If the set $\left\{ y \mid y = x + t \left( x ^ { 2 } - x \right) , 0 \leq t \leq 1, 1 \leq x \leq 2 \right\}$ represents a figure where the maximum distance between two points is $d$ and the area is $S$,
A. $d = 3 , S < 1$
B. $d = 3 , S > 1$
C. $d = \sqrt { 10 } , S < 1$
D. $d = \sqrt { 10 } , S > 1$

The area of the region $$\left\{ (x,y) : 0 \leq x \leq \frac{9}{4}, \quad 0 \leq y \leq 1, \quad x \geq 3y, \quad x + y \geq 2 \right\}$$ is
(A) $\frac{11}{32}$
(B) $\frac{35}{96}$
(C) $\frac{37}{96}$
(D) $\frac{13}{32}$

Let $S = \left\{ ( x , y ) \in \mathbb { R } \times \mathbb { R } : x \geq 0 , y \geq 0 , y ^ { 2 } \leq 4 x , y ^ { 2 } \leq 12 - 2 x \right.$ and $\left. 3 y + \sqrt { 8 } x \leq 5 \sqrt { 8 } \right\}$. If the area of the region $S$ is $\alpha \sqrt { 2 }$, then $\alpha$ is equal to
(A) $\frac { 17 } { 2 }$
(B) $\frac { 17 } { 3 }$
(C) $\frac { 17 } { 4 }$
(D) $\frac { 17 } { 5 }$

The area (in sq. units) of the region described by $\{(x, y) : y^2 \leq 2x \text{ and } y \geq 4x - 1\}$ is:
(1) $\frac{7}{32}$
(2) $\frac{5}{64}$
(3) $\frac{15}{64}$
(4) $\frac{9}{32}$

The area (in sq. units) of the region $\{ ( x , y ) : x \geq 0 , x + y \leq 3 , x ^ { 2 } \leq 4 y$ and $y \leq 1 + \sqrt { x } \}$ is:
(1) $\frac { 59 } { 12 }$
(2) $\frac { 3 } { 2 }$
(3) $\frac { 7 } { 3 }$
(4) $\frac { 5 } { 2 }$

The area (in sq. units) of the region $\left\{ ( x , y ) : 0 \leq y \leq x ^ { 2 } + 1,0 \leq y \leq x + 1 , \frac { 1 } { 2 } \leq x \leq 2 \right\}$ is
(1) $\frac { 23 } { 16 }$
(2) $\frac { 79 } { 24 }$
(3) $\frac { 79 } { 16 }$
(4) $\frac { 23 } { 6 }$

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 (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) $\frac { 9 } { 32 }$

Let the area of the region $\{(x, y): 2y \leq x^2 + 3,\ y + |x| \leq 3,\ y \geq |x-1|\}$ be A. Then $6A$ is equal to:
(1) 16
(2) 12
(3) 14
(4) 18