Let `f prime(x)=(192x^3)/(2+sin^4 pix)` for all `x in RR` with `f(1/2)=0.` If `mlt=int_(1/2)^1f(x)dxlt=M` then the possible values of m and M are (i) `m=13,M= 24` (ii)` m=1/4,M=1/2` (iii) `m=-11,M = 0` (iv) `m=1,M=12`

Options

A. `m=13,M=24`
B. `m=1/4,M=1/2`
C. `m=-11,M=0`
D. `m=1,M=12`

Answer

D

Explanation

`f'(x)=(192x^(3))/2+sin^(4)(pix)AAxepsilonR,f(1/2)=0`
Now, `64x^(3)lef'(x)le96x^(3)AAxepsilon[1/2,1]`
`:.int_(1//2)^(x)64x^(3)dxltint_(1//2)^(x)f'(x)dxleint_(1//2)^(x)96x^(3)dx`
So `16x^(4)-1lef(x)le24x^(4)-3/2AAxepsilon[1/2,1]`
`:.int_(1//2)^(1)(16x^(4)-1)dxleint_(1//2)^(1)f(x)dxleint_(1//2)^(1)f(x)dxleint_(1//2)^(1)(24x^(4)-3/2)dx`
`:. 16/5 . 31/32 - 1/2 le int_(1//2)^(1)f(x)dxle 24/5 . 31/32 -3/4`
`=26/10 le int_(1//2)^(1)f(x)dxle 78/20`