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Let \alpha ,\beta ,\gamma are roots of x^{3}-x^{2}+1= 0  then the equation whose roots are \frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma }  is

  • Option 1)

    x^{3}+x-1= 0

  • Option 2)

    x^{3}-x+1= 0

  • Option 3)

    x^{3}+x+1= 0

  • Option 4)

    x^{3}-x-1= 0

 

Answers (1)

best_answer

Let \frac{1}{\alpha }=y\: \Rightarrow \: \alpha =\frac{1}{y}

Now, \alpha being root of given equation will satisfy it, So

\left (\frac{1}{y} \right )^{3}-\left (\frac{1}{y} \right )^{2}+1= 0\: \Rightarrow \: 1-y+y^{3}=0

\Rightarrow \: y^{3}-y+1=0

\therefore  equation is x^{3}-x+1=0

\therefore Option (B)

 

Transformation of equation -

To find equation whose roots are symmetrical functions of \alpha and \beta , Where \alpha & \beta are roots of some other equation. 

- wherein

Take any of the root to be equal to y  &  calculate \alpha  or  \beta  accordingly in terms of y  & satisfy the given equation to get required equation.

 

 


Option 1)

x^{3}+x-1= 0

This is incorrect

Option 2)

x^{3}-x+1= 0

This is correct

Option 3)

x^{3}+x+1= 0

This is incorrect

Option 4)

x^{3}-x-1= 0

This is incorrect

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divya.saini

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