Ifax3 + 4x − 5 < 0 ∀x∈R then

a<0 and b2 − 4ac < 016 − 4(a)(−5) < 016 + 20a < 020a < −16a < −1620a < −45

Ifax3 + 4x − 5 < 0 ∀x∈R then

a<0 and b2 − 4ac < 016 − 4(a)(−5) < 016 + 20a < 020a < −16a < −1620a < −45

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If $a x^{3} + 4 x - 5 < 0$ $\forall x \in R$ then

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$a < 0$

$a > 0$

$a < \frac{- 4}{5}$

$a > \frac{- 4}{5}$

answer is C.

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$a < 0$ and $b^{2} - 4 a c < 0$

$16 - 4 (a) (- 5) < 0$

$16 + 20 a < 0$

$20 a < - 16$

$a < \frac{- 16}{20}$

$a < \frac{- 4}{5}$

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Ifax3 + 4x − 5 < 0 ∀x∈R then

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If $a x^{3} + 4 x - 5 < 0$ $\forall x \in R$ then