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9th Class Mathematics Polynomials Question Bank Polynomials

  • question_answer
    If\[{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{z}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{{{\mathbf{x}}^{\mathbf{2}}}}\mathbf{+}\frac{\mathbf{1}}{{{\mathbf{y}}^{\mathbf{2}}}}\mathbf{+}\frac{\mathbf{1}}{{{\mathbf{z}}^{\mathbf{2}}}}\mathbf{=6}\], then  the valve of \[{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{z}}^{\mathbf{2}}}\]is

    A)  3 

    B)  4

    C)  8                                

    D)  16

    Correct Answer: A

    Solution :

    (a): \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}-6=0\] \[\Rightarrow {{x}^{2}}+\frac{1}{{{x}^{2}}}-2+{{y}^{2}}+\frac{1}{{{y}^{2}}}-2+{{z}^{2}}+\frac{1}{{{z}^{2}}}-2=0\] \[\Rightarrow {{\left( x-\frac{1}{x} \right)}^{2}}+{{\left( y-\frac{1}{y} \right)}^{2}}+{{\left( z-\frac{1}{z} \right)}^{2}}=0\] \[\Rightarrow \left( x-\frac{1}{x} \right)=0\therefore {{x}^{2}}-1=0\Rightarrow x=\pm 1\] Similarly \[y=\pm 1,z=\pm 1\,\,\therefore {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1+1+1=3\]                               


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