$f(x)$ and $g(x)$ are differentiable functions such that $\frac{f(x)}{g(x)}=$ a non zero constant. If…

$f(x)$ and $g(x)$ are differentiable functions such that $\frac{f(x)}{g(x)}=$ a non zero constant. If $\frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{g}^{\prime}(\mathrm{x})}=\alpha(\mathrm{x})$ and $\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}\right)^{\prime}=\beta(\mathrm{x})$, then $\frac{\alpha(x)-\beta(x)}{\alpha(x)+\beta(x)}=$
  1. 0
  2. $f(x)+g(x)$
  3. 1
  4. $\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})$

Solution

$ \begin{aligned} & \text {} \because \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{g}^{\prime}(\mathrm{x})}=\alpha(\mathrm{x}) \&\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}\right)^{\prime}=\beta(\mathrm{x}) \\ & \because \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\text { a non-zero constant, } \\ & \Rightarrow\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}\right)^{\prime}=0 \Rightarrow \beta(\mathrm{x})=0 \end{aligned} $ Now, $\frac{\alpha(x)-\beta(x)}{\alpha(x)+\beta(x)}=\frac{\alpha(x)-0}{\alpha(x)+0}=1$

Asked in: AP EAMCET 2023 (18 May Shift 2)

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