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 1

First we have to know what is one-one function and onto function. So, an injective function (also known as injection, or one-to-one function) is a function that maps distinct elements of its domain to distinct elements of its codomain. And, a function $f$ from a set $X$ to a set $Y$ is surjective (also known as onto, or a surjection), if for every element $y$ in the codomain $Y$ of $f$, there is at least one element $x$ in the domain $X$ of $f$ such that $f\left( x \right) = y$. So, first we have to check whether the given function is strictly increasing or decreasing in the entire domain, in order to determine whether it’s one- one or not. Next we have to find whether the function ranges within its co-domain or not in order to determine it’s onto function or not.

Complete step by step answer:
Given, $f\left( x \right) = 2{x^3} - 15{x^2} + 36x + 1$ in $f:\left[ {0,3} \right] \to \left[ {1,29} \right]$.
$f\left( x \right)$ is a cubic polynomial.
Now, first derivative of $f\left( x \right)$ is,
$f'\left( x \right) = 6{x^2} - 30x + 36$
So, to find the critical points, we have to solve, $f'\left( x \right) = 0$.
So, $6{x^2} - 30x + 36 = 0$
$\Rightarrow {x^2} - 5x + 6 = 0$
Using middle term factorisation, we get,
$\Rightarrow {x^2} - 3x - 2x + 6 = 0$
$\Rightarrow (x - 3)(x - 2) = 0$
$\Rightarrow x = 2,3$
In the domain, it can be divided into two intervals, $0 \leqslant x < 2$ and $2 < x < 3$.
So, in interval, $0 \leqslant x < 2$, $f'\left( x \right) > 0$, so, $f\left( x \right)$ is increasing.
And, in the interval, $2 < x < 3$, $f'\left( x \right) < 0$, so, $f\left( x \right)$ is decreasing.
Hence, $f\left( x \right)$ is not strictly increasing or strictly decreasing in the entire domain.
So, $f\left( x \right)$ is not a one-one function.
A function has it’s maximum or minimum value in it’s critical points or at the extreme values of it’s domain.
So, $f\left( 2 \right) = 2{\left( 2 \right)^3} - 15{\left( 2 \right)^2} + 36\left( 2 \right) + 1$
$\Rightarrow f\left( 2 \right) = 16 - 60 + 72 + 1$
$\Rightarrow f\left( 2 \right) = 29$
And, $f\left( 3 \right) = 2{\left( 3 \right)^3} - 15{\left( 3 \right)^2} + 36\left( 3 \right) + 1$
$\Rightarrow f\left( 2 \right) = 54 - 135 + 108 + 1$
$\Rightarrow f\left( 2 \right) = 28$
And, at the extremum $0$, $f\left( 0 \right) = 2{\left( 0 \right)^3} - 15{\left( 0 \right)^2} + 36\left( 0 \right) + 1$
$\Rightarrow f\left( 0 \right) = 1$
So, the maximum value of $f\left( x \right)$ is $29$ and the minimum value is $1$.
So, the maximum and minimum value of $f\left( x \right)$ lies on the co-domain of $f$.
Hence, every value in the range of $f$ has a preimage in the domain of $f$
Therefore, we can say that, $f$ is an onto function.
Therefore, $f$ is an onto function but not one-one, correct option is (B).

Note:
We can also check whether a function is one-one or not, by finding that, if $f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$, then, ${x_1} = {x_2}$, if this condition is satisfied, then the function is one-one. And, to check whether a function is onto or not, we have to assume, $y = f\left( x \right) \Rightarrow x = {f^{ - 1}}\left( y \right)$, then substituting x in the function as, $f\left( {{f^{ - 1}}\left( y \right)} \right)$ gives back $y$, then the function is onto function.