Question

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 $\left( {b, + \infty } \right)$

Answer 1

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.

Complete step-by-step answer:
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 \\\$$

Since the discriminant is greater than zero, it has two real roots.
Consider,
$f\left( a \right) = \left( {a - a} \right)\left( {a - b} \right) - 1 = - 1$
$f\left( b \right) = \left( {b - a} \right)\left( {b - b} \right) - 1 = - 1$
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 $\left( {b, + \infty } \right)$

Note: If the discriminant of the function is less than zero then the function has imaginary roots and if the function has discriminant equal to zero then the roots are real and equal.