• # question_answer A function $\mathbf{f}:\mathbf{R}\to \mathbf{R},\mathbf{f}\left( \mathbf{x} \right)={{\mathbf{x}}^{\mathbf{2}}}+\mathbf{x}$, is f A) one-one onto                      B) one-one intoC) Many-one into                    D) Many-one onto

[c] $\because A\text{ }functionf:R\to R$ $\And f\left( x \right)={{x}^{2}}+x.$ $\because f\left( 1 \right)={{1}^{2}}+1=2$ $f\left( -1 \right)={{\left( -1 \right)}^{2}}+\left( -1 \right)=0$ $f\left( 2 \right)={{\left( 2 \right)}^{2}}+\left( +2 \right)=6$ $f\left( -2 \right)={{\left( -2 \right)}^{2}}+\left( -2 \right)=2$ Range of above function, $f=[0,\infty ]={{R}^{+}}$ & we see the above results, 2 is the image of 1 and$-2$. Hence, the function$y=f(x)={{x}^{2}}+x$, be many-one into $\phi$ i.e. option [c] is correct.