If for a posiive integer $n$ , the quadratic equation, $x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right) + .\;.\;.\; + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right) = 10n$ has two consecutive integral solutions, then $n$ is equal to:
$11$
$12$
$9$
$10$
Let $a, b, c, d$ be real numbers between $-5$ and $5$ such that $|a|=\sqrt{4-\sqrt{5-a}},|b|=\sqrt{4+\sqrt{5-b}},|c|=\sqrt{4-\sqrt{5+c}}$ $|d|=\sqrt{4+\sqrt{5+d}}$ Then, the product $a b c d$ is
The number of real solutions of the equation $x\left(x^2+3|x|+5|x-1|+6|x-2|\right)=0$ is
The number of ordered pairs $(x, y)$ of real numbers that satisfy the simultaneous equations $x+y^2=x^2+y=12$ is
If$\frac{{2x}}{{2{x^2} + 5x + 2}} > \frac{1}{{x + 1}}$, then