Let u( x ) and v( x ) are differentiable functions such that ( u ( x )/ v ( x ))=7. If ( u prime( x )/ v prime( x ))= p and (( u ( x )/ v ( x ))) prime= q, then ( p + q / p - q ) has the value equal to

Question

Let u( x ) and v( x ) are differentiable functions such that ( u ( x )/ v ( x ))=7. If ( u prime( x )/ v prime( x ))= p and (( u ( x )/ v ( x ))) prime= q, then ( p + q / p - q ) has the value equal to (A) 1 (B) 0 (C) 7 (D) - 7

Tardigrade Admin · Accepted Answer

u ( x )=7 v ( x ) ⇒ u prime( x )=7 v prime( x ) ⇒ p =7 text (given) again (u(x)/v(x))=7 ⇒((u(x)/v(x))) prime=0 ⇒ q=0; now (p+q/p-q)=(7+0/7-0)=1

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

1

0

7

- 7

$u (x) = 7 v (x) \Rightarrow u^{'} (x) = 7 v^{'} (x) \Rightarrow p = 7 (given)$
again $\frac{u ( x )}{v ( x )} = 7 \Rightarrow (\frac{u ( x )}{v ( x )})^{'} = 0 \Rightarrow q = 0$ ; now $\frac{p + q}{p - q} = \frac{7 + 0}{7 - 0} = 1$

Q. Let u(x) and v(x) are differentiable functions such that v(x)u(x)​=7. If v′(x)u′(x)​=p and (v(x)u(x)​)′=q, then p−qp+q​ has the value equal to

1

0

7

- 7

u(x)=7v(x)⇒u′(x)=7v′(x)⇒p=7 (given) again v(x)u(x)​=7⇒(v(x)u(x)​)′=0⇒q=0; now p−qp+q​=7−07+0​=1

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

$u (x) = 7 v (x) \Rightarrow u^{'} (x) = 7 v^{'} (x) \Rightarrow p = 7 (given)$
again $\frac{u ( x )}{v ( x )} = 7 \Rightarrow (\frac{u ( x )}{v ( x )})^{'} = 0 \Rightarrow q = 0$ ; now $\frac{p + q}{p - q} = \frac{7 + 0}{7 - 0} = 1$