If b a , then the equation ( x - a ) ( x - b ) - 1 = 0 has [2000]

Question

If $b > a$ , then the equation $(x - a) (x - b) - 1 = 0$ has [2000]

Smart Achievers · Accepted Answer

(4)

$Given : (x - a) (x - b) - 1 = 0, b > a$

$or x^{2} - (a + b) x + (a b - 1) = 0$

$Let f (x) = x^{2} - (a + b) x + (a b - 1)$

$D = {(a + b)}^{2} - 4 (a b - 1)$

$= {(a - b)}^{2} + 1 > 0$

$Since coefficient of x^{2} i.e. 1 > 0,$ $∴$ $f (x) represents upward parabola,$

$intersecting x-axis at two points corresponding to two real roots, D being positive.$

$Also f (a) = f (b) = - 1$

$\Rightarrow curve is below x-axis at a and b$

$∴$ $a and b both lie between the roots.$

$Therefore, the graph of given equation is as shown.$

$It is clear from graph, that one root of the equation lies in (- \infty, a) and other in (b, \infty) .$

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