Problem:
The function f defined by f(x)=cx+dax+b, where a,b,c, and d are nonzero real numbers, has the properties f(19)=19,f(97)=97, and f(f(x))=x for all values of x except −d/c. Find the unique number that is not in the range of f.
Solution:
The statement implies that f is its own inverse. The inverse may be found by solving x=cy+day+b for y. This yields f−1(x)=−cx+adx−b. Because the nonzero numbers a,b, c, and d must therefore be proportional to −d,b,c, and −a, respectively, it follows that a=−d, hence that f(x)=cx−aax+b. The conditions f(19)=19 and f(97)=97 lead to the equations
192c=2⋅19a+b972c=2⋅97a+b
Thus (972−192)c=2(97−19)a, from which follows a=58c, which in turn leads to b=−1843c. This determines
f(x)=x−5858x−1843=58+x−581521
which never has the value 58.
OR
The number that is not in the domain of f is x=−d/c, and the number that is not in the range of f is y=a/c. Because f=f−1, it follows that d=−a. The fixed points of f satisfy
ax+b=x(cx+d), or cx2−(a−d)x−b=0
It follows that the sum of the two fixed points is (a−d)/c=2a/c, which is twice the number omitted from the range. Because the fixed points are given as 19 and 97 , the number omitted from the range is (19+97)/2=58.
OR
The equation y=cx+dax+b is equivalent to the equation xy−cax+cdy=cb, hence to the equation (x+cd)(y−ca)=c2bc−ad. Notice that bc−ad cannot be zero, for this would imply that f(x)=ca for all values of x except −cd, contrary to the problem statement. The graph of y=f(x) is therefore a hyperbola, whose center is (c−d,ca). The statement implies that the line y=x is an axis of symmetry. Because the hyperbola intersects this axis at (19,19) and at (97,97), these points are the vertices of the hyperbola, and its center is midway between them. It follows that c−d=ca=219+97=58, which is the value omitted from the range of f (and also from the domain of f).
The problems on this page are the property of the MAA's American Mathematics Competitions