### Question

The equation of a curve is $$y=x^{3}+px^{2}$$ ,where p is a positive constant .

(i)Show that the origin is a stationary point on the curve and find the coordinates of the other stationary point in terms of p.

(ii)Find the nature of each of the stationary points.

Another curve has equation $$y=x^{3}+px^{2}+px$$

(iii)Find the set of values of p for which this curve has n o stationary points.

(i) To determine whether the origin is a stationary point on the curve $$y = x^3 + px^2$$, we need to find the first and second derivatives of the curve and then evaluate them at $$x = 0$$.
The first derivative of $$y$$ with respect to $$x$$ is:
$$y’ = 3x^2 + 2px$$
$$y'(0) = 3(0)^2 + 2p(0) = 0$$
Since $$y'(0) = 0$$, this indicates that the origin ($$x = 0$$) is a stationary point on the curve.
To find the coordinates of the other stationary point, we need to find the value of $$x$$ where $$y’$$ is also equal to zero:
$$0 = 3x^2 + 2px$$
$$0 = x(3x + 2p)$$
$$x = 0$$ (which corresponds to the origin, where we already know there is a stationary point).
$$3x + 2p = 0$$
To find the value of $$x$$ for the second stationary point, solve for $$x$$ in the second equation:
$$3x + 2p = 0$$
$$3x = -2p$$
$$x = -\frac{2p}{3}$$
So, the other stationary point has coordinates $$\left(-\frac{2p}{3}, f\left(-\frac{2p}{3}\right)\right)$$.
(ii) To determine the nature of each of the stationary points, we can use the second derivative test. We need to find the second derivative $$y”$$ and evaluate it at both $$x = 0$$ and $$x = -\frac{2p}{3}$$.
The second derivative of $$y$$ is:
$$y” = 6x + 2p$$
Evaluate $$y”$$ at $$x = 0$$:
$$y”(0) = 6(0) + 2p = 2p$$
Since $$y”(0) = 2p$$, and $$p$$ is positive (given that $$p$$ is a positive constant), this indicates that the stationary point at the origin is a local minimum.
Now, evaluate $$y”$$ at $$x = -\frac{2p}{3}$$:
$$y”\left(-\frac{2p}{3}\right) = 6\left(-\frac{2p}{3}\right) + 2p = -4p + 2p = -2p$$
Since $$y”\left(-\frac{2p}{3}\right) = -2p$$, and $$p$$ is positive, this indicates that the other stationary point at $$\left(-\frac{2p}{3}, f\left(-\frac{2p}{3}\right)\right) = \left(-\frac{2p}{3}, -\frac{4p^3}{27}\right)$$ is also a local minimum.
(iii) For the curve $$y = x^3 + px^2 + px$$, to find the set of values of $$p$$ for which this curve has no stationary points,
$$y’ = 3x^2 + 2px + p$$
For a stationary point, we need $$y’$$ to be equal to zero:
$$0 = 3x^2 + 2px + p$$
The discriminant $$\Delta$$ is given by:
$$\Delta = (2p)^2 – 4(3)(p)$$
$$\Delta = 4p^2 – 12p = 4p(p – 3)$$
To have no stationary points, we need $$\Delta$$ to be negative because it means that the quadratic equation has no real solutions.
$$\therefore$$ $$4p(p – 3) < 0$$
$$\rightarrow 0< p< 3$$

### Question The diagram shows that the function $\mathrm{f}$ defined for $-1 \leq x \leq 4$, where
$$f(x)= \begin{cases}3 x-2 & \text { for }-1 \leqslant x \leqslant 1 \\ \frac{4}{5-x} & \text { for } 1<x \leqslant 4\end{cases}$$
(i)State the range of $f$.
(ii)Copy the diagram and on your copy sketch the graph of $y=f^{-1}(x)$
(iii)Obtain expressions to define the function $f^{-1}$,giving also the set of values for which each expression is valid.

(i) The range of a function represents the set of all possible values that the function can take on for its given domain. In the case of the function $$f(x)$$, it is defined for the domain $$-1 \leq x \leq 4$$, and it consists of two distinct pieces:
$$\Rightarrow$$ For the interval $$-1 \leq x \leq 1$$, the function is defined as $$f(x) = 3x – 2$$. In this interval, as $$x$$ varies from -1 to 1, the function $$f(x)\ varies from \(3(-1) – 2 = -5$$ to $$3(1) – 2 = 1$$. So, for this part of the domain, the range of $$f(x)\ is \(-5 \leq f(x) \leq 1$$.
$$\Rightarrow$$ For the interval $$1 < x \leq 4$$, the function is defined as $$f(x) = \frac{4}{5-x}$$. In this interval, as $$x$$ varies from 1 to 4, the function $$f(x)\ takes on values that change continuously. \(\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} \frac{4}{5-x} = +\infty$$
As $$x$$ approaches 5 from the left, $$f(x)\ becomes unbounded and approaches positive infinity. This means that for \(x$$ values close to, but less than 5, $$f(x)\ can take on arbitrarily large positive values. However, it never reaches 5 because that would make the denominator zero. So, for this part of the domain, the range of \(f(x)\ is \(-\infty < f(x) < +\infty$$, but $$f(x)\ never equals 5. Combining both ranges, the overall range of the function \(f(x$$ is $$-5 \leq f(x) \leq 1$$ for the given domain $$-1 \leq x \leq 4$$, and the function takes all real values within this range except for the value 5, which is excluded in the second piece due to the asymptote at $$x = 5$$.
The range is given as $$-5 \leq f(x) \leq 4$$ for both $$x$$ and $$y$$.
(ii) To sketch the graph of $$y = f^{-1}(x)$$, you can follow these steps:
1. Reflect the graph of $$y = f(x)$$ in the line $$y = x$$. This means that for each point $$(a, b)$$ on the graph of $$y = f(x)\, you plot a corresponding point \((b, a)\ on the graph of \(y = f^{-1}(x)\. 2. Determine the domain of \(f^{-1}(x)\ based on the reflection. Since the original function \(f(x)\ was defined for \(-1 \leq x \leq 4$$, the range of $$f(x)\ will become the domain of \(f^{-1}(x)\ after reflection. 3. Plot the points and sketch the curve for \(y = f^{-1}(x)\ within its valid domain. \(f^{-1}\left ( x \right )$$ approximately correct (independent of f)
Closed region between (1,1)and (4,4) line reaches x-axis.
(iii) To obtain expressions that define the function $$f^{-1}(x)\, we need to find the inverse of each piece of the function \(f(x)\. For the first piece, \(f(x) = 3x – 2\ for \(-1 \leq x \leq 1\, to find the inverse, swap \(x\ and \(y\ and solve for \(y\: \(x = 3y – 2$$
$$3y = x + 2$$
$$y = \frac{x + 2}{3$$
So, the inverse function for this piece is $$f^{-1}(x) = \frac{x + 2}{3\ for \(-1 \leq x \leq 1\. For the second piece, \(f(x) = \frac{4}{5-x\ for \(1 < x \leq 4\, to find the inverse, swap \(x\ and \(y\ and solve for \(y\: \(x = \frac{4}{5-y$$
$$\frac{4}{5-y} = x$$
$$5-y = \frac{4}{x$$
$$-y = \frac{4}{x} – 5$$
$$y = 5 – \frac{4}{x$$
So, the inverse function for this piece is $$f^{-1}(x) = 5 – \frac{4}{x\ for \(1 < x \leq 4\.$$

### 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 the function $$f(x)$$ in the form $$(x+a)^2+b+k$$, you need to complete the square. Here’s how you can do it:
$$f(x) = x^2 – 4x + k$$
$$f(x) = (x^2 – 4x + 4) + k – 4$$
$$f(x) = (x – 2)^2 + (k – 4)$$
So, $$a = -2$$ and $$b = k – 4$$.
(ii) The range of the function $$f(x)$$ in terms of $$k$$ can be determined from the expression $$(x – 2)^2 + (k – 4)$$. The minimum value of $$(x – 2)^2$$ is $$0$$, and it can take any non-negative value. So, the range of $$f(x)$$ is $$(k – 4, +\infty)$$ in terms of $$k$$.
(iii) To make $$f(x)$$ one-one (injective), the function must be strictly increasing or decreasing. For $$f(x) = x^2 – 4x + k$$, it’s a quadratic function, and its behavior depends on the coefficient of the $$x^2$$ term (which is 1 in this case). A quadratic function is strictly increasing if the coefficient of $$x^2$$ is positive. So, for $$f(x)$$ to be one-one, $$1 > 0$$, which implies that $$p$$ must be greater than 2.
(iv) To find the expression for $$f^{-1}(x)$$ and state its domain, we first need to interchange the roles of $$x$$ and $$y$$:
$$x-2=\left ( \pm \right )\sqrt{y+4-k}$$
$$x=2+\sqrt{y+4-k}$$
$$f^{-1}\left ( x \right )=2+\sqrt{x+4-k}$$
Domain is $$x> k-4$$ or $$\left [ k-4,\infty \right ]$$
or $$\left ( k-4,\infty \right )$$

### 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)$$fg(x)=5x$$

Range of fg is$$y\geq 0$$

(ii)$$y=\frac{4}{\left ( 5x+2 \right )}\Rightarrow x=\frac{4-2y}{5y}$$

$$g^{-1}\left ( x \right )=\frac{\left ( 4-2x \right )}{5x}$$

0,2 with no incorrect inequality

$$0< x\leq 2$$

### Question

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

$$g:x\rightarrow 2x-3$$

(i)Find and simplify an expression for g(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) The function $$g(x)$$ is already given as $$g(x) = 2x – 3$$,
$$gg\left ( x \right )=g\left ( 2x-3 \right )=2\left ( 2x-3 \right )-3=4x-9$$
(ii) To find the inverse function $$f^{-1}(x)$$, we need to swap $$x$$ and $$y$$ in the function $$f(x)$$, and then solve for $$y\: \(y = \frac{1}{x^2 – 9}.$$
$$x = \frac{1}{y^2 – 9}.$$
$$y^2 – 9 = \frac{1}{x}.$$
$$y^2 = \frac{1}{x} + 9.$$
$$y = \sqrt{\frac{1}{x} + 9}.$$
$$f^{-1}(x) = \sqrt{\frac{1}{x} + 9}.$$
The domain of $$f^{-1}(x)\ should exclude values that result in a non-real number under the square root. Since the square root of a negative number is not real, the expression \(\frac{1}{x} + 9$$ must greater than or equal to zero:
$$\frac{1}{x} + 9 \geq 0.$$
$$\frac{1}{x} \geq -9.$$
Now, we can see that $$x$$ cannot be zero because dividing by zero is undefined.
So, the domain of $$f^{-1}(x)\ is \(x > 0$$ or $$x < 0$$, which can be expressed as $$x \in (-\infty, 0) \cup (0, +\infty)$$.
(iii)EITHER:
$$\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)$$\vec{CD}=-3\begin{pmatrix}2\\ 1\\ 2\end{pmatrix}+2\begin{pmatrix}4\\ 0\\ 6\end{pmatrix}=\begin{pmatrix}2\\ -3\\ 6\end{pmatrix}$$

$$Unit vector =\frac{1}{7}\begin{pmatrix}2\\ -3\\ 6\end{pmatrix}$$

(ii)$$\vec{OE}=\begin{pmatrix}6\\ 3\\ 6\end{pmatrix}+\begin{pmatrix}1\\-1\tfrac{1}{2}\\ 3\end{pmatrix}=\begin{pmatrix}7\\ 1\tfrac{1}{2}\\ 9\end{pmatrix}$$

$$\vec{OE}.\vec{OD}=56+0+108=164$$

$$\left | \vec{OE} \right |=\sqrt{132.25}=11.5$$

$$\left | \vec{OD} \right |=\sqrt{208}$$

$$164=\sqrt{132.25}\times \sqrt{208}\times \cos \Theta$$

$$\Theta =8.6^{\circ}$$

### 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 define the inverse function $$f^{-1}$$, we first need to swap $$x$$ and $$y$$ in the function $$f(x)$$, and then solve for $$y\: \(y = x^2 + 1.) \(x = y^2 + 1.$$
$$y^2 = x – 1.$$
$$y = \pm \sqrt{x – 1}.$$
So, the inverse function $$f^{-1}(x)\ can be defined as: \(f^{-1}(x) = \pm \sqrt{x – 1}.$$
(ii) To solve the equation $$f(x) = \frac{185}{16}$$, we have the equation:
$$(x^2 + 1)^{2} +1=f(x).$$
$$(x^2 + 1)^{2} +1= \frac{185}{16}$$
$$(x^2 +1)^{2}= \frac{185}{16}-1 .$$
$$x = \pm \sqrt{\frac{169}{16}}.$$
$$x^{2}+1`=\pm \frac{13}{4}$$
$$x = \frac{3}{2}.$$

### Question

(i) Express $$x^{2}-2x-15$$  in the form $$(x+a)^{2}+b$$.

The function f is defined for p ≤ x ≤ q, where p and q are positive constants, by f : x → $$x^{2}$$ − 2x − 15.
The range of f is given by c ≤ fx ≤ d, where c and d are constants.
(ii) State the smallest possible value of c.
For the case where c = 9 and d = 65,
(iii) find p and q,
(iv) find an expression for $$f^{-1}(x)$$.

(i) To express $$x^2 – 2x – 15$$ in the form $$(x+a)^2 + b$$, we complete the square:
\begin{align*}
x^2 – 2x – 15 &= (x^2 – 2x) – 15 \\
&= \left(x^2 – 2x + (-2/2)^2\right) – 15 – (-2/2)^2 \\
&= (x^2 – 2x + 1) – 15 – 1 \\
&= (x – 1)^2 – 16.
\end{align*}
So, $$x^2 – 2x – 15 = (x – 1)^2 – 16$$.
(ii) To find the smallest possible value of $$c$$ in the range $$c \leq fx \leq d$$, we need to consider the range of $$f(x) = x^2 – 2x – 15$$.
The vertex of the parabola $$f(x) = x^2 – 2x – 15$$ is at the point $$(1, -16)$$. Since the coefficient of $$x^2$$ is positive, the parabola opens upward, and its minimum value occurs at the vertex. Therefore, the smallest possible value of $$c$$ is the value of $$f$$ at the vertex, which is $$-16$$.
So, $$c = -16$$.
(iii) $$9\leq (x-1)^{2}-16\leq 65$$ OR $$x^{2}-2x-15=9\rightarrow 6,-4$$
$$25\leq (x-1)^{2}\leq 81$$ $$x^{2}-2x-15=65\rightarrow 10,-8$$
$$5\leq x-1\leq 9$$         $$\Rightarrow p=6$$
$$6\leq x\leq 10$$       $$\Rightarrow q=10$$
(iv) To find an expression for $$f^{-1}(x)$$, we first swap $$x$$ and $$y$$ in the equation for $$f(x)$$ and solve for $$y\: \(x = y^2 – 2y – 15.$$
$$y^2 – 2y = x + 15.$$
$$(y^2 – 2y + 1) = x + 15 + 1.$$
$$(y – 1)^2 = x + 16.$$
$$y – 1 = \pm \sqrt{x + 16}.$$
$$y = 1 \pm \sqrt{x + 16}.$$
So, the expression for $$f^{-1}(x)\ is: \(f^{-1}(x) = 1 \pm \sqrt{x + 16}.$$

### Question

Find the set of values of k for which the line y = 2x − k meets the curve $$y=x^{2}+kx-2$$ at two distinct points.

$$x^{2}+x(k-2)+(k-2)=0$$

$$(k-2)^{2}-4(k-2)> 0$$

$$(k-2)(k-6)> 0$$

k<2 or k>6

Allow $$\left ( -\infty ,2 \right )\cup \left ( 6,\infty \right )$$

### 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.

$$f:x\rightarrow \frac{2}{3-2x}$$ $$g:x\rightarrow 4x+a$$,

(i)$$y=\frac{2}{3-2x}\rightarrow y\left ( 3-2x \right )=2\rightarrow 3-2x=\frac{2}{y}$$

$$\rightarrow 2x=3-\frac{2}{y}\rightarrow f^{-1}\left ( x \right )=\frac{3}{2}-\frac{1}{x}$$

(ii)$$gf(-1)=3f(-1)=\frac{2}{5}$$

$$\frac{8}{5}+a=3\rightarrow a=\frac{7}{5}$$

(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) $$f:x\rightarrow 4\sin x-1$$  for $$-\frac{\pi }{2}\leq x\leq \frac{\pi }{2}$$

Range $$-5\leq f(x)\leq 3$$

(ii) 4s-1=0→$$s=\frac{1}{4}\rightarrow x=0.253$$

$$x=0\rightarrow y=-1$$ (iv)range-$$\frac{1}{2}\pi \leq f^{-1}\left ( x \right )\leq\frac{1}{2}\pi$$

domain $$-5\leq x\leq 3$$

Inverse $$f^{-1}(x)=\sin ^{-1}\left ( \frac{x+1}{4} \right )$$

### Question

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

\begin{aligned}
& \mathrm{f}: x \mapsto 2 x^2+3, \\
& \mathrm{~g}: x \mapsto 3 x+2 .
\end{aligned}

(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) $$gf\left ( x \right )=3\left ( 2x^{2} +3\right )+2=6x^{2}+11$$

$$fg\left ( x \right )=2\left ( 3x +2\right )^{2}+3\0 Allow \(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) $$6\left ( 2x \right )^{2}+11=2\left ( 3x+2 \right )^{2}+3$$

$$6x^{2}-24x=0 oe$$

x=0,4

### 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) Values of $$a$$ and $$b$$:
$$2a + 4b = 8$$
$$2a^2 + 3a + 4b + 14 = 0$$
$$2a^2 + 3a + (8 – 2a) + 14 = 0$$
$$2a^2 + a + 22 = 0$$
$$(a + 2)(2a – 3) = 0$$.
So, $$a = -2$$ or $$a = \frac{3}{2}$$.
Now, substitute $$a$$ back into the equation $$2a + 4b = 8$$ to find $$b$$:
If $$a = -2$$, then $$2(-2) + 4b = 8 \implies -4 + 4b = 8 \implies b = 3$$.
If $$a = \frac{3}{2}$$, then $$2\left(\frac{3}{2}\right) + 4b = 8 \implies 3 + 4b = 8 \implies b = \frac{5}{4}$$.
So, the correct values are $$a = -2$$, $$b = 3$$ or $$a = \frac{3}{2}$$, $$b = \frac{5}{4}$$.
(ii) Expression for $$f^{-1}(x)$$ and Domain:
$$y = \left(x – \frac{1}{2}\right)^2 – \frac{13}{4}$$
$$x – \frac{1}{2} = \pm \sqrt{y + \frac{13}{4}}$$
$$f^{-1}(x) = \frac{1}{2} – \sqrt{x + \frac{13}{4}}$$
The domain is given as $$x \geqslant -\frac{13}{4}$$.

### Question.

Functions f and g are defined for $$x\in R$$ by where a and b are constants.
(a) Given that $$gg(2) = 10$$ and $$f^{−1}(2) = 14$$, find the values of a and b.

(b) Using these values of a and b, find an expression for $$gf(x)$$ in the form $$cx + d$$, where c and d are constants.

(a) Find the values of $$a$$ and $$b$$:
Given the equation $$3(3x + b) + b = 9x + 4b$$,
$$9x + 3b + b = 9x + 4b$$
$$10 = 18 + 4b$$
$$-8 = 4b$$
$$b = -2$$
Now, for the equation $$f^{-1}(2) = 14$$:
$$f^{-1}(x) = 2(x + a)$$
Substitute $$x = 2$$:
$$f^{-1}(2) = 2(2 + a) = 14$$
$$4 + 2a = 14$$
$$2a = 10$$
$$a = 5$$
So, the values are $$a = 5$$ and $$b = -2$$.
(b) Find $$g f(x)$$ in the form $$c x + d$$:
$$g f(x) = 3\left(\frac{1}{2}x – 5\right) – 2$$
$$g f(x) = \frac{3}{2}x – 15 – 2$$
$$g f(x) = \frac{3}{2}x – 17$$
So, the expression for $$g f(x)$$ in the form $$c x + d$$ is $$g f(x) = \frac{3}{2}x – 17$$.

### 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 function $$f(x) = 2 – 3 \cos x$$ for $$0 \leq x \leq 20$$. The range of $$f$$ is the set of all possible values that $$f(x)$$ can take. The cosine function has a range between -1 and 1, so the range of $$f$$ can be found by considering the extreme cases:
$$-3 \cos x \leq 3$$
$$2 – 3 \leq f(x) \leq 2 + 3$$
So, the range of $$f$$ is $$[ -1, 5]$$.
(ii) The function $$f(x) = 2 – 3 \cos x$$ is a cosine function with an amplitude of 3, centered at $$x = 0$$, and shifted vertically by 2 units. The graph oscillates between 5 and -1.
(iii) For a function to have an inverse, it must be one-to-one (injective). The cosine function is not one-to-one over its entire domain, but if we restrict the domain, it becomes one-to-one.
The maximum value for $$p$$ occurs when the cosine function completes a full period. The period of $$\cos x$$ is $$\pi$$, so the largest value for $$p$$ is $$\pi$$.
(iv) For this value of $$p$$, find an expression for $$g^{-1}(x)$$:
If $$g(x) = 2 – 3 \cos x$$ for $$0 \leq x \leq 2\pi$$, then to find $$g^{-1}(x)$$, we switch $$x$$ and $$g(x)$$ and solve for $$x$$:
$$x = 2 – 3 \cos g^{-1}(x)$$
Solve for $$g^{-1}(x)$$:
$$g^{-1}(x) = \cos^{-1}\left(\frac{2 – x}{3}\right)$$
So, for $$p = 2\pi$$, the expression for $$g^{-1}(x)$$ is:
$$g^{-1}(x) = \cos^{-1}\left(\frac{2 – x}{3}\right)$$

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