The function `f:[0,3]vec[1, 29],` defined by `f(x)=2x^3-15 x^2+36 x+1,` is one-one and onto onto but not one-one one-one but not onto neither one-one nor onto

Options

A. one-one and onto
B. onto but not one -one
C. one-one but not onto
D. neither one-one nor on to

Answer

2

Explanation

`f(x)=2x^(3)-15x^(2)+36x+1`
`f(x)=6x^(2)-30x+36`
`=6(x-2)(x-3)`
Thuis f(X) is increasing in [0,2] and decreasing in [2,3]
Therefore f(x) is may one
f`(0)=1`
f(=29
f(3)=28
Range isi [1,29]
Hence f(x) is many one onto