The maximum value of the function $f (x) = {2 x}^{3} - {15 x}^{2} + 36 x - 48$ on the set
$A = (x | x^{2} + 20 \leq 9 x)$ is ...

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Solution

The correct option is A 7
Given,

$x^{2} + 20 \leq 9 x$

$x^{2} - 9 x + 20 \leq 0$

$x^{2} - 4 x - 5 x + 20 \leq 0$

$x (x - 4) - 5 (x - 4) \leq 0$

$(x - 4) (x - 5) \leq 0$

$4 \leq x \leq 5$

now ,

$f (x) = 2 x^{3} - 15 x^{2} + 36 x - 48$

differentiate wrt $x$

$f^{'} (x) = 6 x^{2} - 30 x + 36 = 6 (x^{2} - 5 x + 6)$

$= 6 (x - 2) (x - 3)$

maximum value of $f (x)$ at $x = 5$

$f (5) = 2 (5)^{3} - 15 (5)^{2} + 36 (5) - 48$

$= 250 - 375 + 180 - 48$

$= 430 - 423 = 7$

Suggest Corrections

Similar questions

f (x) = 2 x^{3} - 15 x^{2} + 36 x - 48

A = {x | x^{2} + 20 \leq 9 x}

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f (x) = 2 x^{3} - 15 x^{2} + 36 x + 1

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