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

MCQ

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

A
Both roots in $[a,\,b]$
B
Both roots in $( - \infty ,\,a)$
C
Both roots in $(b,\, + \infty )$
✓
One root in $( - \infty ,\,a)$ and the other in $(b,\, + \infty )$

✓

Answer

Correct option: D.

One root in $( - \infty ,\,a)$ and the other in $(b,\, + \infty )$

d
(d) The equation is ${x^2} - (a + b)\,x + ab - 1 = 0$

$\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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