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

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

Question : Quadratics and Functions

(a) Express \(3x^{2}-12x+14\) in the form \(3(x+a)^{2}+b\) where \(a\) and \(b\) are constants to be found.

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 the function \(f^{-1}\) exists.

For the rest of this question, you should assume that \(k\) has the value found in part (b).

(c) Find an expression for \(f^{-1}(x)\).

(d) Hence or otherwise solve the equation \(ff(x)=29\).

a
Express \(3x^{2}-12x+14\) in the form \(3(x+a)^{2}+b\) where \(a\) and \(b\) are constants to be found.

Working space:

▶️Answer/Explanation

Answer: \(3(x – 2)^2 + 2\)

Working:

Rewrite by completing the square:

\[ 3x^2 – 12x + 14 = 3(x^2 – 4x) + 14 \]

\[ x^2 – 4x = (x – 2)^2 – 4 \]

\[ 3[(x – 2)^2 – 4] + 14 = 3(x – 2)^2 – 12 + 14 = 3(x – 2)^2 + 2 \]

So, \(a = -2\), \(b = 2\).

Key Concept:

Completing the square transforms a quadratic expression into vertex form, revealing the vertex and constants \(a\) and \(b\).

b
Find the least value of \(k\) for which the function \(f^{-1}\) exists.

Working space:

▶️Answer/Explanation

Answer: \(k = 2\)

Working:

\(f(x) = 3(x – 2)^2 + 2\) has a vertex at \(x = 2\). For \(f^{-1}\) to exist, \(f\) must be one-to-one. Since it’s increasing for \(x \geq 2\), the smallest \(k\) is 2.

Key Concept:

A function has an inverse if it is one-to-one, achieved by restricting the domain to where the function is strictly increasing or decreasing.

c
Find an expression for \(f^{-1}(x)\).

Working space:

▶️Answer/Explanation

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

Working:

With domain \(x \geq 2\), set \(y = 3(x – 2)^2 + 2\):

\[ y – 2 = 3(x – 2)^2 \]

\[ (x – 2)^2 = \frac{y – 2}{3} \]

\[ x – 2 = \sqrt{\frac{y – 2}{3}} \quad (\text{positive since } x \geq 2\text{)} \]

\[ x = 2 + \sqrt{\frac{y – 2}{3}} \]

So, \(f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}\), for \(x \geq 2\).

Key Concept:

The inverse function is found by solving \(y = f(x)\) for \(x\), ensuring the domain restriction maintains one-to-one behavior.

d
Hence or otherwise solve the equation \(ff(x)=29\).

Working space:

▶️Answer/Explanation

Answer: \(x = 3\)

Working:

\[ f(x) = 3(x – 2)^2 + 2 \]

\[ f(f(x)) = 3[f(x) – 2]^2 + 2 = 3[3(x – 2)^2]^2 + 2 = 27(x – 2)^4 + 2 \]

\[ 27(x – 2)^4 + 2 = 29 \]

\[ 27(x – 2)^4 = 27 \]

\[ (x – 2)^4 = 1 \]

\[ x – 2 = \pm 1 \]

\[ x = 3 \text{ or } 1 \]

Since \(x \geq 2\), \(x = 3\).

Key Concept:

Composite functions and domain restrictions are used to solve equations involving function iterations.

Syllabus Reference

Quadratics and Functions

  • (a) SL 1.1 – Quadratic functions: completing the square
  • (b) SL 1.2 – Functions: conditions for existence of inverse
  • (c) SL 1.2 – Functions: finding inverse functions
  • (d) SL 1.2 – Functions: composite functions and solving equations

Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)

Question : Functions

Graph of y=f(x) and y=g(x)

In the diagram, the graph with equation \( y=f(x) \) is shown with solid lines and the graph with equation \( y=g(x) \) is shown with broken lines.

a
Describe fully a sequence of three transformations which transforms the graph of \( y=f(x) \) to the graph of \( y=g(x) \).

Working space:

▶️Answer/Explanation

Answer:

1. Reflection in y-axis

2. Translation or shift \( \begin{pmatrix} -1 \\ 0 \end{pmatrix} \)

3. Stretch, factor 2, parallel to y-axis

Working:

Correct order and three correctly named transformations only.

Alternative Solution for first 3 marks:

1. Translation or shift \( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \)

2. Reflection in y-axis

Alternative solutions:

There are alternative solutions which can be marked in the same way, e.g., the given stretch, translation \( \begin{pmatrix} -4 \\ 0 \end{pmatrix} \), reflect in \( x=-2.5 \).

b
Find an expression for \( g(x) \) in the form \( af(bx+c) \) where \( a \), \( b \), and \( c \) are integers.

Working space:

▶️Answer/Explanation

Answer: \( g(x) = 2f(-x-1) \) or \( a=2 \), \( b=-1 \), \( c=-1 \)

Syllabus Reference

Functions

  • (a) SL 1.2 – Transformations of functions
  • (b) SL 1.2 – Composite transformations and function notation

Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)

Question : Functions

The function \( f \) is defined by \( f(x) = \frac{2x + 1}{2x – 1} \) for \( x < \frac{1}{2} \).

The function \( g \) is defined by \( g(x) = 3x + 2 \) for \( x \in \mathbb{R} \).

a(i)
State the value of \( f(-1) \).

Working space:

▶️Answer/Explanation

Correct answer: \( \frac{1}{3} \)

Working:

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

Key Concept:

Evaluate a function by substituting the given value into the expression.

a(ii)
The diagram shows the graph of \( y = f(x) \). Sketch the graph of \( y = f^{-1}(x) \) on this diagram. Show any relevant mirror line.

Graph of y = f(x)

Working space:

▶️Answer/Explanation

Correct answer:

Graph of y = f^{-1}(x)

Working:

The graph of \( y = f^{-1}(x) \) is the reflection of \( y = f(x) \) over the line \( y = x \). The mirror line \( y = x \) is shown on the graph.

Key Concept:

The inverse function’s graph is obtained by reflecting the original function’s graph over the line \( y = x \).

a(iii)
Find an expression for \( f^{-1}(x) \) and state the domain of the function \( f^{-1} \).

Working space:

▶️Answer/Explanation

Correct answer: \( f^{-1}(x) = \frac{x + 1}{2(x – 1)} \), Domain: \( x < 1 \)

Working:

Let \( y = f(x) = \frac{2x + 1}{2x – 1} \).

Solve for \( x \):

\[ y = \frac{2x + 1}{2x – 1} \]

\[ 2x + 1 = y(2x – 1) \]

\[ 2x + 1 = 2xy – y \]

\[ 2xy – 2x = y + 1 \]

\[ 2x(y – 1) = y + 1 \]

\[ x = \frac{y + 1}{2(y – 1)} \]

Thus, \( f^{-1}(x) = \frac{x + 1}{2(x – 1)} \).

Alternative form: \( f^{-1}(x) = \frac{-x – 1}{2 – 2x} \).

Domain of \( f^{-1} \): The denominator \( 2(x – 1) \neq 0 \), so \( x \neq 1 \). Since the range of \( f(x) \) for \( x < \frac{1}{2} \) is \( y < 1 \), the domain of \( f^{-1} \) is \( x < 1 \).

Alternative Method:

\[ y = \frac{2x + 1}{2x – 1} = 1 + \frac{2}{2x – 1} \]

\[ y – 1 = \frac{2}{2x – 1} \]

\[ 2x – 1 = \frac{2}{y – 1} \]

\[ 2x = \frac{2}{y – 1} + 1 \]

This leads to the same expression after simplification.

Key Concept:

The inverse function is found by swapping \( x \) and \( y \) and solving for \( y \). The domain of \( f^{-1} \) is the range of \( f \).

b
Solve the equation \( f(x) = g(f(\frac{1}{4})) \).

Working space:

▶️Answer/Explanation

Correct answer: \( x = \frac{3}{8} \)

Working:

First, compute \( f\left(\frac{1}{4}\right) \):

\[ f\left(\frac{1}{4}\right) = \frac{2 \cdot \frac{1}{4} + 1}{2 \cdot \frac{1}{4} – 1} = \frac{\frac{1}{2} + 1}{\frac{1}{2} – 1} = \frac{\frac{3}{2}}{-\frac{1}{2}} = -3 \]

Then, compute \( g(f(\frac{1}{4})) \):

\[ g(-3) = 3(-3) + 2 = -9 + 2 = -7 \]

The equation is:

\[ f(x) = g(f(\frac{1}{4})) = -7 \]

\[ \frac{2x + 1}{2x – 1} = -7 \]

Solve:

\[ 2x + 1 = -7(2x – 1) \]

\[ 2x + 1 = -14x + 7 \]

\[ 16x = 6 \]

\[ x = \frac{6}{16} = \frac{3}{8} \]

Alternative solution:

Since \( f(x) = -7 \), use the inverse:

\[ x = f^{-1}(-7) \]

\[ f^{-1}(x) = \frac{x + 1}{2(x – 1)} \]

\[ f^{-1}(-7) = \frac{-7 + 1}{2(-7 – 1)} = \frac{-6}{-16} = \frac{3}{8} \]

Key Concept:

Solve composite function equations by evaluating the inner function first, then solving the resulting equation.

Syllabus Reference

Functions

  • (a)(i) ALV: 1.2 – Evaluating functions
  • (a)(ii) ALV: 1.2 – Graphing inverse functions
  • (a)(iii) ALV: 1.2 – Finding the inverse function and its domain
  • (b) ALV: 1.2 – Solving equations involving composite functions

Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)

Question : Functions

The function \( f \) is defined by \( f(x) = 3 + 6x – 2x^2 \) 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 \( f \).

Working space:

▶️Answer/Explanation

Correct answer: \( f(x) = \frac{15}{2} – 2\left(x – \frac{3}{2}\right)^2 \), range \( f(x) \leq \frac{15}{2} \)

Working:

Rewrite \( f(x) = 3 + 6x – 2x^2 \) by completing the square:

\[ f(x) = -2x^2 + 6x + 3 \]

Factor out \(-2\) from the \( x \) terms:

\[ f(x) = -2(x^2 – 3x) + 3 \]

Complete the square:

\[ x^2 – 3x = \left(x – \frac{3}{2}\right)^2 – \frac{9}{4} \]

So:

\[ f(x) = -2\left[\left(x – \frac{3}{2}\right)^2 – \frac{9}{4}\right] + 3 \]

\[ = -2\left(x – \frac{3}{2}\right)^2 + \frac{18}{4} + 3 \]

\[ = -2\left(x – \frac{3}{2}\right)^2 + \frac{18}{4} + \frac{12}{4} \]

\[ = -2\left(x – \frac{3}{2}\right)^2 + \frac{30}{4} \]

\[ = -2\left(x – \frac{3}{2}\right)^2 + \frac{15}{2} \]

Thus, \( a = \frac{15}{2} \), \( b = 2 \), \( c = \frac{3}{2} \).

Range: Since \(-2\left(x – \frac{3}{2}\right)^2 \leq 0\), the maximum occurs at \( x = \frac{3}{2} \):

\[ f\left(\frac{3}{2}\right) = \frac{15}{2} \]

As \( x \to \pm \infty \), \( f(x) \to -\infty \). So, the range is:

\[ f(x) \leq \frac{15}{2} \]

Key Concept:

Completing the square transforms a quadratic into vertex form, revealing the maximum/minimum and range.

b
The graph of \( y = f(x) \) is transformed to the graph of \( y = h(x) \) by a reflection in one of the axes followed by a translation. It is given that the graph of \( y = h(x) \) has a minimum point at the origin. Give details of the reflection and translation involved.

Working space:

▶️Answer/Explanation

Correct answer: Reflect in the \( x \)-axis, then translate left by \( \frac{3}{2} \) and up by \( \frac{15}{2} \).

Working:

\( f(x) = \frac{15}{2} – 2\left(x – \frac{3}{2}\right)^2 \) has a maximum at \( \left(\frac{3}{2}, \frac{15}{2}\right) \). We need \( h(x) \) with a minimum at \( (0, 0) \).

1. Reflection: Reflect in the \( x \)-axis:

\[ y = -f(x) = -\left[3 + 6x – 2x^2\right] = 2x^2 – 6x – 3 \]

Vertex becomes \( \left(\frac{3}{2}, -\frac{15}{2}\right) \), a minimum.

2. Translation: Shift from \( \left(\frac{3}{2}, -\frac{15}{2}\right) \) to \( (0, 0) \):

  • Left by \( \frac{3}{2} \): replace \( x \) with \( x + \frac{3}{2} \)
  • Up by \( \frac{15}{2} \): add \( \frac{15}{2} \)

\[ h(x) = 2\left(x + \frac{3}{2}\right)^2 – 6\left(x + \frac{3}{2}\right) – 3 + \frac{15}{2} \]

Simplify:

\[ = 2\left(x^2 + 3x + \frac{9}{4}\right) – 6x – 9 – 3 + \frac{15}{2} \]

\[ = 2x^2 + 6x + \frac{18}{4} – 6x – 12 + \frac{15}{2} = 2x^2 + \frac{9}{2} – 12 + \frac{15}{2} = 2x^2 \]

So, \( h(x) = 2x^2 \), minimum at \( (0, 0) \).

Key Concept:

Transformations like reflections and translations adjust the position and orientation of a graph’s vertex.

c
Sketch the graph of \( y = g(x) \) and explain why \( g \) is a one-to-one function. You are not required to find the coordinates of any intersections with the axes.

Working space:

▶️Answer/Explanation

Correct answer: \( g(x) \) sketch as parabola; not one-to-one unless restricted.

Working:

Assuming \( g(x) = f(x) = 3 + 6x – 2x^2 \):

Sketch: Downward parabola, vertex at \( \left(\frac{3}{2}, \frac{15}{2}\right) \).

One-to-one: \( f(x) \) is not one-to-one (parabola fails horizontal line test) unless restricted.

Key Concept:

A function is one-to-one if each output corresponds to exactly one input, often requiring domain restriction for quadratics.

d
Sketch the graph of \( y = g^{-1}(x) \) on your diagram in (c), and find an expression for \( g^{-1}(x) \). You should label the two graphs in your diagram appropriately and show any relevant mirror line.

Working space:

▶️Answer/Explanation

Correct answer: \( g^{-1}(x) = \frac{3 + \sqrt{3 + 2x}}{2} \), sketch with \( y = x \) mirror line.

Working:

Restrict \( f(x) \) to \( x \geq \frac{3}{2} \) (decreasing branch):

\[ y = 3 + 6x – 2x^2 \]

Solve for \( x \):

\[ 2x^2 – 6x + (3 – y) = 0 \]

Discriminant: \( \Delta = (-6)^2 – 4 \cdot 2 \cdot (3 – y) = 36 – 24 + 8y = 12 + 8y \)

\[ x = \frac{6 \pm \sqrt{12 + 8y}}{4} \]

Take \( + \) for \( x \geq \frac{3}{2} \):

\[ x = \frac{6 + \sqrt{12 + 8y}}{4} = \frac{6 + 2\sqrt{3 + 2y}}{4} = \frac{3 + \sqrt{3 + 2y}}{2} \]

\[ g^{-1}(x) = \frac{3 + \sqrt{3 + 2x}}{2}, \quad x \leq \frac{15}{2} \]

Sketch: \( y = g(x) \) (right branch), \( y = g^{-1}(x) \) reflected over \( y = x \).

Key Concept:

The inverse function’s graph is a reflection over \( y = x \), requiring a one-to-one function via domain restriction.

Syllabus Reference

Functions

  • (a) Topic-1.2 – Functions
  • (b) Topic-1.2 – Functions
  • (c) Topic-1.2 – Functions
  • (d) Topic-1.2 – Functions

Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)

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