Question

If $f\left( x \right) = {x^3} + 3{x^2} + 12x - 2\sin x$, where $f:R \to R$, then
(a) $f\left( x \right)$ is many-one and onto
(b) $f\left( x \right)$ is one-one and onto
(c) $f\left( x \right)$ is one-one and into
(d) $f\left( x \right)$ is many-one and into

Answer 1

Here, we need to check whether the function is one-one or many-one, and onto or into. A function is one-one if its first derivative is greater than 0. A function is onto if its range is equal to its codomain. When there are at least two domains, which have the same image then it is called many-one function. When there is at least one image in the co-domain which is not an image of any domain then it is called into function.

Complete step by step solution:
We know that $\sin x$ and $\cos x$ lie in the interval $\left[ { - 1,1} \right]$.
Therefore, the minimum and maximum value of $\sin x$ and $\cos x$ is $- 1$ and 1 respectively.
First, we will check whether the given function is one-one or many-one.
We know that if the derivative of a function is greater than 0, then the function is one-one.
Differentiate the function with respect to $x$, we get
$f'\left( x \right) = 3{x^2} + 6x + 12 - 2\cos x$
Rewriting the expression, we get
$\Rightarrow f'\left( x \right) = 3\left( {{x^2} + 2x + 4} \right) - 2\cos x$
We will write 4 as a sum of 1 and 3. Thus, we get
$\Rightarrow f'\left( x \right) = 3\left( {{x^2} + 2x + 1 + 3} \right) - 2\cos x$
Rewriting the terms in the parentheses, we get
$\Rightarrow f'\left( x \right) = 3\left[ {\left( {{x^2} + 2\left( 1 \right)\left( x \right) + {1^2}} \right) + 3} \right] - 2\cos x$
Applying the algebraic identity ${a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}$, we get
$\begin{array}{l} \Rightarrow f'\left( x \right) = 3\left[ {{{\left( {x + 1} \right)}^2} + 3} \right] - 2\cos x\\\ \Rightarrow f'\left( x \right) = 3{\left( {x + 1} \right)^2} + 9 - 2\cos x\end{array}$
Now, we know that the square of a number is always greater than or equal to 0.
Therefore, ${\left( {x + 1} \right)^2} \ge 0$.
Multiplying both sides of the inequality by 3, we get
$3{\left( {x + 1} \right)^2} \ge 0$
Adding 9 to both sides of the inequality, we get
$3{\left( {x + 1} \right)^2} + 9 \ge 9$
The minimum and maximum value of $\cos x$ is $- 1$ and 1 respectively.
Therefore, we get $- 1 \le \cos x \le 1$.
Multiplying both sides by 2, we get
$- 2 \le 2\cos x \le 2$
We can get the minimum value of $f'\left( x \right)$ by substituting the minimum value of the expression $3{\left( {x + 1} \right)^2} + 9$ and the maximum value of the expression $2\cos x$.
Therefore, we get
$\begin{array}{l} \Rightarrow f'\left( x \right) \ge 9 - 2\\\ \Rightarrow f'\left( x \right) \ge 7\end{array}$
We can observe that the derivative of the function is greater than 0. Therefore, the given function is one-one.
Now, we will check whether the given function is into or onto.
A function is onto if its range is equal to its codomain.
The codomain of the given function is $R$.
We will use limits at the extreme values of the codomain to find the range of the function.
Therefore, we get the upper limit of the range as
$\begin{array}{l}\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \left( {{x^3} + 3{x^2} + 12x - 2\sin x} \right)\\\ = \infty \end{array}$
We get the lower limit of the range as
$\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \left( {{x^3} + 3{x^2} + 12x - 2\sin x} \right)\\\ = - \infty \end{array}$
Thus, the range of the function is $\left( { - \infty ,\infty } \right)$, that is all real numbers.
We observe that the range of the function is equal to its codomain. Hence, the function is onto.
We find that the given function is one-one and onto.

$\therefore$ The correct option is option (b).

Note:
We must know how to check whether a function is one-one or onto. A common mistake we can make is to say that if the derivative of the function is greater than 0, then the function is onto. This statement will be incorrect. Another method of checking whether a function is onto or not is by letting $y = f\left( x \right)$, and rewriting the given function to write $x$ in terms of $y$, such that $x = f\left( y \right)$. If $x$ exists for all values of $y$ in the range of $f\left( x \right)$, then the function is onto.