Question 10·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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The function $h$ is defined by $h(x)=a^{x/2}+b$ , where $a$ and $b$ are positive constants. It is known that $h(2)=10$ , $h(6)=16$ , and $h(10)=40$ . Which choice is equal to $h(1)$ ?

A shifted exponential function has a power term plus a fixed amount. If several outputs are given but the base and fixed amount are unknown, divide each input by the number in the exponent’s denominator, then use a Desmos element-list regression to fit all the outputs at once. For the requested input, use the entire exponent: a half-power means a square root, not an extra factor.

Hints

Hint 1
The exponent is $x/2$ , so inputs $2$ , $6$ , and $10$ produce exponents $1$ , $3$ , and $5$ . What three equations do the given outputs make when you keep the same added $b$ in each?
Hint 2
A regression lets Desmos find unknown constants that fit several facts together. Put the three expressions involving $a$ and $b$ in one list, and the three known outputs in another, with $\sim$ between the lists.
Hint 3
To find $h(1)$ , put $1$ in for $x$ . The exponent becomes $1/2$ , and raising a positive number to the one-half power means taking its square root. The added $b$ stays outside the power.

Step-by-step

Approach 1: Fit the constants in Desmos

Step 1
Turn the three outputs into equations
Divide each given input by $2$ to get its exponent: $2/2=1$ , $6/2=3$ , and $10/2=5$ . The shift $b$ is added unchanged to every power, so the three given outputs mean $a+b=10$ , $a^3+b=16$ , and $a^5+b=40$ .
Step 2
Find the base and shift
Type $[a+b,a^3+b,a^5+b]\sim[10,16,40]$ . The $\sim$ tells Desmos to fit all three equations together. Under PARAMETERS, its regression shows $a=2$ and $b=8$ ; both meet the given positive-constant condition.
Step 3
Evaluate the requested input
For $h(1)$ , the exponent is $1/2$ . A half-power means the positive square root of the base. Type $\sqrt a+b$ below the regression; Desmos shows about $9.414$ . Using the fitted constants gives the exact value $\sqrt2+8=8+\sqrt2$ . Choice D.

Approach 2: Cancel the shift by subtracting outputs

Step 1
Find two changes without the shift
Subtract $h(2)$ from $h(6)$ . The same $b$ cancels from both outputs:
$a^3-a=16-10=6.$
Subtract $h(6)$ from $h(10)$ for the next input jump of $4$ :
$a^5-a^3=40-16=24.$
Step 2
Use the changes to find the base
Since $a^5-a^3=a^2(a^3-a)$ , the second change is $a^2$ times the first. Compare the changes:
$24=6a^2.$
Divide by $6$ :
$4=a^2.$
Take the positive root because the problem says $a>0$ :
$a=2.$
Step 3
Recover the fixed shift
The given $h(2)=10$ means $a+b=10$ . Replace $a$ with $2$ :
$2+b=10.$
Subtract $2$ : $b=8$ . This is the amount added after the power.
Step 4
Find the exact output
At $x=1$ , the exponent is $1/2$ . Substitute the constants:
$h(1)=2^{1/2}+8.$
Rewrite the half-power as a square root: $h(1)=8+\sqrt2$ . Choice D.

Question 10·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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