Consider the function `f(x) =27(x^(2/3)-x)/4` How many solutions does the function f(x) =1 have?

Options

A. One
B. Two
C. Three
D. Four

Answer

B

Explanation

For `f(x)=1`
`27/4 ((x^(2))^(1//3) -x)=1`
` (x^(1//3))^(2) -x = 4/27`
Put `x^(1//3)=y`
` x=y^(3)`
` y^(2) - y^(3) = 4/27`
` rArr y^(3) - y^(2) + 4/27 = 0 `
` rArr 27y^(3) - 27y^(2) + 4 =0`
This is a cubic equation.
` " If we put" y=-1/3" then "(3y+1) = 0 " is a factor of cubic equation "`

`(3 y +1) ( 9y^(2) - 12 y + 4)=0`
` (3 y + 1 ) ( 3y-2)^(2) = 0`
Hence `y=-1/3,2/3`
Thus f(x) = 1 has two solutions.