If $b > a$,then the equation $(x - a)(x - b) = 1$ has

Match the following quadratic expressions with their minimum values:

Quadratic expression	The minimum value
i) $x^2 + 4x + 6$	a) $1$
ii) $x^2 - 2x + 5$	b) $2$
iii) $x^2 + 6x + 18$	c) $4$
iv) $x^2 - 4x + 5$	d) $9$

Consider the quadratic equation $(c - 5)x^2 - 2cx + (c - 4) = 0$,where $c \ne 5$. Let $S$ be the set of all integral values of $c$ for which one root of the equation lies in the interval $(0, 2)$ and the other root lies in the interval $(2, 3)$. Then the number of elements in $S$ is

Let $f(x) = (1 + b^2)x^2 + 2bx + 1$ and $m(b)$ be the minimum value of $f(x)$. If $b$ can take any real value,what is the range of $m(b)$?

If $b > a$,then in which interval do the roots of the equation $(x - a)(x - b) - 1 = 0$ lie?

The maximum and minimum values of $\frac{x^2 + x + 1}{x^2 - x + 1}$ are respectively ...........