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Iff(x)=Asin((pix)/2)+b ,f^(prime)(1/2)=s...

`Iff(x)=Asin((pix)/2)+b ,f^(prime)(1/2)=sqrt(2)a n d` `int_0^1f(x)dx=(2A)/pi,t h e ncon s t a n t sAa n dBa r e` `pi/2a n dpi/2` (b) `2/pia n d3/pi` `0a n d-4/pi` (d) `4/pia n d0`

A

`(pi)/(2)and(pi)/(2)`

B

`(2)/(pi)and(3)/(pi)`

C

`0and-(4)/(pi)`

D

`(4)/(pi)and0`

Text Solution

Verified by Experts

The correct Answer is:
D
Given , `f(x)=Asin((pix)/(2))+B,f'((1)/(2))=sqrt(2)`
and `int_(0)^(1)f(x)dx=(2A)/(pi)`
`f'(x)=(Api)/(2)cos(pix)/(2)rArrf'((1)/(2))=(Api)/(2)"cos"(pi)/(4)=(Api)/(2sqrt(2))`
But `int'((1)/(2))=sqrt(2):.(Api)/(2sqrt(2))=sqrt(2)rArrA=(4)/(pi)`
Now , `int_(0)^(1)f(x)dx=(2A)/(pi)rArrint_(0)^(1){Asin((pix)/(2))+B}dx=(2A)/(pi)`
`rArr[-(2A)/(pi)"cos"(pix)/(2)+Bx]_(0)^(1)=(2A)/(pi)rArrB+(2A)/(pi)=(2A)/(pi)`
`rArrB=0`
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