Question

If the functions $f\left(x\right)$ and $g\left(x\right)$ are continuous on [a, b] and differentiable on (a, b), then in the interval (a, b), the equation
$\left|\begin{array}{cc}{f}^{\prime }\left(x\right)& f\left(a\right)\\ g^{\prime }\left(x\right)& g\left(a\right)\end{array}\right|=\frac{1}{a-b}\left|\begin{array}{cc}f\left(a\right)& f\left(b\right)\\ g\left(a\right)& g\left(b\right)\end{array}\right|$

• A. has at least one root
• B. has exactly one root
• C. has at most one root
• D. no root

Answer 1

The correct option is A has at least one root

$\left|\begin{array}{c}\begin{array}{cc}{f}^{\prime }\left(x\right)& f\left(a\right)\end{array}\\ \begin{array}{cc}{g}^{\prime }\left(x\right)& g\left(a\right)\end{array}\end{array}\right|=\frac{1}{a-b}\left|\begin{array}{cc}f\left(a\right)& f\left(b\right)\\ g\left(a\right)& g\left(b\right)\end{array}\right|$
$\Rightarrow f^{\prime }\left(x\right)g\left(a\right)-f\left(a\right)g^{\prime }\left(x\right)=\frac{f\left(a\right)g\left(b\right)-g\left(a\right)f\left(b\right)}{a-b}$

So from lagrange's mean value theorem there exits at least root $\in (a,b)$