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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
PLAN To check nature of function.
(i) One-one To check one-one, we must check whether
`f'(x) gt 0 " or " f'(x) lt 0` in given domain.
(ii) Onto To check onto, we must check
Range = Codomain
Description of Situation To find range in given domain [a, b], put f'(x) = 0 and find `x=a_(1), a_(2), ... a_(n) in [a, b]`
Now, find `{f(a), f(a_(1)),f(a_(2)), ... , f(a_(n)),f(b)}`
its gretest and least values gives you range.
Now, `f:[0, 3] to [1,29] `
`f(x)=2x^(2)-15x^(2)+36x+1`
`therefore f'(x)=6x^(2)-30x+36=6(x^(2)-5x+6)`
`=6(x-2)(x-3)`

For given doamin [0,3], f(x) is increasing as well as decreasing `rArr` many-one
Now, put `f'(x)=0`
`rArr x=2,3`
Thus, for range `f(0)=1, f(2)=29,f(3)=28`
`rArr" Range " in [1,29]`
`therefore` Onto but not one-one.
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