If $x$ is real and satisfies $x + 2 > \sqrt {x + 4} ,$ then
$x < - 2$
$x > 0$
$ - 3 < x < 0$
$ - 3 < x < 4$
If $x$ is real, then the maximum and minimum values of expression $\frac{{{x^2} + 14x + 9}}{{{x^2} + 2x + 3}}$ will be
Number of rational roots of equation $x^{2016} -x^{2015} + x^{1008} + x^{1003} + 1 = 0,$ is equal to
If $x$ is a solution of the equation, $\sqrt {2x + 1} - \sqrt {2x - 1} = 1, \left( {x \ge \frac{1}{2}} \right)$ , then $\sqrt {4{x^2} - 1} $ is equal to
Consider the cubic equation $x^3+c x^2+b x+c=0$ where $a, b, c$ are real numbers. Which of the following statements is correct?
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 ($2$)