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 = \(\frac{x^2}{31} + \frac{1}{31\sqrt{x}}\)
Verified Correct Answer: A.
Step-by-Step Mathematical Solution
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 = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\)
Source: Myschool Past Questions Archive (Record #32455)