JAMB 1983 · UME · Q14

Variation

$y$ varies partly as the square of $x$ and partly as the inverse of the square root of $x$ . Write down the expression for $y$ if $y = 2$ when $x = 1$ and $y = 6$ when $x = 4$ .

Worked solution (try it first)

"Partly … partly" means a sum of two parts with two constants: $y = ax^2 + \dfrac{b}{\sqrt x}$ .
$x = 1$ , $y = 2$ gives $a + b = 2$ .
$x = 4$ , $y = 6$ gives $16a + \frac{b}{2} = 6$ .
Put $a = 2 - b$ into the second equation: $32 - 16b + \frac b2 = 6$ , so $\frac{31b}{2} = 26$ and $b = \frac{52}{31}$ .
Then $a = 2 - \frac{52}{31} = \frac{10}{31}$ .
So $y = \dfrac{10x^2}{31} + \dfrac{52}{31\sqrt x}$ , option A.

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