### Question

(i)It is given that 2 ln(4x − 5 )+ ln(x + 1 )3 ln 3.(i) Show that $$16x^{3}-24x^{2}-15x-2=0$$
(ii) By first using the factor theorem, factorise $$16x^{3}-24x^{2}-15x-2=0$$ completely.

(iii) Hence solve the equation 2 ln (4x − 5 )+ ln(x + 1) = 3 ln 3.

(i) Given the equation $$2 \ln(4x – 5) + \ln(x + 1) = 3 \ln 3$$, use the logarithm laws for product and exponentiation:
Apply the power rule for logarithms:
$$2 \ln(4x – 5) = \ln((4x – 5)^2)$$
$$\ln((4x – 5)^2) + \ln(x + 1) = \ln(27)$$
$$\ln((4x – 5)^2(x + 1)) = \ln(27)$$
Simplify using the property $$\ln a = \ln b \Rightarrow a = b$$:
$$(4x – 5)^2(x + 1) = 27$$
$$16x^3 – 40x^2 + 25x + 16x^2 – 40x + 25 = 27$$
$$16x^3 – 24x^2 – 15x – 2 = 0$$
(ii) Factorize $$16 x^3 – 24 x^2 – 15 x – 2 = 0$$
Test $$x = 2$$ and $$x = -\frac{1}{4}$$ as possible roots.
Confirm that $$x = 2$$ is a root (as $$16(2)^3 – 24(2)^2 – 15(2) – 2 = 0$$).
Divide the polynomial by $$(x – 2)$$ using polynomial division:
$\begin{array}{r|r} 16x^3 – 24x^2 – 15x – 2 & x – 2 \\ \hline 16x^2 & 16x^3 – 32x^2 \\ 8x^2 – 15x & 8x^2 – 16x \\ 8x + 1 & 8x^2 – 16x \\ & 8x + 1 \\ & 0 \\ \end{array}$
This division gives us $$16x^2 + 8x + 1$$ as the quotient. So, the polynomial can be written as:
$$(x – 2)(16x^2 + 8x + 1) = 0$$
(iii) Solve $$2 \ln(4x – 5) + \ln(x + 1) = 3 \ln 3$$
$$2 \ln(4x – 5) = \ln((4x – 5)^2)$$
Using the product rule for logarithms, $$\ln a + \ln b = \ln(ab)$$, the equation becomes:
$$\ln((4x – 5)^2) + \ln(x + 1) = \ln((4x – 5)^2(x + 1))$$
Since $$3 \ln 3 = \ln(3^3) = \ln(27)$$, we can set the inside of the logarithm equal to 27:
$$(4x – 5)^2(x + 1) = 27$$
$$16x^3 – 40x^2 + 25x + 16x^2 – 40x + 25 = 27$$
$$16x^3 – 24x^2 – 15x – 2 = 0$$
We now have a cubic equation to solve: $$16x^3 – 24x^2 – 15x – 2 = 0$$.
Solving cubic equations analytically can be complex, but we can start by looking for rational roots using the Rational Root Theorem. Let’s check for simple rational roots such as 1, -1, 2, -2, etc.
After testing, we find that $$x = 2$$ is a root (since $$16(2)^3 – 24(2)^2 – 15(2) – 2 = 0$$). We can then divide the cubic polynomial by $$(x – 2)$$ to find the other roots.
Therefore, the solution to the equation $$2 \ln(4x – 5) + \ln(x + 1) = 3 \ln 3$$ is $$x = 2$$.

### Question

Find the quotient and remainder when $$x^{4} is divided by x^{2} + 2x − 1$$

To find the quotient and remainder when dividing $$x^4$$ by $$x^2 + 2x – 1$$,
$$\begin{array}{r|r} x^4 & x^2 + 2x – 1 \\ \hline x^2 – 2x + 5 & x^4 + 2x^3 – x^2 \\ & -x^4 – 2x^3 + x^2 \\ & 0x^3 \\ & -2x^3 – 4x^2 + 2x \\ & 2x^3 + 4x^2 – 2x \\ & 0x^2 \\ & -4x^2 + 4x \\ & 4x^2 + 8x – 4 \\ & -12x + 5 \\ \end{array}$$
The division process confirms that the quotient is $$x^2 – 2x + 5$$ and the remainder is $$-12x + 5$$. Therefore, $$x^4$$ divided by $$x^2 + 2x – 1$$ yields a quotient of $$x^2 – 2x + 5$$ and a remainder of $$-12x + 5$$.

### Question

Find the quotient and remainder when 2x2 is divided by x + 2.

To find the quotient and remainder when dividing $$2x^2$$ by $$x + 2$$,
$\begin{array}{r|r} 2x^2 & x + 2 \\ \hline 2x & 2x^2 + 4x \\ -4 & -4x – 8 \\ \end{array}$
From this division, we get:
Quotient: $$2x – 4$$
Remainder: $$8$$
Thus, when $$2x^2$$ is divided by $$x + 2$$, the quotient is $$2x – 4$$ and the remainder is $$8$$.

### Question

(a) Find the quotient and remainder when n 2x3 − x2 + 6x + 3 is divided by x2 + 3.

(b) Using your answer to part (a), find the exact value of $$\int_{1}^{3}\frac{2x^{3}-x^{2}+6x+3}{x^{2}+3}dx$$.

(a) Let’s divide $$2x^3 – x^2 + 6x + 3$$ by $$x^2 + 3$$ again:
$\begin{array}{r|r} 2x^3 – x^2 + 6x + 3 & x^2 + 3 \\ \hline 2x & 2x^3 + 6x \\ -1 & -x^2 – 3 \\ \end{array}$
This gives us:
Quotient: $$2x – 1$$
Remainder: $$6$$
(b) Given the quotient $$2x – 1$$ and the remainder $$6$$, the integral becomes:
$$\int_1^3 (2x – 1 + \frac{6}{x^2 + 3}) dx$$
Integral of the Quotient:
$$\int_1^3 (2x – 1) dx = [x^2 – x]_1^3 = (3^2 – 3) – (1^2 – 1) = 6$$
Integral of the Remainder Term:
For the integral $$\int \frac{6}{x^2 + 3} dx$$, use the fact that $$\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$$.
Here, $$a = \sqrt{3}$$, so:
$$\int \frac{6}{x^2 + 3} dx = \frac{6}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C$$
Evaluate this from 1 to 3:
$$\left[\frac{6}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right)\right]_1^3 = \frac{6}{\sqrt{3}} \left(\tan^{-1}\left(\frac{3}{\sqrt{3}}\right) – \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\right)$$
Since $$\tan^{-1}\left(\sqrt{3}\right) = \frac{\pi}{3}$$ and $$\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$$, we get:
$$= \frac{6}{\sqrt{3}} \left(\frac{\pi}{3} – \frac{\pi}{6}\right) = \frac{6}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{\sqrt{3}}$$
The total integral is $$6 + \frac{\pi}{\sqrt{3}} = \frac{1}{\sqrt{3}} \pi + 6$$.
Therefore, the exact value of $$\int_1^3 \frac{2x^3 – x^2 + 6x + 3}{x^2 + 3} dx$$ is $$\frac{1}{\sqrt{3}} \pi + 6$$.

### Question

The polynomial $$ax^3+5x^2-4x+b$$, where a and b are constants, is denoted by p(x). It is given that (x+2) is a factor of p(x) and that when p(x) is divided by (x+1) the remainder is 2.
Find the values of and b.

Let’s solve it again to find the values of $$a$$ and $$b$$ in the polynomial $$p(x) = ax^3 + 5x^2 – 4x + b$$, given that $$x + 2$$ is a factor (thus, $$p(-2) = 0$$) and the remainder when divided by $$x + 1$$ is 2 ($$p(-1) = 2$$).
Substitute $$x = -2$$ into $$p(x)$$ and set it equal to 0:
$$p(-2) = a(-2)^3 + 5(-2)^2 – 4(-2) + b = 0$$
$$-8a + 20 + 8 + b = 0$$
$$-8a + b = -28 \tag{1}$$
Substitute $$x = -1$$ into $$p(x)$$ and set it equal to 2:
$$p(-1) = a(-1)^3 + 5(-1)^2 – 4(-1) + b = 2$$
$$-a + 5 + 4 + b = 2$$
$$-a + b = -7 \tag{2}$$
Solve the System of Equations
We now have two equations:
1. $$-8a + b = -28$$
2. $$-a + b = -7$$
From equation (2), we can express $$b$$ in terms of $$a$$:
$$b = -7 + a$$
Substitute this into equation (1):
$$-8a + (-7 + a) = -28$$
$$-7a = -21$$
$$a = 3$$
Now substitute $$a = 3$$ into $$b = -7 + a$$:
$$b = -7 + 3$$
$$b = -4$$
Therefore, the values of $$a$$ and $$b$$ are $$a = 3$$ and $$b = -4$$.

### Question

(a)Find the quotient and remainder when 8x3 + 4x2 + 2x + 7 is divided by 4x2 + 1.

(b) Hence find the exact value of $$\int_{0}^{\frac{1}{2}} \frac{8x^{3}+4x^{2}+2x+7}{4x^{2}+1}dx.$$

Ans:

(a)Commence division and reach quotient of the form 2x ± 1

Obtain (quotient) 2x + 1

Obtain (remainder) 6

(b)

Obtain terms x2+x

Obtain term of the form a tan-1 2x

Obtain term 3tan-1 2x

Use x = 0 and x = $$\frac{1}{2}$$ as limits in a solution containing a term of the form a tan-1 2x

Obtain final answer $$\frac{3}{4}\left ( 1+\pi \right )$$ , or exact equivalent

### Question

(a) By sketching a suitable pair of graphs, show that the equation 4 − x 2 = sec $$\frac{1}{2}x$$ has exactly one root in the interval 0 ≤ x < π.

(b)Verify by calculation that this root lies between 1 and 2.

(c)Use the iterative formula $$x_{n+1} = \sqrt{4-sec\frac{1}{2}x_{n}}$$ to determine the root correct to 2 decimal places.
Give the result of each iteration to 4 decimal places.

Ans:

(a)Sketch a relevant graph, e.g.  y = 4 – x2

Sketch a second relevant graph, e.g. y = sec $$\frac{1}{2}x$$ , and justify the give statement

(b)Calculate the value of a relevant expression or values of a pair of relevant expressions at x = 1 and x = 2

Complete the argument with correct calculated values

(c)

Use the iterative process correctly at least twice

Show sufficient iterations to 4 d.p.to justify 1.60 to 2 d.p. or show there is a sign change in the interval (1.595, 1.605)

f

### Question

The polynomial p(x) is defined by$$p(x)=x^{3}-3ax+4a$$,where a is a constant.
(i) Given that (x − 2) is a factor of p(x), find the value of a.
(ii) When a has this value,
(a) factorise p(x)completely,
(b) find all the roots of the equation$$p(x^{2})$$=0

(i) Substitute x = 2 and equate to zero, or divide by x – 2 and equate constant remainder to zero, or equivalent
Obtain a = 4

(ii) (a) Find further (quadratic or linear) factor by division, inspection or factor theorem or equivalent

Obtain x^{2}+ 2x – 8 or x + 4

State (x – 2)^{2}

(x + 4) or equivalent

(b) State any two of the four (or six) roots
State all roots $$(\pm \sqrt{2},\pm 2i)$$provided two are purely imaginary

### Question

The polynomial 4x3 + ax + 2, where a is a constant, is denoted by p(x). It is given that( 2x+ 1) is a factor of p(x).

(i) Find the value of a. [2]
(ii) When a has this value,
(a) factorise p(x), [2]

Ans:

4 (i) Substitute $$x=-\frac{1}{2}$$ and equate to zero, or divide by (2x + 1) and equate constant remainder to zero

Obtain a = 3

(ii) (a) Commence division by (2x + 1) reaching a partial quotient of 2x2 + kx

Obtain factorisation (2x + 1)(2x2x + 2)

[The M1 is earned if inspection reaches an unknown factor 2x2 + Bx + C and an equation in B and/or C, or an unknown factor Ax2 + Bx + 2and an equation in A and/or B.]

(b) State or imply critical value $$x=-\frac{1}{2}$$

Show that 2x2x + 2 is always positive, or that the gradient of 4x3 + 3x + 2 is always positive

Justify final answer $$x>-\frac{1}{2}$$

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