Answer:
f(x) =
![(\frac {1}{2})^((6x))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hfvxmzxxv57qhb08csgffyr385k9hudhsw.png)
Explanation:
The given options are,
a) f(x) =
![2^((6x))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/3bxee1ge4b9gzj3owsgzmyp0ipvb4zpby4.png)
b) f(x) =
![(\frac {1}{2})^((6x))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hfvxmzxxv57qhb08csgffyr385k9hudhsw.png)
c) f(x) =
![2 * (\frac {1}{6})^(x)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/a5461g21hajt5t3h2zm222qcceah63siam.png)
d) f(x) =
![\frac {1}{2} * (\frac {1}{6})^(x)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cc01zke6iy5ey54d3hlemth4krw72w47wy.png)
Now, clearly a) is a monotonically increasing function, hence discarded, and both of c) and d) don't pass through (0, 1) hence they are also discarded.
Only b) is a decay function which does also pass through (0, 1), hence, b) is the correct option.