Let p, q be the roots of the equation $x^{2} - 3 x + 2 = 0.$

The quadratic equation having roots 2p and 2q is:

x² – 3x + 4 = 0

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x² – 6x - 8 = 0

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x² – 6x +4 = 0

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x² – 6x + 8 = 0

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Solution

The correct option is D x² – 6x + 8 = 0
Substitute x by $(\frac{x}{2})$

The equation becomes: ${(\frac{x}{2})}^{2} - 3 (\frac{x}{2}) + 2 = 0, x^{2} - 6 x + 8 = 0.$

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f (x) = (\begin{matrix} x^{2} - 1, if 0 < x < 2 2 x + 3, if 2 \leq x < 3 \end{matrix}

lim x \to 2^{-} f (x) and lim x \to 2^{+} f (x)

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