Question
Let $$f:R \to R$$ be a function such that $$f\left( x \right) = {x^3} - 6{x^2} + 11x - 6.$$ Then :
A.
$$f$$ is one-one and into
B.
$$f$$ is many-one and into
C.
$$f$$ is one-one and onto
D.
$$f$$ is many-one and onto
Answer :
$$f$$ is many-one and onto
Solution :
$$f\left( x \right) = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)$$
$$\therefore f\left( 1 \right) = f\left( 2 \right) = f\left( 3 \right) = 0.$$ So, the function is many-one.
Clearly, for $$x < 0,\,f\left( x \right) < 0$$ and goes on decreasing as $$x$$ decreases.
For $$x > 3,\,f\left( x \right) > 0$$ and goes on increasing as $$x$$ increases.
$$\therefore \,f\left( x \right)$$ can have all real values. So, $$f$$ is onto.
$$f\left( x \right) = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)$$
$$\therefore f\left( 1 \right) = f\left( 2 \right) = f\left( 3 \right) = 0.$$ So, the function is many-one.
Clearly, for $$x < 0,\,f\left( x \right) < 0$$ and goes on decreasing as $$x$$ decreases.
For $$x > 3,\,f\left( x \right) > 0$$ and goes on increasing as $$x$$ increases.
$$\therefore \,f\left( x \right)$$ can have all real values. So, $$f$$ is onto.