The function f:[0,3]→[1,29], defined by f(x)=2x³−5x²+36x+1, is one-one and onto. onto but not one-one. neither one-one nor onto. one-one but not onto.

f(x)=2x³−5x²+36x+1, f:[0,3]→[1,29]

1. Check for one-one mapping-
   f(x) is a cubic polynomial. First derivative of f(x) gives f′(x)=6x²−10x+36. Solve f′(x)=0 to get the critical points.
   x²−(5/3)x+6=0
   (x−2)(x−3)=0 ⇒x∈{2,3}
   f(x) is increasing for x<2 or x>3 (∵f′(x)>0) and f(x) is decreasing for 2<x<3 (∵f′(x)<0). Hence, f(x) is not strictly increasing or strictly decreasing in the entire domain. So f(x) is not one-one.
2. Check for onto mapping-
   f(x) has maximum at x=2
   f(2)=29, which is the maximum value of f (∵in the given domain [0,3], f is increasing in [0,2) and decreasing in (2,3])
   f(0)=1 and f(3)=28
   Hence, range of f(x)=[1,29] which is equal to the co-domain. Hence, f(x) is onto.