If ${log}_{3 x + 5} (a x^{2} + 8 x + 2) > 2$ then $x$ lies in the interval:

(- \frac{4}{3}, - \frac{20}{11})

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(- \frac{4}{3}, - \frac{23}{22})

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(- \frac{5}{3}, - \frac{23}{22})

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None of these

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Solution

The correct option is C $(- \frac{5}{3}, - \frac{23}{22})$
Given,
$l o g_{3 x + 5} (a x^{2} + 8 x + 2) > 2$

$\frac{l o g (a x^{2} + 8 x + 2)}{l o g (3 x + 5)} > 2$

so $3 x + 5 > 0$

$x > - \frac{5}{3}$

Also $l o g (a x^{2} + 8 x + 2) > l o g (3 x + 5)^{2}$

$a x^{2} + 8 x + 2 > 9 x^{2} + 25 + 30 x$

$⟹ x^{2} (a - 9) - 22 x - 23 > 0$

$- 22 x - 23 > 0 ⟹ x < - \frac{23}{22}$

So $x \in (- \frac{5}{3}, - \frac{23}{22})$

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