Search any question & find its solution
Question:
Answered & Verified by Expert
If $b \gt a$, then the equation $(x-a)(x-b)=1$ has
Options:
Solution:
1152 Upvotes
Verified Answer
The correct answer is:
One root in $(-\infty, a)$ and the other in $(b,+\infty)$
The equation is $x^2-(a+b) x+a b-1=0$
$\therefore$ discriminant $=(a+b)^2-4(a b-1)=(b-a)^2+4 \gt 0$
$\therefore$ both roots are real. Let them be $\alpha, \beta$ where
$\alpha=\frac{(a+b)-\sqrt{(b-a)^2+4}}{2}, \beta=\frac{(a+b)+\sqrt{(b-a)^2+4}}{2}$
Clearly, $\alpha \lt \frac{(a+b)-\sqrt{(b-a)^2}}{2}=\frac{(a+b)-(b-a)}{2}=a$
$(\because b\gta)$
and $\beta \gt \frac{(a+b)+\sqrt{(b-a)^2}}{2}=\frac{a+b+b-a}{2}=b$
Hence, one root $\alpha$ is less than $a$ and the other root $\beta$ is greater than $b$.
$\therefore$ discriminant $=(a+b)^2-4(a b-1)=(b-a)^2+4 \gt 0$
$\therefore$ both roots are real. Let them be $\alpha, \beta$ where
$\alpha=\frac{(a+b)-\sqrt{(b-a)^2+4}}{2}, \beta=\frac{(a+b)+\sqrt{(b-a)^2+4}}{2}$
Clearly, $\alpha \lt \frac{(a+b)-\sqrt{(b-a)^2}}{2}=\frac{(a+b)-(b-a)}{2}=a$
$(\because b\gta)$
and $\beta \gt \frac{(a+b)+\sqrt{(b-a)^2}}{2}=\frac{a+b+b-a}{2}=b$
Hence, one root $\alpha$ is less than $a$ and the other root $\beta$ is greater than $b$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.