Question

The area bounded by the curve $y=f\left(x\right)$, the $x-$axis and the ordinates $x=1$ and $x=b$ is $(b-1)\cos (3b+4)$ sq.unit. Then $f\left(x\right)$ is given by:

• A. $(x-1)\sin (3x+4)$
• B. $3(x-1)\sin (3x+4)+\cos (3x+4)$
• C. $\cos (3x+4)-3(x-1)\sin (3x+4)$
• D. none of the above

Answer 1

The correct option is C $\cos (3x+4)-3(x-1)\sin (3x+4)$

${\int }_1^{b}f\left(x\right)dx=(b-1)\cos (3b+4)$
Differentiating both sides w.r.t $b$ we get
$f\left(b\right)=(b-1).3(-\sin (3b+4))+\cos (3b+4).1$
or $f\left(b\right)=\cos (3b+4)-3(b-1)\sin (3b+4)$
By replacing $b\to x$ we get
$\therefore f\left(x\right)=\cos (3x+4)-3(x-1)\sin (3x+4)$