If (f:\mathbb{R}\to\mathbb{R}) is to be defined by (f(x)=\frac{1}{x^2+ax+4}), which condition on (a) makes it a function on all of (\mathbb{R})?

Correct answer: (|a|<4)

For the formula \\(f(x)=\\frac{1}{x^2+ax+4}\\) to define a real-valued function for every real x, its denominator must never be zero. Consider the quadratic denominator \\(x^2+ax+4\\). Since its leading coefficient is positive, it remains strictly positive for all real x only when it has no real zero. The discriminant must therefore be negative: \\(a^2-16<0\\).

Solving this inequality gives \\(a^2<16\\), which is equivalent to \\(-4<a<4\\), or \\(|a|<4\\). At \\(|a|=4\\), the denominator has one real zero, so the reciprocal is undefined at that point. If \\(|a|>4\\), it has two real zeros. Hence option A is correct. The condition does not hold for every real a; it must satisfy the strict inequality.