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# If b is greater than a as $b > a$, then the equation $\left( {x - a} \right)\left( {x - b} \right) - 1 = 0$, has A. both roots in $\left[ {a,b} \right]$B. both roots in $\left( { - \infty ,a} \right)$ C. both roots in $\left( {b, + \infty } \right)$D. one root in $\left( { - \infty ,a} \right)$ and other in $\left( {b, + \infty } \right)$

Last updated date: 27th Mar 2023
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Hint: For a function $f\left( x \right) = a{x^2} + bx + c = 0$ the discriminant is given by $D = {b^2} - 4ac$. If the discriminant of the function is greater than zero the function has two real and distinct values.

Let the given function be $f\left( x \right) = \left( {x - a} \right)\left( {x - b} \right) - 1 = 0$ which can be written as $f\left( x \right) = {x^2} - (a + b)x + ab - 1 = 0$
We know that for the function $f\left( x \right) = a{x^2} + bx + c = 0$ the discriminant is given by $D = {b^2} - 4ac$
So, discriminant of $f\left( x \right) = {x^2} - (a + b)x + ab - 1 = 0$ is
$D = {\left[ { - \left( {a + b} \right)} \right]^2} - 4(ab - 1) \\ D = {a^2} + {b^2} + 2ab - 4ab + 4 \\ D = {a^2} + {b^2} - 2ab + 4 \\ D = {\left( {a - b} \right)^2} + 4 > 0 \\$
$f\left( a \right) = \left( {a - a} \right)\left( {a - b} \right) - 1 = - 1$
$f\left( b \right) = \left( {b - a} \right)\left( {b - b} \right) - 1 = - 1$
But $b > a$ i.e., $a$ and $b$ are distinct as coefficient of ${x^2}$ is positive (it is 1) , minima of $f\left( x \right)$ is between $a$ and $b$.
Hence one root will lie in interval $\left( { - \infty ,a} \right)$ and another root will be in interval $\left( {b, + \infty } \right)$.
Thus, the correct option is D. one root in $\left( { - \infty ,a} \right)$ and other in $\left( {b, + \infty } \right)$