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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$(P)$ If $A \neq I_{2},$ then $|A|=-1$
$(\mathrm{Q})$ If $|\mathrm{A}|=1,$ then $\operatorname{tr}(\mathrm{A})=2$
where $I_{2}$ denotes $2 \times 2$ identity matrix and $\operatorname{tr}(A)$ denotes the sum of the diagonal entries of $A$ Then