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The area bounded by the curves y=f(x), t...

The area bounded by the curves `y=f(x),` the x-axis, and the ordinates `x=1a n dx=b` is `(b-1)sin(3b+4)dot` Then `f(x)` is. `(x-1)cos(3x+4)` `sin(3x+4)` `sin(3x+4)+3(x-1)cos(3x+4)` None of these

A

`(x-1)cos(3x+4)`

B

`8sin(3x+4)`

C

`sin(3x+4)+3(x-1)cos(3x+4)`

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
C
Since, `int_(1)^(b)f(x)dx=(b-1)sin(3b+4)`
On differentiating both sides w.r.t. b, we get
`f(b)=3(b-1)*cos(3b+4)+sin(3b+4) `
` therefore f(x)=sin(3x+4)+3(x-1)cos(3x+4)`
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