Let $f(x)=(a x+b) \cos x+(c x+d) \sin x$ and $f'(x)=x \cos x$ for all $x$. Then,
Answer & explanation
Correct answer: option 1
We have,
$f(x)=(a x+b) \cos x+(c x+d) \sin x$
$\Rightarrow f'(x)=-(a x+b) \sin x+a \cos x+(c x+d) \cos x+c \sin x$
$\Rightarrow f'(x)=(c x+d+a) \cos x+(c-a x-b) \sin x$
But, f'(x) = x cos x for all x
∴ $(c x+d+a) \cos x+(c-a x-b) \sin x=x \cos x$ for all x
$\Rightarrow c=1, d+a=0, a=0$ and $c-b=0$
$\Rightarrow a=0, b=c=1, d=0$