Let $a , b > 0 , a + b = 5$. The maximum value of $\sqrt { a + 1 } + \sqrt { b + 3 }$ is $\_\_\_\_$ .

If the sum of the square of the roots of the equation $x^{2} - (\sin\alpha - 2)x - (1+\sin\alpha) = 0$ is least, then $\alpha$ is equal to
(1) $\frac{\pi}{6}$
(2) $\frac{\pi}{4}$
(3) $\frac{\pi}{3}$
(4) $\frac{\pi}{2}$

If $a \in R$ and the equation $- 3 ( x - [ x ] ) ^ { 2 } + 2 ( x - [ x ] ) + a ^ { 2 } = 0$ (where $[ x ]$ denotes the greatest integer $\leq x )$ has no integral solution, then all possible values of $a$ lie in the interval
(1) $( - 2 , - 1 )$
(2) $( - \infty , - 2 ) \cup ( 2 , \infty )$
(3) $( - 1,0 ) \cup ( 0,1 )$
(4) $( 1,2 )$

If, for a positive integer $n$, the quadratic equation,
$$x(x + 1) + (x + 1)(x + 2) + \ldots + (x + \overline{n-1})(x + n) = 10n$$
has two consecutive integral solutions, then $n$ is equal to:
(1) 12
(2) 9
(3) 10
(4) 11

If $\lambda \in R$ is such that the sum of the cubes of the roots of the equation $x ^ { 2 } + ( 2 - \lambda ) x + ( 10 - \lambda ) = 0$ is minimum, then the magnitude of the difference of the roots of this equation is :
(1) $4 \sqrt { 2 }$
(2) 20
(3) $2 \sqrt { 5 }$
(4) $2 \sqrt { 7 }$

If $m$ is chosen in the quadratic equation $\left( m ^ { 2 } + 1 \right) x ^ { 2 } - 3 x + \left( m ^ { 2 } + 1 \right) ^ { 2 } = 0$ such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is:
(1) $4 \sqrt { 3 }$
(2) $10 \sqrt { 5 }$
(3) $8 \sqrt { 3 }$
(4) $8 \sqrt { 5 }$

The minimum value of the sum of the squares of the roots of $x ^ { 2 } + 3 - a x = 2 a - 1$ is
(1) 6
(2) 4
(3) 5
(4) 8

Let $S = \left\{ \alpha : \log _ { 2 } \left( 9 ^ { 2 \alpha - 4 } + 13 \right) - \log _ { 2 } \left( \frac { 5 } { 2 } \cdot 3 ^ { 2 \alpha - 4 } + 1 \right) = 2 \right\}$. Then the maximum value of $\beta$ for which the equation $\mathrm { x } ^ { 2 } - 2 \left( \sum _ { \alpha \in s } \alpha \right) ^ { 2 } \mathrm { x } + \sum _ { a \in s } ( \alpha + 1 ) ^ { 2 } \beta = 0$ has real roots, is $\_\_\_\_$ .

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

Consider the following quadratic equations in $x$
$$x^2+2x-15=0 \tag{1}$$ $$2x^2+3x+a^2+12a=0 \tag{2}$$
Let us denote the two solutions of (1) by $\alpha$ and $\beta$ ($\alpha < \beta$). We are to find the range of values which $a$ in (2) can take, in order that (2) has two real solutions $\gamma$ and $\delta$ and they satisfy
$$\alpha < \gamma < \beta < \delta.$$
(1) $\alpha = \mathbf{AB}$ and $\beta = \mathbf{C}$.
(2) When we set $b = a^2+12a$, from the condition $\alpha < \gamma$ we have
$$b > \mathbf{DEF}$$
and from the condition $\gamma < \beta < \delta$ we have
$$b < \mathbf{GHI}.$$
Hence the range of the values which $a$ can take is
$$\mathbf{JK} < a < \mathbf{LM}, \quad \mathbf{NO} < a < \mathbf{PQ},$$
where $\mathrm{JK} < \mathrm{NO}$.