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 = 3¹/₃ when x = 3, then the relation between x and y is -

1y = 2x - ⁸/_x
2y = x + ⁴/_x
3y = 2x + ⁴/_x
4y = 2x + ⁸/_x

Answer:- 1

Explanation:-

Solution:

\begin{aligned} y α (x + \frac{1}{x}) \\ \Rightarrow y = kx + \frac{m}{x}, \end{aligned}

$\eqalign{ & {\text{y }}\alpha {\text{ }}\left( {{\text{x}} + \frac{1}{{\text{x}}}} \right) \cr & \Rightarrow {\text{y}} = {\text{kx}} + \frac{{\text{m}}}{{\text{x}}}{\text{,}} \cr}$ Where k and m are constants,
Then,

\begin{aligned} = 4 k + \frac{m}{4} = 6...... (i) \\ a n d \\ = 3 k + \frac{m}{3} = 10...... (i i) \end{aligned}

$\eqalign{ & = 4{\text{k}} + \frac{{\text{m}}}{4} = 6......({\text{i}}) \cr & and \cr & = 3{\text{k}} + \frac{{\text{m}}}{3} = 10......(ii) \cr}$ Multiplying (i) by 3 and (ii) by 4, we get :

\begin{aligned} = 12 k + \frac{3 m}{4} = 18.... (iii) \\ a n d \\ = 12 k + \frac{4 m}{3} = \frac{40}{3} . . . . (iv) \end{aligned}

$\eqalign{ & = 12{\text{k}} + \frac{{3{\text{m}}}}{4} = 18....({\text{iii}}) \cr & and \cr & = 12{\text{k}} + \frac{{4{\text{m}}}}{3} = \frac{{40}}{3}....({\text{iv}}) \cr}$ Subtracting (iv) from (iii), we get :

\begin{aligned} = \frac{3 m}{4} - \frac{4 m}{3} = 18 - \frac{40}{3} \\ \Rightarrow - \frac{7 m}{12} = \frac{14}{3} \\ \Rightarrow m = - 8. \\ Putting m = - 8 in (i), \\ we get : \\ 4 k + \frac{(- 8)}{4} = 6 \\ \Rightarrow 4 k = 8 \\ \Rightarrow k = 2 \\ ∴ y = 2 x - \frac{8}{x} . \end{aligned}

$\eqalign{ & = \frac{{3{\text{m}}}}{4} - \frac{{4{\text{m}}}}{3} = 18 - \frac{{40}}{3} \cr & \Rightarrow - \frac{{7{\text{m}}}}{{12}} = \frac{{14}}{3} \cr & \Rightarrow {\text{m}} = - 8. \cr & {\text{Putting m}} = - {\text{8 in (i),}} \cr & {\text{we get}}: \cr & 4{\text{k}} + \frac{{\left( { - 8} \right)}}{4} = 6 \cr & \Rightarrow 4{\text{k}} = 8 \cr & \Rightarrow {\text{k}} = 2 \cr & \therefore {\text{y}} = 2{\text{x}} - \frac{8}{{\text{x}}}. \cr}$