Functions (Cambridge (CIE) A Level Maths: Pure 1): Exam Questions

The function $\text{f}$ is defined by

$\text{f}\left(x\right)=3x^2+1$

for $x\ge 0$.

Find an expression for ${\text{f}}^{-1}\left(x\right)$.

State the domain and range of ${\text{f}}^{-1}$.

The function $\text{f}$ is defined by

$\text{f}\left(x\right)=x^2+2x-3$

for $x\in \mathrm{ℝ}$.

Sketch the graph of $y=\text{f}\left(x\right)$, giving the coordinates of the points where the graph meets the coordinate axes and the coordinates of the turning point.

Write down the range of $\text{f}$.

The functions $\text{f}$ and $\text{g}$ are defined by

$\text{f}\left(x\right)=x^2$

$\text{g}\left(x\right)=4x-3$

for $x\in \mathrm{ℝ}$.

Write down the range of $\text{f}$.

Find an expression for

(i) $\text{fg}\left(x\right)$,

(ii) $\text{gf}\left(x\right)$.

Solve the equation $\text{f}\left(x\right)=\text{g}\left(x\right)$.

The diagram shows the graph of $y=\text{f}\left(x\right)$ (solid line) together with a dotted line.

Use the graph to write down the domain and range of $\text{f}$.

Given that $(1,1)$ lies on the dotted line, write down the equation of the dotted line.

On the diagram, sketch the graph of $y={\text{f}}^{-1}\left(x\right)$.

The function $\text{f}$ is defined by

$\text{f}:x\mapsto \frac{x^2+1}{x^2}$

for $x\in \mathrm{ℝ},x\ne 0$.

Show that $\text{f}\left(x\right)$ can be written as $1+\frac{1}{x^2}$.

Explain why $\text{f}$ does not have an inverse, and suggest an adaptation to its domain so that the inverse exists.

The domain of $\text{f}$ is changed to $x>0$. Find an expression for ${\text{f}}^{-1}\left(x\right)$, and state its domain and range.

The functions $\text{f}$ and $\text{g}$ are defined by

$\text{f}\left(x\right)=\frac{1}{2}(4x-3)$

$\text{g}\left(x\right)=0.5x+0.75$

for $x\in \mathrm{ℝ}$.

Find an expression for

(i) $\text{fg}\left(x\right)$,

(ii) $\text{gf}\left(x\right)$.

Write down ${\text{f}}^{-1}\left(x\right)$ and state its domain and range.

The function $\text{f}$ is defined by

$\text{f}\left(x\right)=\frac{2}{x}$

Write down the domain of $\text{f}$.

Sketch the graph of $y=\text{f}\left(x\right)$, stating the coordinates of any intersections with the axes and the equations of any asymptotes.

Write down the range of $\text{f}$.

The functions $\text{f}$ and $\text{g}$ are defined by

$\text{f}\left(x\right)=3x^2+2$

$\text{g}\left(x\right)=1-3x$

for $x\in \mathrm{ℝ}$.

Write down the range of $\text{f}$.

Find an expression for

(i) $\text{fg}\left(x\right)$,

(ii) $\text{gf}\left(x\right)$.

Solve the equation $\text{f}\left(x\right)=\text{g}\left(x\right)+1$.

The graph of $y=f\left(x\right)$ is shown below.

The diagram shows the graph of $y=\text{f}\left(x\right)$ (solid curve) together with a dotted line.

(i) Use the graph to write down the domain and range of $\text{f}$.

(ii) Write down the equation of the dotted line.

On the diagram, sketch the graph of $y={\text{f}}^{-1}\left(x\right)$.

The function $\text{f}$ is defined by

$\text{f}\left(x\right)=4x^3+4x^2-7x+2$

Write down the domain and range of $\text{f}$.

Sketch the graph of $y=\text{f}\left(x\right)$, stating the coordinates of any intersections with the axes. (You do not need to give the coordinates of any turning points.)

The functions $\text{f}$ and $\text{g}$ are defined by

$\text{f}\left(x\right)=x^2-2$ for $x\in \mathrm{ℝ}$,

$\text{g}\left(x\right)=1-\frac{2}{x}$ for $x\in \mathrm{ℝ},x\ne 0$.

Write down the range of $\text{f}$.

Leaving your answers as single fractions, find an expression for

(i) $\text{fg}\left(x\right)$,

(ii) $\text{gf}\left(x\right)$.

Given that ${\left(x-1\right)}^2$ is a factor of $x^3-3x+2$, solve the equation $\text{f}\left(x\right)=\text{g}\left(x\right)$.

The function $\text{f}$ is defined by

$\text{f}:x\mapsto \sqrt{25-x^2}$

for $x\in \mathrm{ℝ},-5\le x\le 5$.

Explain why $\text{f}$ does not have an inverse.

Suggest an adaptation to the domain of $\text{f}$ so that:

• the inverse of $\text{f}$ exists,

• the graph of $y=\text{f}\left(x\right)$ lies in the first quadrant only, and

• the domain of $\text{f}$ is as large as possible.

State the range of your adapted $\text{f}$.

The domain of $\text{f}$ is changed to $-5\le x\le 0$. Find an expression for ${\text{f}}^{-1}\left(x\right)$, and state its domain and range.

The function $\text{f}$ is defined by

$\text{f}\left(x\right)=(x-3{)}^2(x-4{)}^2$

for $2\le x\le 5$.

Work out the range of $\text{f}$.

If the domain of $\text{f}$ is changed to $x\le 2$, what is the range of $\text{f}$?

State another domain for $\text{f}$ that would have the same effect as that in part (b).

The diagram shows the graph of $y=\text{f}\left(x\right)$ together with the line $y=x$ (dotted). The function $\text{f}$ has rotational symmetry about the origin, and for $x>0$ there is a vertical line of symmetry at $x=4.5$.

Use the graph to write down the domain and range of $\text{f}$.

On the diagram, sketch the reflection of $\text{f}$ in the line $y=x$, and explain why this cannot be the graph of ${\text{f}}^{-1}\left(x\right)$.

(i) Given that the maximum solution to $\text{f}\left(x\right)=6$ is $x=6$, state the restriction on the domain of $\text{f}$ such that ${\text{f}}^{-1}$ exists.

(ii) Hence write down the domain and range of ${\text{f}}^{-1}$.

The functions $\text{f}$ and $\text{g}$ are defined by

$\text{f}\left(x\right)=(x-1{)}^2-4$ for $x\ge 1$,

$\text{g}\left(x\right)=1+\sqrt{x+4}$ for $x\ge -4$.

Find an expression for

(i) $\text{fg}\left(x\right)$,

(ii) $\text{gf}\left(x\right)$.

Write down ${\text{f}}^{-1}\left(x\right)$ and state its domain and range.

The graphs of $\text{f}$ and ${\text{f}}^{-1}$ are drawn on the same axes. Describe the transformation that maps one graph onto the other.

Find the coordinates of the point where the graphs of $y=\text{f}\left(x\right)$ and $y={\text{f}}^{-1}\left(x\right)$ meet.