Science:Math Exam Resources/Courses/MATH101 C/April 2024/Question 18 (b)/Solution 1

As $q$ gets big, so does $2^{q}$, which means that ${\frac {1}{2^{q}}}$ gets very small. Therefore,

$${\begin{aligned}\lim _{q\to \infty }D(q)=\lim _{q\to \infty }{\frac {1}{2^{q}}}+1=0+1=1.\end{aligned}}$$

Let us use this to solve for ${\tilde {q}}_{e}$:

$${\begin{aligned}S({\tilde {q}}_{e})&=\lim _{q\to \infty }D(q).\\S({\tilde {q}}_{e})&=1.\end{aligned}}$$

Let us use the piecewise formula for $S(q)$ to find the value ${\tilde {q}}_{e}$ such that $S({\tilde {q}}_{e})=1$. If ${\tilde {q}}_{e}\leq 10$, the equation we are trying to solve becomes $0=1$, which of course does not have a solution. If ${\tilde {q}}_{e}>10$, it becomes ${\frac {{\tilde {q}}_{e}}{10}}-1=1$, which does have a solution:

$${\begin{aligned}{\frac {{\tilde {q}}_{e}}{10}}-1&=1.\\{\frac {{\tilde {q}}_{e}}{10}}&=2.\\{\tilde {q}}_{e}&=20.\end{aligned}}$$

Luckily for us, this solution is indeed greater than $10$.

Thus, ${\tilde {q}}_{e}=20$.