Problem: The expression a+yyββaβyaβa+yaβ+aβyyββ, a real, aξ =0, has the value β1 for:
Answer Choices:
A. all but two real values of y
B. only two real values of y
C. all real values of y
D. only one real value of y
E. no real values of y
Solution:
To avoid division by zero, yξ =a and yξ =βa. With these values excluded we may simplify the fraction by multiplying numerator and denominator by (a+y)(aβy), obtaining ayβy2βa2βaya2βay+ay+y2β=β(a2+y2)a2+y2β=β1