Question 130·200 Super-Hard SAT Math Questions·Advanced Math

A point $(x,y)$ on a function means its output at input $x$ is $y$. So both given points give equations: $f(c)=14\quad\Longrightarrow\quad a^c+b=14$, $f(-c)=5\quad\Longrightarrow\quad a^{-c}+b=5$.

The same $b$ is added to each power. Subtract $b$ from both equations: $a^c=14-b,\qquad a^{-c}=5-b$.

Multiplying powers with the same base adds their exponents. Here $c$ and $-c$ add to $0$, so $a^c a^{-c}=a^0=1$. Multiply the expressions from the previous step: $(14-b)(5-b)=1$. This removes both $a$ and $c$. At opposite inputs, the outputs minus the shared shift multiply to $1$.

Because $a$ is positive, $a^{-c}$ must be positive. Since $a^{-c}=5-b$, we need $5-b>0\quad\Longrightarrow\quad b<5$. Keep this condition: two negative factors can multiply to $1$, but neither original power can be negative.

Type $(14-b_1)(5-b_1)\sim1\{b_1<5\}$ in Desmos. The subscript makes $b_1$ a value to fit rather than a slider, and $\sim$ tells Desmos to fit it. Under PARAMETERS, $b_1\approx4.890$. Type each radical choice on a line below it; they evaluate to about $14.110$, $-4.890$, $-14.110$, and $4.890$, in order. The matching value of $b$ is $\frac{19-\sqrt{85}}{2}$. Choice D.