Algebra of real functions

By the definition of the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Since the denominators are the same, add the numerators: \((2x+1)+(x-3)=3x-2\). Therefore, \((f+g)(x)=\frac{3x-2}{x^2+1}\). Option C results from incorrectly treating the constant terms; \(1+(-3)=-2\), not \(+4\). Exam tip: when adding algebraic fractions with the same denominator, add only the numerators and retain the common denominator; here \(x^2+1>0\), so the denominator is never zero.

Substituting \(x=0\), we get \(f(0)=0^2-2(0)+2=2\) and \(g(0)=0^2+2(0)+2=2\). By the definition of the product function, \((fg)(0)=f(0)g(0)=2\times2=4\), so option C is correct. Option B may result from using the value of only one function. In an exam, remember that \((fg)(x)\) denotes the product of \(f(x)\) and \(g(x)\), not their sum.

(\frac{f}{g}=\frac{(x-5)^2+1}{(x-5)^2+5}), which gives (\frac{1}{5}) at ((x-5)^2=0). Set the common square term as (t\ge 0).

Subtract the functions: (f − g)(x) = (x² + ax + 2) − (x² − 2ax + 1) = 3ax + 1. Thus, (f − g)(3) = 9a + 1 = 22, so 9a = 21 and a = 7/3. Therefore, option D is correct.