Themaximum value of the function
Question:

Themaximum value of the function $f(x)=3 x^{3}-18 x^{2}+27 x-40$ on

the set $\mathrm{S}=\left\{x \in R: x^{2}+30 \leq 11 x\right\}$ is :

  1. (1) $-122$

  2. (2) $-222$

  3. (3) 122

  4. (4) 222


Correct Option: 3,

Solution:

Consider the function,

$f(x)=3 x(x-3)^{2}-40$

Now $S=\left\{x \in R: x^{2}+30 \leq 11 x\right\}$

So $x^{2}-11 x+30 \leq 0$

$\Rightarrow \quad x \in[5,6]$

$\therefore f(x)$ will have maximum value

for $x=6$

The maximum value of function is,

$f(6)=3 \times 6 \times 3 \times 3-40=122$

 

 

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