If $f(x)=\frac{4+\sin 2x+A\sin x+B\cos x}{x^2}$ for x ≠ 0 is continuous at x = 0, then A + B + f(0) is:

-4

As f (x) is continuous at x = 0, $f(0)=\underset{x→0}{\lim}f(x)$ & both f(0) and $\underset{x→0}{\lim}f(x)$ are finite.

$f(0)=\underset{x→0}{\lim}\frac{4+\sin 2x+A\sin x+B\cos x}{x^2}$

As D → 0, N should also approach 0 ⇒ 4 + sin 2(0) + A sin 0 + B cos 0 = 0   B = –4

$⇒f(0)=\underset{x→0}{\lim}\frac{4(1-\cos x)+2\sin x\cos x+A\sin x}{x^2}=\underset{x→0}{\lim}2(\frac{\sin x/2}{x/2})^2+\underset{x→0}{\lim}\frac{\sin x}{x}.\underset{x→0}{\lim}\frac{2\cos x+A}{x}$

$=2+\underset{x→0}{\lim}\frac{2\cos x+A}{x}$

As D → 0, N should also apporoach 0 ⇒ A + 2 cos (0) = 0 ⇒ A = -2

$⇒f(0)=2+\underset{x→0}{\lim}\frac{-2\sin^2x/2}{x}=2+0=2⇒f(0)=2$