Let f(x) = { ((1+ax)^(1/x) , x<0), (1+b, x=0), ((x+4)^(1/2)-2 / (x+c)^(1/3)-2, x>0) be continuous at = 0 then e^a bc is equal to

Question

Let f(x) = \(\begin{cases} (1+ax)^\frac{1}{x} &,\ x<0 \\ 1+b &, \ x=0 & \\ \frac{(x+4)^\frac{1}{2}-2}{(x+c)\frac{1}{3}-2} & , x>0 \end{cases} \)be continuous at = 0 then \(e^abc\) is equal to

(1) 64 

(2) 48 

(3) 72 

(4) 36

Answer 1

Correct option is: (2) 48   

\(b+1=3 \Rightarrow b=2\) 

\(e^a \times bc = (b+1)bc\)

\(= 3 \times 2 \times 8=48\)