Question

If $\frac{{(p+\frac{1}{q})}^p{(p-\frac{1}{q})}^q}{{(q+\frac{1}{p})}^p{(q-\frac{1}{p})}^q}-{\left(\frac{p}{x}\right)}^x$, then $x=$

• A. $p-q$
• B. $p+q$
• C. $q-p$
• D. $pq$

Answer 1

Answer: (B)

$p+q$

Given, $\frac{{(p+\frac{1}{q})}^p{(p-\frac{1}{q})}^q}{{(q+\frac{1}{p})}^p{(q-\frac{1}{p})}^q}={\left(\frac{p}{q}\right)}^x$
$\Rightarrow {\left(\frac{p}{q}\right)}^x=\frac{\frac{(pq+1{)}^p}{q^p}\times \frac{(pq-1{)}^q}{q^q}}{\frac{(pq+1{)}^p}{p^p}\times \frac{(pq-1{)}^q}{p^q}}$
$=\frac{p^p\times p^q}{q^p\times q^q}$
$={\left(\frac{p}{q}\right)}^{p+q}$

$\therefore x=p+q$