A) \[\log \left( \frac{1}{2} \right)\]
B) 0
C) 4
D) \[-1+\log 2\]
Correct Answer: D
Solution :
\[\because f(x)=\frac{-{{e}^{x}}+{{2}^{x}}}{x}\] and\[f(x)\]is continuous at x = 0 \[\therefore \] \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0)\] \[\Rightarrow \]\[\underset{x\to 0}{\mathop{\lim }}\,\frac{-{{e}^{x}}+{{2}^{x}}}{x}=f(0)\] \[\Rightarrow \]\[\underset{x\to 0}{\mathop{\lim }}\,\frac{-{{e}^{x}}+{{2}^{x}}\log 2}{1}=f(0)\] (using LHospitals rule) \[\Rightarrow \] \[-{{e}^{0}}+{{2}^{0}}\log 2=f(0)\] \[\Rightarrow \] \[f(0)=-1+log2\]
You need to login to perform this action.
You will be redirected in
3 sec
