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.

(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 $$g^{-1}$$.

(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 $$fg(x)=\frac{1}{7}$$

(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.

(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 $$f(x)=\frac{185}{16}$$.

(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.

(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 $$f^{-1}$$.

(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 $$gf(2x) = fg(x)$$.

(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 $$f^{-1}$$

(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 $$g^{-1}(x)$$

(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).

(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]

(c) Find the area of the shaded region.                                                                                                                         [6]

(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

(c) Area of correct triangle $$\frac{1}{2}(9-3)\times 6$$

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

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

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

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

### 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.

(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

(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.

(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 $$f^{-1}(x)$$:
$$f^{-1}(12) = \sqrt{12 + 4} + 2 = \sqrt{16} + 2 = 4 + 2 = 6$$
$$g(f^{-1}(12)) = a(6) + 2 = 6a + 2$$
$$g(g(f^{-1}(12))) = a(6a + 2)+2= 62$$
$$6a^{2}+2a+2=62$$
$$6a^{2}+2a-60=0$$
$$3a^{2}+a-30=0$$
$$a = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
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.

(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 $$b = -\frac{1}{2}$$.