Question

Let $y\propto p+q$ where $p$ varies directly as $x$ and $q$ varies inversely as $x^2$. If $y=19$ when $x=2$ or $3$ then $y$ in terms of $x$ is

• A. $36x+\frac{5}{x^2}$
• B. $\frac{5}{x}+36x^2$
• C. $5x+\frac{36}{x^2}$
• D. none of these

Answer 1

The correct option is C $5x+\frac{36}{x^2}$

Given : $y\propto p+q$

$\Rightarrow y=k(p+q)$
Also,
$p=ax$ and $q=\frac{b}{x^2}$ where $k,a,b$ are constants of proportionality.
Therefore,
$y=akx+\frac{bk}{x^2}$
Now, $y=19$ at $x=2$
$\Rightarrow 19=2ak+\frac{bk}{4}$ ... $\left(i\right)$
And
$y=19$ at $x=3$
$\Rightarrow 19=3ak+\frac{bk}{9}$ ... $\left(ii\right)$
Multiplying eq $\left(i\right)$ by $3$ and $\left(ii\right)$ by $2$, we get
$6ak+\frac{3bk}{4}=57$ ... $\left(iii\right)$
and
$6ak+\frac{2bk}{9}=38$ ... $\left(iv\right)$
Subtracting $\left(iv\right)$ from $\left(iii\right)$, we get
$\frac{27bk-8bk}{36}=19$
$\Rightarrow \frac{19bk}{36}=19$
$\Rightarrow bk=36$
Substituting in equation $\left(i\right)$, we get
$ak=5$
Hence,
$y=\frac{bk}{x^2}+akx$
$\Rightarrow y=\frac{36}{x^2}+5x$