The maximum value of function $f (x) = 3 x^{3} - 18 x^{2} + 27 x - 40$ on the set $S = {x \in R : x^{2} + 30 \leq 11 x}$ is:

122

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- 222

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- 122

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222

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Solution

The correct option is A $122$
$S = {x \in R, x^{2} + 30 - 11 x \leq 0}$
$= {x \in R, 5 \leq x \leq 6}$
Now $f (x) = 3 x^{3} - 18 x^{2} + 27 x - 40$
$⟹ f^{'} (x) = 9 (x - 1) (x - 3)$ , which is positive in $[5, 6]$
$⟹ f (x)$ increasing in $[5, 6]$
Hence maximum value $= f (6) = 122$

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Similar questions

f (x) = 3 x^{3} - 18 x^{2} + 27 x - 40

S = {x \in R : x^{2} + 30 \leq 11 x}

f (x) = 3 x^{3} - 18 x^{2} + 27 x - 40

S = {x \in R : x^{2} + 30 \leq 11 x}

f (x) = 2 x^{3} - 18 x^{2} + 48 x - 11

S = {x \in R : x^{2} + 42 \leq 13 x}

f (x) = 2 x^{3} - 18 x^{2} + 48 x - 11

S = {x \in R : x^{2} + 42 \leq 13 x}

50^{\circ} C

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