For the equation $|{x^2}| + |x| – 6 = 0$, the roots are

Suppose that $x$ and $y$ are positive number with $xy = \frac{1}{9};\,x\left( {y + 1} \right) = \frac{7}{9};\,y\left( {x + 1} \right) = \frac{5}{{18}}$ . The value of $(x + 1) (y + 1)$ equals

Let $r$ be a real number and $n \in N$ be such that the polynomial $2 x^2+2 x+1$ divides the polynomial $(x+1)^n-r$. Then, $(n, r)$ can be

If $x$ is real, then the maximum and minimum values of the expression $\frac{{{x^2} – 3x + 4}}{{{x^2} + 3x + 4}}$ will be

Let $p, q$ be integers and let $\alpha, \beta$ be the roots of the equation, $x^2-x-1=0$, where $\alpha \neq \beta$. For $n=0,1,2, \ldots$, let $a_n=$ $p \alpha^n+q \beta^n$.

$FACT$ : If $a$ and $b$ are rational numbers and $a+b \sqrt{5}=0$, then $a=0=b$.

($1$) $a_{12}=$

$[A]$ $a_{11}-a_{10}$  $[B]$ $a_{11}+a_{10}$  $[C]$ $2 a_{11}+a_{10}$   $[D]$ $a_{11}+2 a_{10}$

($2$) If $a_4=28$, then $p+2 q=$

$[A] 21$   $[B] 14$   $[C] 7$    $[D] 12$

 answer the quetion ($1$) and ($2$)