Question 77·200 Super-Hard SAT Math Questions·Advanced Math

The first numerator has $x^2-4x+4=(x-2)^2$, and the middle numerator contains $x-2$. So $x=2$ makes both terms zero. It's allowed because $x>0$.

Call the given expression $f(x)$. Type it into Desmos, then type $f(2)$. Desmos shows $2048$; at $x=2$, only the last fraction contributes.

The expression equals $\frac12x^k$ for every allowed $x$, so it must equal that form at $x=2$. Keep the $\frac12$ when you write the equation:

Type $\frac12 2^k\sim2048$. The $\sim$ tells Desmos to fit $k$; under PARAMETERS, it shows $k=12$. The promised equivalent form and the allowed input $2$ pin down the exponent. So the value of $k$ is $12$. Choice B.

The three fractions already have the same denominator, so put their numerators together, keeping the minus sign on the whole middle numerator:

Each numerator term contains $x^{19}$. Factoring it out leaves $x$ from $x^{20}$ and $x^2$ from $x^{21}$:

Because $x^2-4x+4=(x-2)^2$, the bracket becomes

Use the perfect-square identity $A^2-2AB+B^2=(A-B)^2$, with $A=x-2$ and $B=x$:

Simplify inside the square:

Square the result:

The powers of $x$ inside the bracket cancel.

Since $x>0$, the denominator isn't zero. Subtract the denominator's exponent:

Subtract $7$ from $19$:

Divide $4$ by $8$:

So the required exponent is $k=12$. Choice B.