### Question

Solve the inequality $$\left | x-5 \right |< \left | 2x+3 \right |$$ .[4]

Ans:

Either         State or imply non-modular inequality $$\left ( x-5^{2} \right )< \left ( 2x+3 \right )^{2}$$ or corresponding pair of linear equations
Attempt solution of 3-term quadratic equation or of 2 linear equations
Obtain critical values −8 and $$\frac{2}{3}$$
State answer $$x< -8, x> \frac{2}{3}$$

Or                Obtain critical value −8 from graphical method, inspection, equation
Obtain critical value $$\frac{2}{3}$$ similarly
State answer  $$x< -8, x> \frac{2}{3}$$

Question

A piece of wire of length 24cm is bent to form the perimeter of a sector of a circle of radius r cm.

(i)Show that the area of the sector ,$$Acm^{2}$$  is given by $$A=12r-r^{2}$$

(ii)Express A in the form $$a-\left ( r-b \right )^{2}$$,where a and b are constants.

(iii)Given that r can vary,state the greatest value of A and find the corresponding angle of the sector.

(i)$$24=r+r+r\Theta$$

$$\rightarrow \Theta =\frac{24-2r}{r}$$

$$A=\frac{1}{2}r^{2}\Theta =\frac{24r}{2}-r^{2}=12r-r^{2}$$

(ii)$$A=36-(r-6)^{2}$$

(iii)Greatest value of A=36

r=6→$$\Theta =2$$

Question

(i)Express in the form $$4x^{2}-12x$$ in the form $$\left ( 2x+a \right )^{2}+b$$

(ii)Hence ,or otherwise,find the set of values of x satisfying $$4x^{2}-12x> 7$$

(i)(\left ( 2x-2 \right )^{2}-9\)

(ii)$$2x-3> 4$$    $$2x-3< – 4$$

$$x> 3\tfrac{1}{2}$$ (or)  $$x< -\frac{1}{2}$$

Allow $$-\frac{1}{2}> x>3\tfrac{1}{2}$$

OR

$$4x^{2}-12x-7\rightarrow \left ( 2x-7 \right )\left ( 2x+1 \right )$$

$$x> 3\tfrac{1}{2}$$  (or) $$< -\frac{1}{2}$$

Allow $$-\frac{1}{2}> x>3\tfrac{1}{2}$$

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

1(i) Express $$x^{2}+6x+2$$ in the form $$\left ( x+a \right )^{2}+b$$,where a and b are constants.

(ii)Henc3,or otherwise, find the set of values of x for which $$x^{2}+6x+2> 9$$.

(i)$$\left ( x+3 \right )^{2}-7$$

(ii)1,-7 seen

$$x> 1,x< -7$$

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

(i) Express $$-x^{2}+6x-5$$ in the form $$a(x+b)^{2}+c$$, where a, b and c are constants.

The function $$f:x\rightarrow -x^{2}+6x-5$$ defined for $$x\geq m$$ ,where m is constant.

(ii) State the smallest value of m for which f is one-one.
(iii) For the case where m = 5, find an expression for $$f^{-1}(x)$$ and state the domain of $$f^{-1}$$.

(i)$$-1(x-3)^{2}+4$$

(ii)Smallest (m)-is 3

(iii)$$(x-3)^{2}=4-y$$

Correct order of operations

$$f^{-1}(x)=3+\sqrt{4-x}$$

Domain is$$x\leq 0$$

.

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)$$(x-1)^{2}-16$$

(ii)-16

(iii)$$9\leq (x-1)^{2}-16\leq 65$$ OR $$x^{2}-2x-15=9\rightarrow 6,-4$$

$$25\leq (x-1)^{2}\leq 81$$ $$x^{2}-2x-15=65\rightarrow 10,-8$$

$$5\leq x-1\leq 9$$             p=6

$$6\leq x\leq 10$$        q=10

(iv)$$x=\left ( y-1 \right )^{2}-16$$      (interchange x/y)

$$y-1=\pm \sqrt{x+16}$$

f^{-1}\left ( x \right )=1+\sqrt{x+16}

Question

A curve for which $$\frac{\mathrm{d} y}{\mathrm{d} x}=7-x^{2}-6x$$   pases tghrough the p[oint (3,-10)

(i)Find the equation of the curve.

(ii)Express $$7-x^{2}-6x$$ in the form $$a-(x+b)^{2}$$,where a and b are constants.

(iii)Find the set of values of x for which the gradient of the curve is positive.

(i)$$y=7x-\frac{x^{3}}{3}-\frac{6x^{2}}{2}+c$$

Uses (3,-10) →c=5

(ii)$$7-x^{2}-6x=16-(x+3)^{2}$$

(iii)$$16-\left ( x+3 \right )^{2}> 0\rightarrow \left ( x+3 \right )^{2}< 16$$,and solve

End-points x=1 or -7

$$\rightarrow -7< x< 1$$

Question

The diagram shows part of the curve$$y = \frac{3}{\sqrt({1+4x})}$$  and a point P (2, 1)  lying on the curve. The normal to the curve at P intersects the x-axis at Q.

(i) Show that the x-coordinate of Q is  $$\frac{16}{9}$$

(ii) Find, showing all necessary working, the area of the shaded region.

Question

The function f is defined by $$f(x)=-2x^{2}+12x-3 for \varepsilon R$$

(i) Express $$-2x^{2}+12x-3 in the from 2(x+a)^{a}+b,$$where a and b are constants.

(ii) State the greatest value of fx.

The function g is defined by g(x) = 2x + 5 for x ∈ >.

(iii) Find the values of x for which gf(x )+ 1

= 0.

(i) $$-2(x-3) ^{2}$$+15(a=-3,b=15)

(ii) (f(x) ⩽) 15

(iii) gf(x) =$$(-2x^{2}+12x-3)+5=-4x^{2}+24x-6+5 gf(x)+1=0\rightarrow -4x^{2}+24x$$=0 x=0 or 6

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

By using a suitable substitute, solve the equation
$$(2x-3)^2-\frac{4}{(2x-3)^2}-3=0$$

Ans:
u = 2x – 3 leading to u4 – 3u2 – 4 [=0]

(u2 – 4)(u2 + 1) [=0]

2x – 3 = [±]2

$$x=\frac{1}{2}, \frac{5}{2}$$ only

### 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

(a)Express 2x2 − 8x + 14 in the form 2[(x – a)2 + b]

The functions f and g are defined by
f(x) = x2           for x ∈ R >,
g(x) = 2x2 − 8x + 14      for x ∈ R>.

(b) Describe fully a sequence of transformations that maps the graph of y = f(x)  onto the graph of   y= g(x), making clear the order in which the transformations are applied.

(a) $$2\left [ \left \{ \left ( x-2 \right )^{2} \right \}\left \{ +3 \right \} \right ]$$
(b){Translation} $$\binom{\left \{ 2 \right \}}{\left \{ 3 \right \}}$$ OR {Stretch} {y direction} {factor 2}
{Stretch} {y direction} {factor 2} OR {Translation}  $$\binom{\left \{ 2 \right \}}{\left \{ 6 \right \}}$$