$-2+y=6$

$y=8$

So we have two points of intersection: $(4,2)$ and $(-2,8)$ .

Now, to find the area between $x y + 4y = 16$ and $x + y = 6$ , we'll integrate the top function minus the bottom function from $x = -2$ to $x = 4$ . First, we need to express $y$ from both equations in terms of $x$ :

For $x y + 4 y = 16$ , express $y$ in terms of $x$ :

$y(x+4) = 16$

$y = \frac{16}{x+4}$

For $x + y = 6$ , we have:

$y = 6 - x$

The area $A$ is therefore given by:

$A = \int\limits_{-2}^{4}\left( 6 - x - \frac{16}{x+4} \right) dx$

Now, we integrate:

$A = \int\limits_{-2}^{4} (6 - x) dx - \int_{-2}^{4} \frac{16}{x + 4} dx$

$A = \left[6x - \frac{x^2}{2}\right]_{-2}^{4} - \left[16 \ln|x + 4|\right]_{-2}^{4}$

$A = \left( 6 \times 4 - \frac{1}{2} \times 4^2 \right) - \left( 6 \times -2 - \frac{1}{2} \times -2^2 \right) - 16 \left( \ln|4+4| - \ln|-2+4| \right)$

$A = \left( 24 - 8 \right) - \left( -12 - 2 \right) - 16 \left( \ln 8 - \ln 2 \right)$

$A = 16 + 14 - 16 \left( \ln (2^3) - \ln 2 \right)$

$A = 30 - 16 \left( 3 \ln 2 - \ln 2 \right)$

$A = 30 - 16 (2 \ln 2)$

$A = 30- 32 \ln 2$