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JEE Mains · Maths · STD 12 - 6. Application of derivatives

The maximum value of the function \(f\,(x)\, = 3{x^3} - 18{x^2} + 27x\,\, - 40\) on the set  \(S = \{ x\, \in \,R\,:\,{x^2}\, + \,30\, \le \,11x\} \) is

  1. A \(-122\)
  2. B \(-222\)
  3. C \(122\)
  4. D \(222\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(122\)

Step-by-step Solution

Detailed explanation

\(x^{2}-11 x+30 \leq 0\) \((x-6)(x-5) \leq 0\) \(f(x)=3 x^{3}-18 x^{2}+27 x-40\) \(f^{\prime}(x)=9 x^{2}-36 x+27\) \(=9\left(x^{2}-4 x+3\right)=9(x-3)(x-1)\) \(\therefore \) \(f(x)\) will be maximum when \(x=6\) \(f(6)=3(6)^{3}-18(6)^{2}+27(6)-40\) \(=36(18-18)+162-40=122\)