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 x = 0. Then ea bc is equal to

Question

Let \(f(x)=\left\{\begin{array}{lll}(1+a x)^{1 / x} & , & x<0 \\ 1+b & , & x=0 \\ \frac{(x+4)^{1 / 2}-2}{(x+c)^{1 / 3}-2} & , & x>0\end{array}\right.\) 

be continuous at x = 0. Then \(e^{a}\) bc is equal to

(1) 64

(2) 72

(3) 48

(4) 36   

Answer 1

Correct option is: (3) 48 

\(\mathrm{f}\left(0^{-}\right)=\mathrm{e}^{\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\mathrm{ax}}{\mathrm{x}}}=\mathrm{e}^{\mathrm{a}}\)

\(\mathrm{f}(0)=1+\mathrm{b}\)

\(\mathrm{f}\left(0^{+}\right)=\frac{\frac{1}{2 \sqrt{\mathrm{x}+4}}}{\frac{1}{3}(\mathrm{x}+\mathrm{c})^{-\frac{2}{3}}}=\frac{\frac{1}{2(2)}}{\frac{1}{3} \cdot \mathrm{c}^{-\frac{2}{3}}}\)

\(=\frac{3}{4} \mathrm{c}^{2 / 3}\)

Also at \(\mathrm{x}=0;\)

\(c^{1 / 3}=2 \Rightarrow c=8\)

So \(\mathrm{f}\left(0^{+}\right)=\frac{3}{4}(8)^{2 / 3}=3\)

Now, \(\mathrm{e}^{\mathrm{a}}=\mathrm{b}+1=3\)

\(\mathrm{e}^{\mathrm{a}} . \mathrm{b} . \mathrm{c}=3.2 .8=48\)