Question

If b > a. then the equation (x-a) (x-b) -1 = 0 has:

• A. both roots in $[a,b]$
• B. both roots in $(-\infty ,0)$
• C. both roots in $(b,+\infty )$
• D. one root in $(-\infty ,a)$ and other root in $(b,+\infty )$

Answer 1

The correct option is D one root in $(-\infty ,a)$ and other root in $(b,+\infty )$

Since $f\left(x\right)=(x-a)(x-b)-1=0\Rightarrow (x-a)(x-b)=1$


It means x-axis is shifted to -1 unit below the original position. So it is clear from the graph that $\alpha <a$ and $\beta >b$

Alternatively:
Let us assume a = 3, b = 4, given that a < b then the given equation becomes
$\begin{array}{cc}& (x-3)(x-4)-1=0\\ & x^2-7x+11=0\\ \therefore & x=\frac{7\pm \sqrt{49-44}}{2}\Rightarrow x=\frac{7\pm \sqrt{5}}{2}\\ \Rightarrow & x=\frac{7+\sqrt{5}}{2}>4andx=\frac{7-\sqrt{5}}{3}<3\end{array}$
Hence only option (d) is satisfied, hence correct.