Salam Academy
Practise

Question bank

Original exam-style questions, mapped to your syllabus. Answers are checked for mathematical equivalence and every attempt updates your mastery.

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17 questions

SA-9709-P1-FN4-0061 markL1 FoundationCalculatorIntersections
For an increasing function ff, the graphs of y=f(x)y=f(x) and y=f−1(x)y=f^{-1}(x) can only meet
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-0072 marksL1 FoundationCalculatorReflect in y = x
The point (−1,4)(-1,4) is reflected in the line y=xy=x. State the coordinates of its image.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-G012 marksL1 FoundationCalculatorReflect in y = x
The point (2,7)(2,7) lies on the graph of y=f(x)y=f(x). State the coordinates of the corresponding point on the graph of y=f−1(x)y=f^{-1}(x).

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-Q011 markL1 FoundationCalculatorReflect in y = x
The graph of y=f−1(x)y=f^{-1}(x) is the reflection of y=f(x)y=f(x) in
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-Q021 markL1 FoundationCalculatorReflect in y = x
y=f(x)y=f(x) has a horizontal asymptote y=5y=5. The graph of y=f−1(x)y=f^{-1}(x) has
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-Q031 markL1 FoundationCalculatorReflect in y = x
When sketching y=f(x)y=f(x) and y=f−1(x)y=f^{-1}(x) on the same axes, you should
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-0012 marksL2 StandardCalculatorIntersections
f(x)=2x+3f(x)=2x+3. Find the xx-coordinate of the point where the graphs of y=f(x)y=f(x) and y=f−1(x)y=f^{-1}(x) meet.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-0042 marksL2 StandardCalculatorReflect in y = x
f(x)=x+2f(x)=\sqrt{x+2} for x≥−2x\ge-2. The graph of y=f−1(x)y=f^{-1}(x) crosses the yy-axis at (0,c)(0,c). Find cc.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-0051 markL2 StandardCalculatorReflect in y = x
f(x)=1x−2+3f(x)=\dfrac{1}{x-2}+3 for x>2x>2. The graph of y=f−1(x)y=f^{-1}(x) has a vertical asymptote x=kx=k. State kk.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-0081 markL2 StandardCalculatorSelf-inverse
f(x)=3−xf(x)=3-x, x∈Rx\in\mathbb{R}. What can you say about f−1f^{-1}?
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-0101 markL2 StandardCalculatorReflect in y = x
f(x)=x2+3f(x)=x^2+3 for x≥1x\ge1. The domain of f−1f^{-1} is x≥cx\ge c. State cc.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-0122 marksL2 StandardCalculatorIntersections
f(x)=4x−9f(x)=4x-9. Find the coordinates of the point where y=f(x)y=f(x) meets y=f−1(x)y=f^{-1}(x), giving the xx-coordinate.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-0021 markL3 ExamCalculatorIntersections
f(x)=x2+2f(x)=x^2+2 for x≥0x\ge0. Which statement about the graphs of y=f(x)y=f(x) and y=f−1(x)y=f^{-1}(x) is true?
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-0033 marksL3 ExamCalculatorIntersections
f(x)=x2−6f(x)=x^2-6 for x≥0x\ge0. Find the xx-coordinate of the point where the graphs of y=f(x)y=f(x) and y=f−1(x)y=f^{-1}(x) meet.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-0093 marksL3 ExamCalculatorSelf-inverse
f(x)=2xx−2f(x)=\dfrac{2x}{x-2} for x≠2x\ne2. Show that ff is self-inverse, and hence find f−1(6)f^{-1}(6).

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-0112 marksL3 ExamCalculatorReflect in y = x
f(x)=x3+8f(x)=x^3+8, x∈Rx\in\mathbb{R}. State (i) where y=f−1(x)y=f^{-1}(x) crosses the xx-axis (ii) where it crosses the yy-axis.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-FN4-0134 marksL5 ExtensionCalculatorIntersections
f(x)=x2−4x+6f(x)=x^2-4x+6 for x≥2x\ge2. Find the xx-coordinates of all points where the graphs of y=f(x)y=f(x) and y=f−1(x)y=f^{-1}(x) meet.

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Salam Academy Original Practice · Proposed mark scheme