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# Let f(x)={4x-x^3+log(a^2-3a+3),0                 {x-18 >=3 Then the complete set of 'a' such that f(x)has local maxima at x=3

Arun
25763 Points
one year ago

Dear student

f’(x) = 4 – 3x^2
= 1

f(3)

– 15

a^2 – 3a + 3 >= 1

(a-1)(a-2) >= 0

a belongs to ( – infinity, 1) U (2, infinity)
Vikas TU
14149 Points
one year ago
Dear student
F’(x) = 4x – 3x^2 +  $\frac{1}{(a^2-3a+3)}$
So , the value of x is x>= 15
put x = 15
for this t be local maximum
The value of (a^2-3a+3) must be less than 0
So , (a-1)(a-2) must be less than 0
So , value of a must be (-inf , -1) U (1,2)
Because there is log and value can not be between (-1 , 1)
Hope this helps
Good Luck