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Monotonicity

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Question

The least positive integral value of ‘a’ such that 2x+ $\frac{a}{x^{2}} \geq 6, \forall x \in R i s$

Moderate

Solution

$Let f(x) =2x+ \frac{a}{x^{2}} - 6$
$W e h a v e t o f i n d' a' w h e n 2 x + \frac{a}{x^{2}} - 6 \geq 0, \forall x \in R$
$\Rightarrow w e h a v e t o e n s u r e t h a t m i n i m u m v a l u e (a) o f f (x) ⩾ 0$
$∴ f^{'} (x) = 2 - \frac{2 a}{x^{3}} a n d f^{"} (x) = \frac{6 a}{x^{4}}$
$A l s o f^{"} (a^{\frac{1}{3}}) > 0$
$f^{'} (x) = 0 \Rightarrow x = a^{\frac{1}{3}}$
$h e n c e f (x) i s m a x i m u m a t x = a^{\frac{1}{3}}$
$∴ f (x) \geq 0 i f 2 a^{\frac{1}{3}} + \frac{a}{a^{\frac{2}{3}}} - 6 \geq 0, \forall x \in R$
$∴ a^{\frac{1}{3}} \geq \frac{6}{3} a \geq 8$

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