MHT CET20225 Aug 2022Morning ShiftMathematicsApplication of DerivativesActual
The maximum value of the function f(x)=3 x^3-18 x 2+27 x-40 on the set S= x R x^2+30 11 x is
Options
- A122
- B222
- C810
- D162
Correct answer
A. 122
Step-by-step solution
aligned & f(x)=3 x^3-18 x^2+27 x-40 & f^ (x)=9 x^2-36 x+27 & =9(x-1)(x-3) aligned Sign scheme for f^ (x) Now, x^2+30 11 x aligned & (x-5)(x-6) 0 & x [5,6] aligned For the interval [5,6] f(x) is increasing Hence f_ =f(6) x [5,6] i.e. on the set S =3 6^3-18 6^2+27 6-40=122