Question

If the area bounded by the line y=f( x ) $,$x-axis and the line ordinate $x=1$ and $x=b$ is $(b-1)\sin (3b+4),$ then

• A. $f\left(x\right)=\cos (3x+4)+3(x-1)\sin (3x+4)$
• B. $f\left(x\right)=\sin (3x+4)+3(x-1)\cos (3x+4)$
• C. $f\left(x\right)=\sin (3x+4)-3(x-1)\cos (3x+4)$
• D. none of these

Answer 1

Answer: (B)

$f\left(x\right)=\sin (3x+4)+3(x-1)\cos (3x+4)$

We are given ${\int }_1^{b}f\left(x\right)dx=(b-1)\sin (3b+4)$

Differentiating with respect to $b$, we get

${\left[f\left(x\right)\right]}_{x=b}=\sin (3b+4)+(b-1).3\cos (3b+4)\therefore f\left(x\right)=\sin (3x+4)+3(x-1)\cos (3x+4)$