The equation of the ellipse is $x^2 + 4y^2 = 4$. Dividing by 4 gives the standard form: $ \frac{x^2}{4} + \frac{y^2}{1} = 1 $ This ellipse has semi-axes $a=2$ and $b=1$. The area of the ellipse ($A_E$) is $A_E = \pi ab$. $ A_E = \pi (2)(1) = 2\pi $
The region is bounded by $y = |x| - 1$ and $y = 1 - |x|$. This defines a rhombus with vertices at $(0, 1)$, $(1, 0)$, $(0, -1)$, and $(-1, 0)$. The diagonals are $d_1 = 2$ and $d_2 = 2$. The area of the rhombus ($A_R$) is: $ A_R = \frac{1}{2} d_1 d_2 = \frac{1}{2} (2)(2) = 2 $
The vertices of the rhombus $(\pm 1, 0)$ and $(0, \pm 1)$ satisfy $x^2 + 4y^2 \le 4$. For $(\pm 1, 0)$: $1^2 + 4(0)^2 = 1 < 4$. For $(0, \pm 1)$: $0^2 + 4(\pm 1)^2 = 4$. The rhombus is contained within or on the boundary of the ellipse.
The required area is inside the ellipse and outside the rhombus. $ \text{Area} = \text{Area of Ellipse} - \text{Area of Rhombus} $ $ \text{Area} = A_E - A_R $ $ \text{Area} = 2\pi - 2 $ $ \text{Area} = 2(\pi - 1) $
If $y = y(x)$ satisfies the differential equation
$16(\sqrt{x+ 9\sqrt{x}})(4 + \sqrt{9 + \sqrt{x}}) \cos y \, dy = (1 + 2 \sin y) dx, x > 0$ and $y(256) = \frac{\pi}{2}, y(49) = \alpha$, then $2 \sin \alpha$ is equal to :
Let $f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} \,dx$, $x > 0$, $\lim_{x \rightarrow 0} f(x) = 0$ and $f(1) = \frac{1}{4}$.
If $A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$ and $B = \text{adj}(\text{adj } A)$ be such that $|B| = 81$, then $\alpha^2$ is equal to
Let $f$ be a differentiable function satisfying $f(x) = 1 - 2x + \int_{0}^{x} e^{(x - t)} f(t) dt$, $x \in \mathbf{R}$ and let $g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$, $x \in \mathbf{R}$. If $\text{p}$ and $\text{q}$ are respectively the points of local minima and local maxima of $g$, then the value of $|\text{p} + \text{q}|$ is equal to _________.
If $y = y(x)$ satisfies the differential equation
$16(\sqrt{x+ 9\sqrt{x}})(4 + \sqrt{9 + \sqrt{x}}) \cos y \, dy = (1 + 2 \sin y) dx, x > 0$ and $y(256) = \frac{\pi}{2}, y(49) = \alpha$, then $2 \sin \alpha$ is equal to :