Suppose y varies as the sum of two quantities of which one varies directly as x and the other inversely as x. If y = 6 when x = 4 and y = 3 1 3 when x = 3, then the relation between x and y is -

Answer: Option A

$$\begin{array}{rl}& \text{y }\alpha \left(\text{x}+\frac{1}{\text{x}}\right)\\ & \Rightarrow \text{y}=\text{kx}+\frac{\text{m}}{\text{x}}\text{,}\end{array}$$
Where k and m are constants,
Then,
$$\begin{array}{rl}& =4\text{k}+\frac{\text{m}}{4}=\mathrm{6......}(\text{i})\\ & and\\ & =3\text{k}+\frac{\text{m}}{3}=\mathrm{10......}(ii)\end{array}$$
Multiplying (i) by 3 and (ii) by 4, we get :
$$\begin{array}{rl}& =12\text{k}+\frac{3\text{m}}{4}=\mathrm{18....}(\text{iii})\\ & and\\ & =12\text{k}+\frac{4\text{m}}{3}=\frac{40}{3}....(\text{iv})\end{array}$$
Subtracting (iv) from (iii), we get :
$$\begin{array}{rl}& =\frac{3\text{m}}{4}-\frac{4\text{m}}{3}=18-\frac{40}{3}\\ & \Rightarrow -\frac{7\text{m}}{12}=\frac{14}{3}\\ & \Rightarrow \text{m}=-8.\\ & \text{Putting m}=-\text{8 in (i),}\\ & \text{we get}:\\ & 4\text{k}+\frac{\left(-8\right)}{4}=6\\ & \Rightarrow 4\text{k}=8\\ & \Rightarrow \text{k}=2\\ & \therefore \text{y}=2\text{x}-\frac{8}{\text{x}}\end{array}$$