Question

The area bounded by the curve y=f(x), x-axis and ordinates x = 1 and x=b is $(b-1)sin(3b+4)$, then f(x) is

• A. $3(x-1)cos(3x+4)+sin(3x+4)$
• B. $(b-1)sin(3x+4)+3cos(3x+4)$
  
• C. $(b-1)cos(3x+4)+3sin(3x+4)$
• D. $Noneofthese$

Answer 1

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

Area bounded by the curve y=f(x), x-axis and the ordinates x=1 and x=b is ${\int }_1^{b}f\left(x\right)dx$
$\therefore$ From the question ${\int }_1^{b}f\left(x\right)dx=(b-1)sin(3b+4)$
Differentiate with respect to b, we get
$f\left(b\right).1=3(b-1)cos(3b+4)+sin(3b+4)$
$f\left(x\right)=3(x-1)cos(3x+4)+sin(3x+4)$.