SATHEE: Application of Derivatives 4 Question 8

Application of Derivatives 4 Question 8

####9. The maximum value of the function

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

on the set $Missing or unrecognized delimiter for \left$ is (2019 Main, 11 Jan I)

(a) 122

(b) -122

(d) 222

Show Answer

Answer:

Correct Answer: 9. (a)

Solution:

We have,

$\begin{aligned} f (x) & = 3 x^{3} - 18 x^{2} + 27 x - 40 \\ \Rightarrow f^{'} (x) & = 9 x^{2} - 36 x + 27 \\ = & 9 (x^{2} - 4 x + 3) = 9 (x - 1) (x - 3) \end{aligned}$

Also, we have $Missing or unrecognized delimiter for \left$

Clearly, $x^{2} + 30 \leq 11 x$

$\begin{array}{llrl} \Rightarrow & x^{2} - 11 x + 30 & \leq 0 \\ \Rightarrow & (x - 5) (x - 6) & \leq 0 \Rightarrow x \in [5, 6] \\ So, & S & = [5, 6] \end{array}$

Note that $f (x)$ is increasing in $[5, 6]$

$[∵ f^{'} (x) > 0$ for $x \in [5, 6]$

$∴ f (6)$ is maximum, where

$f (6) = 3 (6)^{3} - 18 (6)^{2} + 27 (6) - 40 = 122$

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Application of Derivatives 4 Question 8

Answer:

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