**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.

**▶️Answer/Explanation**

(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 \(x^{4} is divided by x^{2} + 2x − 1\)

**▶️Answer/Explanation**

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 2x^{2} is divided by x + 2.

**▶️Answer/Explanation**

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 2x ^{3} − x^{2} + 6x + 3 is divided by x^{2} + 3. **

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

**▶️Answer/Explanation**

(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.

**▶️Answer/Explanation**

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 8x ^{3} + 4x^{2} + 2x + 7 is divided by 4x^{2} + 1.**

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

**Answer/Explanation**

Ans:

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

Obtain (quotient) 2x + 1

Obtain (remainder) 6

(b)

Obtain terms x^{2}+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.

**Answer/Explanation**

Ans:

(a)Sketch a relevant graph, e.g. y = 4 – x^{2}

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

Obtain final answer 1.60

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

**Answer/Explanation**

**(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 4*x*^{3} + *ax* + 2, where *a* is a constant, is denoted by p(*x)*. It is given that( 2*x*+ 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]** (b)** solve the inequality p(*x*) > 0, justifying your answer. [3]

**Answer/Explanation**

Ans:

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

Obtain *a* = 3

**(ii) (a)** Commence division by (2*x* + 1) reaching a partial quotient of 2*x*^{2} + *kx*

Obtain factorisation (2*x* + 1)(2*x ^{2}* –

*x*+ 2)

[The M1 is earned if inspection reaches an unknown factor 2*x ^{2}* +

*Bx*+

*C*and an equation in

*B*and/or

*C*, or an unknown factor

*Ax*+

^{2}*Bx*+ 2and an equation in A and/or B.]

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

Show that 2*x ^{2}* –

*x*+ 2 is always positive, or that the gradient of 4

*x*+ 3

^{3}*x*+ 2 is always positive

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