Question

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

• A. Both roots in $[a,b]$
• B. Both roots in $[-\infty ,a]$
• C. Both roots in $[b,\infty ]$
• D. One root in $[-\alpha ,a]$ and other in $[b,\alpha ]$

Answer 1

Answer: (D)

One root in $[-\alpha ,a]$ and other in $[b,\alpha ]$

$(x-a)(x-b)-1=0$

$\Rightarrow x^2-(a+b)x-ab-1=0$

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

it has two real roots further $f\left(a\right)=-1$ and $f\left(b\right)=-1$ but $b>a$ i.e $a$ and $b$ are distinct as coefficient of $x^2$ is position (it is $1$), minima of $f\left(x\right)$ is between $a$ and $b$

Hence, one root will be in the interval $(-\alpha ,a)$ and the other root will be in the interval $(b,\alpha )$.