Super-Hard SAT Math Question 140 | Free SAT Prep | Aniko

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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 $x\ge 0$ . C‍ont‍еnt by Аniко.аі

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

For nonlinear function “must be true” questions, focus on qualitative behavior instead of heavy algebra: first analyze how the key nonlinear part changes (here, $r^x$ with $0<r<1$ ), then translate that change through the function (denominator smaller means fraction larger). For maximum/minimum claims, look for bounds and asymptotes: show the output stays below some value and whether it can ever equal that value. For existence claims like “there exists $a$ ,” set up the target equation and check whether the required intermediate value is always possible given the parameter restrictions. aniko‌.aі

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Think about $r^x$ when $0<r<1$

As $x$ increases, does $r^x$ get larger or smaller when the base is between 0 and 1?

Compare the denominator at two different $x$ -values

If $r^{x_1}>r^{x_2}$ , what can you say about $1+n r^{x_1}$ compared to $1+n r^{x_2}$ ? Then what happens to the fraction?

Check the “maximum value” idea using an upper bound

Show that $1+n r^x>1$ for all $x\ge 0$ . What does that imply about how large $\frac{m}{1+n r^x}$ can be? (А‍ni‍ко.аі)

For Statement III, set up an equation for $a$

Write $\frac{m}{1+n r^a}+d=d+\frac{m}{2}$ and simplify until you get something like $r^a=$ (an expression in $n$ ). Then decide if that expression must always be between 0 and 1.

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Graph the function with sliders

Enter So‍ur⁠ce: аnі‌kо‍.аі

$f(x)=m/(1+n\,r^x)+d$

Create sliders for $m$ , $n$ , $r$ , and $d$ , and set $0<r<1$ , $m>0$ , $n>0$ , $d>0$ .

Check Statement I (increasing) with a table

Make a table for $x$ values like $0,1,2,3,4$ . Observe whether the $f(x)$ values increase as $x$ increases for several different slider settings (still keeping $0<r<1$ ).

Check Statement II (no maximum) by looking at the long-run behavior

Add a line $y=m+d$ . Increase $x$ (zoom out or extend the table to larger $x$ ). Observe that the graph gets closer and closer to $y=m+d$ but stays below it.

Test Statement III by changing $n$

Add a line $y=d+m/2$ . Try values with $n>1$ and values with $0<n<1$ . Look for whether the graph intersects $y=d+m/2$ when $n<1$ ; if it does not, then Statement III is not guaranteed.

Step-by-step Explanation

Use how $r^x$ changes when $0<r<1$

If $0<r<1$ and $x_1<x_2$ , then $r^{x_1}>r^{x_2}$ . So as $x$ increases, $r^x$ gets smaller.

Decide whether $f$ is increasing (Statement I)

As $x$ increases, $r^x$ decreases, so $n r^x$ decreases, so the denominator $1+n r^x$ decreases.

Since $m>0$ , dividing by a smaller positive number gives a larger value:

\frac{m}{1+n r^x}

increases as $x$ increases. Adding $d$ does not change increasing/decreasing behavior, so $f$ is increasing on $x\ge 0$ . Thus, Statement I is true.

Check whether a maximum exists (Statement II)

That means

\frac{m}{1+n r^x}<\frac{m}{1}=m

So for all $x\ge 0$ ,

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

As $x$ gets very large, $r^x\to 0$ , so $f(x)$ approaches $m+d$ but never equals $m+d$ for any finite $x$ . Therefore, $f$ has no maximum value on $x\ge 0$ , and Statement II is true.

Test whether Statement III must be true

Statement III claims there must be some $a>0$ such that

f(a)=d+\frac{m}{2}

Solve:

\begin{aligned} \frac{m}{1+n r^a}+d &= d+\frac{m}{2}\\ \frac{m}{1+n r^a} &= \frac{m}{2}\\ 1+n r^a &= 2\\ r^a &= \frac{1}{n} \end{aligned}

A solution $a>0$ exists only if $\frac{1}{n}$ is a possible value of $r^a$ . Since $0<r<1$ , the range of $r^a$ for $a>0$ is $(0,1)$ .

If $0<n<1$ , then $\frac{1}{n}>1$ , which is impossible for $r^a$ . So Statement III is not guaranteed.

Therefore, the statements that must be true are I and II only, so the correct choice is I and II only.

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Step-by-step Explanation

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