If `g(x) =2f(2x^3-3x^2)+f(6x^2-4x^3-3) AA x in R` and `f''(x) gt 0 AA x in R` then g(x) is increasing in the interval

Options

A. `(-oo,-(1)/(2))uu(0,1)`
B. `(-(1)/(2),0)uu(1,oo)`
C. `(0,oo)`
D. `(-oo,1)`

Answer

B

Explanation

Since `f''(x)gt0`
`rArr" "f'(x)` is always increasing.
`g'(x)=2f'(2x^(3)-3x^(2))xx(6x^(2)-6x)+f'(6x^(2)-4x^(3)-3)(12x-12x^(2))`
`=12(x^(2)-4).(f'(2x^(3)-3x^(2))-f'(6x^(2)-4x^(3)-3))`
`=12x(x-1)[f'(2x^(3)-3x^(2))-f'(6x^(2)-4x^(3)-3)]`
For increasing `g'(x)gt0`
Case I
`xlt0 or xgt1 rArr f'(2x^(3)-3x^(2))gtf'(6x^(2)-4x^(3)-3)`
`" "{because f'(x)" is increasing"}`
`rArr" "(x-1)^(2)(x+(1)/(2))gt0`
`rArr" "xgt-(1)/(2) therefore in (-(1)/(2),0)uu(1,oo)`
Case II
`0ltxlt1`
`rArr" "(x-1)^(2)(x+1//2)lt0`
`rArr" "xlt-1//2`
So, no solution for this case.