Question

Let $u\left(x\right)$ and $v\left(x\right)$ are differentiable function such that $\frac{u \left(x\right)}{v \left(x\right)}=7.$ If $\frac{u^{'} \left(x\right)}{v^{'} \left(x\right)}=p$ and $\left(\frac{u \left(x\right)}{v \left(x\right)}\right)^{'}=q$ , then $\frac{p + q}{p - q}$ has the value equal to

Answer 1

Given,
$\frac{u \left(x\right)}{v \left(x\right)}=7$
$\Rightarrow u\left(x\right)=7v\left(x\right)$
Differentiating both sides w.r.t. $x$ , we get,
$\Rightarrow u^{'}\left(x\right)=7v^{'}\left(x\right)$
$\Rightarrow \frac{u^{'} \left(\right. x \left.\right)}{v^{'} \left(\right. x \left.\right)}=7$
So, $p=7$
Again,
$\frac{u \left(x\right)}{v \left(x\right)}=7$
Differentiating both sides w.r.t. $x$ , we get,
$\Rightarrow \left(\frac{u \left(x\right)}{v \left(x\right)}\right)^{'}=0$
So, $q=0$
Now,
$\frac{p + q}{p - q}=\frac{7 + 0}{7 - 0}$
$\Rightarrow \frac{p + q}{p - q}=1$