If `S_(1) and S_(2)` are respectively the sets of local minimum and local maximum point of the function, `f(x)=9x^(4)+12x^(3)-36x^(2)+25, x in R`, then (a) `S_(1)={-2}:S_(2)={0,1}` (b) `S_(1)={-2,0}:S_(2)={1}` (c) `S_(1)={-2,1}:S_(2)={0}` (d) `S_(1)={-1}:S_(2)={0,2}`

Options

A. `S_(1)={-2}:S_(2)={0,1}`
B. `S_(1)={-2,0}:S_(2)={1}`
C. `S_(1)={-2,1}:S_(2)={0}`
D. `S_(1)={-1}:S_(2)={0,2}`

Answer

C

Explanation

Given function is
`f(x)=9x^(4)+12x^(3)-36x^(2)+25=y ("let")`
For maxima or minima put `(dy)/(dx)=0`
`rArr (dx)/(dx)=36x^(2)=-72x=0`
`rArr x^(3)+x^(2)-2x=0`
`rArr x[x^(2)+ +x-2]=0`
`rArr x[x^(2)+2x-x-2]=0`
`rAr x[x(x+2)-1(x+2)]=0`
`rArr x(x-1)(x+2)=0`
`rArr x=-2,0,1`
By sin method, we have following

Since, `(dy)/(dx)` changes its' sign from negative at `x='-2' and '1'`, so x=-2 1 are points of local minima.
Also, `(dy)/(dx)` changes it's sign from positive to negative at `x=0, so x =0` is point of local maxima.
`:. S_(1)={-2,1} and S_(2)={0}`