The function f is defined by $$f(x) = \frac{x^2 + 10}{2x + 5}$$ where f has its maximum possible domain - AQA - A-Level Maths Pure - Question 10 - 2022 - Paper 3

To show that points P and Q are stationary points, we need to find the derivative of the function and set it to zero:

1. Differentiate f(x): Using the quotient rule, we have:

$f'(x) = \frac{(2x + 5)(2x) - (x^2 + 10)(2)}{(2x + 5)^2}$

2. Set the derivative to zero: Setting the numerator equal to zero:

$(2x + 5)(2x) - (x^2 + 10)(2) = 0$

This simplifies to:

$2x^2 + 5(2x) - 2(x^2 + 10) = 0$

Which leads to:

$2x^2 + 10x - 2x^2 - 20 = 0 \Rightarrow 10x - 20 = 0 \Rightarrow x = 2$

Now substituting back to find y:

$f(2) = \frac{2^2 + 10}{2(2) + 5} = \frac{14}{9}$

Hence, points P and Q intersect concisely at these x-values, which implies they are stationary points.

To determine the range of f, we analyze the behavior of the function. Since the function is a rational function and employs a polynomial for the numerator and denominator, we can postulate that the range will exclude values outside the limits prescribed by the horizontal asymptotes and the stationary points found previously.

Thus, the range of f can be expressed in set notation as:

$R(f) = \{ y \in \mathbb{R} : y \neq \frac{5}{2} \}$

This excludes the horizontal asymptote that approaches (y = \frac{5}{2}) as (x) approaches infinity.