f(x) = 2x^3 – 15x^2 + 36x + 1
⇒ f′(x) = 6(x^2 − 5x + 6) = 6(x-2)(x-3)
f(x) has minima and maxima at x=3 & x=2 respectively.
Since f(x) is non monotonic in ∈ [0,3]
⇒ f(x) is not one-one.
And, f(x) is increasing in x ∈[0,2) and decreasing in ∈(2,3]
f(0) = 1, f(2) = 29 & f(3) = 28
since f(x) is continuous, so range of f(x) is minimum and maximum value b/w [0,3].
minimum value=1 at x=0 & maximum value=29 at x=2.
Therefore Range of f(x) is [1,29]
⇒ f(x) is onto.
Hence, f(x) is surjective(onto) but not injective(one-one). is the answer.