<p>f : N <span class="math-tex">\(\rightarrow\)</span> N : f(x) = x<sup>2</sup> + x + 1 is</p>
<p style="display:inline">one-one and into</p>
<p style="display:inline">many-one and onto</p>
<p style="display:inline">one-one and onto</p>
<p style="display:inline">many-one and into</p>
Step-by-Step Solution
Key Concept: To determine if a function is one-one and onto, solve f(p)=f(q) for injectivity within the domain N and verify if the set of outputs exactly matches the codomain N.
<p>f(x) = x<sup>2</sup> + x + 1<br />
One-one function<br />
Let p, q be two arbitrary elements in N<br />
Then, f(p) = f(q)<br />
<span class="math-tex">\(\Rightarrow\)</span> p<sup>2</sup> + p + 1 = q<sup>2</sup> + q + 1<br />
<span class="math-tex">\(\Rightarrow\)</span> p<sup>2</sup> - q<sup>2</sup> + p - q = 0<br />
<span class="math-tex">\(\Rightarrow\)</span> (p - q) (p + q + 1) = 0<br />
<span class="math-tex">\(\Rightarrow\)</span> p = q, p + q + 1 <span class="math-tex">\(\ne\)</span> 0 (<span class="math-tex">\(\because\)</span> p, q <span class="math-tex">\(\in\)</span> N)<br />
When f(p) = f(q), p = q<br />
thus, f(x) is one-one function.<br />
Onto function<br />
For x = 1, f(x) assumes value 3.<br />
As, f(x) cannot assume value less than 3, for x <span class="math-tex">\(\in\)</span> N<br />
Thus, f(x) is not onto function. It is into function.</p>
Correct Answer: A