If $f\left(x\right)$ and $g\left(x\right)$ are continuous functions in $\left[a,b\right]$ and they are differentiable in $\left(a,b\right)$, then in the interval $\left(a,b\right)$, the equation

$\left|\begin{array}{cc}f\text{'}\left(x\right)& f\left(a\right)\\ g\text{'}\left(x\right)& g\left(a\right)\end{array}\right|=\frac{1}{a-b}\left|\begin{array}{cc}f\left(a\right)& f\left(b\right)\\ g\left(a\right)& g\left(b\right)\end{array}\right|$

Given equation is $\left|\begin{array}{cc}f\text{'}\left(x\right)& f\left(a\right)\\ g\text{'}\left(x\right)& g\left(a\right)\end{array}\right|=\frac{1}{a-b}\left|\begin{array}{cc}f\left(a\right)& f\left(b\right)\\ g\left(a\right)& g\left(b\right)\end{array}\right|$.

$f\left(x\right), g\left(x\right)$ are continuous and differentiable in $\left[a,b\right],\left(a,b\right)$ respectively.

Let us consider Lagrange's mean value theorem for $f\left(x\right)$ and $g\left(x\right).$

Lagrange's mean value theorem: If a function is continuous and differentiable in a given interval $\left(a,b\right)$, then there exists

                                                      a point $c$ in the interval such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

$\Rightarrow f\text{'}\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ and $g\text{'}\left(x\right)=\frac{g\left(b\right)-g\left(a\right)}{b-a}$.

And both the functions have at least one real solution $\Rightarrow$the combination of both functions will have at least one real solution in the interval $\left(a,b\right).$

$\Rightarrow f\left(a\right)g\text{'}\left(x\right)-g\left(a\right)f\text{'}\left(x\right)$ will have at least one solution.

$$\begin{gathered} \Rightarrow f\left(a\right)g\text{'}\left(x\right)-f\text{'}\left(x\right)g\left(a\right)=f\left(a\right)\left(\frac{g\left(b\right)-g\left(a\right)}{b-a}\right)-g\left(a\right)\left(\frac{f\left(b\right)-f\left(a\right)}{b-a}\right) \\ =\left(\frac{1}{b-a}\right)\left[f\left(a\right)g\left(b\right)-f\left(a\right)g\left(a\right)-g\left(a\right)f\left(b\right)+g\left(a\right)f\left(a\right)\right] \\ \Rightarrow f\left(a\right)g\text{'}\left(x\right)-f\text{'}\left(x\right)g\left(a\right)=\frac{f\left(a\right)g\left(b\right)-f\left(b\right)g\left(a\right)}{b-a} ......\left(i\right) \end{gathered}$$

Now, by solving the given equation, we get

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)g\left(a\right)-g\text{'}\left(x\right)f\left(a\right)=\frac{f\left(a\right)g\left(b\right)-g\left(a\right)f\left(b\right)}{b-a} \\ \Rightarrow f\left(a\right)g\text{'}\left(x\right)-f\text{'}\left(x\right)g\left(a\right)=\frac{f\left(a\right)g\left(b\right)-g\left(a\right)f\left(b\right)}{a-b}......\left(ii\right) \end{gathered}$$

Hence, we got $\left(i\right)=\left(ii\right)$

As $\left(i\right)$ having at least one root says $\left(ii\right)$ i.e. the given equation have at least one root.

Thus, the equation have at least one root.

 Therefore, the correct answer is option $1.$