11th Class Mathematics Sample Paper Mathematics Olympiad - Sample Paper-1

  • 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 into

    C) Many-one into                    

    D) Many-one onto

    Correct Answer: C

    Solution :

    [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.


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