The area bounded by the curve $y = \left| x ^ { 2 } - 9 \right|$ and the line $y = 3$ is
(1) $8 \sqrt { 6 } - 16 \sqrt { 12 } - 72$
(2) $8 \sqrt { 6 } + 8 \sqrt { 12 } - 72$
(3) $16 \sqrt { 6 } + 16 \sqrt { 12 } - 72$
(4) $16 \sqrt { 6 } - 16 \sqrt { 12 } - 64$

If the area of the region $\left\{ ( x , y ) : x ^ { \frac { 2 } { 3 } } + y ^ { \frac { 2 } { 3 } } \leq 1 , x + y \geq 0 , y \geq 0 \right\}$ is $A$, then $\frac { 256 A } { \pi }$ is

The area of the region $\{(x, y): x^2 \leq y \leq |x^2 - 4|, y \geq 1\}$ is
(1) $\frac{4(\sqrt{5}-1)}{3} + 4$
(2) $\frac{4(\sqrt{5}-1)}{3} + 2$
(3) $\frac{2(\sqrt{5}-1)}{3} + 4$
(4) $\frac{2(\sqrt{5}-1)}{3} + 2$

The area of the region $\{ ( x , y ) : x ^ { 2 } \leq y \leq | x ^ { 2 } - 4 | , y \geq 1 \}$ is
(1) $\frac { 4 } { 3 } ( 4 \sqrt { 2 } - 1 )$
(2) $\frac { 4 } { 3 } ( 4 \sqrt { 2 } + 1 )$
(3) $\frac { 3 } { 4 } ( 4 \sqrt { 2 } + 1 )$
(4) $\frac { 3 } { 4 } ( 4 \sqrt { 2 } - 1 )$

Let T and C respectively, be the transverse and conjugate axes of the hyperbola $16 x ^ { 2 } - y ^ { 2 } + 64 x + 4 y + 44 = 0$. Then the area of the region above the parabola $x ^ { 2 } = y + 4$, below the transverse axis T and on the right of the conjugate axis C is:
(1) $4 \sqrt { 6 } + \frac { 44 } { 3 }$
(2) $4 \sqrt { 6 } + \frac { 28 } { 3 }$
(3) $4 \sqrt { 6 } - \frac { 44 } { 3 }$
(4) $4 \sqrt { 6 } - \frac { 28 } { 3 }$

Let $A = \left\{ ( x , y ) \in \mathbb { R } ^ { 2 } : y \geq 0,2 x \leq y \leq \sqrt { 4 - ( x - 1 ) ^ { 2 } } \right\}$ and $B = \left\{ ( x , y ) \in \mathbb { R } \times \mathbb { R } : 0 \leq y \leq \min \left\{ 2 x , \sqrt { 4 - ( x - 1 ) ^ { 2 } } \right\} \right\}$. Then the ratio of the area of $A$ to the area of $B$ is
(1) $\frac { \pi - 1 } { \pi + 1 }$
(2) $\frac { \pi } { \pi - 1 }$
(3) $\frac { \pi } { \pi + 1 }$
(4) $\frac { \pi + 1 } { \pi - 1 }$

Let $q$ be the maximum integral value of $p$ in $[0, 10]$ for which the roots of the equation $x^{2} - px + \frac{5}{4}p = 0$ are rational. Then the area of the region $\left\{(x, y): 0 \leq y \leq (x - q)^{2},\, 0 \leq x \leq q\right\}$ is
(1) 243
(2) 25
(3) $\frac{125}{3}$
(4) 164

Let $\Delta$ be the area of the region $\left\{ ( x , y ) \in \mathbb { R } ^ { 2 } : x ^ { 2 } + y ^ { 2 } \leq 21 , y ^ { 2 } \leq 4 x , x \geq 1 \right\}$. Then $\frac { 1 } { 2 } \left( \Delta - 21 \sin ^ { - 1 } \frac { 2 } { \sqrt { 7 } } \right)$ is equal to
(1) $2 \sqrt { 3 } - \frac { 1 } { 3 }$
(2) $\sqrt { 3 } - \frac { 2 } { 3 }$
(3) $2 \sqrt { 3 } - \frac { 2 } { 3 }$
(4) $\sqrt { 3 } - \frac { 4 } { 3 }$

The area of the region $A = \left\{ ( x , y ) : | \cos x - \sin x | \leq y \leq \sin x , 0 \leq x \leq \frac { \pi } { 2 } \right\}$ is: (1) $1 - \frac { 3 } { \sqrt { 2 } } + \frac { 4 } { \sqrt { 5 } }$ (2) $\sqrt { 5 } + 2 \sqrt { 2 } - 4.5$ (3) $\frac { 3 } { \sqrt { 5 } } - \frac { 3 } { \sqrt { 2 } } + 1$ (4) $\sqrt { 5 } - 2 \sqrt { 2 } + 1$

The area of the region $\left\{ ( x , y ) : x ^ { 2 } \leq y \leq 8 - x ^ { 2 } , y \leq 7 \right\}$ is
(1) 27
(2) 18
(3) 20
(4) 21

Let $A$ be the area bounded by the curve $y = x(x - 3)$, the $x$-axis and the ordinates $x = -1$ and $x = 2$. Then $12A$ is equal to $\_\_\_\_$.

Let for $x \in \mathbb{R}$, $f(x) = \frac{x + |x|}{2}$ and $g(x) = \begin{cases} x, & x < 0 \\ x^2, & x \geq 0 \end{cases}$. Then area bounded by the curve $y = f(g(x))$ and the lines $y = 0$, $2y - x = 15$ is equal to $\underline{\hspace{1cm}}$.

Let $y = p(x)$ be the parabola passing through the points $(-1, 0)$, $(0, 1)$ and $(1, 0)$. If the area of the region $\{(x, y) : (x+1)^2 + (y-1)^2 \leq 1,\; y \leq p(x)\}$ is $A$, then $12\pi - 4A$ is equal to $\_\_\_\_$.

The area of the region $\left\{(x, y) : y ^ { 2 } \leq 4 x , x < 4 , \frac { x y (x - 1)(x - 2) } { (x - 3)(x - 4) } > 0 , x \neq 3 \right\}$ is
(1) $\frac { 16 } { 3 }$
(2) $\frac { 64 } { 3 }$
(3) $\frac { 8 } { 3 }$
(4) $\frac { 32 } { 3 }$

The area enclosed between the curves $y = x | x |$ and $y = x - | x |$ is :
(1) $\frac { 4 } { 3 }$
(2) 1
(3) $\frac { 2 } { 3 }$
(4) $\frac { 8 } { 3 }$

If the area of the region $\left\{ ( x , y ) : \frac { \mathrm { a } } { x ^ { 2 } } \leq y \leq \frac { 1 } { x } , 1 \leq x \leq 2,0 < \mathrm { a } < 1 \right\}$ is $\left( \log _ { \mathrm { e } } 2 \right) - \frac { 1 } { 7 }$ then the value of $7 \mathrm { a } - 3$ is equal to:
(1) 0
(2) 2
(3) $- 1$
(4) 1

The area of the region in the first quadrant inside the circle $x ^ { 2 } + y ^ { 2 } = 8$ and outside the parabola $y ^ { 2 } = 2 x$ is equal to : (1) $\frac { \pi } { 2 } - \frac { 1 } { 3 }$ (2) $\pi - \frac { 1 } { 3 }$ (3) $\frac { \pi } { 2 } - \frac { 2 } { 3 }$ (4) $\pi - \frac { 2 } { 3 }$

Let the area of the region enclosed by the curves $y = 3 x , 2 y = 27 - 3 x$ and $y = 3 x - x \sqrt { x }$ be $A$. Then $10 A$ is equal to
(1) 172
(2) 162
(3) 154
(4) 184

The area (in sq. units) of the region described by $\left\{ ( x , y ) : y ^ { 2 } \leq 2 x \right.$, and $\left. y \geq 4 x - 1 \right\}$ is
(1) $\frac { 11 } { 32 }$
(2) $\frac { 8 } { 9 }$
(3) $\frac { 11 } { 12 }$
(4) $\frac { 9 } { 32 }$

The area (in square units) of the region enclosed by the ellipse $x ^ { 2 } + 3 y ^ { 2 } = 18$ in the first quadrant below the line $y = x$ is
(1) $\sqrt { 3 } \pi - \frac { 3 } { 4 }$
(2) $\sqrt { 3 } \pi + 1$
(3) $\sqrt { 3 } \pi$
(4) $\sqrt { 3 } \pi + \frac { 3 } { 4 }$

Let the area of the region $\left\{ ( x , y ) : x - 2 y + 4 \geq 0 , x + 2 y ^ { 2 } \geq 0 , x + 4 y ^ { 2 } \leq 8 , y \geq 0 \right\}$ be $\frac { m } { n }$, where $m$ and $n$ are coprime numbers. Then $\mathrm { m } + \mathrm { n }$ is equal to $\_\_\_\_$.

Let $\mathrm { fx } = \int _ { 0 } ^ { \mathrm { x } } \mathrm { gt } \log _ { \mathrm { e } } \frac { 1 - \mathrm { t } } { 1 + \mathrm { t } } \mathrm { dt }$, where g is a continuous odd function. If $\int _ { - \frac { \pi } { 2 } } ^ { \frac { \pi } { 2 } } \mathrm { fx } + \frac { \mathrm { x } ^ { 2 } \cos \mathrm { x } } { 1 + \mathrm { e } ^ { \mathrm { x } } } \mathrm { dx } = \frac { \pi ^ { 2 } } { \alpha } - \alpha$, then $\alpha$ is equal to $\_\_\_\_$.

If the area of the region $( x , y ) : 0 \leq y \leq \min 2 x , 6 x - x ^ { 2 }$ is $A$, then $12 A$ is equal to $\_\_\_\_$ .

The area of the region enclosed by the parabola $(y-2)^2 = x - 1$, the line $x - 2y + 4 = 0$ and the positive coordinate axes is $\underline{\hspace{1cm}}$.

Let the area of the region enclosed by the curve $y = \min \{ \sin x , \cos x \}$ and the $x$ axis between $x = - \pi$ to $x = \pi$ be $A$. Then $A ^ { 2 }$ is equal to $\_\_\_\_$