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

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Both roots in $(- \infty, a)$

Both roots in $(b, + \infty)$

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

Both roots in $[a, b]$

answer is D.

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Detailed Solution

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

$∴$ discriminant $= {(a + b)}^{2} - 4 (ab - 1) = {(b - a)}^{2} + 4 > 0$

$∴$ both roots are real. Let them be $α, β$ where

$α = \frac{(a + b) - \sqrt{{(b -)}^{2} + 4}}{2}, β = \frac{(a + b) + \sqrt{{(b - a)}^{2} + 4}}{2}$

Clearly, $α < \frac{(a + b) - \sqrt{{(b - a)}^{2}}}{2} = \frac{(a + b) - (b - a)}{2} = a$

$(∵ b > a)$

and $β > \frac{(a + b) + \sqrt{{(b - a)}^{2}}}{2} = \frac{a + b + b - a}{2} = b$

Hence, one root $α$ is less than a and the other root $β$ is greater than b.

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