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if   \alpha ,\beta \neq 0,and\; f\left ( n \right )=\alpha ^{n}+\beta ^{n}\; and

=K\left ( 1-\alpha \right )^{2}\; \left ( 1-\beta \right )^{2}\; \left ( \alpha -\beta \right )^{2},then\; K\; is\; equal\; to\; :

  • Option 1)

    1

  • Option 2)

    -1

  • Option 3)

    \alpha \beta \;

  • Option 4)

    \frac{1}{\alpha \beta }

 

Answers (2)

As we learnt in 

Multiplication of two determinants -

- wherein

Determinant multiplication is row to row , row to column , column to row or column to column

 

\Rightarrow \begin{vmatrix} 1+1+1 & 1+\alpha +\beta & 1+\alpha ^{2}+\beta ^{2}\\ 1+\alpha +\beta & 1+\alpha^{2} +\beta ^{2} & 1+\alpha ^{3}+\beta ^{3}\\ 1+\alpha ^{2}+\beta ^{2} & 1+\alpha ^{3}+\beta ^{3} & 1+\alpha ^{4}+\beta ^{4} \end{vmatrix}

\Rightarrow \begin{vmatrix} 1 & 1 & 1\\ 1 & \alpha & \beta \\ 1 & \alpha ^{2} & \beta ^{2} \end{vmatrix} \begin{vmatrix} 1 & 1 & 1\\ 1 & \alpha & \alpha ^{2}\\ 1 & \beta & \beta ^{2} \end{vmatrix}

\Rightarrow \left ( 1-\alpha \right )^{2}\left ( 1-\beta \right )^{2}\left ( \alpha -\beta \right )^{2}

\therefore K=1


Option 1)

1

This option is correct.

Option 2)

-1

This option is incorrect.

Option 3)

\alpha \beta \;

This option is incorrect.

Option 4)

\frac{1}{\alpha \beta }

This option is incorrect.

Posted by

Sabhrant Ambastha

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