MHT CET · Maths · Application of Derivatives
The minimum value of the function \(\mathrm{f}(x)=2 x^3-15 x^2+36 x-48 \quad\) on the set \(\mathrm{A}=\left\{x \mid x^2+20 \leqslant 9 x\right\}\) is
- A -16
- B -7
- C 16
- D 7
Answer & Solution
Correct Answer
(A) -16
Step-by-step Solution
Detailed explanation
\(\mathrm{A} =\left\{x \mid x^2+20 \leq 9 x\right\} \)
\( =\left\{x \mid x^2-9 x+20 \leq 0\right\} \)
\( =\{x \mid(x-4)(x-5) \leq 0\} \)
\( \mathrm{A} =\{4,5\} \)
\( \mathrm{f}(x) =2 x^3-15 x^2+36 x-48 \)
\( \therefore \quad \mathrm{f}^{\prime}(x) =6 x^2-30 x+36 \)
\( =6\left(x^2-5 x+6\right) \)
\( =6(x-2)(x-3) \lt 0 \forall x \in(4,5)\)
\(\therefore \mathrm{f}(x)\) is strictly increasing in the interval \((4,5)\).
\(\therefore \) Minimum value of \(\mathrm{f}(x)\) when \(x \in(4,5)\) is
\(f(4)=-16\)
\( =\left\{x \mid x^2-9 x+20 \leq 0\right\} \)
\( =\{x \mid(x-4)(x-5) \leq 0\} \)
\( \mathrm{A} =\{4,5\} \)
\( \mathrm{f}(x) =2 x^3-15 x^2+36 x-48 \)
\( \therefore \quad \mathrm{f}^{\prime}(x) =6 x^2-30 x+36 \)
\( =6\left(x^2-5 x+6\right) \)
\( =6(x-2)(x-3) \lt 0 \forall x \in(4,5)\)
\(\therefore \mathrm{f}(x)\) is strictly increasing in the interval \((4,5)\).
\(\therefore \) Minimum value of \(\mathrm{f}(x)\) when \(x \in(4,5)\) is
\(f(4)=-16\)
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