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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

  1. A -16
  2. B -7
  3. C 16
  4. D 7
Verified Solution

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
\(f(4)=-16\)
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