If linear function f(x) and g(x) satisfy ∫[(3x-1)cosx+(1-2x)sinx]dx=f(x)cosx+g(x)sinx+c, then
fx=3x-1
fx=3x-5
gx=3x-1
gx=3+x
Explanation for the correct option.
Step 1: Solve L.H.S of the given equation.
∫[(3x-1)cosx+(1-2x)sinx]dx=∫3xcosxdx-∫cosxdx+∫sinxdx-∫2xsinxdx=3∫xcosxdx-∫cosxdx+∫sinxdx-2∫xsinxdx=3(cosx+xsinx)-sinx+-cosx-2sinx-xcosx+c=3cosx+3xsinx-sinx-cosx-2sinx+2xcosx+c=cosx(3-1+2x)+sinx3x-1-2+c=2+2xcosx+3x-3sinx+c=2x+1cosx+3x-1sinx+c
Step 2: Find the value of fxandgx.
The given equation is ∫[(3x-1)cosx+(1-2x)sinx]dx=f(x)cosx+g(x)sinx+c
By solving L.H.S, we get
2x+1cosx+3x-1sinx+c=f(x)cosx+g(x)sinx+c
From here we get
fx=2(x+1)andgx=3x-1
Hence, option C is correct.
Use the factor theorem to determine whether g(x) is a factor of f(x)
f(x)=22x2+5x+2;g(x)=x+2