If f(x) $= a x^{3} - 9 x^{2} + 9 x + 3$ is increasing on R, then

a < 3

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a \geq 3

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a \leq 3

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none of these

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Solution

The correct option is D $a \geq 3$
$f (x) = a x^{3} - 9 x^{2} + 9 x + 3$
$\Rightarrow f^{'} (x) = 3 a x^{2} - 18 x + 9$
For increasing, $f^{'} (x) \geq 0$
$\Rightarrow 3 a x^{2} - 18 x + 9 \geq 0$
$\Rightarrow a > 0$ and $D \leq 0$
$\Rightarrow 18^{2} - 108 a \leq 0$
$\Rightarrow a \geq 3$

Suggest Corrections

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a

f (x) = a x^{3} - 9 a x^{2} + 9 x + 3

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f (x) = k x^{3} - 9 x^{2} + 9 x + 3

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f (x) = k x^{3} - 9 x^{2} + 9 x + 3

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