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 P and Q are stationary points, we need to find the derivative of the function f(x) and set it to zero. First, we differentiate f(x):

Using the quotient rule:

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

Simplifying the derivative:

1. Expand the numerator: $= \frac{4x^2 + 10x - (2x^2 + 20)}{(2x + 5)^2}$ $= \frac{2x^2 + 10x - 20}{(2x + 5)^2}$

2. Set the numerator to zero to find stationary points: $2x^2 + 10x - 20 = 0$ $x^2 + 5x - 10 = 0$

3. Use the quadratic formula to solve for x: $x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-10)}}{2(1)}$ $= \frac{-5 \pm \sqrt{25 + 40}}{2}$ $= \frac{-5 \pm \sqrt{65}}{2}$

This leads to two stationary points: $x = \frac{-5 + \sqrt{65}}{2}, \quad x = \frac{-5 - \sqrt{65}}{2}$

Thus, points P and Q are stationary points of the curve.

To find the range of f, we need to analyze the behavior of the function:

1. As the domain of f excludes $x = -\frac{5}{2}$, we check the limits:

   • As $x$ approaches $-\frac{5}{2}$ from the left, $f(x) \to -\infty$.
   • As $x$ approaches $-\frac{5}{2}$ from the right, $f(x) \to +\infty$.

2. Lastly, we check for horizontal asymptotes as $x \to \infty$ or $x \to -\infty$:

   • As $x \to \infty$, (f(x) \to 1).
   • As $x \to -\infty$, (f(x) \to 1).

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

$R(f) = (-\infty, -5) \cup (1, +\infty)$