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Let f prime(x)=(192x^3)/(2+sin^4 pix) fo...

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 for `x in [(1/2),1]` the possible values of m and M are

A

a. `m=13,M=24`

B

b. `m=1/4,M=1/2`

C

c. `m=-11,M=0`

D

d. `m=1,M=12`

Text Solution

Verified by Experts

The correct Answer is:
D
`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`
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