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



    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  \(x< -8, x> \frac{2}{3}\)


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.



\(\rightarrow \Theta =\frac{24-2r}{r}\)

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


(iii)Greatest value of A=36

r=6→\(\Theta =2\)


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


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

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

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


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


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

Domain is \(x> k-4\) or \(\left [ k-4,\infty \right ]\) 

or \(\left ( k-4,\infty \right )\)


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

\(x> 1,x< -7\)


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


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

\(\left | \vec{OD} \right |=\sqrt{208}\)

\(164=\sqrt{132.25}\times \sqrt{208}\times \cos \Theta\)

\(\Theta =8.6^{\circ}\)


(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 \(f^{-1}\).



(ii)Smallest (m)-is 3


Correct order of operations 


Domain is\( x\leq 0\)



(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 \(f^{-1}(x)\).




(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}


 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.



Uses (3,-10) →c=5


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

End-points x=1 or -7

\(\rightarrow -7< x< 1\)


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.




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


(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).




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


Largest k is 2 Accept \(k\leq 2\)


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

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


By using a suitable substitute, solve the equation


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


(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


(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}  \( \binom{\left \{ 2 \right \}}{\left \{ 6 \right \}}\)