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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 onto

Text Solution

Verified by Experts

The correct Answer is:
B

We have, `f(x)=2x^(3)-15x^(2)+36x+1`
`f'(x)=6x^(2)-30x+36=6(x-3)(x-2)`
`"Clearly", f'(x) gt 0" for x in (0,2)and f'(x) lt 0"For x" in (2,3)"`
So, f'(x) is increasing on (0,2) and decreasing (2,3).
Also, f(0)=1,f(2)=29 and f(3)=28
As f is a continous function. So, it attains every value between its minimum value 1 and maximum value `f(2)=29`, So, f is onto and it is not one-one
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