The function $f:[0,3]\rightarrow [1,29]$ defined by $f(x)=2x^{3}-15x^{2}+36x+1$ is
the graph of function looks like,
it is clearly onto(surjective) since every value of codomain[1,29] is mapped with some value of x, but it is not one-one(injective) because for every value of f(x)$\epsilon$(28,29) there are two corresponding value of x
so option 2 is correct
@joshi_nitish
f(1)=24
f(2)=29
f(3)=28
So, it is first increasing, then decreasing function
but how do u got again increasing?
I mean increasing-decreasing then again increasing?
Injective means for every x there is unique value of f(x)
So here for x= 0 we have f(x) = 1
x= 1 f(x)= 24
x= 2 f(x)= 29
x= 2 f(x)= 28
So,every x has unique value of f(x)
Thus,it should be injective
Now coming to surjective part..if every element of set B(HERE [1,29]) is image of set A(here [0,3]) tehn it will be surjective but it is not the case.
So, answer should be C
But, according to below link answer is B
Please check if my logic or understanding is wrong
https://books.google.co.in/books?id=_RAwDwAAQBAJ&pg=PA32&lpg=PA32&dq=the%20function%20f%3A[0%2C3]%20-%3E%20[1%2C29]%20defined&source=bl&ots=2g8axUMuzW&sig=EZzVx5E7L-uP6rXmRZaLUXc4Lig&hl=en&sa=X&ved=0ahUKEwjz48Puw5rYAhXJsI8KHYgED3YQ6AEISTAF#v=onepage&q&f=false
one to one (injective)----->A function for which every element of the range of the function corresponds to exactly one element of the domain.
onto (surjective)------>The function is surjective (onto) if each element of the range is mapped to by at least one element of the domain.
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