(1) The denominator can be factorised:
x³-ax²-x+a=x³-x-(ax²-a)=x(x²-1)-a(x²-1)=(x-a)(x²-1)=(x-a)(x-1)(x+1).
The numerator can be factorised:
(x+2)(x+1).
The common factor x+1 can be cancelled (for x≠-1) so the expression is (x+2)/((x-1)(x-a).
We can use partial fractions:
(x+2)/((x-1)(x-a))≡A/(x-1)+B/(x-a) where A and B are constants:
x+2≡A(x-a)+B(x-1)=Ax-Aa+Bx-B.
Matching coefficients:
A+B=1, -Aa-B=2, add these equations: A-Aa=3, A(1-a)=3, A=3/(1-a).
B=1-A=1-3/(1-a)=(1-a-3)/(1-a)=-(2+a)/(1-a)
So the expression becomes 3/((x-1)(1-a))-(2+a)/((x-a)(1-a)).
(2) I read this as y=log₂(tan(√x)).
The argument of square root must be positive, and the argument of log must be strictly positive. The argument of tangent must be less than π/2. Therefore 0<x<π²/4, and the range is (-∞,∞). However, √x can be in the third quadrant where tangent is positive. π<√x<3π/2, meaning π²<x<9π²/4. This pattern repeats every cycle, without altering the range, but the domain is discontinuous. There is no valid value of x between π²/4 and π². Each domain region gets progressively wider.