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QJEEM19Ma7e10: JEE Main 2019 Mathematics - Class 12 JEEM19Ma7e10 Solution: The maximum value of the function f(x) = 3x 3 – 18x 2 + 27x –…
The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = is :
Multiple Choice Options
- A 222
- B 122
- C
- D 222
Step-by-Step Verified Solution
S = {x R, x2 + 30 11x 0}
= {x R, 5 x 6}
Now f(x) = 3x3 18x2 + 27x 40
f '(x) = 9(x 1)(x 3),
which is positive in [5, 6]
f(x) increasing in [5, 6]
Hence maximum value = f(6) = 122
= {x R, 5 x 6}
Now f(x) = 3x3 18x2 + 27x 40
f '(x) = 9(x 1)(x 3),
which is positive in [5, 6]
f(x) increasing in [5, 6]
Hence maximum value = f(6) = 122