MHT CET · Maths · Application of Derivatives
The maximum value of the function \(f(x)=3 x^3-18 x 2+27 x-40\) on the set \(S=\left\{x \in R x^2+30 \leq 11 x\right\}\) is
- A \(122\)
- B \(222\)
- C \(810\)
- D \(162\)
Answer & Solution
Correct Answer
(A) \(122\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & f(x)=3 x^3-18 x^2+27 x-40 \\ & \Rightarrow f^{\prime}(x)=9 x^2-36 x+27 \\ & =9(x-1)(x-3)\end{aligned}\)
Sign scheme for \(f^{\prime}(x)\)
Now, \(x^2+30 \leq 11 x\)
\(\begin{aligned} & \Rightarrow(x-5)(x-6) \leq 0 \\ & \Rightarrow x \in[5,6]\end{aligned}\)
For the interval \([5,6] f(x)\) is increasing
Hence \(f_{\max }=f(6)\)
\(\forall x \in[5,6]\) i.e. on the set S
\(=3 \times 6^3-18 \times 6^2+27 \times 6-40=122\)
Sign scheme for \(f^{\prime}(x)\)
Now, \(x^2+30 \leq 11 x\)
\(\begin{aligned} & \Rightarrow(x-5)(x-6) \leq 0 \\ & \Rightarrow x \in[5,6]\end{aligned}\)
For the interval \([5,6] f(x)\) is increasing
Hence \(f_{\max }=f(6)\)
\(\forall x \in[5,6]\) i.e. on the set S
\(=3 \times 6^3-18 \times 6^2+27 \times 6-40=122\)
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