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In a quadratic function, the value of the product of the roots $(\alpha, \beta)$ is $4$. Find the value of

                        $\dfrac{\alpha^{n}+\beta^{n}}{\alpha^{-n}+\beta^{-n}}$

  1. $n^{4}$
  2. $4^{n}$
  3. $2^{2n-1}$
  4. $4^{n-1}$
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Best answer
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$\dfrac{\alpha^n+\beta^n}{\alpha^{-n}+\beta^{-n}}$

$=\dfrac{\alpha^n+\beta^n}{ \left ( \dfrac{1}{\alpha^{n}}+\frac{1}{\beta^{n}} \right ) }$

$=\dfrac{\alpha^n+\beta^n}{ \left ( \dfrac{ \alpha^n + \beta^n}{\alpha^n \beta^n} \right )}$

$=(\alpha \beta)^n$

=$4^n$

as product of roots, $ \alpha \beta = 4 $

Correct Answer: $B$

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Assume roots (α,β) = (2,2) because Product of roots α*β = 2*2 = 4 is satisfying as per question.

 

Let n=3

(α^n+β^n)/(α^−n+β^−n) = (2^3+2^3)/(2^-3+2^-3) = (8+8)/(1/8)+(1/8) = 16/(1/4) = 16*4 = 64

Now trying option elimination with n=3 

Option B = 4^n = 4^4= 64

Matching with our answer 64 which was calculated above

 

Answer:

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