Question
The function $$f:\left[ {0,\,3} \right] \to \left[ {1,\,29} \right],$$ defined by $$f\left( x \right) = 2{x^3} - 15{x^2} + 36x + 1,$$ is :
A.
one-one and onto
B.
onto but not one-one
C.
one-one but not onto
D.
neither one-one nor onto
Answer :
onto but not one-one
Solution :
$$\eqalign{
& f\left( x \right) = 2{x^3} - 15{x^2} + 36x + 1 \cr
& f'\left( x \right) = 6{x^2} - 30x + 36 = 6\left( {x - 2} \right)\left( {x - 3} \right) \cr} $$
Thus, $$f\left( x \right)$$ is increasing in $$\left[ {0,\,2} \right]$$ and decreasing in $$\left[ {2,\,3} \right].$$ Therefore $$f\left( x \right)$$ is many-one
$$f\left( 0 \right) = 1\,;\,\,f\left( 2 \right) = 29\,;\,\,f\left( 3 \right) = 28$$
Range is $$\left[ {1,\,29} \right]$$
Hence, $$f\left( x \right)$$ is many-one-onto.