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
A. y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\) B. y = x2 + \(\frac{1}{\sqrt{x}}\) C. y = x2 + \(\frac{1}{x}\) D. y = \(\frac{x^2}{31} + \frac{1}{31\sqrt{x}}\)
Correct Answer: A
Explanation
y = kx2 + \(\frac{c}{\sqrt{x}}\) y = 2when x = 1 2 = k + \(\frac{c}{1}\) k + c = 2 y = 6 when x = 4 6 = 16k + \(\frac{c}{2}\) 12 = 32k + c k + c = 2 32k + c = 12 = 31k + 10 k = \(\frac{10}{31}\) c = 2 - \(\frac{10}{31}\) = \(\frac{62 - 10}{31}\) = \(\frac{52}{31}\) y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\)
