JEE Main2019Math
Q.
The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = {xR: x2 + 30 11x} is
- A
122
- B
–122
- C
222
- D
–222
Solution
The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = {xR: x2 + 30 11x} is
122
–122
222
–222