Final answer:
To determine the positve values of 'a' for which the function f(x) is continuous, we need to ensure that the numerator and denominator do not have common factors that would cause a discontinuity. The denominator x²-a must not be zero, but if 'a' is either 1 or 4, the numerator will cancel with the denominator at the corresponding values, ensuring continuity. Thus, the correct values of 'a' are 1 and 4.
Step-by-step explanation:
The question asks for the positive values of a for which the function f(x) is continuous for all real numbers x. The function is given by f(x) = (x-1)(x²-4)/x²-a. To find the values of a where the function is continuous, we need to ensure that the denominator does not become zero as this would cause a discontinuity.
First, notice that the numerator (x-1)(x²-4) factors further into (x-1)(x-2)(x+2). Now, let's look at the denominator x²-a. For the function to be continuous, we must not have any values of x where the denominator becomes zero. This means we need to set x²-a ≠ 0, which implies a cannot equal any real squared number.
However, we have a factor of (x-1) in the numerator which would cancel out with (x-1) if it were in the denominator. Thus, if a was 1, we would have continuity since the (x-1) term would cancel and there would not be a problem at x = 1. If a were 4, the same would be true because (x-2) would cancel the zero of the denominator at x = 2. Therefore, the function f(x) is continuous for a being positive values 1 and 4, which corresponds to option (E) 1 and 4 only.