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 \(\mathrm{S}=\left\{x \in \mathrm{R} / x^2+30 \leq 11 x\right\}\) is
- A \(122\)
- B \(-122\)
- C \(-222\)
- D \(222\)
Answer & Solution
Correct Answer
(A) \(122\)
Step-by-step Solution
Detailed explanation
\(\mathrm{S} =\left\{x \in \mathrm{R} / x^2+30 \leq 11 x\right\} \)
\( =\left\{x \in \mathrm{R} / x^2-11 x+30 \leq 0\right\} \)
\( =\{x \in \mathrm{R} /(x-5)(x-6) \leq 0\} \)
\( =\{x \in \mathrm{R} / x \in[5,6]\} \)
\( \mathrm{f}(x) =3 x^3-18 x^2+27 x-40 \)
\( \therefore \quad \mathrm{f}^{\prime}(x) =9 x^2-36 x+27 \)
\( =9(x-1)(x-3)>0 \quad \forall x \in[5,6]\)
\(\Rightarrow \mathrm{f}(x)\) is increasing in \([5,6]\).
\(\therefore \text { Maximum value } =\mathrm{f}(6) \)
\( =3(6)^3-18(6)^2+27(6)-40 \)
\( =122\)
\( =\left\{x \in \mathrm{R} / x^2-11 x+30 \leq 0\right\} \)
\( =\{x \in \mathrm{R} /(x-5)(x-6) \leq 0\} \)
\( =\{x \in \mathrm{R} / x \in[5,6]\} \)
\( \mathrm{f}(x) =3 x^3-18 x^2+27 x-40 \)
\( \therefore \quad \mathrm{f}^{\prime}(x) =9 x^2-36 x+27 \)
\( =9(x-1)(x-3)>0 \quad \forall x \in[5,6]\)
\(\Rightarrow \mathrm{f}(x)\) is increasing in \([5,6]\).
\(\therefore \text { Maximum value } =\mathrm{f}(6) \)
\( =3(6)^3-18(6)^2+27(6)-40 \)
\( =122\)
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