Question 140·200 Super-Hard SAT Math Questions·Advanced Math

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f(x)=\frac{m}{1+n\,r^x}+d

The function $f$ is defined by the equation above for all $x\ge 0$ . The constants $m$ , $n$ , and $d$ are positive, and $0<r<1$ .

Which choice correctly identifies all statements that must be true?

I. The function $f$ is increasing on $x\ge 0$ .

II. The function $f$ has no maximum value on $x\ge 0$ .

III. There exists a value $a>0$ such that $f(a)=d+\frac{m}{2}$ .

The cue is a decaying exponential inside a denominator, with $0<r<1$ . Track how the denominator changes as $x$ grows, then ask whether any allowed input gives the largest output. A statement that must be true has to work for every allowed set of constants, so one counterexample can disprove it. Don't mistake a height the function approaches for a maximum it reaches.

Hints

Hint 1
Because $0<r<1$ , raising $r$ to a larger power makes $r^x$ smaller, though it stays positive. What does that do to the denominator $1+nr^x$ ?
Hint 2
A maximum is an output no allowed input beats. The domain has no last input: whenever $x\ge0$ is allowed, $x+1$ is allowed too. Once you know which way $f$ moves, can it have a maximum?
Hint 3
To disprove a claim that must hold, try one allowed set of constants. If $n=1$ , the denominator at $x=0$ is $2$ . Compare $f(0)$ with the output in III, but keep its condition $a>0$ in view.

Step-by-step

Track the denominator, then test III
Step 1
Find how the power changes
Because $0<r<1$ , $r^x$ decreases as $x$ increases but stays positive. For any increase $h>0$ , $r^{x+h}=r^x r^h<r^x$ because $r^h<1$ .
Step 2
Follow the denominator
Since $n>0$ , multiplying $r^x$ by $n$ keeps it decreasing. Adding $1$ doesn't change that direction. So the positive denominator $1+nr^x$ gets smaller as $x$ grows.
Step 3
Decide whether the function increases
With $m>0$ , dividing $m$ by a smaller positive denominator gives a larger fraction. Adding the fixed $d$ doesn't change the comparison. So $f(x)$ rises whenever $x$ rises, and statement I holds.
Step 4
Test for a maximum
A maximum is an output no allowed input beats. For every $x\ge0$ , the larger input $x+1$ is also allowed, and Step 3 gives $f(x+1)>f(x)$ . A strictly increasing function with no last input has no maximum. So statement II holds.
Step 5
Evaluate an allowed counterexample
To test whether III must hold, choose the allowed value $n=1$ . At $x=0$ , $r^0=1$ , making the denominator $2$ . Choose $m=2$ , $d=1$ , and $r=\frac12$ as other allowed values. Type $f(x)=\frac{2}{1+(\frac12)^x}+1$ , then $f(0)$ . Desmos prints $2$ for $f(0)$ .
Step 6
Check the positive-input condition
Type $1+\frac22$ for the target $d+\frac m2$ . Desmos prints $2$ , the same output as $f(0)$ . But $f$ strictly increases, so every $a>0$ gives $f(a)>f(0)$ ; none gives that target. Statement III need not hold, leaving I and II only. Choice C.

Question 140·200 Super-Hard SAT Math Questions·Advanced Math

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