Question-9

Consider the function \(f(x)= x+\cfrac{1}{x}\). Which of the following options are correct?

• \(f\) is an injective function, if the domain of \(f\) is \(\mathbb{R}\setminus\{0\}\).
• \(f\) is an injective function if the domain of \(f\) is \(\mathbb{N}\).
• \(f\) is an injective function if the domain of \(f\) is \(\mathbb{Q}\setminus\{0\}\).
• \(f\) is an injective function if the domain of \(f\) is \(\mathbb{Z}\setminus\{0\}\).
• \(f\) is a surjective function if the domain of \(f\) is \(\mathbb{N}\) and co-domain of \(f\) is \(\mathbb{Q}\).
• \(f\) is a surjective function if the domain of \(f\) is \(\mathbb{Z}\setminus\{0\}\) and co-domain of \(f\) is \(\mathbb{Q}\).

Answer

• \(f\) is an injective function, if the domain of \(f\) is \(\mathbb{R}\setminus\{0\}\).
• \(f\) is an injective function if the domain of \(f\) is \(\mathbb{N}\).
• \(f\) is an injective function if the domain of \(f\) is \(\mathbb{Q}\setminus\{0\}\).
• \(f\) is an injective function if the domain of \(f\) is \(\mathbb{Z}\setminus\{0\}\).
• \(f\) is a surjective function if the domain of \(f\) is \(\mathbb{N}\) and co-domain of \(f\) is \(\mathbb{Q}\).
• \(f\) is a surjective function if the domain of \(f\) is \(\mathbb{Z}\setminus\{0\}\) and co-domain of \(f\) is \(\mathbb{Q}\).

Solution

We have: \[ f( x) =x+\frac{1}{x} \]

Let us first try to see what all values this function can take for its natural domain, \(\displaystyle \mathbb{R} \backslash \{0\}\):

\[ \begin{aligned} x+\frac{1}{x} & =y\\ x^{2} -yx+1 & =0 \end{aligned} \]

For \(\displaystyle x\in \mathbb{R} \backslash \{0\}\), the discriminant of the above equation has to be non-negative:

\[ y^{2} -4\geqslant 0\Longrightarrow y\in ( \infty ,-2] \cup [ 2,\infty ) \]

Thus the range of \(\displaystyle f\) is \(\displaystyle ( \infty ,-2] \cup [ 2,\infty )\).

Let us now probe the injectivity of \(\displaystyle f\):

\[ \begin{aligned} f( x_{1}) & =f( x_{2})\\ x_{1} +\frac{1}{x_{1}} & =x_{2} +\frac{1}{x_{2}}\\ x_{1} -x_{2} +\frac{x_{2} -x_{1}}{x_{1} x_{2}} & =0\\ \frac{( x_{1} -x_{2})( x_{1} x_{2} -1)}{x_{1} x_{2}} & =0\\ ( x_{1} -x_{2})( x_{1} x_{2} -1) & =0 \end{aligned} \]

This suggests one of two possibilities:

\[ x_{1} =x_{2} \ \ \text{OR} \ \ x_{1} x_{2} =1 \]

With this background work, let us now jump into the options:

Option-1: Domain of \(\displaystyle f\) is \(\displaystyle \mathbb{R} \backslash \{0\}\)

We see that \(\displaystyle f\) is not injective. One simple example is \(\displaystyle f( 2) =f\left(\frac{1}{2}\right)\).

Option-2: Domain of \(\displaystyle f\) is \(\displaystyle \mathbb{N}\)

\(\displaystyle f\) has to be injective. Recall the condition \(\displaystyle x_{1} x_{2} =1\). If \(\displaystyle x_{1} ,x_{2} \in \mathbb{N}\), then \(\displaystyle x_{1} =x_{2} =1\).

Option-3: Domain of \(\displaystyle f\) is \(\displaystyle \mathbb{Q} \backslash \{0\}\)

Not injective. Similar to option-1

Option-4: Domain of \(\displaystyle f\) is \(\displaystyle \mathbb{Z} \backslash \{0\}\)

\(\displaystyle f\) is injective. Similar to option-2

Option-5: Domain of \(\displaystyle f\) is \(\displaystyle \mathbb{N}\) and co-domain is \(\displaystyle \mathbb{Q}\)

\(\displaystyle f\) cannot be surjective as \(\displaystyle 1\in \mathbb{Q}\), but \(\displaystyle 1\) is not in the range of \(\displaystyle f\).

Option-6: Domain of \(\displaystyle f\) is \(\displaystyle \mathbb{Z}\) and co-domain is \(\displaystyle \mathbb{Q}\)

\(\displaystyle f\) cannot be surjective. Similar to option-5