If $α$ and $β$ are the zeroes of polynomial, $f (x) = x^{2} - 2 x - 3,$ then find new quadratic polynomial having zeroes $\frac{1}{α} a n d \frac{1}{β} .$

k (3 x^{2} + 2 x + 1)

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k (3 x^{2} + 2 x - 1)

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k (3 x^{2} + 3 x - 1)

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k (2 x^{2} + 2 x - 1)

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Solution

The correct option is B
$k (3 x^{2} + 2 x - 1)$
$f (x) = x^{2} - 2 x - 3 ∵ α a n d β a r e z e r o e s o f f (x) α + β = \frac{- b}{a} = \frac{- (- 2)}{1} = 2 α β = \frac{c}{a} = \frac{- 3}{1} = - 3 For new quadratic polynomial zeroes are \frac{1}{α} a n d \frac{1}{β} S = \frac{1}{α} + \frac{1}{β} = \frac{β + α}{α β} = \frac{2}{(- 3)} = \frac{- 2}{3} P = \frac{1}{α} . \frac{1}{β} = \frac{1}{α β} = \frac{1}{(- 3)} = \frac{- 1}{3} ∴ Required new quadratic polynomial is = k [x^{2} - S x + P] = k [x^{2} - (\frac{- 2}{3}) x + (\frac{- 1}{3})] = k [x^{2} + \frac{2 x}{3} - \frac{1}{3}] = k [3 x^{2} + 2 x - 1]$
So, option b is correct.

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