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The function f:[0,3]vec[1, 29], defined ...

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

A

one-one and onto

B

onto but not one -one

C

one-one but not onto

D

neither one-one nor on to

Text Solution

Verified by Experts

The correct Answer is:
2
`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
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