The area bounded by the curve y=f(x), the x−axis and the ordinates x=1 and x=b is (b−1)cos(3b+4) sq.unit. Then f(x) is given by:
A
(x−1)sin(3x+4)
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B
3(x−1)sin(3x+4)+cos(3x+4)
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C
cos(3x+4)−3(x−1)sin(3x+4)
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D
none of the above
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Solution
The correct option is Ccos(3x+4)−3(x−1)sin(3x+4) ∫b1f(x)dx=(b−1)cos(3b+4) Differentiating both sides w.r.t b we get f(b)=(b−1).3(−sin(3b+4))+cos(3b+4).1 or f(b)=cos(3b+4)−3(b−1)sin(3b+4) By replacing b→x we get ∴f(x)=cos(3x+4)−3(x−1)sin(3x+4)