Problem: If k is a positive number and f is a function such that, for every positive number x,
[f(x2+1)]x​=k
then, for every positive number y,
[f(y29+y2​)]y12​​
is equal to
Answer Choices:
A. k​
B. 2k
C. kk​
D. k2
E. yk​
Solution:
[f(y29+y2​)]y2​​​=⎣⎢⎢⎡​[f((y3​)2+1)]y3​​⎦⎥⎥⎤​2=k2.