(a) Show that a function f: R → R defined as f ( x )= x^2+ x +1 is neither one-one nor onto. Also, find all the values of x for which f ( x )=3.

$$\begin{aligned} & \text { Let } \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right) \text { for some } \mathrm{x}_{1}, \mathrm{x}_{2} \in R \\ & \text { Then } \mathrm{x}_{1}^{2}+\mathrm{x}_{1}+1=\mathrm{x}_{2}^{2}+\mathrm{x}_{2}+1 \\ & \Rightarrow\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)\left(1+\mathrm{x}_{1}+\mathrm{x}_{2}\right)=0 \\ & \Rightarrow \mathrm{x}_{1}-\mathrm{x}_{2}=0 \text { or } \mathrm{x}_{1}+\mathrm{x}_{2}=-1 \\ & \Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2} \text { or } \mathrm{x}_{1}+\mathrm{x}_{2}=-1 \text { so if } \mathrm{x}_{1}+\mathrm{x}_{2}=-1, \mathrm{x}_{1} \neq \mathrm{x}_{2} \end{aligned} $$ Hence f is not one -one Let $\displaystyle \mathrm{y}=\mathrm{f}(\mathrm{x})$ where $\displaystyle \mathrm{x} \in R$ Then $\displaystyle \mathrm{y}=\mathrm{x}^{2}+\mathrm{x}+1$. $$\begin{aligned} & \Rightarrow \mathrm{x}^{2}+\mathrm{x}+1-\mathrm{y}=0 \\ & \Rightarrow \mathrm{x}=\frac{-1 \pm \sqrt{4 \mathrm{y}-3}}{2} \end{aligned} $$ For x to be real, $\displaystyle 4 \mathrm{y}-3 \geq 0$ $$\Rightarrow \mathrm{y} \geq \frac{3}{4} $$ Hence, range $\displaystyle =\left[\frac{3}{4}, \infty\right) \neq$ codomain Hence, f is not onto. $$\begin{aligned} & \mathrm{f}(\mathrm{x})=3 \Rightarrow \mathrm{x}^{2}+\mathrm{x}+1=3 \Rightarrow \mathrm{x}^{2}+\mathrm{x}-2=0 \\ & \Rightarrow \mathrm{x}=\frac{-1 \pm \sqrt{9}}{2}=-2,1 \end{aligned} $$ Let $\displaystyle (a, b) \in \mathrm{N} \times \mathrm{N}$