The number of ordered pairs $(x, y)$ of positive integers satisfying $2^x + 3^y = 5^{xy}$ is

Let $[r]$ denote the greatest integer not exceeding $r$. The roots of the equation $3 x^2 + 6 x + 5 + \alpha (x^2 + 2 x + 2) = 0$ are complex numbers whenever $\alpha > L$ or $\alpha < M$. If $(L - M)$ is minimum,then the greatest value of $[r]$ such that $L y^2 + M y + r < 0$ for all $y \in R$ is:

Let $a, b, c, d$ be real numbers between $-5$ and $5$ such that $|a|=\sqrt{4-\sqrt{5-a}}$,$|b|=\sqrt{4+\sqrt{5-b}}$,$|c|=\sqrt{4-\sqrt{5+c}}$,and $|d|=\sqrt{4+\sqrt{5+d}}$. Then,the product $abcd$ is

If $\alpha, \beta$ are the real roots of $x^2+p x+q=0$ and $\alpha^4, \beta^4$ are the roots of $x^2-r x+s=0$,then the equation $x^2-4 q x+2 q^2-r=0$ has always

$\alpha$ is the maximum value of $1-2x-5x^2$ and $\beta$ is the minimum value of $x^2-2x+r$. If $5\alpha x^2+\beta x+6>0$ for all real values of $x$,then the interval in which $r$ lies is

Let $p, q$ be real numbers. If $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$, $\beta$ is a root of $x^{2}+9 p^{2} x+15 q^{2}=0$ and $0 < \alpha < \beta$, then the equation $x^{2}+6 p^{2} x+10 q^{2}=0$ has a root $\gamma$ that always satisfies: