If $x+\frac{1}{x}=a, x^2+\frac{1}{x^3}=b$, then $x^3+\frac{1}{x^2}$ is

Solution of the equation $\sqrt {x + 3 - 4\sqrt {x - 1} }  + \sqrt {x + 8 - 6\sqrt {x - 1} }  = 1$ is

The sum of all the real roots of the equation $\left( e ^{2 x }-4\right)\left(6 e ^{2 x }-5 e ^{ x }+1\right)=0$ is

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 ($2$)

Let $\alpha $ and $\beta $ are roots of $5{x^2} - 3x - 1 = 0$ , then $\left[ {\left( {\alpha  + \beta } \right)x - \left( {\frac{{{\alpha ^2} + {\beta ^2}}}{2}} \right){x^2} + \left( {\frac{{{\alpha ^3} + {\beta ^3}}}{3}} \right){x^3} -......} \right]$ is

Let $f: R \rightarrow R$ be the function $f(x)=\left(x-a_1\right)\left(x-a_2\right)$ $+\left(x-a_2\right)\left(x-a_3\right)+\left(x-a_3\right)\left(x-a_1\right)$ with $a_1, a_2, a_3 \in R$.Then, $f(x) \geq 0$ if and only if