Question

Let $u\left(x\right)$ and $v\left(x\right)$ are differentiable functions 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 $\left(\frac{p+q}{p-q}\right)$ has the value equal to

• A. $1$
• B. $0$
• C. $7$
• D. $-7$

Answer 1

The correct option is A $1$

Given 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 ${\left(\frac{u\left(x\right)}{v\left(x\right)}\right)}^{{}^{\prime }}=q$
$u\left(x\right)=7v\left(x\right)$
Differentiating both the sides, we get
$u^{{}^{\prime }}\left(x\right)=7v^{{}^{\prime }}\left(x\right)$
$\therefore \frac{u^{{}^{\prime }}\left(x\right)}{v^{{}^{\prime }}\left(x\right)}=7=p$
$\therefore p=7$
We know,
$\frac{u\left(x\right)}{v\left(x\right)}=7$
Again differentiating both the sides, we get
${\left(\frac{u\left(x\right)}{v\left(x\right)}\right)}^{{}^{\prime }}=0=q$
$\therefore q=0$
We have to find the value of $\frac{p+q}{p-q}$
$\left(\frac{p+q}{p-q}\right)=\frac{7+0}{7-0}=1$