Let $f(x)=\left\{\begin{matrix}\frac{a(1-x\sin x)+b\cos x+5}{x^2},&x<0\\3,&x=0\\\{1+(\frac{cx+dx^3}{x^2})\}^{1/x},&x>0\end{matrix}\right.$

If f is continuous at x = 0, then the value of a + b + c + d is:

-5 + log3

f (x) is continuous at x = 0 ⇒  L.H.L. = R.H.L. = f (0) = 3

Consider L.H.L.

$\underset{x→0^-}{\lim}\frac{a[1-x(x\frac{-x^3}{3!}+\frac{x^5}{5!}-+...)]+b(\frac{1-x^2}{2!}+\frac{x^4}{4!}-+...)+5}{x^2}$

Since $\left.\begin{matrix}coeff.(x^0)=0⇒a+b+5=0\\coeff.(x)=0⇒x\\coeff.(x^2)=3⇒-a-\frac{b}{2}=3\end{matrix}\right\}a=-1,b=-4$

Consider R.H.L.

$\underset{x→0^+}{\lim}(1+\frac{x+dx^3}{x^2})^{1/x}=3$ for limit to exist : c = 0 and $\underset{x→0}{\lim}(1+dx)^{1/x}=e^d=3⇒d=ln\,3$