Question
The area bounded by the curves $$y = f\left( x \right),$$ the $$x$$-axis and the ordinates $$x = 1$$ and $$x = b$$ is $$\left( {b - 1} \right)\sin \left( {3b + 4} \right).$$ Then $$f\left( x \right)$$ is-
A.
$$\left( {x - 1} \right)\cos \left( {3x + 4} \right)$$
B.
$$\sin \,\left( {3x + 4} \right)$$
C.
$$\sin \,\left( {3x + 4} \right) + 3\left( {x - 1} \right)\cos \left( {3x + 4} \right)$$
D.
none of these
Answer :
$$\sin \,\left( {3x + 4} \right) + 3\left( {x - 1} \right)\cos \left( {3x + 4} \right)$$
Solution :
$$ATQ\,\int_1^b {f\left( x \right)dx} = \left( {b - 1} \right)\sin \left( {3b + 4} \right)$$
Differentiating both sides w.r.t $$b,$$ we get
$$\eqalign{
& f\left( b \right) = 3\left( {b - 1} \right)\cos \left( {3b + 4} \right) + \sin \left( {3b + 4} \right) \cr
& \Rightarrow f\left( x \right) = 3\left( {x - 1} \right)\cos \left( {3x + 4} \right) + \sin \left( {3x + 4} \right) \cr} $$