Solve $|x - 2| + |x - 1| = x - 3$

Let $a, b, c, d \in R^+$ such that $256abcd \geq (a+b+c+d)^4$ and $3a + b + 2c + 5d = 11$. Then $a^3 + b + c^2 + 5d$ is equal to:

If $a, b, c \in \mathbb{R}$ and $1$ is a root of the equation $ax^2 + bx + c = 0$,then the curve $y = 4ax^2 + 3bx + 2c$ $(a \neq 0)$ intersects the $x$-axis at

If the roots of the equation $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{5}{2}$ are $p$ and $q$ $(p > q)$ and the roots of the equation $(p+q)x^4 - pqx^2 + \frac{p}{q} = 0$ are $\alpha, \beta, \gamma, \delta$,then $(\Sigma \alpha)^2 - \Sigma \alpha \beta + \alpha \beta \gamma \delta = $

If $x$ is real,the function $f(x) = \frac{(x - a)(x - b)}{(x - c)}$ will assume all real values,provided

The number of ordered pairs $(x, y)$ of real numbers that satisfy the simultaneous equations $x+y^2=x^2+y=12$ is