Question

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

• A. $1$
• B. $0$
• C. $7$
• D. $-7$

Answer 1

Answer: (A)

$1$

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