FPM 2023 November01

Question 1

Figure 1 shows the triangle $ABC$
$\angle ABC = 90^\circ$ $AB = (2 + 4\sqrt{5}) \mathrm{cm}$ $BC = (a + b\sqrt{5}) \mathrm{cm}$ where $a$ and $b$ are integers.
The area of triangle $ABC = (34 + 11\sqrt{5}) \mathrm{cm}^2$

Without using a calculator, find the value of $a$ and the value of $b$ (4)

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Solution

We are given a triangle $\triangle ABC$ with the following properties:

$\angle ABC = 90^\circ$ (right triangle).
$AB = (2 + 4\sqrt{5}) \ \mathrm{cm}$.
$BC = (a + b\sqrt{5}) \ \mathrm{cm}$, where $a$ and $b$ are integers.
The area of the triangle $\triangle ABC = (34 + 11\sqrt{5}) \ \mathrm{cm}^2$.

We aim to find the values of $a$ and $b$. Using the formula for the area of a right triangle:

$$ \text{Area} = \frac{1}{2} \times AB \times BC, $$

we substitute the known values:

$$ (34 + 11\sqrt{5}) = \frac{1}{2} \times (2 + 4\sqrt{5}) \times (a + b\sqrt{5}). $$

Step 1: Simplify the Equation

Multiply through by $2$ to eliminate the fraction:

$$ 68 + 22\sqrt{5} = (2 + 4\sqrt{5})(a + b\sqrt{5}). $$

Expand the right-hand side:

$$ (2 + 4\sqrt{5})(a + b\sqrt{5}) = 2a + 2b\sqrt{5} + 4a\sqrt{5} + 20b. $$

Combine like terms:

$$ = (2a + 20b) + (2b + 4a)\sqrt{5}. $$

Equating the coefficients of $1$ and $\sqrt{5}$ on both sides:

$$ 2a + 20b = 68, $$ $$ 2b + 4a = 22. $$

Step 2: Solve the System of Equations

From the first equation:

$$ 2a + 20b = 68 \implies a + 10b = 34. \tag{1} $$

From the second equation:

$$ 2b + 4a = 22 \implies b + 2a = 11. \tag{2} $$

Substitute $a$ from equation (1) into equation (2):

$$ a = 34 - 10b. \tag{3} $$

Substitute equation (3) into equation (2):

$$ b + 2(34 - 10b) = 11. $$

Expand and simplify:

$$ b + 68 - 20b = 11, $$ $$ -19b + 68 = 11 \implies -19b = -57 \implies b = 3. $$

Substitute $b = 3$ into equation (3):

$$ a = 34 - 10(3) = 34 - 30 = 4. $$

Step 3: Final Answer

The values of $a$ and $b$ are:

$$ a = 4, \quad b = 3. $$

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Question 2

$$ g(x)=2x^2+\frac{1}{2}x-3 $$

(a) Express $g(x)$ in the form $p(x+q)^2+r$ where $p$, $q$ and $r$ are rational numbers to be found. (3)

(b) Find
(i) the minimum value of $g(x)$
(ii) the value of $x$ at which this minimum occurs. (2)

$$ h(x)=2x^6+\frac{1}{2}x^3-3 $$

(c) Hence, or otherwise, write down
(i) the minimum value of $h(x)$
(ii) the value of $x$ at which this minimum occurs. (2)

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Solution:

(a) Express $g(x)$ in the form $p(x+q)^2+r$

Given:

$$ g(x)=2x^2+\frac{1}{2}x-3 $$

We complete the square:

$$ \begin{aligned} g(x) &= 2\left(x^2+\frac{1}{4}x\right)-3 \\ &= 2\left( x^2+\frac{1}{4}x+\frac{1}{64}-\frac{1}{64} \right)-3 \\ &= 2\left( \left(x+\frac{1}{8}\right)^2-\frac{1}{64} \right)-3 \\ &= 2\left(x+\frac{1}{8}\right)^2-\frac{2}{64}-3 \\ &= 2\left(x+\frac{1}{8}\right)^2-\frac{1}{32}-3 \\ &= 2\left(x+\frac{1}{8}\right)^2-\frac{97}{32}. \end{aligned} $$

Thus, $g(x)$ in the form $p(x+q)^2+r$ is:

$$ g(x)=2\left(x+\frac{1}{8}\right)^2-\frac{97}{32}, $$

where $p=2$, $q=\frac{1}{8}$, and $r=-\frac{97}{32}$.

(b) Find the minimum value of $g(x)$ and the value of $x$ at which it occurs

(i) The minimum value of $g(x)$ occurs when the square term is zero:

$$ \left(x+\frac{1}{8}\right)^2=0 \implies x=-\frac{1}{8}. $$

Substituting $x=-\frac{1}{8}$ into $g(x)$:

$$ g\left(-\frac{1}{8}\right) = 2(0)^2-\frac{97}{32} = -\frac{97}{32}. $$

Thus, the minimum value of $g(x)$ is $-\frac{97}{32}$, and it occurs at $x=-\frac{1}{8}$.

(c) Determine the minimum value of $h(x)$ and the value of $x$ at which it occurs

Given:

$$ h(x)=2x^6+\frac{1}{2}x^3-3. $$

Notice that $h(x)$ is related to $g(x)$:

$$ h(x)=g(x^3), $$

where

$$ g(x)=2x^2+\frac{1}{2}x-3. $$

From part (b), the minimum value of $g(x)$ is $-\frac{97}{32}$, and it occurs at $x=-\frac{1}{8}$. Replacing $x$ in $g(x)$ with $x^3$ in $h(x)$, the minimum value of $h(x)$ occurs when

$$ x^3=-\frac{1}{8}. $$

Solving for $x$:

$$ x=\sqrt[3]{-\frac{1}{8}}=-\frac{1}{2}. $$

(i) The minimum value of $h(x)$ is:

$$ h(x)=-\frac{97}{32}. $$

(ii) The value of $x$ at which this minimum occurs is:

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

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Question 3

$$ g'(x)=mx^2-10x-37 \text{ where } m \text{ is an integer} $$

The curve $y=g(x)$ passes through the point with coordinates $(1,20)$
Given that $(x-5)$ is a factor of $g(x)$

(a) show that $g(x)=2x^3-5x^2-37x+60$ (5)

(b) Hence, or otherwise, use algebra to solve the equation $g(x)=0$ (3)

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Solution

We are given:

$$ g'(x)=mx^2-10x-37 \quad \text{where } m \text{ is an integer.} $$

The curve $y=g(x)$ passes through the point $(1,20)$, and $(x-5)$ is a factor of $g(x)$.

(a) Show that $g(x)=2x^3-5x^2-37x+60$.

Step 1: Integrate $g'(x)$

Integrating $g'(x)$ with respect to $x$ gives:

$$ g(x) = \int (mx^2-10x-37)\,dx = \frac{m}{3}x^3-5x^2-37x+C, $$

where $C$ is the constant of integration.

Step 2: Use the given point

Since $g(x)$ passes through $(1,20)$, substitute $x=1$ and $g(1)=20$:

$$ 20 = \frac{m}{3}(1)^3 - 5(1)^2 - 37(1) + C. $$

Simplify:

$$ 20 = \frac{m}{3} - 5 - 37 + C. $$ $$ 20 = \frac{m}{3} - 42 + C. $$ $$ C = 62-\frac{m}{3}. \tag{1} $$

Step 3: Use the factor theorem

Since $(x-5)$ is a factor of $g(x)$, substituting $x=5$ into $g(x)$ must yield $0$:

$$ g(5) = \frac{m}{3}(5)^3 - 5(5)^2 - 37(5) + C = 0. $$

Simplify:

$$ \frac{m}{3}(125)-5(25)-37(5)+C=0. $$ $$ \frac{125m}{3}-125-185+C=0. $$ $$ \frac{125m}{3}-310+C=0. \tag{2} $$

Step 4: Solve for $m$

Substitute $C=62-\frac{m}{3}$ from equation (1) into equation (2):

$$ \frac{125m}{3}-310+62-\frac{m}{3}=0. $$

Simplify:

$$ \frac{124m}{3}-248=0. $$ $$ \frac{124m}{3}=248. $$ $$ m=6. $$

Step 5: Substitute $m$ back into $g(x)$

Substituting $m=6$ into $g(x)$:

$$ g(x) = \frac{6}{3}x^3-5x^2-37x+C. $$ $$ g(x)=2x^3-5x^2-37x+C. $$

Using $C=62-\frac{6}{3}=60$:

$$ g(x)=2x^3-5x^2-37x+60. $$

Thus, we have shown that:

$$ g(x)=2x^3-5x^2-37x+60. $$

(b) Solve the equation $g(x)=0$

We need to solve:

$$ 2x^3-5x^2-37x+60=0. $$

Step 1: Use the factor theorem

Since $(x-5)$ is a factor, divide $g(x)$ by $(x-5)$.

Performing synthetic division: \[\begin{array}{r|rrrr} 5& 2&-5&-37&60\\ & &10&25&-60\\ \hline &2& 5&-12&0 \end{array}\]

Thus:

$$ g(x)=(x-5)(2x^2+5x-12). $$

Step 2: Solve the quadratic

Solve $2x^2+5x-12=0$ using the quadratic formula:

$$ x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \quad \text{where } a=2,\; b=5,\; c=-12. $$ $$ x = \frac{-5\pm\sqrt{5^2-4(2)(-12)}}{2(2)}. $$ $$ x = \frac{-5\pm\sqrt{25+96}}{4}. $$ $$ x = \frac{-5\pm\sqrt{121}}{4}. $$ $$ x = \frac{-5\pm11}{4}. $$ $$ x=\frac{6}{4}=\frac{3}{2}, \quad x=\frac{-16}{4}=-4. $$

Final Solution

The solutions to $g(x)=0$ are:

$$ x=5, \quad x=\frac{3}{2}, \quad x=-4. $$

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Question 4

The point $A$ with coordinates $(12,14)$ and the point $B$ with coordinates $(q,2)$ where $q$ is a constant, lie on the straight line with equation $3y-2x-p=0$ where $p$ is a constant.

(a) Find the value of $p$ and the value of $q$ (3)

The line $L$ is perpendicular to $AB$ and passes through the point $X$, which lies on $AB$ such that $AX:XB=1:2$

(b) Find an equation for $L$ in the form $ax+by+c=0$ where $a$, $b$ and $c$ are integers to be found. (6)

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Solution:

Part (a): Find the value of $p$ and the value of $q$

We are given the points $A(12,14)$ and $B(q,2)$, and the line equation

$$ 3y-2x-p=0, $$

where $p$ is a constant. Both points lie on this line, so we substitute their coordinates into the equation.

Step 1: Substitute the coordinates of point $A(12,14)$ into the equation.

$$ 3(14)-2(12)-p=0 $$ $$ 42-24-p=0 $$ $$ 18=p $$

Thus, the value of $p$ is $18$.

Step 2: Substitute the coordinates of point $B(q,2)$ into the equation.

$$ 3(2)-2(q)-p=0 $$

Substitute $p=18$:

$$ 6-2q-18=0 $$ $$ -2q-12=0 $$ $$ -2q=12 $$ $$ q=-6 $$

Thus, the value of $q$ is $-6$.

Part (b): Find an equation for the line $L$

The line $L$ is perpendicular to the line $AB$ and passes through a point $X$ on $AB$, such that the ratio

$$ AX:XB=1:2. $$

We need to find the equation of line $L$.

Step 1: Find the coordinates of point $X$ using the section formula.

The coordinates of point $X$ dividing the line segment $AB$ in the ratio $m:n$ are:

$$ x_X=\frac{mx_B+nx_A}{m+n}, \qquad y_X=\frac{my_B+ny_A}{m+n} $$

Here,

$$ A(12,14), \qquad B(-6,2), \qquad m:n=1:2. $$

Substituting the values:

$$ x_X = \frac{1(-6)+2(12)}{1+2} = \frac{-6+24}{3} = \frac{18}{3} = 6 $$ $$ y_X = \frac{1(2)+2(14)}{1+2} = \frac{2+28}{3} = \frac{30}{3} = 10 $$

Thus, the coordinates of point $X$ are:

$$ (6,10). $$

Step 2: Find the slope of line $AB$.

The slope of line $AB$ is:

$$ m_{AB} = \frac{y_B-y_A}{x_B-x_A} = \frac{2-14}{-6-12} = \frac{-12}{-18} = \frac{2}{3} $$

Step 3: Find the slope of line $L$.

Since line $L$ is perpendicular to $AB$, its slope is the negative reciprocal of $\frac{2}{3}$:

$$ m_L=-\frac{3}{2} $$

Step 4: Use the point-slope form of the line equation.

The point-slope form is:

$$ y-y_1=m(x-x_1) $$

Substituting $m_L=-\frac{3}{2}$ and the point $(6,10)$:

$$ y-10=-\frac{3}{2}(x-6) $$

Step 5: Simplify the equation.

$$ y-10=-\frac{3}{2}x+9 $$ $$ y=-\frac{3}{2}x+19 $$

Multiply through by $2$:

$$ 2y=-3x+38 $$

Rearranging into the form $ax+by+c=0$:

$$ 3x+2y-38=0 $$

Thus, the equation of line $L$ is:

$$ \boxed{3x+2y-38=0} $$

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Question 5

Figure 2 shows the graph of part of the curve $C$ with equation $y=\sqrt{2x+6}$. The finite region enclosed by the curve $C$ and the straight line with equation $3y-x=3$ is rotated through $360^\circ$ about the $x$-axis.

Use algebraic integration to find the exact volume of the solid generated.
Give your answer in terms of $\pi$ (8)

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Solution:

We are given the curve $C$ with equation $y = \sqrt{2x + 6}$ and the straight line with equation $3y - x = 3$. The finite region enclosed by these two curves is rotated through $360^\circ$ about the $x$-axis. We are tasked with finding the exact volume of the solid generated by this rotation using algebraic integration.

Step 1: Express the equation of the line in terms of $y$

The given straight line equation is:

$$ 3y - x = 3 $$

Solving for $y$:

$$ 3y = x + 3 $$ $$ y = \frac{x + 3}{3} $$

Thus, the equation of the line is $y = \frac{x + 3}{3}$.

Step 2: Find the points of intersection between the curve and the line

To find the points of intersection, we set the expressions for $y$ from the curve and the line equal to each other:

$$ \sqrt{2x + 6} = \frac{x + 3}{3} $$

Now, square both sides to eliminate the square root:

$$ 2x + 6 = \left( \frac{x + 3}{3} \right)^2 $$ $$ 2x + 6 = \frac{(x + 3)^2}{9} $$

Multiply both sides by 9 to clear the denominator:

$$ 9(2x + 6) = (x + 3)^2 $$ $$ 18x + 54 = x^2 + 6x + 9 $$

Rearrange the equation:

$$ x^2 + 6x + 9 - 18x - 54 = 0 $$ $$ x^2 - 12x - 45 = 0 $$

Now, solve this quadratic equation using the quadratic formula:

$$ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(-45)}}{2(1)} $$ $$ x = \frac{12 \pm \sqrt{144 + 180}}{2} $$ $$ x = \frac{12 \pm \sqrt{324}}{2} $$ $$ x = \frac{12 \pm 18}{2} $$

Thus, the solutions for $x$ are:

$$ x = \frac{12 + 18}{2} = 15 \quad \text{and} \quad x = \frac{12 - 18}{2} = -3 $$

So, the points of intersection occur at $x = -3$ and $x = 15$.

Step 3: Set up the volume integral

The volume of the solid generated by rotating a region around the $x$-axis is given by the formula:

$$ V = \pi \int_{a}^{b} \left( f(x)^2 - g(x)^2 \right) \, dx $$

where $f(x)$ and $g(x)$ represent the functions describing the outer and inner radii of the solid, respectively. In this case:

$f(x) = \sqrt{2x + 6}$ is the curve (outer radius),
$g(x) = \frac{x + 3}{3}$ is the line (inner radius).

Thus, the volume integral becomes:

$$ V = \pi \int_{-3}^{15} \left( \left( \sqrt{2x + 6} \right)^2 - \left( \frac{x + 3}{3} \right)^2 \right) \, dx $$ $$ V = \pi \int_{-3}^{15} \left( (2x + 6) - \frac{(x + 3)^2}{9} \right) \, dx $$

Step 4: Simplify the integrand

First, expand $\frac{(x + 3)^2}{9}$:

$$ \frac{(x + 3)^2}{9} = \frac{x^2 + 6x + 9}{9} $$

So, the integrand becomes:

$$ 2x + 6 - \frac{x^2 + 6x + 9}{9} $$

To simplify, rewrite the expression with a common denominator:

$$ 2x + 6 = \frac{18x + 54}{9} $$

Thus, the integrand is:

$$ \frac{18x + 54}{9} - \frac{x^2 + 6x + 9}{9} = \frac{18x + 54 - x^2 - 6x - 9}{9} $$ $$ = \frac{-x^2 + 12x + 45}{9} $$

The volume integral is now:

$$ V = \pi \int_{-3}^{15} \frac{-x^2 + 12x + 45}{9} \, dx $$

Step 5: Integrate the expression

We can factor out $\frac{1}{9}$ from the integral:

$$ V = \frac{\pi}{9} \int_{-3}^{15} (-x^2 + 12x + 45) \, dx $$

Now, integrate term by term:

$$ \int -x^2 \, dx = -\frac{x^3}{3}, \quad \int 12x \, dx = 6x^2, \quad \int 45 \, dx = 45x $$

Thus, the integral becomes:

$$ V = \frac{\pi}{9} \left[ -\frac{x^3}{3} + 6x^2 + 45x \right]_{-3}^{15} $$

Step 6: Evaluate the definite integral

First, evaluate the expression at $x = 15$:

$$ -\frac{15^3}{3} + 6(15^2) + 45(15) = -\frac{3375}{3} + 6(225) + 675 = -1125 + 1350 + 675 = 900 $$

Next, evaluate the expression at $x = -3$:

$$ -\frac{(-3)^3}{3} + 6(-3)^2 + 45(-3) = -\frac{-27}{3} + 6(9) - 135 = 9 + 54 - 135 = -72 $$

Thus, the definite integral is:

$$ 900 - (-72) = 900 + 72 = 972 $$

Step 7: Final volume calculation

Now, multiply by $\frac{\pi}{9}$:

$$ V = \frac{\pi}{9} \times 972 = 108\pi $$

Thus, the exact volume of the solid generated is:

$$ \boxed{108\pi} $$

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Question 6

Figure 3 shows a right pyramid with a horizontal square base.

$$ \begin{aligned} &AB = BC = CD = DA = x \mathrm{~cm} \\ &AV = BV = CV = DV = x \mathrm{~cm} \end{aligned} $$

$O$ is the point of intersection of the diagonals of the base.
The vertex $V$ of the pyramid is vertically above $O$

(a) Show that $VO=\frac{\sqrt{2}}{2}x \mathrm{~cm}$ (3)

(b) Find, in degrees, the size of the angle $AVC$ (2)

The volume of the pyramid is $200 \mathrm{~cm}^3$
Given that the volume of a pyramid $=\frac{1}{3}\times$ base area $\times$ height

(d) Find to 3 significant figures, the value of $x$ (3)

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Solution

Figure 3 shows a right pyramid with a horizontal square base. The following are given:

$$ AB = BC = CD = DA = x \text{ cm}, $$ $$ AV = BV = CV = DV = x \text{ cm}. $$

$O$ is the point of intersection of the diagonals of the square base, and the vertex $V$ is vertically above $O$.

(a) Show that $VO = \frac{\sqrt{2}}{2}x$

The diagonals of the square base intersect at $O$, dividing each diagonal into two equal parts. Let $AC$ and $BD$ be the diagonals.

The length of a diagonal of the square can be found using the Pythagorean theorem:

$$ AC = \sqrt{x^2 + x^2} = \sqrt{2x^2} = x\sqrt{2}. $$

Since $O$ is the midpoint of $AC$, we have:

$$ AO = \frac{AC}{2} = \frac{x\sqrt{2}}{2}. $$

The vertex $V$ is vertically above $O$, so $VO$ forms the height of the right triangle $AVO$, where $AV = x$ and $AO = \frac{x\sqrt{2}}{2}$. Using the Pythagorean theorem:

$$ AV^2 = AO^2 + VO^2 $$ $$ x^2 = \left(\frac{x\sqrt{2}}{2}\right)^2 + VO^2 $$ $$ x^2 = \frac{2x^2}{4} + VO^2 $$ $$ VO^2 = x^2 - \frac{x^2}{2} = \frac{x^2}{2} $$ $$ VO = \sqrt{\frac{x^2}{2}} = \frac{x\sqrt{2}}{2}. $$

Thus, $VO = \frac{\sqrt{2}}{2}x$ cm.

(b) Find the size of $\angle AVC$

Consider the triangle $AVC$. The sides of the triangle are:

$$ AV = x, \quad VC = x, \quad AC = x\sqrt{2}. $$

Using the cosine rule:

$$ \cos \angle AVC = \frac{AV^2 + VC^2 - AC^2}{2 \cdot AV \cdot VC} $$ $$ \cos \angle AVC = \frac{x^2 + x^2 - (x\sqrt{2})^2}{2 \cdot x \cdot x} $$ $$ \cos \angle AVC = \frac{x^2 + x^2 - 2x^2}{2x^2} $$ $$ \cos \angle AVC = \frac{0}{2x^2} = 0. $$

Since $\cos \angle AVC = 0$, $\angle AVC = 90^\circ$.

(c) Find the angle between the planes $VAB$ and $VDC$

Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. The line $MN$ is parallel to the square base and lies in the plane $ABCD$. The angle between the planes $VAB$ and $VDC$ is equal to the angle $\angle MVN$.

From the geometry of the pyramid:

$VM = VN = \sqrt{VO^2 + OM^2}$, where $OM = \frac{AB}{2} = \frac{x}{2}$.
$VO = \frac{x\sqrt{2}}{2}$ (from part (a)).

Using the Pythagorean theorem in $\triangle VOM$:

$$ VM = \sqrt{VO^2 + OM^2} = \sqrt{\left(\frac{x\sqrt{2}}{2}\right)^2 + \left(\frac{x}{2}\right)^2} $$ $$ VM = \sqrt{\frac{x^2}{2} + \frac{x^2}{4}} = \sqrt{\frac{2x^2 + x^2}{4}} = \sqrt{\frac{3x^2}{4}} = \frac{x\sqrt{3}}{2}. $$

The length $MN = AB = x$. The angle $\angle MVN$ can be found using the cosine rule:

$$ \cos \angle MVN = \frac{VM^2 + VN^2 - MN^2}{2 \cdot VM \cdot VN} $$ $$ \cos \angle MVN = \frac{\left(\frac{x\sqrt{3}}{2}\right)^2 + \left(\frac{x\sqrt{3}}{2}\right)^2 - x^2}{2 \cdot \frac{x\sqrt{3}}{2} \cdot \frac{x\sqrt{3}}{2}} $$ $$ \cos \angle MVN = \frac{\frac{3x^2}{4} + \frac{3x^2}{4} - x^2}{\frac{3x^2}{2}} $$ $$ \cos \angle MVN = \frac{\frac{6x^2}{4} - x^2}{\frac{3x^2}{2}} = \frac{\frac{2x^2}{4}}{\frac{3x^2}{2}} = \frac{1}{3}. $$

Thus:

$$ \angle MVN = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.5^\circ. $$

(d) Find the value of $x$

The volume of the pyramid is given as $200$ cm$^3$. The formula for the volume of a pyramid is:

$$ \text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height}. $$

The base area is the area of the square:

$$ \text{Base Area} = x^2. $$

The height is $VO = \frac{x\sqrt{2}}{2}$ (from part (a)). Substituting these values:

$$ 200 = \frac{1}{3} \times x^2 \times \frac{x\sqrt{2}}{2} $$ $$ 200 = \frac{x^3\sqrt{2}}{6} $$ $$ x^3 = \frac{200 \cdot 6}{\sqrt{2}} = \frac{1200}{\sqrt{2}} = 600\sqrt{2} $$ $$ x = \sqrt[3]{600\sqrt{2}} = 9.467 \approx 9.47 \text{ cm} \, \text{(to 3 significant figures)}. $$

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Question 7

A geometric series $G$ with common ratio $r$, has first term $16$ and third term $\frac{2704}{625}$

(a) Find the two possible values of $r$ (2)

Given that $r \gt 0$

(b) find the sum to infinity of $G$ (2)

The sum to $n$ terms of $G$ is greater than $33$

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Solution

A geometric series $G$ with common ratio $r$, has first term 16 and third term $\frac{2704}{625}$.

Given:

First term $a = 16$
Third term $ar^2 = \frac{2704}{625}$

(a) Find the two possible values of $r$

From the formula for the $n$-th term of a geometric series:

$$ T_n = ar^{n-1} $$

For the third term, $T_3$:

$$ ar^2 = \frac{2704}{625} $$ $$ 16r^2 = \frac{2704}{625} $$ $$ r^2 = \frac{2704}{625 \times 16} $$ $$ r^2 = \frac{169}{625} $$ $$ r = \pm \frac{13}{25} $$

Thus, the two possible values of $r$ are:

$$ r = \frac{13}{25} \quad \text{or} \quad r = -\frac{13}{25} $$

(b) Find the sum to infinity of $G$ (Given that $r > 0$)

The sum to infinity of a geometric series is given by:

$$ S_\infty = \frac{a}{1-r}, \quad \text{for } |r| \lt 1 $$

Substituting $a = 16$ and $r = \frac{13}{25}$:

$$ S_\infty = \frac{16}{1 - \frac{13}{25}} $$ $$ = \frac{16}{\frac{25 - 13}{25}} $$ $$ = \frac{16 \times 25}{12} $$ $$ = \frac{400}{12} $$ $$ = \frac{100}{3} $$

Thus, the sum to infinity is:

$$ S_\infty = \frac{100}{3} $$

(c) Find the least possible value of $n$ (using logarithms)

The sum to $n$ terms of a geometric series is given by:

$$ S_n = \frac{a(1 - r^n)}{1 - r}, \quad \text{for } |r| \lt 1 $$

We are given that $S_n > 33$, and $r = \frac{13}{25}$. Substituting into the formula:

$$ \frac{16(1 - r^n)}{1 - \frac{13}{25}} > 33 $$ $$ \frac{16(1 - r^n)}{\frac{12}{25}} > 33 $$ $$ 16(1 - r^n) > 33 \times \frac{12}{25} $$ $$ 16(1 - r^n) > \frac{396}{25} $$ $$ 1 - r^n > \frac{396}{25 \times 16} $$ $$ 1 - r^n > \frac{99}{100} $$ $$ r^n \lt \frac{1}{100} $$

Taking logarithms (base 10) on both sides:

$$ \log_{10}(r^n) \lt \log_{10}\left(\frac{1}{100}\right) $$ $$ n \log_{10}\left(\frac{13}{25}\right) \lt -2 $$ $$ n > \frac{-2}{\log_{10}\left(\frac{13}{25}\right)} $$

Using $\log_{10}\left(\frac{13}{25}\right) = \log_{10}(13) - \log_{10}(25)$:

$$ \log_{10}(13) \approx 1.11394, \quad \log_{10}(25) \approx 1.39794 $$ $$ \log_{10}\left(\frac{13}{25}\right) \approx 1.11394 - 1.39794 = -0.284 $$ $$ n > \frac{-2}{-0.284} $$ $$ n > 7.04 $$

Thus, the least possible value of $n$ is:

$$ n = 8 $$

===========================

Question 8

$$ y=\frac{2e^{3x+1}}{5x^2} $$

(a) Find $\frac{dy}{dx}$

Give your answer in the form $\frac{Ae^{3x+1}(Bx-A)}{Cx^3}$ where $A$, $B$ and $C$ are prime numbers to be found. (5)

The value of $x$ increases by $2\%$

(b) Use your answer to part (a) to find an estimate, in terms of $x$, for the percentage change in $y$

Give your answer in the form $(Px-Q)$ where $P$ and $Q$ are integers. (3)

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Solution

We are given:

$$ y = \frac{2 e^{3x+1}}{5x^2} $$

(a) Find $ \frac{\mathrm{d}y}{\mathrm{d}x} $

The function $ y $ is a product of $ 2e^{3x+1} $ and $ \frac{1}{5x^2} $. To differentiate, we use the product rule:

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}\left(2e^{3x+1}\right) \cdot \frac{1}{5x^2} + 2e^{3x+1} \cdot \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{5x^2}\right). $$

Step 1: Differentiate $ 2e^{3x+1} $:

$$ \frac{\mathrm{d}}{\mathrm{d}x}\left(2e^{3x+1}\right) = 2 \cdot e^{3x+1} \cdot \frac{\mathrm{d}}{\mathrm{d}x}(3x+1) = 6e^{3x+1}. $$

Step 2: Differentiate $ \frac{1}{5x^2} $:

$$ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{5x^2}\right) = \frac{1}{5} \cdot \frac{\mathrm{d}}{\mathrm{d}x}(x^{-2}) = \frac{1}{5} \cdot (-2x^{-3}) = -\frac{2}{5x^3}. $$

Step 3: Substitute into the product rule:

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \left(6e^{3x+1}\right) \cdot \frac{1}{5x^2} + \left(2e^{3x+1}\right) \cdot \frac{-2}{5x^3}. $$

Simplify each term:

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6e^{3x+1}}{5x^2} - \frac{4e^{3x+1}}{5x^3}. $$

Combine terms over a common denominator:

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6e^{3x+1}x - 4e^{3x+1}}{5x^3}. $$

Factorize the numerator:

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2e^{3x+1}(3x-2)}{5x^3}. $$

Thus, the answer is:

$$ \boxed{\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2e^{3x+1}(3x-2)}{5x^3}}. $$

Here, $ A = 2 $, $ B = 3 $, and $ C = 5 $ are the prime numbers.

(b) Estimate the percentage change in $ y $

We know that the percentage change in $ y $ can be approximated by:

$$ \text{Percentage change in } y \approx \frac{\mathrm{d}y}{\mathrm{d}x} \cdot \frac{\Delta x}{y} \times 100\%, $$

where $ \Delta x $ is the percentage change in $ x $ (in decimal form). For a $ 2\% $ increase, $ \Delta x = 0.02x $.

Substitute $ \frac{\mathrm{d}y}{\mathrm{d}x} $:

$$ \text{Percentage change in } y \approx \frac{2e^{3x+1}(3x-2)}{5x^3} \cdot \frac{0.02x}{\frac{2e^{3x+1}}{5x^2}} \times 100\%. $$

Simplify the expression:

$$ \text{Percentage change in } y \approx \frac{2e^{3x+1}(3x-2)}{5x^3} \cdot \frac{0.02x \cdot 5x^2}{2e^{3x+1}} \times 100\%. $$ $$ \text{Percentage change in } y \approx \frac{(3x-2) \cdot 0.02 \cdot 5x^3}{5x^3} \times 100\%. $$ $$ \text{Percentage change in } y \approx (3x-2) \cdot 0.02 \cdot \frac{x^3}{x^3} \times 100\%. $$ $$ \text{Percentage change in } y \approx (3x-2) \cdot 0.02 \times 100\%. $$ $$ \text{Percentage change in } y \approx 2(3x-2)\%. $$

Expand and simplify:

$$ \text{Percentage change in } y \approx 6x - 4\%. $$

Thus, the answer is:

$$ \boxed{6x - 4\%} $$

where $ P = 6 $ and $ Q = 4 $.

=================================

Question 9

(a) Expand $\left(1-8x^2\right)^{-\frac{1}{2}}$ in ascending powers of $x$, up to and including the term in $x^6$ giving each coefficient as an integer. (3)

$$ g(x)=\frac{a+bx}{\sqrt{1-8x^2}} \text{ where } a \text{ and } b \text{ are prime numbers} $$

Given that the fourth and fifth terms, in ascending powers of $x$, in the series expansion of $g(x)$ are $20x^3$ and $48x^4$ respectively,

(b) find the value of $a$ and the value of $b$ (4)

Using the first five terms, in ascending powers of $x$, in the series expansion of $g(x)$

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Solution

Part (a)

We aim to expand $\left(1-8x^2\right)^{-\frac{1}{2}}$ in ascending powers of $x$ up to and including the term in $x^6$. Using the binomial expansion for $(1 + u)^n$:

$$ (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots, $$

where $n = -\frac{1}{2}$ and $u = -8x^2$, we proceed as follows:

Step 1: Expand the series

Substitute $n = -\frac{1}{2}$ and $u = -8x^2$ into the formula:

$$ \left(1 - 8x^2\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-8x^2) + \frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2!}(-8x^2)^2 $$ $$ \quad + \frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3!}(-8x^2)^3 + \cdots. $$

Step 2: Simplify each term

Zeroth term: $1$
First term: $\left(-\frac{1}{2}\right)(-8x^2) = 4x^2$

Second term:

$$ \frac{-\frac{1}{2}(-\frac{1}{2}-1)}{2!}(-8x^2)^2 = \frac{-\frac{1}{2}\left(-\frac{3}{2}\right)}{2}(64x^4) $$ $$ = \frac{3}{8}(64x^4) = 24x^4 $$

Third term:

$$ \frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3!}(-8x^2)^3 $$ $$ = \frac{-\frac{1}{2}\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}(512x^6) $$ $$ = \frac{1 \cdot 3 \cdot 5}{2 \cdot 2 \cdot 2 \cdot 6}(512x^6) = 160x^6 $$

Step 3: Final expansion

Collecting terms, we have:

$$ \left(1 - 8x^2\right)^{-\frac{1}{2}} = 1 + 4x^2 + 24x^4 + 160x^6 + \cdots $$

Part (b)

We are given $\mathrm{g}(x) = \frac{a + bx}{\sqrt{1 - 8x^2}}$, where $a$ and $b$ are prime numbers. The fourth and fifth terms in the expansion of $\mathrm{g}(x)$ are $20x^3$ and $48x^4$, respectively. We determine $a$ and $b$ step-by-step.

Step 1: Expand $\mathrm{g}(x)$

Substitute the expansion of $\left(1 - 8x^2\right)^{-\frac{1}{2}}$ into $\mathrm{g}(x)$:

$$ \mathrm{g}(x) = (a + bx)\left(1 + 4x^2 + 24x^4 + 160x^6 + \cdots\right) $$ $$ = a(1 + 4x^2 + 24x^4 + 160x^6) + bx(1 + 4x^2 + 24x^4 + 160x^6). $$

Expand and collect terms:

$$ \mathrm{g}(x) = a + 4ax^2 + 24ax^4 + 160ax^6 + bx + 4bx^3 + 24bx^5 + 160bx^7 + \cdots. $$

Step 2: Match coefficients

The fourth term in the expansion is $20x^3$, and the fifth term is $48x^4$. Match coefficients for $x^3$ and $x^4$:

Coefficient of $x^3$: $4b = 20 \implies b = 5$
Coefficient of $x^4$: $24a = 48 \implies a = 2$

Step 3: Verify the values of $a$ and $b$

Substitute $a = 2$ and $b = 5$ back into the expansion of $\mathrm{g}(x)$:

$$ \mathrm{g}(x) = 2 + 5x + 8x^2 + 20x^3 + 48x^4 + \cdots. $$

The coefficients match the given conditions, confirming that $a = 2$ and $b = 5$.

Part (c) $\int_0^{0.2} \mathrm{g}(x) \, \mathrm{d}x$

We now estimate $\int_0^{0.2} \mathrm{g}(x) \, \mathrm{d}x$ using the first five terms in the expansion of $\mathrm{g}(x)$.

Substitute $\mathrm{g}(x) = 2 + 5x + 8x^2 + 20x^3 + 48x^4$ into the integral:

$$ \int_0^{0.2} \mathrm{g}(x) \, \mathrm{d}x = \int_0^{0.2} \left(2 + 5x + 8x^2 + 20x^3 + 48x^4\right) \, \mathrm{d}x $$

First, find the indefinite integral:

$$ \int \left( 2 + 5x + 8x^2 + 20x^3 + 48x^4 \right) \, dx = 2x + \frac{5}{2}x^2 + \frac{8}{3}x^3 + 5x^4 + \frac{48}{5}x^5 + C $$

Now, evaluate this expression at the limits $x = 0.2$ and $x = 0$:

$$ I = \left[ 2x + \frac{5}{2}x^2 + \frac{8}{3}x^3 + 5x^4 + \frac{48}{5}x^5 \right]_0^{0.2} $$

Substitute the upper limit $x = 0.2$ into the expression:

$$ = \left( 2(0.2) + \frac{5}{2}(0.2)^2 + \frac{8}{3}(0.2)^3 + 5(0.2)^4 + \frac{48}{5}(0.2)^5 \right) $$

Now, compute each term:

$$ = 0.4 + \frac{5}{2}(0.04) + \frac{8}{3}(0.008) + 5(0.0016) + \frac{48}{5}(0.00032) $$ $$ = 0.4 + 0.1 + 0.0213333 + 0.008 + 0.003072 $$ $$ = 0.5324053 $$

Therefore, the value of the definite integral is approximately:

$$ I \approx 0.5324 $$

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Question 10

(a) Using formulae on page 2, show that

(i) $\sin 2A = 2\sin A \cos A$
(ii) $\cos 2A = 2\cos^2 A - 1$ (3)

$$ f(\theta)=\frac{2\tan\theta}{1+\tan^2\theta} $$

(b) Show that $f(\theta)=\sin 2\theta$ (4)

$$ 5\tan\left(x+\frac{\pi}{6}\right) = \left[1+\tan^2\left(x+\frac{\pi}{6}\right)\right] \left[1-2\cos^2\left(x+\frac{\pi}{6}\right)\right] $$

(d) Using calculus, find the exact value of

$$ \int_0^{\frac{\pi}{2}} \left( \frac{4\tan\theta}{1+\tan^2\theta} -\cos 5\theta +2 \right) \,d\theta $$

(4)

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Solution

(a) Using compound angle formulae, show that:

(i) $\sin 2A = 2 \sin A \cos A$

To derive the identity for $\sin 2A$, we use the compound angle formula for sine:

$$ \sin(2A) = \sin(A + A) $$

Using the sum formula for sine, $\sin(x + y) = \sin x \cos y + \cos x \sin y$, we get:

$$ \sin(2A) = \sin A \cos A + \cos A \sin A = 2 \sin A \cos A $$

Thus, we have shown that:

$$ \sin 2A = 2 \sin A \cos A $$

(ii) $\cos 2A = 2 \cos^2 A - 1$

Now, to derive the identity for $\cos 2A$, we use the compound angle formula for cosine:

$$ \cos(2A) = \cos(A + A) $$

Using the sum formula for cosine, $\cos(x + y) = \cos x \cos y - \sin x \sin y$, we get:

$$ \cos(2A) = \cos A \cos A - \sin A \sin A = \cos^2 A - \sin^2 A $$

We now use the Pythagorean identity $\sin^2 A + \cos^2 A = 1$, which implies that $\sin^2 A = 1 - \cos^2 A$. Substituting this into the above equation:

$$ \cos(2A) = \cos^2 A - (1 - \cos^2 A) = 2 \cos^2 A - 1 $$

Thus, we have shown that:

$$ \cos 2A = 2 \cos^2 A - 1 $$

(iii) $\mathrm{f}(\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}$

The given function is:

$$ \mathrm{f}(\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} $$

This is a standard trigonometric identity known as the double angle formula for tangent. We can recognize that this expression represents the following identity for $\tan 2\theta$:

$$ \tan(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} $$

Thus, we have shown that:

$$ \mathrm{f}(\theta) = \tan 2\theta $$

(b) Show that $\mathrm{f}(\theta) = \sin 2\theta$

We are tasked with showing that $\mathrm{f}(\theta) = \sin 2\theta$. We start by recalling the double angle formula for tangent:

$$ \mathrm{f}(\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} $$

Next, substitute $\tan \theta = \frac{\sin \theta}{\cos \theta}$ into this expression:

$$ \mathrm{f}(\theta) = \frac{2 \left(\frac{\sin \theta}{\cos \theta}\right)}{1 + \left(\frac{\sin \theta}{\cos \theta}\right)^2} $$

Simplifying the numerator:

$$ \mathrm{f}(\theta) = \frac{2 \sin \theta}{\cos \theta} $$

Now, simplify the denominator:

$$ 1 + \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} $$

So the expression for $\mathrm{f}(\theta)$ becomes:

$$ \mathrm{f}(\theta) = \frac{2 \sin \theta}{\cos \theta} \cdot \cos^2 \theta = 2 \sin \theta \cos \theta $$

Finally, we recognize that:

$$ 2 \sin \theta \cos \theta = \sin 2\theta $$

Thus, we have shown that:

$$ \mathrm{f}(\theta) = \sin 2\theta $$

Solve the equation: $\tan 2\left(x + \frac{\pi}{6}\right) = -\frac{2}{5}$

Step 1: Solve for $2\left(x + \frac{\pi}{6}\right)$

The general solution for $\tan \theta = k$ is given by:

$$ \theta = \tan^{-1}(k) + n\pi, \quad n \in \mathbb{Z}. $$

Here, $k = -\frac{2}{5}$. Thus:

$$ 2\left(x + \frac{\pi}{6}\right) = \tan^{-1}\left(-\frac{2}{5}\right) + n\pi. $$

Using a calculator to compute $\tan^{-1}\left(-\frac{2}{5}\right)$:

$$ \tan^{-1}\left(-\frac{2}{5}\right) \approx -0.3805. $$

So the equation becomes:

$$ 2\left(x + \frac{\pi}{6}\right) = -0.3805 + n\pi. $$

Step 2: Solve for $x$

Divide through by 2:

$$ x + \frac{\pi}{6} = \frac{-0.3805 + n\pi}{2}. $$

Subtract $\frac{\pi}{6}$ from both sides:

$$ x = \frac{-0.3805 + n\pi}{2} - \frac{\pi}{6}. $$

Simplify the expression for $x$:

$$ x = \frac{-0.3805}{2} + \frac{n\pi}{2} - \frac{\pi}{6}. $$

Combine terms:

$$ x = -0.1903 + \frac{n\pi}{2} - \frac{\pi}{6}. $$

Finding a common denominator for the terms involving $\pi$:

$$ \frac{n\pi}{2} - \frac{\pi}{6} = \frac{3n\pi}{6} - \frac{\pi}{6} = \frac{(3n-1)\pi}{6}. $$

Thus:

$$ x = -0.1903 + \frac{(3n-1)\pi}{6}. $$

Step 3: Determine solutions within the given interval

The interval for $x$ is $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$, or approximately $-1.571 \leq x \leq 1.571$.

Substitute values of $n$ to find valid solutions:

$$ \text{For } n = 0: \quad x = -0.1903 + \frac{(3(0)-1)\pi}{6} = -0.1903 - \frac{\pi}{6} \approx -0.714. $$ $$ \text{For } n = 1: \quad x = -0.1903 + \frac{(3(1)-1)\pi}{6} = -0.1903 + \frac{2\pi}{6} = -0.1903 + \frac{\pi}{3} \approx 0.8568. $$

No other values of $n$ will yield solutions within the interval.

Final Answer

The solutions to $\tan 2\left(x + \frac{\pi}{6}\right) = -\frac{2}{5}$ in the interval $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ are:

$$ x \approx -0.714 \quad \text{and} \quad x \approx 0.857 \quad \text{(to 3 significant figures)}. $$

(d) Use calculus to find the exact value of:

$$ \int_0^{\frac{\pi}{2}} \left( \frac{4 \tan \theta}{1 + \tan^2 \theta} - \cos 5\theta + 2 \right) \, d\theta $$

We begin by breaking the integral into three parts:

$$ \int_0^{\frac{\pi}{2}} \left( \frac{4 \tan \theta}{1 + \tan^2 \theta} \right) \, d\theta - \int_0^{\frac{\pi}{2}} \cos 5\theta \, d\theta + \int_0^{\frac{\pi}{2}} 2 \, d\theta $$

1. Simplifying the first term:

The first term is:

$$ \frac{4 \tan \theta}{1 + \tan^2 \theta} $$

Using the identity:

$$ \frac{2 \tan \theta}{1 + \tan^2 \theta} = \sin 2\theta $$

we have:

$$ \frac{4 \tan \theta}{1 + \tan^2 \theta} = 2 \sin 2\theta $$

Thus, the integral becomes:

$$ \int_0^{\frac{\pi}{2}} 2 \sin 2\theta \, d\theta $$

Using the integral of $\sin x$, which is $-\cos x$, we get:

$$ \int_0^{\frac{\pi}{2}} 2 \sin 2\theta \, d\theta = -\cos 2\theta \Big|_0^{\frac{\pi}{2}} = -\cos \pi + \cos 0 = -(-1) + 1 = 2 $$

2. Simplifying the second term:

The second term is:

$$ \int_0^{\frac{\pi}{2}} \cos 5\theta \, d\theta $$

The integral of $\cos k\theta$ is:

$$ \frac{\sin k\theta}{k} $$

Thus:

$$ \int_0^{\frac{\pi}{2}} \cos 5\theta \, d\theta = \frac{1}{5} \sin 5\theta \Big|_0^{\frac{\pi}{2}} = \frac{1}{5} \left(\sin \frac{5\pi}{2} - \sin 0\right) $$

Since $\sin \frac{5\pi}{2} = \sin \frac{\pi}{2} = 1$, we get:

$$ \frac{1}{5}(1 - 0) = \frac{1}{5} $$

3. Simplifying the third term:

The third term is:

$$ \int_0^{\frac{\pi}{2}} 2 \, d\theta $$

This is a simple linear integral:

$$ \int_0^{\frac{\pi}{2}} 2 \, d\theta = 2\theta \Big|_0^{\frac{\pi}{2}} = 2 \times \frac{\pi}{2} - 2 \times 0 = \pi $$

Final Calculation:

Now, combining all the results from the three integrals:

$$ 2 - \frac{1}{5} + \pi = \pi + 2 - \frac{1}{5} $$

To combine the terms:

$$ \pi + 2 - \frac{1}{5} = \pi + \frac{10}{5} - \frac{1}{5} = \pi + \frac{9}{5} $$

Thus, the exact value of the integral is:

$$ \boxed{\pi + \frac{9}{5}} $$

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Question 11

Solve the simultaneous equations

$$ \begin{aligned} 2\log_4 x &= \log_3 3y^2 \\ \log_2 x^3 + 8\log_9 y &= 13 \end{aligned} $$

Show your working clearly. (8)

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Solution

Solution to the Simultaneous Equations

We are solving the simultaneous equations:

$$ \begin{aligned} 2 \log_4 x &= \log_3 (3y^2), \\ \log_2 x^3 + 8 \log_9 y &= 13. \end{aligned} $$

Step 1: Substitutions for Simpler Variables

Let $ \log_2 x = a $ and $ \log_3 y = b $. Using the properties of logarithms, we rewrite the equations:

1. From the first equation:

$$ 2 \log_4 x = \log_3 (3y^2). $$

Using $ \log_4 x = \frac{a}{2} $ and $ \log_3 (3y^2) = \log_3 3 + \log_3 y^2 = 1 + 2b $:

$$ 2 \cdot \frac{a}{2} = 1 + 2b \quad \Rightarrow \quad a = 1 + 2b. $$

2. From the second equation:

$$ \log_2 x^3 + 8 \log_9 y = 13. $$

Using $ \log_2 x^3 = 3a $ and $ \log_9 y = \frac{b}{2} $:

$$ 3a + 8 \cdot \frac{b}{2} = 13 \quad \Rightarrow \quad 3a + 4b = 13. $$

Step 2: Solve the System of Equations

We now have the system of linear equations:

$$ \begin{aligned} 1. & \quad a = 1 + 2b, \\ 2. & \quad 3a + 4b = 13. \end{aligned} $$

Substitute $ a = 1 + 2b $ into the second equation:

$$ 3(1 + 2b) + 4b = 13. $$

Simplify:

$$ 3 + 6b + 4b = 13 \quad \Rightarrow \quad 10b = 10 \quad \Rightarrow \quad b = 1. $$

Substitute $ b = 1 $ into $ a = 1 + 2b $:

$$ a = 1 + 2(1) = 3. $$

Step 3: Final Values

From $ a = \log_2 x $ and $ b = \log_3 y $:

$$ \log_2 x = 3 \quad \Rightarrow \quad x = 2^3 = 8, $$ $$ \log_3 y = 1 \quad \Rightarrow \quad y = 3^1 = 3. $$

Conclusion

The solutions to the simultaneous equations are:

$$ \boxed{x = 8, \quad y = 3.} $$

Further Pure Mathematics (Summary)

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FPM 2023 November01

Question 1

Solution

Step 1: Simplify the Equation

Step 2: Solve the System of Equations

Step 3: Final Answer

Question 2

Solution:

(a) Express $g(x)$ in the form $p(x+q)^2+r$

(b) Find the minimum value of $g(x)$ and the value of $x$ at which it occurs

(c) Determine the minimum value of $h(x)$ and the value of $x$ at which it occurs

Question 3

Solution

Step 1: Integrate $g'(x)$

Step 2: Use the given point

Step 3: Use the factor theorem

Step 4: Solve for $m$

Step 5: Substitute $m$ back into $g(x)$

Step 1: Use the factor theorem

Step 2: Solve the quadratic

Final Solution

Question 4

Solution:

Part (a): Find the value of $p$ and the value of $q$

Part (b): Find an equation for the line $L$

Question 5

Solution:

Step 1: Express the equation of the line in terms of $y$

Step 2: Find the points of intersection between the curve and the line

Step 3: Set up the volume integral

Step 4: Simplify the integrand

Step 5: Integrate the expression

Step 6: Evaluate the definite integral

Step 7: Final volume calculation

Question 6

Solution

(a) Show that $VO = \frac{\sqrt{2}}{2}x$

(b) Find the size of $\angle AVC$

(c) Find the angle between the planes $VAB$ and $VDC$

(d) Find the value of $x$

Question 7

Solution

(a) Find the two possible values of $r$

(b) Find the sum to infinity of $G$ (Given that $r > 0$)

(c) Find the least possible value of $n$ (using logarithms)

Question 8

Solution

(a) Find \( \frac{\mathrm{d}y}{\mathrm{d}x} \)

(b) Estimate the percentage change in \( y \)

Question 9

Solution

Part (a)

Step 1: Expand the series

Step 2: Simplify each term

Step 3: Final expansion

Part (b)

Step 1: Expand $\mathrm{g}(x)$

Step 2: Match coefficients

Step 3: Verify the values of $a$ and $b$

Part (c) $\int_0^{0.2} \mathrm{g}(x) \, \mathrm{d}x$

Question 10

Solution

(a) Using compound angle formulae, show that:

(i) $\sin 2A = 2 \sin A \cos A$

(ii) $\cos 2A = 2 \cos^2 A - 1$

(iii) $\mathrm{f}(\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}$

(b) Show that $\mathrm{f}(\theta) = \sin 2\theta$

Solve the equation: $\tan 2\left(x + \frac{\pi}{6}\right) = -\frac{2}{5}$

Step 1: Solve for $2\left(x + \frac{\pi}{6}\right)$

Step 2: Solve for $x$

Step 3: Determine solutions within the given interval

Final Answer

(d) Use calculus to find the exact value of:

1. Simplifying the first term:

2. Simplifying the second term:

3. Simplifying the third term:

Final Calculation:

Question 11

Solution

Solution to the Simultaneous Equations