### Question

The sequence of values given by the iterative formula $$x_{n+1}=\frac{6+8x_{n}}{8+x^{2}_{n}}$$ with initial value x1= 2 converges
to $$\alpha$$.
(a) Use the iterative formula to find the value of $$\alpha$$ correct to 4 significant figures. Give the result of
each iteration to 6 significant figures.

(b) State an equation satisfied by $$\alpha$$ and hence determine the exact value of $$\alpha$$.

(a) Iteration 1:
$$x_2 = \frac{6 + 8x_1}{8 + x_1^2} = \frac{6 + 8(2)}{8 + 2^2} = \frac{22}{12} = 1.833333$$
Iteration 2:
$$x_3 = \frac{6 + 8x_2}{8 + x_2^2} = \frac{6 + 8(1.833333)}{8 + (1.833333)^2} = \frac{20.666664}{11.361112} = 1.819638$$
Iteration 3:
$$x_4 = \frac{6 + 8x_3}{8 + x_3^2} = \frac{6 + 8(1.819638)}{8 + (1.819638)^2} = \frac{20.557104}{11.275174} = 1.817927$$
Iteration 4: $$x_4 \approx 1.817927$$
Iteration 5: $$x_5 \approx 1.817923$$
Iteration 6: $$x_6 \approx 1.817923$$
Iteration 7: $$x_7 \approx 1.817923$$
After several iterations, the value of $$x_n$$ converges to approximately $$1.817923$$ (rounded to 6 significant figures).
(b) To determine an equation satisfied by α, let’s consider what happens in the limit as $$n$$ approaches infinity:
$$\lim_{{n \to \infty}} x_n = \alpha$$
$$\alpha = \frac{6 + 8\alpha}{8 + \alpha^2}$$
$$\alpha(8 + \alpha^2) = 6 + 8\alpha$$
$$8\alpha + \alpha^3 = 6 + 8\alpha$$
$$\alpha^3 = 6$$
$$\alpha = \sqrt[3]{6}$$

### Question

(a) Solve the equation |2x − 5| = |x + 6|.

(b) Hence find the value of y such that |21−y − 5|=|2−y + 6|. Give your answer correct to 3 significant
figures.

(a) To solve the equation $$|2x – 5| = |x + 6|$$, we can consider two cases: one where the expression inside the absolute value on the left is positive and one where it’s negative.
Case 1: $$2x – 5$$ is positive (i.e., $$2x – 5 > 0$$):
$$2x – 5 = x + 6$$
$$2x – x = 6 + 5$$
$$x = 11$$
Case 2: $$2x – 5$$ is negative (i.e., $$2x – 5 < 0$$):
$$-(2x – 5) = x + 6$$
$$-2x + 5 = x + 6$$
$$5 = 3x + 6$$
$$-1 = 3x$$
$$x = -\frac{1}{3}$$
$$\therefore$$ $$x = 11$$ and $$x = -\frac{1}{3}$$.
(b) $$\left|2^{1-y} – 5\right| = \left|2^{-y} + 6\right|$$.
For $$x = 11$$:
$$\left|2^{1-y} – 5\right| = \left|2^{-y} + 6\right|$$
$$\left|2^{1-y} – 5\right| = \left|2^{-y} + 6\right|$$
Since $$x = 11$$ is a solution for $$|2x – 5| = |x + 6|$$, we can use this value for $$x$$ in our equation. Solving for $$y$$:
$$\left|2^{1-y} – 5\right| = \left|2^{-y} + 6\right|$$
$$\left|2^{1-y} – 5\right| = \left|2^{-y} + 6\right|$$
For $$x = -\frac{1}{3}$$:
$$\left|2^{1-y} – 5\right| = \left|2^{-y} + 6\right|$$
$$\left|2^{1-y} – 5\right| = \left|2^{-y} + 6\right|$$
Since $$x = -\frac{1}{3}$$ is a solution for $$|2x – 5| = |x + 6|$$, we can use this value for $$x$$ in our equation. Solving for $$y$$:
$$\left|2^{1-y} – 5\right| = \left|2^{-y} + 6\right|$$
$$\left|2^{1-y} – 5\right| = \left|2^{-y} + 6\right|$$
For $$x = 11$$:
$$y \approx 0.362$$
For $$x = -\frac{1}{3}$$:
$$y \approx 1.147$$
So, the solutions for $$y$$ are approximately $$0.362$$ and $$1.147$$, correct to 3 significant figures.

### Question

(a) Sketch, on the same diagram, the graphs of y = |x + 2k| and y = |2x – 3k|, where k is a positive constant.
Give, in terms of k, the coordinates of the points where each graph meets the axes.
(b) Find, in terms of k, the coordinates of each of the two points where the graphs intersect.
(c) Find, in terms of k, the largest value of t satisfying the inequality
$$|2^t+2k|\geq|2^{t+1}-3k|$$.

Ans:

(a) Draw two V-shaped graphs with one vertex on negative x-axis and one vertex on positive x-axis
Draw correct graphs related correctly to each other
State correct coordinates -2k, 25, $$\frac{3}{2}k$$. 3k
(b) State or imply non-modulus equation $$(x+2k)^2=(2x-3k)^2$$ or pair of linear equations
Attempt solution of 3-term quadratic equation or pair of linear equations
Obtain $$x=\frac{1}{3}k$$,   x = 5k
Obtain \)y=\frac{7}{3}k\), y = 7k
(c) Relate $$2^t$$ to larger value of x from part (b)
Apply logarithms to obtain $$t=\frac{In(5k)}{In2}$$

### Question

Solve the equation $$sec^2\theta cot\theta =8$$ for $$0<\theta<\pi$$.

Ans:

State $$\frac{1}{cos^2\theta} \times \frac{cos \theta}{sin \theta} =8$$
Attempt use of $$sin 2\theta$$ identity to obtain $$sin 2\theta = k$$
Obtain $$sin 2\theta = \frac{1}{4}$$
Use correct process to find two values of $$\theta$$ between 0 and $$\pi$$.
Obtain 0.126 and 1.44

Alternative method for question 2

State $$\frac{1+tan^2 \theta}{tan \theta} =8$$
Attempt solution of 3-term quadratic equation to find values of $$tan\theta$$
Obtain $$tan \theta = \frac{8± \sqrt{60}}{2}$$
Solve $$tan \theta = ….$$ to find two values of $$\theta$$ between 0 and $$\pi$$.
Obtain 0.126 and 1.44

### Question

(a) Sketch, on the same diagram, the graphs of y=|3x-5| and y=x+2.
(b) Solve the equation |3x-5|=x+2.

Ans:

(a) Draw V-shaped graph with vertex on positive x-axis
Draw correct graph of y=x+2 with smaller positive gradient

(b) Solve 3x-5=x+2 to obtain $$x=\frac{7}{2}$$
Attempt solution of linear equation where signs of 3x and x are different.
Obtain $$x=\frac{3}{4}$$
Alternative method for question 1(b)
State or imply non-modulus equation $$(3x-5)^2=(x+2)^2$$
Attempt solution of 3-term quadratic equation
Obtain $$\frac{3}{4}$$ and $$\frac{7}{2}$$

### Question

(a) Solve the equation |2x − 5| = |x + 6|.                                                                                                                             [3]

(b) Hence find the value of y such that |21−y − 5|=|2−y +6|. Give your answer correct to 3 significant
figures.                                                                                                                                                                                  [2]

Ans

(a) State or imply non-modulus equation   (2x – 5)2 = ( x + 6)2  or pair of
linear equations

Attempt solution of 3-term quadratic equation or of pair of linear
equations

Obtain $$-\frac{1}{3}\ and \ 11$$

(b) Apply logarithms and use power law for 2-y = k  where k > 0
from (a)
Obtain −3.46

### Question

Solve the inequality |3x − 7| < |4x + 5|

Ans

1  State or imply non-modulus inequality  (3x –  7)2 < (4x +  5)2
corresponding equation or pair of linear equations

Attempt solution of 3-term quadratic equation/inequality or of two
linear equations

Obtain critical values 12 − and  $$\frac{2}{7}$$

State answer $$x< -12, x> \frac{2}{7}$$

or

$$(-\infty .-12)\cup \left ( \frac{2}{7},\infty \right )\ or \(-\infty .-12),\left ( \frac{2}{7},\infty \right )$$

### Question

(a)Sketch, on the same diagram, the graphs of y = 3x and y = |x − 3|.

(b)Find the coordinates of the point where the two graphs intersect.

(c)Deduce the solution of the inequality 3x < |x − 3|.

(a)Draw V-shaped graph with vertex on positive x-axis

Draw straight line through origin with positive gradient greater than gradient of first graph, together with a V shaped graph.

(b)Solve linear equation with signs of 3x and x different or solve non-modulus equation  (3x)2 = (x- 3)2

Obtain x = $$\frac{3}{4}$$

Obtain y = $$\frac{9}{4}$$

(c)State x < $$\frac{3}{4}$$

### Question

Solve the equation [5x − 2] = [4x + 9].

Ans:

Solve 5x – 2 = 4x + 9  to obtain x =11

Attempt solution of linear equation where signs of 5x and 4x are different

Obtain final value x = $$-\frac{7}{9}$$

Alternative method for question 1

State or imply non-modulus equation $$\left ( 5x-2 \right )^{2} = \left ( 4x+9 \right )^{2}$$

Attempt solution of 3-term quadratic equation

Obtain x = $$-\frac{7}{9}$$   and x =11

### Question

(i) Solve the inequality |3x − 5|< |x + 3|.

(ii) Hence find the greatest integer n satisfying the inequality$$\left | 3^{0.1n+1}-5 \right |$$  < $$\left | 3^{0.1n+1} +3\right |$$

<p(1)State or imply non-modular inequality $$\left ( 3x-5 \right )^{2}$$  or corresponding equation or pair of different linear equations/inequalities

Attempt solution of 3-term quadratic equation/inequality or of two different linear equations/inequalities.

Obtain critical values $$\frac{1}{2}$$ and 4.

State answer $$\frac{1}{2}< x< 4$$ or equivalent.

(2) Attempt to find n ( not necessarily an integer so far) from $$3^{0.1n}=$$ 0r<their positive upper value from part (1) or $$3^{0.1n+1}=$$ or<3×their positive upper value from part (1)

### Question

(i) Solve the equation| 9x − 2 |= |3x + 2|.

(ii) Hence, using logarithms, solve the equation $$|3^{y+2}-2|=|3^{y+2}-2|$$,giving your answer correct to 3 significant figures

1(i) State or imply non-modular equation$$(9x-2)^{2}=(3x+2)^{2}$$or pair of linear  equations

Attempt solution of quadratic equation or of 2 linear equations

Obtain 0 and $$\frac{2}{3}$$

1(ii) Apply logarithms and use power law for$$3^{y}$$ =k where k >0

Obtain −0.369

### Question

(i) Solve the inequality $$2x − 7 < 2x − 9$$.

(ii) Hence find the largest integer n satisfying the inequality $$2 ln n − 7 < 2 ln n − 9.$$

Attempt to solve quadratic equation in $$e^{x}$$

Obtain$$e^{x}=\frac{1}{3},e^{x}=27$$

Use correct process at least once for solving$$e^{x}$$=

c where c >0

Obtain −ln 3 from a correct solution

Obtain 3 ln3 from a correct solution

### Question

Given that x satisfies the equation| 2x + 3 |= |2x − 1|, find the value of

|4x − 3| −| 6x|.                   [4]

Solve non-modular equation $$(2x+3)^{2}=(2x-1)^{2}$$or linear equation with signs of 2x different

Obtain$$x=-\frac{1}{2}$$

Substitute negative value into expression and show correct evaluation of modulus at least once

Obtain 5-3=2 with no errors seen

### Question

(i) Solve the inequality |2x − 5| < |x + 3 |.

(ii) Hence find the largest integer y satisfying the inequality| 2 ln y − 5 |<| ln y + 3 |.

(i) State or imply non-modulus inequality $$(2x-5)^{2}<(x+3)^{2}$$or
corresponding equation or pair of linear equations Attempt solution of 3-term quadratic inequality or equation or of 2 linear equations

Obtain critical values $$\frac{2}{3}$$ and 8

(ii) Attempt to find y from ln y = upper limit of answer to part (i)

Obtain 2980

### Question

1 (i) Solve the equation $$\left | 3x+4 \right |=\left | 3x+11 \right |$$. [3]

(ii) Hence, using logarithms, solve the equation $$\left | 3×2^{y}+4 \right |=\left | 3×2^{y}-11 \right |$$ , giving the answer correct to 3 significant figures. [2]

Ans:

1 (i) State or imply equation (3x + 4)2 = (3x – 11)2 or 3x + 4 = – (3x – 11)
Attempt solution of ‘quadratic’ equation or linear equation M1
Obtain $$x=\frac{7}{6}$$ or equivalent (and no other solutions) A1 [3]

(ii) Use logarithms to solve equation of form 2y = their answer to (i) ( must be + ve)
Obtain 0.222 (and no other solutions)

Question

Solve the inequality $$\left | x+1 \right |< \left | 3x+5 \right |$$.

Either

State or imply non-modular inequality $$(x+1)^{2}< (3x+5)^{2}$$, or

corresponding equation or pair of linear equations
Make reasonable solution attempt at a 3-term quadratic, or solve
two linear equations
Obtain critical values −2 and $$-\frac{3}{2}$$

State correct answer x < −2 or $$x> -\frac{3}{2}$$.

Or

Obtain one critical value, e.g. x = −2, by solving a linear equation (or inequality)
or from a graphical method or by inspection B1
Obtain the other critical value similarly B2
State correct answer x < −2 or $$x> -\frac{3}{2}$$.

### Question

Solve the equation $$\left | x^{3}-14 \right |=13$$,showing all your working .

We will consider two cases: one where $$x^3 – 14$$ is positive, and one where $$x^3 – 14$$ is negative.
Case 1: $$x^3 – 14$$ is positive (i.e., $$x^3 > 14$$):
In this case, the equation becomes:
$x^3 – 14 = 13$
$x^3 = 13 + 14$
$x^3 = 27$
$x = \sqrt[3]{27}$
$x = 3$
So, one solution is $$x = 3$$.
Case 2: $$x^3 – 14$$ is negative (i.e., $$x^3 < 14$$):
In this case, the equation becomes:
$-(x^3 – 14) = 13$
$-x^3 + 14 = 13$
$-x^3 = 13 – 14$
$-x^3 = -1$
$x^3 = 1$
$x = \sqrt[3]{1}$
$x = 1$
So, the other solution is $$x = 1$$.
Therefore, the solutions to the equation $$|x^3 – 14| = 13$$ are $$x = 3$$ and $$x = 1$$.

### Question

Solve  the inequality $$\left | x-2 \right |\geq \left | x+5 \right |$$.

We can consider different cases based on the sign of the expressions inside the absolute values:
Case 1: When $$x – 2$$ and $$x + 5$$ are both positive (i.e., $$x > 2$$ and $$x > -5$$):
In this case, the inequality becomes:
$$x – 2 \geq x + 5$$
$$-2 \geq 5$$
This inequality is not possible; there are no solutions in this case.
Case 2: When $$x – 2$$ is positive and $$x + 5$$ is negative (i.e., $$x > 2$$ and $$x < -5$$):
In this case, the inequality becomes:
$$x – 2 \geq -(x + 5)$$
$$x – 2 \geq -x – 5$$
$$2x – 2 \geq -5$$
$$2x \geq -5 + 2$$
$$2x \geq -3$$
$$x \geq -\frac{3}{2}$$
So, the solution in this case is $$x \geq -\frac{3}{2}$$.
Case 3: When $$x – 2$$ is negative and $$x + 5$$ is positive (i.e., $$x < 2$$ and $$x > -5$$):
In this case, the inequality becomes:
$$-(x – 2) \geq x + 5$$
$$-x + 2 \geq x + 5$$
$$2 \geq 2x + 5$$
$$-3 \geq 2x$$
$$x \leq -\frac{3}{2}$$
So, the solution in this case is $$x \leq -\frac{3}{2}$$.
Case 4: When $$x – 2$$ is negative and $$x + 5$$ is negative (i.e., $$x < 2$$ and $$x < -5$$):
In this case, the inequality becomes:
$$-(x – 2) \geq -(x + 5)$$
$$-x + 2 \geq -x – 5$$
$$2 \geq -5$$
This inequality is always true.
Now, let’s combine the solutions from all three cases:
1. No solution when $$x > 2$$ and $$x > -5$$.
2. Solution when $$x \geq -\frac{3}{2}$$.
3. Solution when $$x \leq -\frac{3}{2}$$.
4. Always true when $$x < 2$$ and $$x < -5$$.
So, the solution to the inequality $$|x – 2| \geq |x + 5|$$ is:
$$-\frac{3}{2} \leq x \leq 2$$
This means $$x$$ can take any value between $$-\frac{3}{2}$$ and 2, inclusive.

### Question

Solve the equation $$\left | 2^{x} -7\right |=1$$, giving answers correct to 2 decimal places where appropriate.

To solve the equation $$|2^x – 7| = 1$$, we can consider two cases:
Case 1: $$2^x – 7$$ is positive:
In this case, the equation becomes:
$$2^x – 7 = 1$$
$$2^x = 1 + 7$$
$$2^x = 8$$
$$x = \log_2(8)$$
$$x = 3$$
\So, one solution is $$x = 3$$.
Case 2: $$2^x – 7$$ is negative:
$$-(2^x – 7) = 1$$
$$-(2^x – 7) = 1$$
$$-2^x + 7 = 1$$
$$-2^x = -6$$
$$2^x = 6$$
$$x = \log_2(6)$$
$$x \approx 2.585$$
So, the other solution is $$x \approx 2.59$$ (rounded to 2 decimal places).
Therefore, the solutions to the equation $$|2^x – 7| = 1$$ are $$x = 3$$ and $$x \approx 2.59$$ (rounded to 2 decimal places).

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