5.9 Inverse functions (exercise)  (Page 4/3)

Working rules

A. Onto function

• A function is onto if every image in the co-domain has a pre-image in the domain set.
• Range = Co-domain

B. Bijection

• If a function is both one-one and onto, then the function is a bijection.

C. Inverse function

• Solve given function for “x”.
• Substitute “x” by inverse symbol “ $f^{-1}\left(x\right)$ ” and “y” by “x” to find the rule of inverse function.
• Exchange the domain and range of the given function with that of inverse function.

Problem 1: A function $f:R\to R$ is given by :

$$f\left(x\right)=\sin \left(3x+2\right)$$

Is the function surjective?

Solution :

Statement of the problem : Surjection is another name for onto function. Here, given function is a trigonometric function. We need to check whether its range is equal to co-domain.

According to question :

$$\text{Domain}=R$$

$$\text{Co-domain}=R$$

Now, let

$$y=f\left(x\right)=\sin \left(3x+2\right)$$

Solving for “x”, we have :

$$\Rightarrow 5x+2={\sin }^{-1}y$$

$$\Rightarrow x=\frac{{\sin }^{-1}y-2}{5}$$

The inverse trigonometric function is valid in its domain of [-1,1]. Hence, values of “y”, for which “x” is real, is [-1,1]. Therefore,

$$\text{Range}=[-1,1]$$

Clearly, range is not equal to co-domain. Thus, given function is not an onto function.

Problem 2: A function $f:R\to R$ is given by :

$$f\left(x\right)=\frac{a{x}^2+6a-8}{a+6x-8x^2}$$

Find the interval of values of “a” for which function is onto.

Solution :

Statement of the problem : It is given that the rational function is valid for all values of “x” as its domain is “R”. It is also given that co-domain of the function is “R”. In order to find the interval of "a", we consider that the function is an onto function. This implies that range of the function is equal to co-domain of the function i.e. “R”.

Let

$$y=\frac{a{x}^2+6a-8}{a+6x-8x^2}$$

$$\Rightarrow a{x}^2+6x-8-ay-6xy+8x^{2}y=0$$

Rearranging as quadratic equation in variable “x”, we have :

$$\Rightarrow \left(a+8y\right)x^2+6\left(1-y\right)x-\left(8+ay\right)=0$$

Since “x” is real, discreminant of the quadratic equation should be non-negative,

$$\Rightarrow 36\left(1-y\right){}^2-4X\left(a+8y\right)X-\left(8+ay\right)\ge 0$$

$$\Rightarrow 9{\left(1-y\right)}^2+\left(a+8y\right)X\left(8+ay\right)\ge 0$$

Expanding and rearranging as quadratic equation in variable “y”, we have :

$$\Rightarrow 9\left(1+y^2-2y\right)+\left(8a+a^{2}y+64y+8a{y}^2\right)\ge 0$$

$$\Rightarrow \left(9+8a\right)y^2+\left(a^2+46\right)y+\left(9+8a\right)\ge 0$$

But, we have assumed that function is an onto function. As such, range of the function is equal to co-domain i.e “R”. It means that above inequality holds for all real values of “y”. Now, remember (as explained for sign scheme of quadratic equation) that a quadratic expression evaluates to non-negative number if (i) coefficient of “y2” is a positive number and (ii) descreminant is a non-positive number. Hence,

$$\Rightarrow 9+8a>0\Rightarrow a>-9/8$$

and

$$\Rightarrow {\left(a^2+46\right)}^2-\{2\left(9+8a\right){\}}^2\le 0$$

$$\Rightarrow \left(a^2+46+18+16a\right)\left(a^2+46-18-16a\right)\le 0$$

$$\Rightarrow \left(a^2+16a+64\right)\left(a^2-16a+28\right)\le 0$$

$$\Rightarrow {\left(a+8\right)}^2\left(a-2a-14a+28\right)\le 0$$

$${\left(a+8\right)}^2\left(a-2\right)\left(a-14\right)\le 0$$

$$\Rightarrow \left(a-2\right)\left(a-14\right)\le 0$$

$$\Rightarrow 2\le a\le 14$$

The first condition earlier yielded as a>= 9/8. The required interval, therefore, is intersection of two intervals, which is [2,14].

Problem 3: Let “f” be a one-one function with domain [x,y,z] and range [1,2,3]. It is given that only one of three conditions given below are true and remaining two are false :

$$f\left(x\right)=1;f\left(y\right)\ne 1;f\left(z\right)\ne 2$$

Determine $f^{-1}\left(1\right)$ .

Solution :

Statement of the problem : With three conditions, there are three different possibilities for one being true and other two being false. We shall check each such possibility to see whether function is onto function. If the function is onto, then the function is a bijection as it is already given that function is an injection.