If `f(x)=(ax + b)sinx+(cx +d)cos x`, then the values of a b, c and d such that `f"(x)=xcos x` for all x, then (a + b + c + d)is

Options

A. a=0, b=c1, d=0
B. a=b=c=d=1
C. a=d=0, b=c=-1
D. none of these

Answer

A

Explanation

We have,
`f(x)=(ax+b)cosx+(cx-d)sinx`
`implies" "f'(x)=-(ax+b)sinx+acosx+(cx+d)cosx+csinx`
`implies" "f'(x)=(cx+d+a)cosx+(c-ax-b)sinx`
But, `f'(x)=xcosx` for all x.
`:." "(cx+d+a)cosx+(c-ax-b)sinx=xcosx` for all x
`implies" "c=a,d+a=0,a=0" and "c-b=0`
`implies" "a=0, b=c=1, d=0.`