Set of values of b for which local extrema of the function f(x) are positive where `f(x)=(2)/(3)a^(2)x^(3)-(5a)/(2)x^(2)+3x+b` and maximum occurs at `x(1)/(3)` is -

Options

A. `(-4,oo)`
B. `(-3//8,oo)`
C. `(-10,3//8)`
D. non of these

Answer

B

Explanation

We have
`f(x)=(2)/(3)a^2x^3-(5a)/(2)x^2+3x+b`
`f(x)=2a^2x^2-5ax +3 and f''(x)=4a^2x-5a`
It is given that f(x) attains maximum at x=1/3. Therefore
`f(1)/(3)=0 and f''(1/3)lt 0 `
`rArr f(1/3)=0 " and "f''(1/3)lt 0`
`rArr (2a^2)/(9)-(5a)/(3)+3=0 and 4a^2-15a lt 0 `
`rArr (2a-9)(a-3)=0 and a(4a -15) lt 0`
`rArr a=3, 9/2 and 0 lt a lt (15)/(4)rArr a=3`
`therefore f(x)=18x^2 -15x+3 and f(x)=36x-15 `
`rArr a=3,9/2 and lt a lt (15)/(4) rArr a=3`
`therefore f(x) = 18x^2-15x+3 and f''(x) =36 x-15`
At extereme points , we have
`f(x)=0 rArr 6x^2 x = 1/2,1/3`
Clearly `f''(1/2) gt 0 " and " f(1/3) lt 0`
So `x=1/2` the points of local minimum and `x=1/3` is the points of local maximum
Puting a=3 in (i) we get
`f(x)=6x^3-(15)/(2)x^2+3x+b`
The exterme values of f(x) are
`f(1/2)=6/8-15/8+3/2b=(3)/(8)+b`
and `f(1/3)=(6)/(27)-(15)/(18)+1+b=7/18+b`
For these exterme values to be positive , we must have
`3/8+b gt 0 and 7/18 +b gt 0 `
For theser exterme values to be positive, we must have
`3/8+b gt " and b gt -7/18 +b gt 0 `
`rArr b gt - 3/8 and b gt -7/18 rArr b gt -3/8 rArr b in (-3/8,oo)`