JEE Main201911 Jan 2019Morning ShiftMathematicsApplication of DerivativesActual

Themaximum value of the finction f(x)=3 x³-18 x²+27 x-40 on the set S = x R: x²+30 11 x is :

Options

A-122
B-222
C122
D222

Correct answer

C. 122

Step-by-step solution

Consider the function, f(x)=3 x(x-3)²-40 Now S= x k: x²+30 11 x So x²-11 x+30 0 x [5,6] f(x) will have maximum value for x=6 The maximum value of function is, f(6)=3 6 3 3-40=122

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