CIE A level -Pure Mathematics 1 : Topic : 1.2 Function: function, domain, range : Exam Questions Paper 1

Question

The function $ f:x\rightarrow x^{2}-4x+k$ is defined for the domain $x\geq p$ ,where k and p are constants.

(i)Express $f\left ( x \right )$ in the form ,$\left ( x+a \right )^{2}+b+k$ where a and b are constants.

(ii)State the range of f in terms of k.

(iii)State the smallest value of p for which f is one-one.

(iv)For the value of p found in part (iii),find an expression for $f^{-1}\left ( x \right )$ and state the domain of $f^{-1}$ giving your answer in terms of k.

▶️Answer/Explanation

(i) To express $f(x)$ in the form $(x + a)^2 + b + k$, we expand the square:
$f(x) = x^2 – 4x + k = (x^2 – 4x + 4) – 4 + k = (x – 2)^2 – 4 + k = (x – 2)^2 + (k – 4)$
Therefore, $f(x)$ can be expressed as $(x + a)^2 + b + k$ with $a = -2$ and $b = k – 4$.
(ii) The range of $f$ represents all possible values of $f(x)$ as $x$ varies within its domain. Since $f(x) = (x – 2)^2 + (k – 4)$, we can observe that the square term $(x – 2)^2$ is always non-negative.
Thus, the smallest value $f(x)$ can attain is when $(x – 2)^2 = 0$, which occurs when $x = 2$. In this case, $f(x) = (2 – 2)^2 + (k – 4) = k – 4$. Hence, the smallest value in the range is $k – 4$.
Therefore, the range of $f$ is given by $f(x) \geq k – 4$ for all $x$ in the domain $x \geq p$.
(iii) For $f$ to be one-to-one (injective), each distinct value in the domain must correspond to a distinct value in the range. This means that no two different $x$ values can produce the same $f(x)$ value.
To ensure this, we need to set the smallest value of the range $k – 4$ to be greater than any other value in the range. In other words, $k – 4$ must be the minimum value of $f(x)$ for all $x$ in the domain $x \geq p$.
Therefore, the smallest value of $p$ for which $f$ is one-to-one is $p = 2$.
(iv) To find the inverse function $f^{-1}(x)$, we swap the roles of $x$ and $f(x)$ in the equation $f(x) = (x – 2)^2 + (k – 4)$ and solve for $x$:
$x = (f^{-1}(x) – 2)^2 + (k – 4)$
Let’s solve for $f^{-1}(x)$:
$x – (k – 4) = (f^{-1}(x) – 2)^2$
$\sqrt{x – (k – 4)} = f^{-1}(x) – 2$
$f^{-1}(x) = \sqrt{x – (k – 4)} + 2$
The domain of $f^{-1}(x)$ is determined by the range of $f(x)$. From part (ii), we know that $f(x) \geq k – 4$. Therefore, for $f^{-1}(x)$ to be defined, we need $x – (k – 4) \geq k – 4$, which simplifies to $x \geq 2k – 4$.
Thus, the domain of $f^{-1}(x)$ is $x \geq 2k – 4$.

Question

The functions f and g are defined by

$f(x)=\frac{4}{x}-2$ for$ x> 0$

$g(x)=\frac{4}{5x+2}$ for $x\geq 0$

(i)Find and simplify an expression for fg(x) and state the range of fg.

(ii)Find an expression for $g^{-1}(x)$ and find the domain of $g^{-1}$.

▶️Answer/Explanation

(i) To find the expression for $f \circ g$ or $fg(x)$, we substitute the definition of $g(x)$ into $f(x)$:
$fg(x) = f(g(x)) = f\left(\frac{4}{5x+2}\right) = \frac{4}{\frac{4}{5x+2}} – 2 = 5x.$
Range of f(x) is greater than zero.
So the simplified expression for $fg(x)$ is $5x$. The range of $fg$ is all real numbers since there are no restrictions on $x$ in the given expressions.
(ii) To find the expression for $g^{-1}(x)$ and its domain, we follow these steps:
Replace $g(x)$ with $y$:
$y=\frac{4}{5x+2}.$
$5x+2 = \frac{4}{y}.$
$5x = \frac{4}{y} – 2.$
$x = \frac{4-2y}{5y}.$
Therefore, the expression for $g^{-1}(x)$ is:
$g^{-1}(x) = \frac{4-2x}{5x}.$
The domain of $g^{-1}$ consists of 5x>0 \cup 4-2x\neq \geq 0 \Rightarrow x\leq 2.
Theefore,domain of $g^{-1}$ is 0<x\leq 2.

Question

$f:x\rightarrow \frac{1}{x^{2}-9}$

$g:x\rightarrow 2x-3$

(i)Find and simplify an expression for g(g(x)).

(ii)Find an expression for $f^{-1}\left ( x \right )$ and state the domain of $ f^{-1}$

(iii)Solve the equation $fg(x)=\frac{1}{7}$

▶️Answer/Explanation

(i) Expression for $g(g(x))$:
$g(x) = 2x – 3$
$gg(x) = g(2x – 3) = 2(2x – 3) – 3 = 4x – 9.$
(ii) Expression for $f^{-1}(x)$ and its domain:
To find the inverse of a function $f(x)$, we need to swap the roles of $x$ and $y$ and solve for $x$. So, for $f(x) = \frac{1}{x^2 – 9}$:
Swap $x$ and $y$: $x = \frac{1}{y^2 – 9}$
Solve for $y$: $y^2 – 9 = \frac{1}{x}$
Take the square root of both sides: $y = \pm \sqrt{\frac{1}{x} + 9}$
The domain of $f^{-1}(x)$ is determined by the original function $f(x)$.
In this case, the denominator of $f(x)$ cannot be zero, so $x^2 – 9 \neq 0$, which means $x \neq \pm 3$. Additionally, since we’re taking the square root of $\frac{1}{x} + 9$, we need $\frac{1}{x} + 9 \geq 0$, which implies $x > 0$. Therefore, the domain of $f^{-1}(x)$ is $x > 0$ and $x \neq 3$.
(iii)EITHER:
$\frac{1}{\left ( 2x-3 \right )^{2}-9}=\frac{1}{7}$
$\left ( 2x-3 \right )^{2}=16$ or $4x^{2}-12x-7=0$
$x=\frac{7}{2}$ or $-\frac{1}{2}$\
$x=\frac{7}{2}$

OR:

$g(x)=f^{-1}\left ( \frac{1}{7} \right )$
$g(x)=4$
$2x-3=4$
$x=\frac{7}{2}$

Question

The position vectors of points A and B relative to an origin O are a and b respectively. The position vectors of points C and D relative to O are 3a and 2b respectively. It is given that
$a=\begin{pmatrix}2\\ 1\\ 2\end{pmatrix}$ and $b=\begin{pmatrix}4\\ 0\\ 6\end{pmatrix}$(i) Find the unit vector in the direction of $\vec{CD}$
(ii) The point E is the mid-point of CD. Find angle EOD.

▶️Answer/Explanation

(i)To find the unit vector in the direction of $\vec{CD}$,we need to calculate the vector $\vec{CD}$ first.
The position vector of point C relative to O is 3A , and the position vector of point D relative to O is 2b. Therefore,$\vec{CD}=2b-3a$
Substituting the given values of a and b, we have:
$\vec{CD}=2\begin{pmatrix}4\\ 0\\ 6\end{pmatrix}-3\begin{pmatrix}2\\ 1\\ 2\end{pmatrix}$
Expanding the calculation, we get:
$\vec{CD}=\begin{pmatrix}8\\ 0\\ 12\end{pmatrix}-\begin{pmatrix}6\\ 3\\ 6\end{pmatrix}$
$\vec{CD}=\begin{pmatrix}2\\ -3\\ 6\end{pmatrix}$
To find the unit vector in the direction of $\vec{CD}$,we divide $\vec{CD} $by its magnitude
$\left \| \vec{CD} \right \|=\sqrt{2^{2}+(-3)^{2}+6^{2}}=\sqrt{49}=7$
Therefore, the unit vector$ \vec{CD}$ in the direction of $\vec{CD}$ is :
$\hat{CD}=\frac{1}{7}\begin{pmatrix}2\\ -3\\ 6\end{pmatrix}$
$=\begin{pmatrix}\frac{2}{7}\\ -\frac{3}{7}\\ \frac{6}{7}\end{pmatrix}$
(ii) The point E is the midpoint of CD. To find the position vector of point E, we can use the midpoint formula:
$\vec{OE}=\frac{1}{2}\left ( \vec{OC}+\vec{OD} \right )$.The position vector of point C relative to O is
3a, and the position vector of point D relative to O is 2b. Therefore:
$\vec{OC}=3a $ and $\vec{OD}=2b$
Substituting the given values of a and b, we have:
$\vec{OC}=3\begin{pmatrix}2\\ 1\\ 2\end{pmatrix}=\begin{pmatrix}6\\ 3\\ 6\end{pmatrix}$
$\vec{OD}=2\begin{pmatrix}4\\ 0\\ 6\end{pmatrix}=\begin{pmatrix}8\\ 0\\ 12\end{pmatrix}$
Using the midpoint formula:
$\vec{OE}=\frac{1}{2}\left ( \vec{OC}+\vec{OD} \right )$
$=\frac{1}{2}\left ( \begin{pmatrix}6\\ 3\\6 \end{pmatrix} +\begin{pmatrix}8\\ 0\\ 12\end{pmatrix}\right )$
$=\frac{1}{2}\begin{pmatrix}14\\ 3\\ 18\end{pmatrix}=\begin{pmatrix}7\\ \frac{3}{2}\\ 9\end{pmatrix}$
To find the angle EOD, we can use the dot product of vectors. The dot product between two vectors is given by:
$\vec{a}.\vec{b}=\left | \vec{a} \right |\left | \vec{b} \right |\cos \Theta$
where $\vec{a} $ and $\vec{b}$ are vectors, $\left | \vec{a} \right | and \left | \vec{b} \right | $are their magnitudes, and $\Theta$ is the angle between them.
$\vec{EO}=\begin{pmatrix}7\\ \frac{3}{2}\\ 9\end{pmatrix}$ (position vector of point E relative to O)
$\vec{OD}=2b=2\begin{pmatrix}4\\ 0\\ 6\end{pmatrix}=\begin{pmatrix}8\\ 0\\ 12\end{pmatrix}$ (position vector of point D relative to O)
To find the angle between these vectors, we can use the dot product formula:
$\vec{EO}.\vec{OD}=\left | \vec{EO} \right |.\left | \vec{OD} \right |\cos \Theta$
Let’s calculate the dot product and magnitudes:
$\vec{EO}.\vec{OD}=\begin{pmatrix}7\\ \frac{3}{2}\\ 9 \end{pmatrix}.\begin{pmatrix}8\\ 0\\ 12\end{pmatrix}$
$=(7)(8)+(\frac{3}{2})(0)+(9)(12)=56+0+108 =164$
$\left | \vec{EO}\right |=\sqrt{7^{2}+\left ( \frac{3}{2} \right )^{2}+9^{2}}=\sqrt{49+\frac{9}{4}+81}=\sqrt{\frac{289}{4}}=\frac{17}{2}$
$\left | \vec{OD} \right |=\sqrt{8^{2}+0^{2}+12^{2}}=\sqrt{64+0+144}=\sqrt{208}=4\sqrt{13}$
Substituting the values into the formula:
$\vec{EO}.\vec{OD}=\left | \vec{EO} \right |\left |\vec{OD} \right |\cos \Theta$
$164=\frac{17}{2}.4\sqrt{13}.\cos \Theta$
$\cos \Theta =\frac{164}{\frac{17}{2}.4\sqrt{13}}$
Now, we can find the value of the angle \Theta using the inverse cosine function:
$\Theta =\cos ^{-1}\left ( \frac{328}{17\sqrt{13}} \right )$
$\Theta \approx 8.6^{\circ}$
Therefore, the angle EOD is approximately $8.6^{\circ}.$

Question

The function f is defined by $f:x\rightarrow x^{2}+1$ for $x\geq $

(i) Define in a similar way the inverse function $f^{-1}$.

(ii) Solve the equation $f(x)=\frac{185}{16}$.

▶️Answer/Explanation

(i) To find the inverse function $f^{-1}(x)$, we first need to switch the roles of $x$ and $f(x)$ and then solve for the new variable. So, let $y = x^2 + 1$. Now, we want to solve for $x$, which is the inverse function, so we have:

$x^2 + 1 = y$

$x^2 = y – 1$

$x = \sqrt{y – 1}$

So, the inverse function $f^{-1}(y)$ is:

$f^{-1}(y) = \sqrt{y – 1}$

(ii) To solve the equation $f(x) = \frac{185}{16}$, we’ll substitute $f(x)$ with its expression $x^2 + 1$:

$x^2 + 1 = \frac{185}{16}$

$x^2 = \frac{185}{16} – 1$

$x^2 = \frac{185}{16} – \frac{16}{16}$

$x^2 = \frac{185 – 16}{16}$

$x^2 = \frac{169}{16}$

To solve for $x$, take the square root of both sides:

$x = \pm \frac{\sqrt{169}}{\sqrt{16}}$

$x = \pm \frac{13}{4}$

So, the solutions to the equation $f(x) = \frac{185}{16}$ are $x = \frac{13}{4}$ and $x = -\frac{13}{4}$.

Question

The function f is defined by $f:x\rightarrow \frac{2}{3-2x}$ for $x\epsilon R$ ,$x\neq \frac{3}{2}$

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

The function g is defined by $g:x\rightarrow 4x+a$ for $x\epsilon R$ ,where a is ac constant.

(ii)Find the value of a for which gf(-1)=3

(iii) Find the possible values of a given that the equation$ f^{-1}\left ( x \right )=g^{-1}\left ( x \right )$ has two equal roots.

▶️Answer/Explanation

(i) To find the inverse function $f^{-1}(x)$, we first swap $x$ and $f(x)$ and then solve for $x$. Starting with the original function $f(x) = \frac{2}{3 – 2x}$, we have:

$y = \frac{2}{3 – 2x}$

$3 – 2x = \frac{2}{y}$

$2x = 3 – \frac{2}{y}$

$x = \frac{3 – \frac{2}{y}}{2}$

So, the inverse function $f^{-1}(y)$ is:

$f^{-1}(y) = \frac{3 – \frac{2}{y}}{2}$

(ii) We are given the function $g(x) = 4x + a$. First, find $f(-1)$ and then substitute it into $g(x)$:

$f(-1) = \frac{2}{3 – 2(-1)} = \frac{2}{3 + 2} = \frac{2}{5}$

Now, substitute $f(-1)$ into $g(x)$:

$g(f(-1)) = 4 \cdot \frac{2}{5} + a = \frac{8}{5} + a$

$\frac{8}{5} + a = 3$

$a = 3 – \frac{8}{5} = \frac{15}{5} – \frac{8}{5} = \frac{7}{5}$

So, the value of $a$ that makes $g(f(-1)) = 3$ is $a = \frac{7}{5}$.

(iii)$g^{-1}\left ( x \right )=\frac{x-a}{4}=f^{-1}\left ( x \right )$

$\rightarrow x^{2}-x\left ( a+6 \right )+4(=0)$

Solving \left ( a+6 \right )^{2}=16 or $a^{2}+12a+20(=0)$

$\rightarrow a=-2$ or -10

Question

The function f is defined by $f: x\rightarrow 4 sin x – 1$ for $\frac{-\Pi}{2}\leq x\leq\frac{\Pi}{2}$.
(i) State the range of f.
(ii) Find the coordinates of the points at which the curve y = f(x) intersects the coordinate axes.
(iii) Sketch the graph of y = f(x).
(iv) Obtain an expression for $f^{-1}(x)$ , stating both the domain and range of $f^{-1}$.

▶️Answer/Explanation

(i) State the range of f:
The range of a function is the set of all possible values of the function. In this case, we have the function $f(x) = 4\sin(x) – 1$ for $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$. To find the range, we need to consider the possible values of $f(x)$.
The sine function has a range of [-1, 1]. Therefore, $4\sin(x)$ can take values between -4 and 4. Subtracting 1 from this range, we get the range of $f(x)$ as:
Range of f: [-5, 3]
(ii) Find the coordinates of the points at which the curve $y = f(x)$ intersects the coordinate axes:
To find the points where the curve intersects the coordinate axes, we need to find the values of $x$ for which $f(x) = 0$.
For the x-axis (y = 0):
$4\sin(x) – 1 = 0$
$4\sin(x) = 1$
$\sin(x) = \frac{1}{4}$
Taking the inverse sine of both sides:
$x = \arcsin\left(\frac{1}{4}\right)$
For the y-axis (x = 0):
$f(0) = 4\sin(0) – 1 = 0 – 1 = -1$
So, the points of intersection with the coordinate axes are:
For the x-axis: $\left(\arcsin\left(\frac{1}{4}\right), 0\right)$
For the y-axis: $(0, -1)$

Sketch the graph of $y = f(x)$:
To sketch the graph of $y = f(x)$, we can plot points and observe the behavior of the sine function within the given range. The points we found in part (ii) will be useful.
$\Rightarrow$ At $x = 0$, $f(x) = -1$, as calculated in part (ii).
$\Rightarrow$ As $x$ increases from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, $f(x)$ will oscillate between -5 and 3 due to the sine function.
The graph will be a sinusoidal curve oscillating between -5 and 3, with a phase shift of 0 and an amplitude of 4. It intersects the x-axis at $\left(\arcsin\left(\frac{1}{4}\right), 0\right)$ and the y-axis at $(0, -1)$.
(iv) Obtain an expression for $f^{-1}(x)$, stating both the domain and range of $f^{-1}$:
To find the inverse of $f(x)$, we need to interchange the roles of $x$ and $y$ and solve for $x$. So, we have:
$y = 4\sin(x) – 1$
$4\sin(x) = y + 1$
$\sin(x) = \frac{y + 1}{4}$
$x = \arcsin\left(\frac{y + 1}{4}\right)$
The domain of $f^{-1}(x)$ is the range of $f(x)$, which is [-5, 3], and the range of $f^{-1}(x)$ is the domain of $f(x)$, which is $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$.
So, the expression for $f^{-1}(x)$ is:
$f^{-1}(x) = \arcsin\left(\frac{x + 1}{4}\right)$ for $-5 \leq x \leq 3$

Question

The functions f and g are defined for x ≥ 0 by

$f:x\rightarrow 2x^{2}+3,$
$g:x\rightarrow 3x+2$

(i) Show that $gf(x) = 6x^2 + 11$ and obtain an unsimplified expression for \(fg(x)|).

(ii) Find an expression for $(fg)^{−1}x$ and determine the domain of $(fg)^{−1}$.

(iii) Solve the equation $gf(2x) = fg(x)$.

▶️Answer/Explanation

(i) Show that $gf(x) = 6x^2 + 11$ and obtain an unsimplified expression for $fg(x)$:
To find $gf(x)$, we first need to compute $g(f(x))$. Given that $f(x)$ and $g(x)$ are defined, we can proceed as follows:
$f(x) = 2x^2 + 3$ (given)
$g(x) = 3x + 2$ (given)
$g(f(x)) = g(2x^2 + 3) = 3(2x^2 + 3) + 2 = 6x^2 + 9 + 2 = 6x^2 + 11$
So, $gf(x) = 6x^2 + 11$.
$fg(x) = f(3x + 2) = 2(3x + 2)^2 + 3 = 2(9x^2 + 12x + 4) + 3 = 18x^2 + 24x + 8 + 3 = 18x^2 + 24x + 11$
An unsimplified expression for $fg(x)$ is $18x^2 + 24x + 11$.
(ii)$y=2\left ( 3x+2 \right )^{2}+3 \Rightarrow 3x+2=\left ( \pm \right )\sqrt{\left ( \frac{y-3}{2} \right )}oe$
$\Rightarrow x=\left ( \pm \right )\frac{1}{3}\sqrt{\frac{\left ( y-3 \right )}{2}}-\frac{2}{3}oe$
$\Rightarrow \left ( fg \right )^{-1}\left ( x \right )=\frac{1}{3}\sqrt{\frac{\left ( y-3 \right )}{2}}-\frac{2}{3}oe$
Solve their $\left ( fg \right )^{-1}\left ( x \right )\geq 0$ or attempt range of fg
Domain is $x\geq 11$
(iii) Solve the equation $gf(2x) = fg(x)$:
$gf(x) = 6x^2 + 11$ and $fg(x) = 18x^2 + 24x + 11$.
$gf(2x) = fg(x)$
$6(2x)^2 + 11 = 18x^2 + 24x + 11$
$6(4x^2) + 11 = 18x^2 + 24x + 11$
$24x^2 + 11 = 18x^2 + 24x + 11$
$24x^2 – 18x^2 – 24x = 0$
$6x^2 – 24x = 0$
$6x(x – 4) = 0$
Now,we have two solutions,
$6x = 0$ ⟹ $x = 0$
$x – 4 = 0$ ⟹ $x = 4$
So, the solutions to the equation $gf(2x) = fg(x)$ are $x = 0$ and $x = 4$.

Question

The function f is such that $f(x) =a^2 x^2-ax=3b$ for $x\leq\frac{1}{2a}$, where a and b are constants.

(i) For the case where $f(-2) = 4a^2 – b + 8$ and $f(-3) = 7a^2 – b + 14$, find the possible values of a
and b.

(ii) For the case where a = 1 and b = -1, find an expression for $f^-{1}(x)$ and give the domain of $f^{-1}$

▶️Answer/Explanation

(i) $2a + 4b = 8$
$2a^2 + 3a + 4b + 14 = 0$
$(a + 2)(2a – 3) = 0$
This gives us two possible values for $a$:
$\Rightarrow a + 2 = 0 \Rightarrow a = -2$
$\Rightarrow 2a – 3 = 0 \Rightarrow 2a = 3 \Rightarrow a = \frac{3}{2}$
For $a = -2$:
$2(-2) + 4b = 8$
$-4 + 4b = 8$
$4b = 12$
$b = 3$
For $a = \frac{3}{2}$:
$2\left(\frac{3}{2}\right) + 4b = 8$
$3 + 4b = 8$
$4b = 5$
$b = \frac{5}{4}$
So, the possible values of $a$ are $a = -2$ or $a = \frac{3}{2}$, and the corresponding values of $b$ are $b = 3$ or $b = \frac{5}{4}$.
(ii) For the case where $a = 1$ and $b = -1$, we can find an expression for $f^{-1}(x)$ and determine its domain.
Given $a = 1$ and $b = -1$, we have the equation:
$f(x) = x^2 – x – 3$
To find the inverse, interchange $x$ and $f(x)$:
$x = f^{-1}(f(x)) = (f^{-1}(x))^2 – f^{-1}(x) – 3$
Let $y = f^{-1}(x)$:
$x = y^2 – y – 3$
$y^2 – y – (x + 3) = 0$
$y = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$
In this case, $a = 1$, $b = -1$, and $c = -(x + 3)$. Plugging these values into the quadratic formula:
$y = \frac{1 \pm \sqrt{1 – 4(-x – 3)}}{2}$
Now, we need to determine the domain of $f^{-1}(x)$. Since $f^{-1}(x)$ involves a square root, the expression inside the square root, $1 – 4(-x – 3)$, must be greater than or equal to 0:
$1 – 4(-x – 3) \geq 0$
$1 + 4x + 12 \geq 0$
$4x + 13 \geq 0$
$4x \geq -13$
$x \geq -\frac{13}{4}$
So, the domain of $f^{-1}(x)$ is $x \geq -\frac{13}{4}$, which means that $f^{-1}(x)$ is defined for all $x$ greater than or equal to $-\frac{13}{4}$.

Question

The function f is defined by f(x) = 2 − 3 cos x for 0 ≤ x ≤ 20.

(i) State the range of f.

(ii) Sketch the graph of y=f(x).

The function g is defined by g(x)= 2 − 3 cos x for 0 ≤ x ≤ p, where p is a constant.

(iii) State the largest value of p for which g has an inverse.

(iv) For this value of p, find an expression for $g^{-1}(x)$

▶️Answer/Explanation

(i) State the range of f:
The range of a function is the set of all possible values of the function. In this case, $f(x) = 2 – 3\cos(x)$ for $0 \leq x \leq 20$. To find the range, we need to consider the possible values of $f(x)$.
The cosine function $\cos(x)$ varies between -1 and 1. Therefore, $3\cos(x)$ varies between -3 and 3. When you subtract this from 2, you get:
$2 – 3\cos(x)$ ranges between $2 – 3 = -1$ and $2 + 3 = 5$.
So, the range of $f$ is $-1 \leq f(x) \leq 5$.
(ii) Sketch the graph of $y = f(x)$:
To sketch the graph of $y = f(x)$, you can use the range we found in part (i). The graph of $y = f(x)$ is a cosine curve that has been shifted vertically downward by 3 units and upward by 2 units.
The highest point on the graph occurs at $x = 0$, where $f(x) = 5$, and the lowest point occurs at $x = \pi$, where $f(x) = -1$. The graph oscillates between these points over the interval $0 \leq x \leq 20$.

(iii) State the largest value of $p$ for which $g$ has an inverse:
For a function to have an inverse, it must be one-to-one (injective), which means that each output value (y-value) must correspond to a unique input value (x-value).
The function $g(x) = 2 – 3\cos(x)$ for $0 \leq x \leq p$ is essentially the same as $f(x)$, except for the restriction on the domain ($0 \leq x \leq p$).
In this case, the cosine function $\cos(x)$ has a period of $2\pi$. To ensure that $g(x)$ is one-to-one over the given interval, $0 \leq x \leq p$, the value of $p$ should be less than or equal to one period of $\cos(x)$, which is $\pi$.
Therefore, the largest value of $p$ for which $g$ has an inverse is $p = \pi$.
(iv) For $p = 2\pi$, find an expression for $g^{-1}(x)$:
When $p = 2\pi$, the interval of $g$ is $0 \leq x \leq 2\pi$. In this case, $g(x) = 2 – 3\cos(x)$ for $0 \leq x \leq 2\pi$.
To find $g^{-1}(x)$, we need to interchange the roles of $x$ and $g(x)$ and solve for $x$:
$x = 2 – 3\cos(g^{-1}(x))$
$\cos(g^{-1}(x)) = \frac{2 – x}{3}$
$g^{-1}(x) = \arccos\left(\frac{2 – x}{3}\right)$
The domain of $g^{-1}(x)$ will be the range of $g(x)$ over the interval $0 \leq x \leq 2\pi$, which is $2 – 3\cos(x)$ for $-1 \leq x \leq 5$. So, the domain of $g^{-1}(x)$ is $-1 \leq x \leq 5$.

Question

(i) Express $x^{2}-4x+7$ in the form$(x+a)^{2}+b.$

The function f is defined by $f(x) = x^{2} − 4x + 7 $ for x < k, where k is a constant.

(ii) State the largest value of k for which f is a decreasing function.

The value of k is now given to be 1.

(iii) Find an expression for $f^{-1}(x ) $and state the domain of $ f^{-1}(x) $

(iv) The function g is defined by $g(x )= \frac{2}{x-1} $for x > 1. Find an expression for gof(x) and state the range of gof(x).

▶️Answer/Explanation

(i) To express $x^2 – 4x + 7$ in the desired form, we complete the square:
$x^2 – 4x + 7 = (x^2 – 4x + ?) + 7$
$x^2 – 4x + 7 = (x^2 – 4x + 4) – 4 + 7$
$x^2 – 4x + 7 = (x – 2)^2 – 4 + 7$
$x^2 – 4x + 7 = (x – 2)^2 + 3$
So, $x^2 – 4x + 7$ in the desired form is $(x – 2)^2 + 3$.
(ii) State the largest value of $k$ for which $f$ is a decreasing function:
To find where $f(x) = x^2 – 4x + 7$ is decreasing, we need to determine where its derivative $f'(x) = 2x – 4$ is negative.
Set $f'(x) < 0$:
$2x – 4 < 0$
$2x < 4$
$x < 2$
So, $f(x)$ is decreasing for $x < 2$.
(iii) Find an expression for $f^{-1}(x)$ and state the domain of $f^{-1}(x)$:
To find the inverse $f^{-1}(x)$, swap $x$ and $y$ and solve for $y$:
$x = y^2 – 4y + 7$
$y^2 – 4y = x – 7$
$y^2 – 4y + 4 = x – 7 + 4$
$(y – 2)^2 = x – 3$
$y – 2 = \pm \sqrt{x – 3}$
$y = 2 \pm \sqrt{x – 3}$
The domain of $f^{-1}(x)$ is the range of $f(x)$, which is $f(x) \geq 3$. So, the domain of $f^{-1}(x)$ is $x \geq 3$

Question

The diagram shows a curve with equation $y=4x^{\frac{1}{2}}-2x\ for \ x\geqslant 0,$ and a straight line with equation
y = 3 − x. The curve crosses the x-axis at A (4, 0) and crosses the straight line at B and C.
(a) Find, by calculation, the x-coordinates of B and C. [4]

(b) Show that B is a stationary point on the curve. [2]

▶️Answer/Explanation
(a) $y=4x^{\frac{1}{2}}-2x=3-x\rightarrow x-4x^{\frac{1}{2}}+3(=0)$

$\left ( x^{\frac{1}{2}}-1 \right )\left ( x^{\frac{1}{2}}-3 \right )(=0)\ or \ (u-1)(u-3)(=0)$

$x^{\frac{1}{2}}=1,3$

x =1, 9
Alternative method for question 12(a)

$\left ( 4x^{\frac{1}{2}} \right )^{2}=(3+x)^{2}$

$16x=9+6x+x^{2}\rightarrow x^{2}-10x+9(=0)$

(x – 1 )( x – 9 ) ( =0 )

x = 1, 9

(b) $\frac{dy}{dx}2x^{1/2}-2$

$\frac{dy}{dx}2x^{1/2}-2=0$ when x =1 hence B is a stationary point

$\int (4x^{\frac{1}{2}}-2x)(dx)=\left [ \frac{4x^{\frac{3}{2}}}{\frac{3}{2}}-x^{2} \right ]$

$(72-81)-\left ( \frac{64}{3}-16 \right )$

$-14\frac{1}{3}$

$Shaded \ region=18-14\frac{1}{3}=3\frac{2}{3}$

Question

(a) Express −3x² + 12x + 2 in the form −3 (x-a)² + b, where a and b are constants.

The one-one function f is defined by f : x → −3x² + 12x + 2 for x ≤ k.

(b) State the largest possible value of the constant k.

It is now given that k = −1.

(c) State the range of f.

(d) Find an expression for f ⁻¹ (x).

The result of translating the graph of y = f (x) by $\binom{-3}{1}$ is the graph of y = g (x).

(e) Express g(x) in the form px² + qx + r, where p, q and r are constants.

▶️Answer/Explanation

(a) To express the quadratic function $-3x^2 + 12x + 2$ in the form $-3(x-a)^2 + b$, we can complete the square. First, factor out the common factor of -3:
$-3(x^2 – 4x) + 2$
$-3(x^2 – 4x + 4 – 4) + 2$
$-3((x-2)^2 – 4) + 2$
$-3(x-2)^2 + 12 + 2$
$-3(x-2)^2 + 14$
So, the expression in the desired form is $-3(x-2)^2 + 14$, where $a = 2$ and $b = 14$.
(b) The largest possible value of the constant $k$ is determined by the vertex of the parabola $-3x^2 + 12x + 2$, which occurs at the value of $x$ that maximizes the function. The vertex of a parabola in the form $a(x-h)^2 + k$ is at the point $(h, k)$. In this case, the vertex occurs at $x = 2$ (from part a), so $k$ is the maximum value of the function.
So, the largest possible value of the constant $k$ is $k = -3(2-2)^2 + 14 = 14$.
(c) The range of the function $f(x) = -3x^2 + 12x + 2$ can be determined by analyzing its vertex and the direction of its opening. The function is a downward-facing parabola (since the coefficient of $x^2$ is negative), and we found in part (a) that its vertex is at $(2, 14)$.
Since it’s a downward-facing parabola, the maximum value occurs at the vertex, which is $14$.
Therefore, the range of $f(x)$ is $(-\infty, 14]$.
(d) To find the inverse function $f^{-1}(x)$, we switch the roles of $x$ and $y$ and solve for $y$:
$y = -3x^2 + 12x + 2$
$x = -3y^2 + 12y + 2$
$3y^2 – 12y = -x + 2$
$3(y^2 – 4y) = -x + 2$
$3(y^2 – 4y + 4 – 4) = -x + 2$
$3((y-2)^2 – 4) = -x + 2$
$3(y-2)^2 – 12 = -x +2\]
\(3(y-2)^2 = -x + 14\]
\((y-2)^2 = -\frac{1}{3}x + \frac{14}{3}$
$y – 2 = \pm \sqrt{-\frac{1}{3}x + \frac{14}{3}}$
$y = 2 \pm \sqrt{-\frac{1}{3}x + \frac{10}{3}}$
So, $f^{-1}(x)$ has two branches:
$f^{-1}_1(x) = 2 + \sqrt{-\frac{1}{3}x + \frac{10}{3}}$
$f^{-1}_2(x) = 2 – \sqrt{-\frac{1}{3}x + \frac{10}{3}}$
(e) To express $g(x)$ in the form $px^2 + qx + r$, where $p$, $q$, and $r$ are constants, we need to find the equation for $g(x)$ after translating the graph of $f(x)$ by the vector $\begin{bmatrix}-3 \\ 1\end{bmatrix}$. This translation involves shifting the graph three units to the left (negative 3 in the x-direction) and one unit up (positive 1 in the y-direction).
The original equation for $f(x)$ was $f(x) = -3(x-2)^2 + 14$. After the translation, $g(x)$ becomes:
$g(x) = -3(x-2+3)^2 + 14+1$
$g(x) = -3(x+1)^2 + 15$
$g(x) = -3(x^2+2x+1) + 15$
$g(x) = -3x^2 -6x -3 + 15$
$g(x) = -3x^2 -6x+12$
So, the expression for $g(x)$ in the desired form is $g(x) = -3x^2 -6x+12$.

Question

Functions f and g are defined as follows:
$f:x\rightarrow x^{2}+2x+3$ for $x\leq -1$
$g:x\rightarrow 2x+1$ for $x\geq -1$

(a) Express f(x) in the form $(x+a)^{2}+b$ and state the range of f.

(b) Find the expression for $f^{-1}(x)$.
(c) Solve the equation gf(x)=13

▶️Answer/Explanation

(a) To express $f(x) = x^2 + 2x + 3$ in the form $(x + a)^2 + b$,
$f(x) = x^2 + 2x + 3$
$f(x) = (x^2 + 2x + 1 – 1) + 3$
$f(x) = (x + 1)^2 – 1 + 3$
$f(x) = (x + 1)^2 + 2$
So, $f(x)$ can be expressed in the form $(x + a)^2 + b$ with $a = 1$ and $b = 2$.
The range of $f$ is the set of all real numbers greater than or equal to the minimum value of the quadratic function, which occurs at its vertex. The vertex of the parabola $f(x) = (x + 1)^2 + 2$ is at $(-1, 2)$. Therefore, the range of $f$ is $[2, +\infty)$.
(b) To find the expression for $f^{-1}(x)$, we swap the roles of $x$ and $y$ in the equation and solve for $y\:
\(x = (y + 1)^2 + 2$
$(y + 1)^2 = x – 2$
$y + 1 = \pm \sqrt{x – 2}$
$y = -1 \pm \sqrt{x – 2}$
So, the expression for $f^{-1}(x)$ has two branches:
$f^{-1}_1(x) = -1 + \sqrt{x – 2}$
$f^{-1}_2(x) = -1 – \sqrt{x – 2}$
(c) To solve the equation $g(f(x)) = 13$, we need to find $x$ such that:
$2(x^2 + 2x + 3) + 1 = 13$
$2x^2 + 4x + 7 = 13$
$2x^2 + 4x – 6 = 0$
$x^2 + 2x – 3 = 0$
$(x + 3)(x – 1) = 0$
1. $x + 3 = 0 \implies x = -3$
2. $x – 1 = 0 \implies x = 1$
x=-1 (rejected)
$\therefore$ x=-3 only

Question

Functions f and g are defined as follows:
$f(x)=(x-2)^{2}-4$ for$ \ x\geqslant 2$
$g(x)=ax+2$ for$ \ x\epsilon \mathbb{R}$
where a is a constant.
(a) State the range of f.
(b) Find $f^{-1}(x)$
(c) Given that a=$-\frac{5}{3}$, solve the equation f(x) = g(x)
(d) Given instead that $g(gf^{-1}(12)=62$), find the possible values of a.

▶️Answer/Explanation

(a) To find the range of the function $f(x) = (x-2)^2 – 4$ for $x \geq 2$, we need to determine the possible values that $f(x)$ can take. First, note that for $x \geq 2$, $(x-2)^2$ is always non-negative. Subtracting 4 from a non-negative number will not change the sign, so the range of $f(x)$ is all real numbers greater than or equal to $-4$. In interval notation, this is $(-4, \infty)$.
(b) To find $f^{-1}(x)$, we first swap $x$ and $y$ and solve for $y$:
$x = (y-2)^2 – 4$
$x + 4 = (y-2)^2$
$\sqrt{x + 4} = |y – 2|$
$\sqrt{x + 4} = y – 2$
$y = \sqrt{x + 4} + 2$
So, $f^{-1}(x) = \sqrt{x + 4} + 2$.
(c) Given that $a = -\frac{5}{3}$, solve the equation $f(x) = g(x)$:
$(x-2)^2 – 4 = -\frac{5}{3}x + 2$
$(x-2)^2 – 4 + \frac{5}{3}x – 2 = 0$
$(x-2)^2 + \frac{5}{3}x – 6 = 0$
$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$
In this case, $a = 1$, $b = \frac{5}{3}$, and $c = -6$. Plug these values into the formula and solve for $x$:
$x = \frac{-\frac{5}{3} \pm \sqrt{\left(\frac{5}{3}\right)^2 – 4(1)(-6)}}{2(1)}$
$x = \frac{-\frac{5}{3} \pm \sqrt{\frac{25}{9} + 24}}{2}$
$x = \frac{-\frac{5}{3} \pm \sqrt{\frac{25}{9} + \frac{216}{9}}}{2}$
$x = \frac{-\frac{5}{3} \pm \sqrt{\frac{25 + 216}{9}}}{2}$
$x = \frac{-\frac{5}{3} \pm \sqrt{\frac{241}{9}}}{2}$
$x = \frac{-\frac{5}{3} \pm \frac{\sqrt{241}}{3}}{2}$
$x = \frac{-\frac{5}{6} \pm \frac{\sqrt{241}}{6}}{1}$
$x_1 = \frac{-5 + \sqrt{241}}{6}$
$x_2 = \frac{-5 – \sqrt{241}}{6}$
These are the correct solutions for the equation $(x-2)^2 + \frac{5}{3}x – 6 = 0$.
(d) Given $g(g(f^{-1}(12))) = 62$, we need to find the possible values of $a$. First, let’s find $f^{-1}(12)$ by plugging $x = 12$ into $f^{-1}(x)$:
$f^{-1}(12) = \sqrt{12 + 4} + 2 = \sqrt{16} + 2 = 4 + 2 = 6$
$g(f^{-1}(12)) = a(6) + 2 = 6a + 2$
$g(g(f^{-1}(12))) = a(6a + 2)+2= 62$
$6a^{2}+2a+2=62$
$6a^{2}+2a-60=0$
$3a^{2}+a-30=0$
The quadratic formula is :
$a = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$
In this equation, $a = 3$, $b = 1$, and $c = -30$. Plug these values into the formula:
$a = \frac{-1 \pm \sqrt{1^2 – 4(3)(-30)}}{2(3)}$
Now, discriminant ($b^2 – 4ac$:
$1^2 – 4(3)(-30) = 1 + 360 = 361$
$a = \frac{-1 \pm \sqrt{361}}{2(3)}$
$a = \frac{-1 \pm 19}{6}$
$a_1 = \frac{-1 + 19}{6} = \frac{18}{6} = 3$
$a_2 = \frac{-1 – 19}{6} = \frac{-20}{6} = -\frac{10}{3}$
So, the solutions for $a$ are $a_1 = 3$ and $a_2 = -\frac{10}{3}$.

Question

The diagram shows the graph of $y = f(x)$, where $\frac{3}{2} \cos 2x+\frac{1}{2}$ for 0 ≤ x ≤ π.

(a) State the range of f.

A function g is such that $g(x) = f(x) + k$ , where k is a positive constant. The x-axis is a tangent to the curve $y = g(x)$.

(b) State the value of k and hence describe fully the transformation that maps the curve $y = f(x)$ onto $y = g(x)$.

(c) State the equation of the curve which is the reflection of $y = f(x)$ in the x-axis. Give your answer in the form $y = a cos 2x + b$, where a and b are constants.

▶️Answer/Explanation

(a) To find the range of the function $f(x) = \frac{3}{2} \cos(2x) + \frac{1}{2}$ for $0 \leq x \leq \pi$, we need to consider the range of the cosine function. The range of the cosine function is $[-1, 1]$. When you multiply this range by $\frac{3}{2}$, you get $[- \frac{3}{2}, \frac{3}{2}]$, and then when you add $\frac{1}{2}$, the range becomes $[-1, 2]$. So, the range of $f(x)$ is $[-1, 2]$.
(b) Since the $x$-axis is a tangent to the curve $y = g(x)$, this means that the minimum value of $g(x)$ occurs when it touches the $x$-axis. We have already established that the minimum value of $f(x)$ is $-1$, so we want $g(x)$ to touch the $x$-axis at $y = -1$.
To achieve this, we add a positive constant $k$ to $f(x)$, making $g(x) = f(x) + k$. Since we want $g(x)$ to touch the $x$-axis at $y = -1$, we set $k = -1$. Therefore, $k = -1$, and the transformation that maps the curve $y = f(x)$ onto $y = g(x)$ is a vertical translation of $1$ unit downwards.
(c) To find the equation of the curve which is the reflection of $y = f(x)$ in the $x$-axis, you can simply negate the entire function, changing the sign of $f(x)$. So, the equation of the reflected curve is:$y = -\left(\frac{3}{2} \cos(2x) + \frac{1}{2}\right) = -\frac{3}{2} \cos(2x) – \frac{1}{2}$
This can be written in the form $y = a \cos(2x) + b$, where $a = -\frac{3}{2}$ and $b = -\frac{1}{2}$.

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