Mark (√) against the correct answer in the following: f : N → N : f(x) = $x^2$ + x + 1 is A. one - one and onto B. one - one and into C. many - one and onto D. many - one and into

Home » Mark (√) against the correct answer in the following: f : N → N : f(x) = + x + 1 is A. one – one and onto B. one – one and into C. many – one and onto D. many – one and into

Mark (√) against the correct answer in the following:
f : N → N : f(x) = $x^2$ + x + 1 is
A. one – one and onto
B. one – one and into
C. many – one and onto
D. many – one and into

CBSE | Class 12 | Functions | Maths | RS Aggarwal

Mark (√) against the correct answer in the following:
f : N → N : f(x) = $x^2$ + x + 1 is
A. one – one and onto
B. one – one and into
C. many – one and onto
D. many – one and into

Solution:

Option (B) is correct.
One-One function
Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{N}$
Therefore, $f(p)=f(q)$
$\begin{array}{l} \Rightarrow \mathrm{p}_{2}+\mathrm{p}+1=\mathrm{q}_{2}+\mathrm{q}+1 \\ \Rightarrow \mathrm{p}_{2}-\mathrm{q}_{2}+\mathrm{p}-\mathrm{q}=0 \\ \Rightarrow(\mathrm{p}-\mathrm{q})(\mathrm{p}+\mathrm{q}+1)=0 \\ \Rightarrow \mathrm{p}=\mathrm{q} \quad \mathrm{p}+\mathrm{q}+1 \neq 0(\because \mathrm{p}, \mathrm{q} \in \mathrm{N}) \end{array}$
When $f(p)=f(q), p=q$
Therefore, $\mathrm{f}(\mathrm{x})$ is one-one function.
Onto function
For $\mathrm{x}=1, \mathrm{f}(\mathrm{x})$ assumes value 3 .
Since, $f(x)$ cannot assume value less than 3 , for $x \in N$
Hence, $\mathrm{f}(\mathrm{x})$ is not onto function. It is into function.