MCQ
If $b > a$, then the equation $(x - a)\,(x - b) = 1$ has
- ABoth roots in $[a,\,b]$
- BBoth roots in $( - \infty ,\,a)$
- CBoth roots in $(b,\, + \infty )$
- ✓One root in $( - \infty ,\,a)$ and the other in $(b,\, + \infty )$
$\therefore $ discriminant $ = {(a + b)^2} - 4(ab - 1) = {(b - a)^2} + 4 > 0$
$\therefore $ both roots are real. Let them be $\alpha ,\beta $ where
$\alpha = \frac{{(a + b) - \sqrt {{{(b - a)}^2} + 4} }}{2}$, $\beta = \frac{{(a + b) + \sqrt {{{(b - a)}^2} + 4} }}{2}$
Clearly, $\alpha < \frac{{(a + b) - \sqrt {{{(b - a)}^2}} }}{2} = \frac{{(a + b) - (b - a)}}{2} = a\,\,$
$(\because b > a)$
and $\beta > \frac{{(a + b) + \sqrt {{{(b - a)}^2}} }}{2} = \frac{{a + b + b - a}}{2} = b$
Hence, one root $\alpha $ is less than $a$ and the other root $\beta$ is greater than $b$.
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[Note: Here $z$ takes the values in the complex plane and $\operatorname{Im} z$ and $\operatorname{Re} z$ denote, respectively, the imaginary part and the real part of $z]$
| column-$I$ | column-$II$ |
| $(A)$ The set of points $z$ satisfying $|z-i| z||=|z+i| z||$ is contained in or equal to | $(p)$ an ellipse with eccentricity $\frac{4}{5}$ |
| $(B)$ The set of points $z$ satisfying $|z+4|+|z-4|=10$ is contained in or equal to | $(q)$ the set of points $z$ satisfying $\operatorname{Im} z=0$ |
| $(C)$ If $|\omega|=2$, then the set of points $z=\omega-1 / \omega$ is contained in or equal to | $(r)$ the set of points $z$ satisfying $|\operatorname{Im} z| \leq 1$ |
| $(D)$ If $|\omega|=1$, then the set of points $z=\omega+1 / \omega$ is contained in or equal to | $(s)$ the set of points $z$ satisfying $|\operatorname{Re} z| \leq 1$ |
| $(t)$ the set of points $z$ satisfying $|z| \leq 3$ |