CIE A level -Pure Mathematics 1 : Topic : 1.2 Function: inverse of a one-one function : Exam Questions Paper 1

Question

(a) Express $ 3x^2 – 12x + 14 $ in the form $ 3(x + a)^2 + b $, where $ a $ and $ b $ are constants.
The function $ f(x) = 3x^2 – 12x + 14 $ is defined for $ x \geq k $, where $ k $ is a constant.
(b) Find the least value of $ k $ for which $ f^{-1} $ exists.
Assume $ k $ is the value from part (b).
(c) Find an expression for $ f^{-1}(x) $.
(d) Solve the equation $ f(f(x)) = 29 $.

▶️ Answer/Explanation

Solution

Part (a): Express in vertex form

Rewrite: $ 3x^2 – 12x + 14 = 3(x^2 – 4x) + 14 $
Complete square: $ x^2 – 4x = (x – 2)^2 – 4 $
So: $ 3[(x – 2)^2 – 4] + 14 = 3(x – 2)^2 + 2 $
Answer: $ 3(x – 2)^2 + 2 $

Part (b): Least value of $ k $

$ f(x) = 3(x – 2)^2 + 2 $, vertex at $ x = 2 $
For $ f^{-1} $ to exist, $ f $ must be one-to-one. Increasing for $ x \geq 2 $.
Answer: $ k = 2 $

Part (c): Inverse function

Set $ y = 3(x – 2)^2 + 2 $:
$ y – 2 = 3(x – 2)^2 $
$ (x – 2)^2 = \frac{y – 2}{3} $
$ x = 2 + \sqrt{\frac{y – 2}{3}} $ (positive since $ x \geq 2 $)
Answer: $ f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}} $

Part (d): Solve $ f(f(x)) = 29 $

$ f(x) = 3(x – 2)^2 + 2 $
$ f(f(x)) = 3[3(x – 2)^2]^2 + 2 = 27(x – 2)^4 + 2 $
Set: $ 27(x – 2)^4 + 2 = 29 $
$ (x – 2)^4 = 1 $
$ x – 2 = \pm 1 $
$ x = 3 $ (since $ x \geq 2 $)
Answer: $ x = 3 $

Question

The diagram shows the graph of $ y = f(x) $ with solid lines and $ y = g(x) $ with broken lines.
(a) Describe a sequence of three transformations that transforms the graph of $ y = f(x) $ to $ y = g(x) $.
(b) Find an expression for $ g(x) $ in the form $ af(bx + c) $, where $ a $, $ b $, and $ c $ are integers.

▶️ Answer/Explanation

Solution

Part (a): Sequence of transformations

1. Reflection in y-axis
2. Translation $ \begin{pmatrix} -1 \\ 0 \end{pmatrix} $
3. Vertical stretch, scale factor 2
Answer: Reflection in y-axis, translate left by 1 unit, stretch vertically by factor 2

Part (b): Expression for $ g(x) $

$ g(x) = af(bx + c) $
From transformations: $ a = 2 $ (vertical stretch), $ b = -1 $ (reflection in y-axis), $ c = -1 $ (translation left by 1)
Answer: $ g(x) = 2f(-x – 1) $

Question

The function $ f(x) = \frac{2x + 1}{2x – 1} $ for $ x < \frac{1}{2} $.
The function $ g(x) = 3x + 2 $ for $ x \in \mathbb{R} $.
(a)(i) State the value of $ f(-1) $.
(a)(ii) Given the graph of $ y = f(x) $ below, sketch the graph of $ y = f^{-1}(x) $, showing the mirror line.

(a)(iii) Find an expression for $ f^{-1}(x) $ and state its domain.
(b) Solve the equation $ f(x) = g(f(\frac{1}{4})) $.

▶️ Answer/Explanation

Solution

Part (a)(i): Value of $ f(-1) $

$ f(-1) = \frac{2(-1) + 1}{2(-1) – 1} = \frac{-1}{-3} $
Answer: $ \frac{1}{3} $

Part (a)(ii): Sketch graph of $ y = f^{-1}(x) $

Reflect $ y = f(x) $ over $ y = x $.
$Graph of y = f^{-1}(x)$
Answer: Graph of $ y = f^{-1}(x) $ is the reflection of $ y = f(x) $ over the line $ y = x $, as shown.

Part (a)(iii): Expression and domain of $ f^{-1}(x) $

Set $ y = \frac{2x + 1}{2x – 1} $:
$ 2x + 1 = y(2x – 1) $
$ 2xy – y = 2x + 1 $
$ x(2y – 2) = y + 1 $
$ x = \frac{y + 1}{2(y – 1)} $
Domain: $ x \neq 1 $, range of $ f $ is $ x < 1 $.
Answer: $ f^{-1}(x) = \frac{x + 1}{2(x – 1)} $, domain: $ x < 1 $

Part (b): Solve $ f(x) = g(f(\frac{1}{4})) $

Compute: $ f\left(\frac{1}{4}\right) = \frac{2 \cdot \frac{1}{4} + 1}{2 \cdot \frac{1}{4} – 1} = \frac{\frac{3}{2}}{-\frac{1}{2}} = -3 $
$ g(-3) = 3(-3) + 2 = -7 $
Solve: $ \frac{2x + 1}{2x – 1} = -7 $
$ 2x + 1 = -7(2x – 1) $
$ 2x + 1 = -14x + 7 $
$ 16x = 6 $
Answer: $ x = \frac{3}{8} $

Question

The function $ f(x) = 3 + 6x – 2x^2 $ for $ x \in \mathbb{R} $.
(a) Express $ f(x) $ in the form $ a – b(x – c)^2 $, where $ a $, $ b $, and $ c $ are constants, and state the range of $ f $.
(b) The graph of $ y = f(x) $ is transformed to $ y = h(x) $ by a reflection in one axis followed by a translation, with $ y = h(x) $ having a minimum at the origin. Give details of the reflection and translation.
(c) Sketch the graph of $ y = g(x) $ and explain why $ g $ is a one-to-one function.
(d) Sketch the graph of $ y = g^{-1}(x) $ on the diagram in (c), label both graphs, show the mirror line, and find an expression for $ g^{-1}(x) $.

▶️ Answer/Explanation

Solution

Part (a): Express $ f(x) $ and state range

$ f(x) = -2x^2 + 6x + 3 $
Complete square: $ -2(x^2 – 3x) + 3 = -2\left[\left(x – \frac{3}{2}\right)^2 – \frac{9}{4}\right] + 3 $
$ = -2\left(x – \frac{3}{2}\right)^2 + \frac{18}{4} + 3 = -2\left(x – \frac{3}{2}\right)^2 + \frac{15}{2} $
Range: Maximum at $ x = \frac{3}{2} $, $ f\left(\frac{3}{2}\right) = \frac{15}{2} $, so $ f(x) \leq \frac{15}{2} $.
Answer: $ f(x) = \frac{15}{2} – 2\left(x – \frac{3}{2}\right)^2 $, range: $ f(x) \leq \frac{15}{2} $

Part (b): Reflection and translation for $ h(x) $

$ f(x) $ has maximum at $ \left(\frac{3}{2}, \frac{15}{2}\right) $.
Reflect in x-axis: $ y = -f(x) = 2x^2 – 6x – 3 $, minimum at $ \left(\frac{3}{2}, -\frac{15}{2}\right) $.
Translate: Left by $ \frac{3}{2} $, up by $ \frac{15}{2} $.
Answer: Reflect in x-axis, translate left by $ \frac{3}{2} $ and up by $ \frac{15}{2} $.

Part (c): Sketch $ y = g(x) $ and one-to-one explanation

Assume $ g(x) = f(x) $.
Sketch: Downward parabola, vertex at $ \left(\frac{3}{2}, \frac{15}{2}\right) $.
Graph of y = g(x)
One-to-one: Not one-to-one unless restricted (e.g., $ x \geq \frac{3}{2} $ or $ x \leq \frac{3}{2} $) due to parabola’s symmetry.
Answer: Downward parabola, not one-to-one without domain restriction.

Part (d): Sketch $ y = g^{-1}(x) $ and find expression

Restrict $ g(x) $ to $ x \geq \frac{3}{2} $:
$ y = 3 + 6x – 2x^2 $
Solve: $ 2x^2 – 6x + 3 – y = 0 $
Discriminant: $ \Delta = 36 – 8(3 – y) = 12 + 8y $
$ x = \frac{6 + \sqrt{12 + 8y}}{4} = \frac{3 + \sqrt{3 + 2y}}{2} $
Sketch: Reflect right branch over $ y = x $, label graphs, show $ y = x $.
Answer: $ g^{-1}(x) = \frac{3 + \sqrt{3 + 2x}}{2} $, domain: $ x \leq \frac{15}{2} $, sketch with mirror line $ y = x $.

CIE A level -Pure Mathematics 1 : Topic : 1.2 Function: inverse of a one-one function : Exam Questions Paper 1

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