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(C) Given that $\frac{f(x)}{g(x)} = c$,where $c$ is a non-zero constant. Taking the derivative of both sides with respect to $x$,we get $\left(\frac{f

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(C) Given that $\frac{f(x)}{g(x)} = c$,where $c$ is a non-zero constant. Taking the derivative of both sides with respect to $x$,we get $\left(\frac{f(x)}{g(x)}\right)^{\prime} = 0$. Since $\beta(x) = \left(\frac{f(x)}{g(x)}\right)^{\prime}$,it follows that $\beta(x) = 0$. Also,from $\frac{f(x)}{g(x)} = c$,we have $f(x) = c \cdot g(x)$. Differentiating both sides,$f^{\prime}(x) = c \cdot g^{\prime}(x)$,which implies $\frac{f^{\prime}(x)}{g^{\prime}(x)} = c$. Since $\alpha(x) = \frac{f^{\prime}(x)}{g^{\prime}(x)}$,we have $\alpha(x) = c$. Now,substituting these values into the expression $\frac{\alpha(x) - \beta(x)}{\alpha(x) + \beta(x)}$,we get $\frac{c - 0}{c + 0} = \frac{c}{c} = 1$.

$f(x)$ and $g(x)$ are differentiable functions such that $\frac{f(x)}{g(x)} = c$,where $c$ is a non-zero constant. If $\frac{f^{\prime}(x)}{g^{\prime}(x)} = \alpha(x)$ and $\left(\frac{f(x)}{g(x)}\right)^{\prime} = \beta(x)$,then $\frac{\alpha(x) - \beta(x)}{\alpha(x) + \beta(x)} = $

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