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Question

(c) Since $\alpha ,\;\beta ,\;\gamma ,\;\delta $ form an increasing $G.P.$,

so $\alpha \delta = \beta \gamma $where $\alpha < \beta < \gamma

Vedclass · Accepted Answer

$(2, 32)$

Vedclass · Answer

(c) Since $\alpha ,\;\beta ,\;\gamma ,\;\delta $ form an increasing $G.P.$,

so $\alpha \delta = \beta \gamma $where $\alpha < \beta < \gamma < \delta $.

On solving ${x^2} - 3x + a = 0$,

we get $x = \frac{1}{2}(3 \pm \sqrt {9 - 4a} )$.

Also $\alpha < \beta $.

Hence $\alpha = \frac{1}{2}(3 - \sqrt {9 - 4a} ),\;$

$\beta = \frac{1}{2}(3 + \sqrt {9 - 4a} )$

Similarly from ${x^2} - 12x + b = 0$, we get

$\gamma = \frac{1}{2}(12 - \sqrt {144 - 4b} ),\;$

$\delta = \frac{1}{2}(12 + \sqrt {144 - 4b} )$

Substituting these values of $\alpha ,\;\beta ,\;\gamma ,\;\delta $ in $\alpha \delta = \beta \gamma $

and simplifying, we get $(a,\;b) = (2,\;32)$.

Trick : Check the alternates; only $(c)$ satisfies the condition.

$\alpha ,\;\beta $ are the roots of the equation ${x^2} - 3x + a = 0$ and $\gamma ,\;\delta $ are the roots of the equation ${x^2} - 12x + b = 0$. If $\alpha ,\;\beta ,\;\gamma ,\;\delta $ form an increasing $G.P.$, then $(a,\;b) = $

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