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BCECE Engineering BCECE Engineering Solved Paper-2006

  • question_answer
    If \[\left| \begin{matrix}   {{x}^{n}} & {{x}^{n+2}} & {{x}^{n+3}}  \\    {{y}^{n}} & {{y}^{n+2}} & {{y}^{n+3}}  \\    {{z}^{n}} & {{z}^{n+2}} & {{z}^{n+3}}  \\ \end{matrix} \right|\]\[=(y-z)(z-x)(x-y)\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right),\]then \[n\]is equal to:

    A)  2                            

    B)        \[-2\]                    

    C)        \[-1\]                    

    D)         1

    Correct Answer: C

    Solution :

                    \[\left| \begin{matrix}    {{x}^{n}} & {{x}^{n+2}} & {{x}^{n+3}}  \\    {{y}^{n}} & {{y}^{n+2}} & {{y}^{n+3}}  \\    {{z}^{n}} & {{z}^{n+2}} & {{z}^{n+3}}  \\ \end{matrix} \right|\]                 \[=(y-z)(z-x)(x-y)\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\] The degree of determinant                                 \[=n+(n+2)+(n+3)\]                                 \[=3n+5\] and the degree of \[\text{RHS = 2}\]                 \[\therefore \]  \[3n+5=2\]                 \[\Rightarrow \]               \[n=-1.\]


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