If $\alpha$ and $\beta$ are the roots of the equation $x ^ { 2 } + p x + 2 = 0$ and $\frac { 1 } { \alpha }$ and $\frac { 1 } { \beta }$ are the roots of the equation $2 x ^ { 2 } + 2 q x + 1 = 0$, then $\left( \alpha - \frac { 1 } { \alpha } \right) \left( \beta - \frac { 1 } { \beta } \right) \left( \alpha + \frac { 1 } { \beta } \right) \left( \beta + \frac { 1 } { \alpha } \right)$ is equal to:
(1) $\frac { 9 } { 4 } \left( 9 + q ^ { 2 } \right)$
(2) $\frac { 9 } { 4 } \left( 9 - q ^ { 2 } \right)$
(3) $\frac { 9 } { 4 } \left( 9 + p ^ { 2 } \right)$
(4) $\frac { 9 } { 4 } \left( 9 - p ^ { 2 } \right)$

Let $a , b , c$ be in arithmetic progression. Let the centroid of the triangle with vertices $( a , c ) , ( 2 , b )$ and $( a , b )$ be $\left( \frac { 10 } { 3 } , \frac { 7 } { 3 } \right)$. If $\alpha , \beta$ are the roots of the equation $a x ^ { 2 } + b x + 1 = 0$, then the value of $\alpha ^ { 2 } + \beta ^ { 2 } - \alpha \beta$ is:
(1) $- \frac { 71 } { 256 }$
(2) $\frac { 69 } { 256 }$
(3) $\frac { 71 } { 256 }$
(4) $- \frac { 69 } { 256 }$

Let $p$ and $q$ be two positive numbers such that $p + q = 2$ and $p ^ { 4 } + q ^ { 4 } = 272$. Then $p$ and $q$ are roots of the equation:
(1) $x ^ { 2 } - 2 x + 2 = 0$
(2) $x ^ { 2 } - 2 x + 8 = 0$
(3) $x ^ { 2 } - 2 x + 136 = 0$
(4) $x ^ { 2 } - 2 x + 16 = 0$

The number of real solutions of the equation, $x ^ { 2 } - | x | - 12 = 0$ is:
(1) 2
(2) 3
(3) 1
(4) 4

The number of pairs $a , b$ of real numbers, such that whenever $\alpha$ is a root of the equation $x ^ { 2 } + a x + b = 0 , \quad \alpha ^ { 2 } - 2$ is also a root of this equation, is :
(1) 6
(2) 8
(3) 4
(4) 2

The set of all values of $k > - 1$, for which the equation $\left( 3 x ^ { 2 } + 4 x + 3 \right) ^ { 2 } - ( k + 1 ) \left( 3 x ^ { 2 } + 4 x + 3 \right) \left( 3 x ^ { 2 } + 4 x + 2 \right) + k \left( 3 x ^ { 2 } + 4 x + 2 \right) ^ { 2 } = 0$ has real roots, is: (1) $\left[ - \frac { 1 } { 2 } , 1 \right)$ (2) $\left( 1 , \frac { 5 } { 2 } \right]$ (3) $\left( \frac { 1 } { 2 } , \frac { 3 } { 2 } \right] - \{ 1 \}$ (4) $[ 2,3 )$

$\operatorname { cosec } 18 ^ { \circ }$ is a root of the equation:
(1) $x ^ { 2 } - 2 x - 4 = 0$
(2) $4 x ^ { 2 } + 2 x - 1 = 0$
(3) $x ^ { 2 } + 2 x - 4 = 0$
(4) $x ^ { 2 } - 2 x + 4 = 0$

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

$\alpha = \sin 36 ^ { \circ }$ is a root of which of the following equation
(1) $16 x ^ { 4 } - 20 x ^ { 2 } + 5 = 0$
(2) $16 x ^ { 4 } + 20 x ^ { 2 } + 5 = 0$
(3) $10 x ^ { 4 } - 10 x ^ { 2 } - 5 = 0$
(4) $16 x ^ { 4 } - 10 x ^ { 2 } + 5 = 0$

Let $f ( x ) = a x ^ { 2 } + b x + c$ be such that $f ( 1 ) = 3 , f ( - 2 ) = \lambda$ and $f ( 3 ) = 4$. If $f ( 0 ) + f ( 1 ) + f ( - 2 ) + f ( 3 ) = 14$, then $\lambda$ is equal to
(1) $- 4$
(2) $\frac { 13 } { 2 }$
(3) $\frac { 23 } { 2 }$
(4) $4$

If for some $p , q , r \in R$, all have positive sign, one of the roots of the equation $\left( p ^ { 2 } + q ^ { 2 } \right) x ^ { 2 } - 2 q ( p + r ) x + q ^ { 2 } + r ^ { 2 } = 0$ is also a root of the equation $x ^ { 2 } + 2 x - 8 = 0$, then $\frac { q ^ { 2 } + r ^ { 2 } } { p ^ { 2 } }$ is equal to $\_\_\_\_$.

The number of real solutions of the equation $3 \left( \mathrm { x } ^ { 2 } + \frac { 1 } { \mathrm { x } ^ { 2 } } \right) - 2 \left( \mathrm { x } + \frac { 1 } { \mathrm { x } } \right) + 5 = 0$, is
(1) 4
(2) 0
(3) 3
(4) 2

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 $\_\_\_\_$ .

The number of real solutions of the equation $3\left(x^2 + \frac{1}{x^2}\right) - 2\left(x + \frac{1}{x}\right) + 5 = 0$ is
(1) 4
(2) 0
(3) 3
(4) 2

Let $x = \frac { m } { n } \left( m , n \right.$ are co-prime natural numbers) be a solution of the equation $\cos \left( 2 \sin ^ { - 1 } x \right) = \frac { 1 } { 9 }$ and let $\alpha , \beta ( \alpha > \beta )$ be the roots of the equation $m x ^ { 2 } - n x - m + n = 0$. Then the point $( \alpha , \beta )$ lies on the line
(1) $3 x + 2 y = 2$
(2) $5 x - 8 y = - 9$
(3) $3 x - 2 y = - 2$
(4) $5 x + 8 y = 9$

Let $f : \mathbf { R } - \{ 0 \} \rightarrow ( - \infty , 1 )$ be a polynomial of degree 2, satisfying $f ( x ) f \left( \frac { 1 } { x } \right) = f ( x ) + f \left( \frac { 1 } { x } \right)$. If $f ( K ) = - 2 K$, then the sum of squares of all possible values of $K$ is :
(1) 7
(2) 6
(3) 1
(4) 9

Find sum of the roots of given equation $(x - 1)^{2} - 5|x - 1| + 6 = 0$ for $x \in \mathbb{R}$

If $\alpha , \beta$ are roots of quadratic equation $\lambda \mathrm { x } ^ { 2 } - ( \lambda + 3 ) \mathrm { x } + 3 = 0$ and $\alpha < \beta$ such that $\frac { 1 } { \alpha } - \frac { 1 } { \beta } = \frac { 1 } { 3 }$ ，then find sum of all possible values of $\lambda$ ． （A） 3 （B） 2 （C） 1 （D） 4

If person A and person B can finish together whole work in 22.5 days. If B alone takes 24 days more to complete the work than A alone, find the number of days taken by A alone to finish the given work.\ (A) 18\ (B) 36\ (C) 60\ (D) 24

Q1 Consider the equation
$$(x-1)^2 = |3x-5|.$$
(1) Among all solutions of equation (1), the solutions satisfying $x \geqq \frac{5}{3}$ are $x = \mathbf{A}$ and $x = \mathbf{B}$, where $\mathbf{A} < \mathbf{B}$.
(2) Equation (1) has a total of $\mathbf{C}$ solutions. When the minimum one is denoted by $\alpha$, the integer $m$ satisfying $m-1 < \alpha \leqq m$ is $\mathbf{DE}$.

Consider the quadratic function of $x$
$$x ^ { 2 } + ( 4 a - 6 ) x + 2 a + b + 5 = 0 . \tag{1}$$
We are to find the conditions under which one solution is $-1$ and the other solution satisfies the inequality
$$| x + 2 a | < a + 1 . \tag{2}$$
(1) The condition for equation (1) to have the solution $-1$ is
$$b = \mathbf { L } a - \mathbf { M N } .$$
Denote the other solution by $\alpha$. When $\alpha$ is expressed in terms of $a$, we have
$$\alpha = \mathbf { O P } a + \mathbf { Q } .$$
(2) When $a > \mathbf{RS}$, the inequality (2) has a solution, and its solution is
$$\mathbf { TU } a - \mathbf { VV } < x < - a + \mathbf { W } .$$
Hence the conditions are: that $a$ and $b$ satisfy (3) and that $a$ satisfies
$$\mathbf { X } < a < \mathbf { Y } .$$

For the real numbers $a$ and $b$ satisfying
$$a ^ { 3 } = \frac { 1 } { \sqrt { 5 } - 2 } , \quad b ^ { 3 } = 2 - \sqrt { 5 }$$
we are to find the value of $a + b$.
When we set $x = a + b$, we have
$$x ^ { 3 } = ( a + b ) ^ { 3 } = a ^ { 3 } + b ^ { 3 } + \mathbf { A } a b ( a + b ) .$$
Since $a b = \mathbf { B C }$, we know that this $x$ satisfies
$$x ^ { 3 } + \mathbf { D } x - \mathbf { E } = 0 .$$
The left side of this equation can be factorized as follows:
$$\begin{aligned} x ^ { 3 } + \mathbf { D } x - \mathbf { E } & = \left( x ^ { 3 } - \mathbf { F } \right) + \mathbf { D } \left( x - \frac { \mathbf { F } } { \mathbf { E } } \right) \\ & = ( x - \mathbf { F } ) \left( x ^ { 2 } + x + \mathbf { G } \right) . \end{aligned}$$
Since
$$x ^ { 2 } + x + \frac { \mathbf { G } } { \mathbf { H } } = \left( x + \frac { \mathbf { H } } { \mathbf { I } } \right) ^ { 2 } + \frac { \mathbf { J K } } { \mathbf { L } } > 0 ,$$
we obtain $x = a + b = \mathbf { M }$.

Q1 Let $a$ and $b$ be rational numbers and let $p$ be a real number. Consider the quadratic equation
$$x ^ { 2 } + a x + b = 0 \tag{1}$$
which has a solution $x = \frac { \sqrt { 5 } + 3 } { \sqrt { 5 } + 2 }$, and consider the inequality
$$x + 1 < 2 x + p + 3 . \tag{2}$$
(1) First, we are to find the values of $a$ and $b$.
When we rationalize the denominator of $x = \frac { \sqrt { 5 } + 3 } { \sqrt { 5 } + 2 }$, we have
$$x = \sqrt { \mathbf { A } } - \mathbf { B }$$
Since this is a solution of equation (1), by substituting this in (1) we have
$$- a + b + \mathbf { C } + ( a - \mathbf { D } ) \sqrt { \mathbf { E } } = 0 .$$
Hence we see that
$$a = \mathbf { F } \text { and } b = \mathbf { G H } .$$
(2) Next, we are to find the smallest integer $p$ such that both solutions of equation (1) satisfy inequality (2).
When we solve inequality (2), we have
$$x > - p - 1 .$$
Since both solutions of equation (1) satisfy this, we see that
$$p > \sqrt { \mathbf { J } } - \mathbf { K } .$$
Hence the smallest integer $p$ is $\mathbf { L }$.

Let $a$ be a constant other than 0 . Let
$$\begin{aligned} & f ( x ) = x ^ { 2 } + 2 a x - 4 a - 12 , \\ & g ( x ) = a x ^ { 2 } + 2 x - 4 a + 4 . \end{aligned}$$
(1) When the solutions of $f ( x ) = 0$ and the solutions of $g ( x ) = 0$ coincide, $a$ is $\mathbf { M N }$, and their solutions are $x = \mathbf { O P }$ and $x = \mathbf { Q }$.
(2) $g ( x ) = 0$ has just one solution when $a = \frac { \mathbf { R } } { \mathbf { S } }$, and in this case the solution is $x =$ $\mathbf{TU}$.
(3) The range of $a$ such that $f ( x ) < g ( x )$ for all $x$ is $\mathbf{VW}$.

Consider the two quadratic functions
$$f ( x ) = - 2 x ^ { 2 } , \quad g ( x ) = x ^ { 2 } + a x + b$$
Function $g ( x )$ satisfies the following two conditions:
(i) the value of $g ( x )$ is minimized at $x = 3$;
(ii) $\quad g ( 4 ) = f ( 4 )$.
(1) From condition (i) we see that $a = -$ A . Further, from condition (ii) we see that $b = - \mathbf { B C }$. Hence the minimum value of function $g ( x )$ is $- \mathbf { D E }$.
(2) Let us find the value of $x$ such that $f ( x ) = g ( x )$ and $x$ is not 4 . Since $x$ satisfies
$$x ^ { 2 } - \mathbf { F } x - \mathbf { G } \mathbf { G } = 0 \text {, }$$
we obtain $x = - \mathbf { H }$.
(3) The value of $f ( x ) - g ( x )$ on $- \mathrm { H } \leqq x \leqq 4$ is maximized at $x = \square$, and its maximum value is JK.