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 :
$$\eqalign{
& {\text{We}}\,{\text{have}}\,f\left( x \right) = 2{x^3} - 15{x^2} + 36x + 1 \cr
& \Rightarrow f'\left( x \right) = 6{x^2} - 30x + 36 \cr
& = 6\left( {{x^2} - 5x + 6} \right) \cr
& = 6\left( {x - 2} \right)\left( {x - 3} \right) \cr
& \because f'\left( x \right) > 0\forall x \in \left[ {0,2} \right]\,{\text{and}}\,f'\left( x \right) < 0\,\forall \,x \in \left[ {2,3} \right] \cr} $$
$$\therefore f\left( x \right)$$ is increasing on [0, 2] and decreasing on [2,3]
$$\therefore f\left( x \right)$$ is many one on [0, 3]
$${\text{Also}}\,f\left( 0 \right) = 1,f\left( 2 \right) = 29,f\left( 3 \right) = 28$$
$$\therefore $$ Global min = 1 and Global max = 29
i.e., Range of $$f$$ = [1, 29] = codomain
$$\therefore f$$ is onto.