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)$$\left ( x-2 \right )^{2}-4+k$$

(ii)$$f\left ( x \right )> k-4$$ or $$\left [ k-4,\infty \right ]$$ or $$\left ( k-4,\infty \right )$$

(ii)smallest value of p=2

(iv) $$x-2=\left ( \pm \right )\sqrt{y+4-k}$$

$$x=2+\sqrt{y+4-k}$$

$$f^{-1}\left ( x \right )=2+\sqrt{x+4-k}$$

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}\0. Answer/Explanation (i)\(fg(x)=5x$$

Range of fg is$$y\geq$$

(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)$$gg\left ( x \right )=g\left ( 2x-3 \right )=2\left ( 2x-3 \right )-3=4x-9$$

(ii)$$y=\frac{1}{\left ( x^{2} -9\right )}\rightarrow x^{2}=\frac{1}{y}+9$$

$$f^{-1}\left ( c \right )=\sqrt{\frac{1}{x}+9}$$

Attempt solution of $$\sqrt{\frac{1}{x}+9}> 3$$ or attempt to find range of f $$\left ( y> 0 \right )$$

Domain is $$x> 0$$

(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)$$x=\pm \sqrt{y-1}$$

$$f^{-1}:x\rightarrow \sqrt{x-1}$$ for $$x> 1$$

(ii) $$f(x)=(x^{2}+1)^{2}+1$$

$$x^{2}+1=\pm \frac{13}{4}$$

$$x=\frac{3}{2}$$

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 (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 $$f^{-1}$$

Ans:(i) $$2a+4b=8$$
$$2a^2+3a+4b+14$$
$$2a^2+3a+(8-2a)=14\rightarrow (a+2)(2a-3)=0$$
$$a=-2$$ or $$\frac{3}{2}$$
$$b=3$$ or $$\frac{5}{4}$$

(ii) 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)
−1 ⩽ f(x) ⩽ 5 or [–1, 5] (may use y or f instead of f(x)) (iii) (greatest value of p =)π

9(iv)
$$x = 2− 3cosx → cosx =1⁄3(2-x)$$

$$g^{-1}=cos^{-1}\frac{2-x}{3}$$

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)
−1 ⩽ f(x) ⩽ 5 or [–1, 5] (may use y or f instead of f(x)) (iii) (greatest value of p =)π

9(iv)
$$x = 2− 3cosx → cosx =1⁄3(2-x)$$

$$g^{-1}=cos^{-1}\frac{2-x}{3}$$

Queastion

(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)

$$\left [ (x-2^{2}) \right ]+\left [ 3 \right ]$$

(ii)

Largest k is 2 Accept $$k\leq 2$$

(iii)

$$y=(x-2^{2})+3\Rightarrow x-2=\left ( \pm \right )\sqrt{y-3}$$

$$\Rightarrow F^{-1}(x)=2-\sqrt{x-3}forx>4$$ ### 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.                                                                                             

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

(c) Find the area of the shaded region.                                                                                                                         

Ans

12 (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

12(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

12 (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)  $$\left \{ -3\left ( x – 2 \right )^{2} \right \}$$    {+14}

(b)   [k =] 2

(c) [Range is y] ⩽ –13

(d) y = -3 (x-2)2 + 14 leading to (x-2)2 =  $$\frac{14-y}{3}$$

x = 2 (±) $$\sqrt{\frac{14-y}{3}}$$

[f-1 (x)] = 2 – $$\sqrt{\frac{14-y}{3}}$$

(e)  $$\left [ g(x) =\right ] \left \{ -3\left ( x+3-2 \right )^{2} \right \} + \left \{ 14+1 \right \}$$

g(x) = -3x2 – 6x + 12

### Question

Functions f and g are defined as follows:
f:x  →x2 + 2x + 3 for x ≤ -1,
g:x →2x + 1 for x ≥ -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

Ans:

(a) $$[f(x)=](x+1)^2+2$$
Range [of f is (y)] ≥ 2
(b) $$y=(x+1)^2+2$$ leading to $$x=[±]=-\sqrt{y-2}-1$$
$$f^{-1}=-\sqrt{x-2}-1$$
(c) $$2(x^2+2x+3)+1=13$$
$$2x^2+4x-6[=0]$$ leading to (2)(x-1)(x+3)[=0]
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 ggf −1 (12) = 62, find the possible values of a.                                   

Ans

9 (a) Range of f is f(x) ⩾−4

9 (b) $$y=(x-2)^{2}-4\Rightarrow (x-2)^{2}=y+4\Rightarrow x-2=+\sqrt{y+4}\ or \ \pm \sqrt{y+4}$$

$$(3x+2)(x-3)[=0] \ or \ \frac{7\pm \sqrt{7^{2}-4(3)(-6)}}{6}\ OE\$$

x = 3 only

9(d) f-1 (12) = 6

g(f–1 (12)) = 6a + 2

g(g(f–1 (12))) =a(6a + 2) + 2 = 62

6a2 + 2a – 60 [= 0]

$$a=-\frac{10}{3}\ or \ 3$$

Alternative method for Question 9(d)

$$g(f^{-1}(x))=a\left ( \sqrt{x+4+2} \right )+2 \ or \ gg(x)=a(ax+2)+2$$

$$g(g(f^{-1}(12)))=a(6a+2)+2=62$$

6a2 + 2a – 60 [= 0]

$$a=-\frac{10}{3}\ or \ 3$$

### Question. The diagram shows the graph of $$y = f(x)$$, where $$f(x) =3/2cos 2x +1/2$$ 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.

(c)  $$y=-\frac{3}{2}\cos 2x-\frac{1}{2}$$