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

Let `f^(prime)(x)=(192 x^3)/(2+sin^4pix)fora l lx in Rw i t hf(1/2)=0.Ifmlt=int_(1/2)^1f(x)dxlt=M ,` then the possible values of `ma n dM` are `m=13 ,M=24` (b) `m=1/4,M=1/2` `m=-11 ,M=0` (d) `m=1,M=12`

A

m = 13 , M = 24

B

`m=(1)/(4),M=(1)/(2)`

C

m =- 11 , M = 0

D

m = 1 , M = 12

Text Solution

Verified by Experts

Here , `f'(x) = (192)/(2+sin^(4)pix):. (192x^(3))/(3)le(192x^(3))/(2)`
On integrating between the limits `(1)/(2)` to x , we get
`int_(1//2)^(x)(192x^(3))/(3)dxleint_(1//2)^(x)f'(x)dx leint_(1//2)^(x)(192x^(3))/(2)dx`
`rArr (192)/(12)(x^(4)-(1)/(16))lef(x)-f(0)le24x^(4)-(3)/(2)`
`rArr 16x^(4)-1lef(x)le24x^(4)-(3)/(2)` Again integrating between the limits `(1)/(2)` to 1 , we get
`int_(1//2)^(1)(16x^(4)-1)dx le int_(1//2)^(1)f(x)dxle int_(1//2)^(1)(24x^(4)-(3)/(2))dx`
`rArr [(16x^(5))/(5)-x]_(1//2)^(1)le int_(1//2)^(1) f (x) le [ (24x^(5))/(5)-(3)/(2)x]_(1//2)^(1)`
`rArr ((11)/(5)+(2)/(5))le int_(1//2)^(1) f(x) dx le ((33)/(10)+(6)/(10))`
`rArr2.6le int_(1//2)^(1) f(x)dx le 3.9` `rArr2.6le int_(1//2)^(1) f(x)dx le 3.9` `("*")` None of the option is correct .
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