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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

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The function f is defined by $$f(x) = \frac{x^2 + 10}{2x + 5}$$ where f has its maximum possible domain. The curve $y = f(x)$ intersects the line $y = x$ at the p... show full transcript

Worked Solution & Example Answer: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

Step 1

State the value of x which is not in the domain of f.

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Answer

To determine the value of x that is not in the domain of f, we need to identify the points where the denominator equals zero. The denominator is given by:

2x+5=02x + 5 = 0

Solving this equation:

2x=−52x = -5

x=−52x = -\frac{5}{2}

Thus, the value of x which is not in the domain of f is −52-\frac{5}{2}.

Step 2

Show that P and Q are stationary points of the curve.

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Answer

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)=(2x+5)(2x)−(x2+10)(2)(2x+5)2f'(x) = \frac{(2x + 5)(2x) - (x^2 + 10)(2)}{(2x + 5)^2}

Simplifying the derivative:

  1. Expand the numerator: =4x2+10x−(2x2+20)(2x+5)2= \frac{4x^2 + 10x - (2x^2 + 20)}{(2x + 5)^2} =2x2+10x−20(2x+5)2= \frac{2x^2 + 10x - 20}{(2x + 5)^2}

  2. Set the numerator to zero to find stationary points: 2x2+10x−20=02x^2 + 10x - 20 = 0 x2+5x−10=0x^2 + 5x - 10 = 0

  3. Use the quadratic formula to solve for x: x=−5±52−4(1)(−10)2(1)x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-10)}}{2(1)} =−5±25+402= \frac{-5 \pm \sqrt{25 + 40}}{2} =−5±652= \frac{-5 \pm \sqrt{65}}{2}

This leads to two stationary points: x=−5+652,x=−5−652x = \frac{-5 + \sqrt{65}}{2}, \quad x = \frac{-5 - \sqrt{65}}{2}

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

Step 3

Using set notation, state the range of f.

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Answer

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

  1. As the domain of f excludes x=−52x = -\frac{5}{2}, we check the limits:

    • As xx approaches −52-\frac{5}{2} from the left, f(x)→−∞f(x) \to -\infty.
    • As xx approaches −52-\frac{5}{2} from the right, f(x)→+∞f(x) \to +\infty.
  2. Lastly, we check for horizontal asymptotes as x→∞x \to \infty or x→−∞x \to -\infty:

    • As x→∞x \to \infty, (f(x) \to 1).
    • As x→−∞x \to -\infty, (f(x) \to 1).

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

R(f)=(−∞,−5)∪(1,+∞)R(f) = (-\infty, -5) \cup (1, +\infty)

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