AQA A-Level Mathematics Paper 1, June 2025: Question 12

📐 Step-by-Step Solution

1. Analyze the function: For x ∈ ℝ , the square of any real number satisfies x² ≥ 0 .
2. Add the constant: x² + 5 ≥ 0 + 5 ⇒ f(x) ≥ 5 .
3. Convert to set notation: The question specifically demands set notation, not just an inequality.

✅ Correct Answers

Any standard formal set notation representing values greater than or equal to 5:

• {y : y ≥ 5} or {f(x) : f(x) ≥ 5}
• {x : x ≥ 5}
• Interval notation: [5, ∞)

🧠 Exam Technique

• M1: Awarded for finding the critical bound ≥ 5 or > 5 .
• A1: Awarded strictly for wrapping the inequality in valid set builder brackets { ... } or valid interval brackets [5, ∞) .
• Notice the square bracket at 5 in [5, ∞) because 5 is attainable ( f(0) = 5 ).

❌ Common Errors

• Writing just f(x) ≥ 5 or y ≥ 5 without braces (loses the A1 mark for failing to use set notation).
• Using round brackets for 5 in interval notation: (5, ∞) is incorrect because 5 is included.
• Confusing range with domain and writing {x : x ∈ ℝ} .

💡 Key Knowledge

A function must be one-to-one (bijective) over its defined domain to have an inverse function.

If a function is many-to-one, an inverse cannot exist because each input to the inverse would have more than one output, violating the definition of a function.

✅ Full-Mark Response

Step 1: Demonstrate that f is many-to-one using a counterexample:

f(1) = (1)² + 5 = 6 and f(-1) = (-1)² + 5 = 6

Step 2: State the clear conclusion:

"Since two different inputs produce the same output, f is a many-to-one function (or not one-to-one), and therefore f does not have an inverse."

🧠 Alternative Acceptable Justifications

• Algebraic symmetry: State that f(-x) = f(x) for all x , meaning it is an even / many-to-one function.
• Horizontal Line Test: Draw a sketch of the parabola y = x² + 5 with vertex at (0, 5), draw a horizontal line intersecting it twice, and state that it fails the horizontal line test.

❌ Common Errors

• Simply stating "No" without a mathematical justification (scores 0/2).
• Saying "You can't take the square root of a negative" — this confuses the process of rearranging for the inverse with the mathematical criterion for an inverse to exist.
• Failing to explicitly link the demonstration to the term "many-to-one" or "not one-to-one".

💡 Visualizing the Transformation

• The graph of y = g⁻¹(x) is the reflection of y = g(x) in the line y = x.
• Since g(x) = √x has domain x ≥ 0 and range y ≥ 0 , its inverse is g⁻¹(x) = x² with restricted domain x ≥ 0 .
• The curves intersect at the origin (0, 0) and the fixed point (1, 1) on y = x .

✅ What Must Be Drawn on the Diagram

1. Start the curve at the origin (0, 0) .
2. Between x = 0 and x = 1 , draw a curve that is convex (curving upwards), lying underneath the line y = x .
3. Cross the line y = x at exactly the same point where y = g(x) crosses it (the intersection point (1, 1) ).
4. For values beyond (1, 1) , continue the curve upwards steeply above the line y = x .
5. Crucial: Keep the curve entirely in Quadrant 1 only!

🧠 Mark Scheme Requirements

• M1: Draws a convex curve between the origin and the intersection point (1, 1) with no doubt that it crosses (1, 1) .
• A1: A fully correct, smooth parabolic branch for y = x² restricted strictly to x ≥ 0 .

❌ Common Errors

• Extending the parabola into Quadrant 2 (for negative x ). Because the original function had range y ≥ 0 , the domain of the inverse is x ≥ 0 . Any branch for x < 0 loses the A1 mark!
• Drawing the curve above y = x near the origin. Near 0, x² < x < √x .
• Missing the mutual intersection point on the line of symmetry.

Part (d)(i) — Expression for h(x) [1 Mark]

Substitute the inner function f(x) into the outer function g(x) :

h(x) = g(f(x)) = g(x² + 5)

Note: "No ISW" (Ignore Subsequent Working) applies — do not attempt to simplify this further into x + √5 , as that is mathematically invalid and will lose the mark!

Part (d)(ii) — Range of h(x) [1 Mark]

Step-by-step deduction:

1. From part (a), the range of the inner function is f(x) ≥ 5 .
2. The outer function g(t) = √t is strictly increasing for t ≥ 0 .
3. Therefore, the minimum value occurs when x² + 5 is at its minimum (at x = 0 ):

h(0) = √(0² + 5) = √5

🧠 Exam Technique for Composite Ranges

Always track the range of the inner function first:

Domain of h (x ∈ ℝ) → Range of f ([5, ∞)) → fed into g → Range of h ([√5, ∞))

Students who directly substitute without considering the domain of the inner function often write h(x) ≥ 0 by mistake.

❌ Common Errors

• Writing h(x) ≥ 0 assuming all square root functions have a minimum of 0 (forgetting that the expression inside can never drop below 5).
• Evaluating fg(x) instead of gf(x) : fg(x) = (√x)² + 5 = x + 5 , which is the wrong order of composition.
• Incorrect algebraic expansion: writing √(x² + 5) = x + √5 .