**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 **

**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]

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

**▶️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

(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 −3x^{2} + 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 → −3x^{2} + 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 px^{2} + 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 \(b = -\frac{1}{2}\).