At what values of $a$ , the curve $x^{4} + 3 a x^{3} + 6 x^{2} + 5$ is not situated below any of its tangent lines

| a | > \frac{4}{3}

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| a | < \frac{4}{3}

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| a | > 1

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| a | < \frac{1}{3}

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Solution

The correct option is B $| a | < \frac{4}{3}$
$y = 4 x^{3} + 3 a x^{3} + 6 x^{2} + 5$
For given situation curve must be concave so, $\frac{d^{2} y}{d x^{2}} > 0$
$y^{'} = 4 x^{3} + 9 a x^{2} + 12 x$
$y^{''} = 12 x^{2} + 18 a x + 12 > 0 = 6 (2 x^{2} + 3 a x + 2) > 0$
$D < 0 \Rightarrow 9 a^{2} - 16 < 0$
$\Rightarrow | a | < \frac{4}{3}$

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Similar questions

a,

f (x) = x^{4} + 3 a x^{3} + 6 x^{2} + 5

a,

f (x) = x^{4} + 3 a x^{3} + 6 x^{2} + 5

x^{4} + y^{4} = c^{4}

a

b

a^{- 4 / 3} + b^{- 4 / 3}

y = \frac{x^{4}}{4} + \frac{a x^{3}}{3} + \frac{a x^{2}}{2} + x + 1, x ϵ R

a

x^{4} + y^{4} = a^{4}

p

q

p^{- 4 / 3} + q^{- 4 / 3}

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