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 $\frac{u^{{}^{\prime }}\left(x\right)}{v^{{}^{\prime }}\left(x\right)}=p$ and ${\left(\frac{u\left(x\right)}{v\left(x\right)}\right)}^{{}^{\prime }}=q$ , then $\frac{p+q}{p-q}$ has the value equal to

Answer 1

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,
$\Rightarrow u^{{}^{\prime }}\left(x\right)=7v^{{}^{\prime }}\left(x\right)$
$\Rightarrow \frac{u^{{}^{\prime }}(x)}{v^{{}^{\prime }}(x)}=7$
So, $p=7$
Again,
$\frac{u\left(x\right)}{v\left(x\right)}=7$
Differentiating both sides w.r.t. $x$ , we get,
$\Rightarrow {\left(\frac{u\left(x\right)}{v\left(x\right)}\right)}^{{}^{\prime }}=0$
So, $q=0$
Now,
$\frac{p+q}{p-q}=\frac{7+0}{7-0}$
$\Rightarrow \frac{p+q}{p-q}=1$