FPM 2023 November02

Question 1

The equation $k x^2 + 8x + 3k = 0$ where $k$ is a constant, has real unequal roots.

Find the set of values of $k$ giving your answer in an exact simplified form. (5)

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Solution

Step 1: Condition for real and unequal roots

For a quadratic equation $ax^2 + bx + c = 0$, the roots are real and unequal if and only if the discriminant $\Delta$ satisfies:

$ \Delta > 0. $

The discriminant is given by:

$ \Delta = b^2 - 4ac. $

Step 2: Identifying coefficients

In the given equation $kx^2 + 8x + 3k = 0$, the coefficients are:

$ \begin{aligned} a &= k, \\ b &= 8, \\ c &= 3k. \end{aligned} $

Step 3: Compute the discriminant

Substitute the coefficients into the formula for the discriminant:

$ \begin{aligned} \Delta &= b^2 - 4ac \\ &= 8^2 - 4(k)(3k) \\ &= 64 - 12k^2. \end{aligned} $

Step 4: Solve the inequality $\Delta > 0$

For the roots to be real and unequal, we require $\Delta > 0$:

$ \begin{aligned} 64 - 12k^2 &> 0 \\ 12k^2 &< 64 \\ k^2 &< \frac{64}{12} \\ k^2 &< \frac{16}{3}. \end{aligned} $

Step 5: Solve for $k$

Taking the square root on both sides, we obtain:

$ -\sqrt{\frac{16}{3}} < k < \sqrt{\frac{16}{3}}. $

$ -\frac{4\sqrt{3}}{3} < k < \frac{4\sqrt{3}}{3}. $

Step 6: Final answer

The set of values of $k$ for which the equation has real and unequal roots is:

$ k \in \left(-\frac{4\sqrt{3}}{3}, \frac{4\sqrt{3}}{3}\right). $

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Question 2

In triangle $ABC$, $AB = 3x$ cm, $BC = 5x$ cm and $\angle ABC = 110^{\circ}$.

(a) Find, in degrees to one decimal place, the size of $\angle BCA$. (4)

The area of triangle $ABC$ is $24$ cm$^2$.

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

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Solution

Step 1: Understanding the problem

We are given a triangle $ABC$ where:

$AB = 3x\,\mathrm{cm}$,
$BC = 5x\,\mathrm{cm}$,
$\angle ABC = 110^{\circ}$.

Additionally, the area of $\triangle ABC$ is given as $24\,\mathrm{cm}^2$. The goal is to:

Find the size of $\angle BCA$ (in degrees, to one decimal place).
Determine the value of $x$ (to 3 significant figures).

Step 2: Using the Sine Rule to find $\angle BCA$

To find $\angle BCA$, we first determine the length of $AC$ using the Law of Cosines:

$ AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC). $

Substituting the known values:

$ \begin{aligned} AC^2 &= (3x)^2 + (5x)^2 - 2 \cdot (3x) \cdot (5x) \cdot \cos(110^{\circ}) \\ &= 9x^2 + 25x^2 - 30x^2 \cdot \cos(110^{\circ}). \end{aligned} $

Simplify to:

$ AC^2 = 34x^2 - 30x^2 \cdot \cos(110^{\circ}). $

$ AC = \sqrt{34x^2 - 30x^2 \cdot \cos(110^{\circ})}. $

Next, we apply the Sine Rule:

$ \frac{AB}{\sin(\angle BCA)} = \frac{AC}{\sin(\angle ABC)}. $

Let $\angle BCA = \theta$. Substituting the known values:

$ \frac{3x}{\sin(\theta)} = \frac{\sqrt{34x^2 - 30x^2 \cdot \cos(110^{\circ})}}{\sin(110^{\circ})}. $

Rearranging for $\sin(\theta)$:

$ \sin(\theta) = \frac{3x \cdot \sin(110^{\circ})} {\sqrt{34x^2 - 30x^2 \cdot \cos(110^{\circ})}}. $

Using numerical values:

$\cos(110^{\circ}) \approx -0.3420$,
$\sin(110^{\circ}) \approx 0.9397$.

Substitute these into the equation for $\sin(\theta)$:

$ \sin(\theta) = \frac{3x \cdot 0.9397} {\sqrt{34x^2 - 30x^2 \cdot (-0.3420)}}. $

Simplify the denominator:

$ \sqrt{34x^2 - 30x^2 \cdot (-0.3420)} = \sqrt{34x^2 + 10.26x^2} = \sqrt{44.26x^2} = \sqrt{44.26} \cdot x \approx 6.65x. $

Thus:

$ \sin(\theta) = \frac{3x \cdot 0.9397}{6.65x} = \frac{2.8191}{6.65} \approx 0.4239. $

Now, find $\theta$:

$ \theta = \arcsin(0.4239) \approx 25.1^{\circ}. $

Step 3: Using the area formula to find $x$

The area of $\triangle ABC$ can also be expressed using the formula:

$ \text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(\angle ABC). $

Substituting the known values:

$ \begin{aligned} 24 &= \frac{1}{2} \cdot (3x) \cdot (5x) \cdot \sin(110^{\circ}) \\ 24 &= \frac{1}{2} \cdot 15x^2 \cdot 0.9397. \end{aligned} $

Simplify to:

$ 15x^2 \cdot 0.9397 = 48. $

Divide through by $15 \cdot 0.9397$:

$ x^2 = \frac{48}{15 \cdot 0.9397} \approx \frac{48}{14.0955} \approx 3.406. $

Take the square root to find $x$:

$ x = \sqrt{3.406} \approx 1.85. $

Step 4: Final answers

The size of $\angle BCA$ is $25.1^{\circ}$ (to one decimal place).
The value of $x$ is $1.85$ (to 3 significant figures).

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Question 3

A particle $P$ moves along the $x$-axis.

At time $t$ seconds ($t \geqslant 0$) the acceleration, $a$ m/s$^2$, of $P$ is given by $a = 6t - 16$. When $t = 0$, $P$ is at the origin and is moving with velocity $12$ m/s.

(a) Find an expression in terms of $t$ for

(i) the velocity of $P$ at time $t$ seconds

(ii) the displacement of $P$ at time $t$ seconds. (4)

(b) Hence find the time at which $P$ first returns to the origin. (3)

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Solution

Step 1: Expression for velocity of $P$

The acceleration of the particle is given as:

$ a = 6t - 16. $

Velocity $v$ is obtained by integrating acceleration with respect to time:

$ v = \int a \, dt = \int (6t - 16) \, dt. $

Perform the integration:

$ v = 3t^2 - 16t + C, $

where $C$ is the constant of integration.

At $t = 0$, the velocity $v = 12 \, \mathrm{m/s}$. Substitute these values to find $C$:

$ 12 = 3(0)^2 - 16(0) + C \implies C = 12. $

Thus, the velocity of $P$ at time $t$ seconds is:

$ v = 3t^2 - 16t + 12. $

Step 2: Expression for displacement of $P$

The displacement $x$ is obtained by integrating velocity with respect to time:

$ x = \int v \, dt = \int (3t^2 - 16t + 12) \, dt. $

Perform the integration:

$ x = t^3 - 8t^2 + 12t + C_1, $

where $C_1$ is the constant of integration.

At $t = 0$, the particle is at the origin, so $x = 0$. Substitute these values to find $C_1$:

$ 0 = (0)^3 - 8(0)^2 + 12(0) + C_1 \implies C_1 = 0. $

Thus, the displacement of $P$ at time $t$ seconds is:

$ x = t^3 - 8t^2 + 12t. $

Step 3: Time at which $P$ first returns to the origin

The particle returns to the origin when $x = 0$. Solve the equation:

$ t^3 - 8t^2 + 12t = 0. $

Factorize the equation:

$ t(t^2 - 8t + 12) = 0. $

Solve for $t$:

$ t = 0 \quad \text{or} \quad t^2 - 8t + 12 = 0. $

Solve the quadratic equation $t^2 - 8t + 12 = 0$ using factorization:

$ t^2 - 8t + 12 = (t - 6)(t - 2) = 0. $

Thus:

$ t = 6 \quad \text{or} \quad t = 2. $

Since $t = 0$ corresponds to the initial position, the particle first returns to the origin at:

$ t = 2 \quad \text{seconds}. $

Final Answers

(a) (i) The velocity of $P$ at time $t$ seconds is:

$ v = 3t^2 - 16t + 12. $

(ii) The displacement of $P$ at time $t$ seconds is:

$ x = t^3 - 8t^2 + 12t. $

(b) The particle first returns to the origin at:

$ t = 2 \quad \text{seconds}. $

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Question 4

(a) On the axes opposite, draw the line with equation

(i) $y = -x - 1$
(ii) $y - 3x + 8 = 0$
(iii) $2y = x + 8$

(b) Show, by shading on your graph, the region $R$ defined by the inequalities

$y \geqslant -x - 1$ and $y \geqslant 3x - 8$ and $2y \leqslant x + 8$. (1)

For all points in $R$ with coordinates $(x, y)$,

$$P = 2y - 3x.$$

(i) the greatest value of $P$

(ii) the least value of $P$. (4)

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Solution

Step 1: Draw the lines

(a) On the axes, we draw the following lines:

(i) The line $y = -x - 1$

This line has a gradient of $-1$ and a $y$-intercept of $-1$. To plot this:

Point 1: $(0, -1)$ (intercept)
Point 2: $(1, -2)$ (substitute $x = 1$)

Join these points with a straight line.

(ii) The line $y - 3x + 8 = 0$

Rewrite as $y = 3x - 8$. This line has a gradient of $3$ and a $y$-intercept of $-8$. To plot this:

Point 1: $(0, -8)$ (intercept)
Point 2: $(1, -5)$ (substitute $x = 1$)

Join these points with a straight line.

(iii) The line $2y = x + 8$

Rewrite as $y = \frac{x}{2} + 4$. This line has a gradient of $\frac{1}{2}$ and a $y$-intercept of $4$. To plot this:

Point 1: $(0, 4)$ (intercept)
Point 2: $(2, 5)$ (substitute $x = 2$)

Join these points with a straight line.

Step 2: Identify the region $R$

The region $R$ is defined by the following inequalities:

$y \geqslant -x - 1$ (above the blue line)
$y \geqslant 3x - 8$ (above the red line)
$2y \leqslant x + 8 \implies y \leqslant \frac{x}{2} + 4$ (below the green line)

Shade the intersection of these regions to represent $R$.

Step 3: Evaluate $P$ at vertices of $R$

The vertices of $R$ are determined by the intersections of the lines:

Intersection of $y = -x - 1$ and $y = 3x - 8$
Intersection of $y = -x - 1$ and $y = \frac{x}{2} + 4$
Intersection of $y = 3x - 8$ and $y = \frac{x}{2} + 4$

Solve these equations to find the coordinates of the vertices.

Intersection 1: $y = -x - 1$ and $y = 3x - 8$

Equating the two equations:

$ -x - 1 = 3x - 8 $

$ -4x = -7 $

$ x = \frac{7}{4} $

$ y = -\frac{7}{4} - 1 = -\frac{11}{4} $

Vertex: $ \left(\frac{7}{4}, -\frac{11}{4}\right) $

Intersection 2: $y = -x - 1$ and $y = \frac{x}{2} + 4$

Equating the two equations:

$ -x - 1 = \frac{x}{2} + 4 $

$ -\frac{3x}{2} = 5 $

$ x = -\frac{10}{3} $

$ y = -\left(-\frac{10}{3}\right) - 1 = \frac{10}{3} - 1 = \frac{7}{3} $

Vertex: $ \left(-\frac{10}{3}, \frac{7}{3}\right) $

Intersection 3: $y = 3x - 8$ and $y = \frac{x}{2} + 4$

Equating the two equations:

$ 3x - 8 = \frac{x}{2} + 4 $

$ \frac{5x}{2} = 12 $

$ x = \frac{24}{5} $

$ y = \frac{24}{10} + 4 = \frac{12}{5} + 4 = \frac{32}{5} $

Vertex: $ \left(\frac{24}{5}, \frac{32}{5}\right) $

Step 4: Calculate $P = 2y - 3x$

Evaluate $P$ at each vertex:

At $ \left(\frac{7}{4}, -\frac{11}{4}\right) $:
$ P = 2\left(-\frac{11}{4}\right) - 3\left(\frac{7}{4}\right) = -\frac{22}{4} - \frac{21}{4} = -\frac{43}{4} = -10.75 $
At $ \left(-\frac{10}{3}, \frac{7}{3}\right) $:
$ P = 2\left(\frac{7}{3}\right) - 3\left(-\frac{10}{3}\right) = \frac{14}{3} + 10 = \frac{44}{3} $
At $ \left(\frac{24}{5}, \frac{32}{5}\right) $:
$ P = 2\left(\frac{32}{5}\right) - 3\left(\frac{24}{5}\right) = \frac{64}{5} - \frac{72}{5} = -\frac{8}{5} = -1.6 $

Therefore:

The greatest value of $P$ is $ \boxed{\frac{44}{3}}. $

The least value of $P$ is $ \boxed{-10.75}. $

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Question 5

Figure 1 shows part of the curve $S$ with equation $y = px^2 + qx + r$ where $p, q, r$ are constants.

The points $A, B$ and $P$ with coordinates $(-2,0)$, $(6,0)$ and $(4,-6)$ respectively lie on $S$.

(a) Show that an equation of $S$ is $y = \frac{x^2}{2} - 2x - 6$. (3)

The line $l$ is the normal to $S$ at the point $P$.

(b) Show that an equation of $l$ is $2y + x + 8 = 0$. (5)

The finite region shown shaded in Figure 1 is bounded by $S$ and $l$.

(c) Use algebraic integration to find the exact area of the shaded region. (7)

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Solution

Step 1: Finding the equation of $S$

The equation of $S$ is $y = px^2 + qx + r$. Substituting the coordinates of points $A$, $B$, and $P$ into this equation:

For $A(-2, 0)$:

$$ 0 = p(-2)^2 + q(-2) + r \Rightarrow 4p - 2q + r = 0 \tag{1} $$

For $B(6, 0)$:

$$ 0 = p(6)^2 + q(6) + r \Rightarrow 36p + 6q + r = 0 \tag{2} $$

For $P(4, -6)$:

$$ -6 = p(4)^2 + q(4) + r \Rightarrow 16p + 4q + r = -6 \tag{3} $$

Subtracting equations:

$$ (2) - (1): \quad 36p + 6q + r - (4p - 2q + r) = 0 \Rightarrow 32p + 8q = 0 \Rightarrow 4p + q = 0 \tag{4} $$ $$ (3) - (1): \quad 16p + 4q + r - (4p - 2q + r) = -6 \Rightarrow 12p + 6q = -6 \Rightarrow 2p + q = -1 \tag{5} $$

From equations (4) and (5):

$$ q = -4p \tag{6} $$ $$ 2p + (-4p) = -1 \Rightarrow -2p = -1 \Rightarrow p = \frac{1}{2}, \quad q = -2, \quad r = -6. $$

Thus, the equation of $S$ is:

$$ y = \frac{x^2}{2} - 2x - 6. $$

Step 2: Finding the equation of $l$ (the normal to $S$ at $P$)

The derivative of $y = \frac{x^2}{2} - 2x - 6$ is:

$$ \frac{dy}{dx} = x - 2. $$

At $P(4, -6)$:

$$ \frac{dy}{dx} = 4 - 2 = 2. $$

The gradient of the normal line is the negative reciprocal of the tangent's gradient:

$$ m_{\text{normal}} = -\frac{1}{2}. $$

Using the point-slope form of a line equation at $P(4, -6)$:

$$ y - (-6) = -\frac{1}{2}(x - 4). $$

Simplifying:

$$ y + 6 = -\frac{1}{2}x + 2 \Rightarrow 2y + 12 = -x + 4 \Rightarrow 2y + x + 8 = 0. $$

Thus, the equation of $l$ is:

$$ 2y + x + 8 = 0. $$

Step 3: Finding the area of the shaded region

The shaded region is bounded by $S$ and $l$. Solving for the intersection points:

$$ \text{Substitute } y = \frac{x^2}{2} - 2x - 6 \text{ into } 2y + x + 8 = 0: $$ $$ 2\left(\frac{x^2}{2} - 2x - 6\right) + x + 8 = 0 \Rightarrow x^2 - 4x - 12 + x + 8 = 0 \Rightarrow x^2 - 3x - 4 = 0. $$

Factoring:

$$ (x - 4)(x + 1) = 0 \Rightarrow x = 4 \text{ or } x = -1. $$

Since the straight line is above the curve between $x=-1$ and $x=4$, the area is:

$$ \text{Area} = \int_{-1}^{4} \left( -\frac{x}{2} - 4 \right) dx - \int_{-1}^{4} \left( \frac{x^2}{2} - 2x - 6 \right) dx . $$

Step 4: Evaluate the curve integral

First, consider the integral:

$$ I_1 = \int_{-1}^{4} \left( \frac{x^2}{2} - 2x - 6 \right) dx. $$

We integrate term by term:

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

Thus, the first integral becomes:

$$ I_1 = \left[ \frac{x^3}{6} - x^2 - 6x \right]_{-1}^{4}. $$

Now, evaluate at the bounds:

At $x = 4$:

$$ \frac{4^3}{6} - 4^2 - 6 \cdot 4 = \frac{64}{6} - 16 - 24 = \frac{64}{6} - 40 = \frac{-176}{6} = -\frac{88}{3}. $$

At $x = -1$:

$$ \frac{(-1)^3}{6} - (-1)^2 - 6 \cdot (-1) = \frac{-1}{6} - 1 + 6 = \frac{29}{6}. $$

Thus, the value of the first integral is:

$$ I_1 = \left( -\frac{88}{3} \right) - \left( \frac{29}{6} \right) = -\frac{88}{3} - \frac{29}{6}. $$

Combining these with a common denominator:

$$ -\frac{88}{3} = -\frac{176}{6}, \quad I_1 = -\frac{176}{6} - \frac{29}{6} = -\frac{205}{6}. $$

Step 5: Evaluate the second integral

Next, we evaluate the second integral:

$$ I_2 = \int_{-1}^{4} \left( -\frac{x}{2} - 4 \right) dx. $$

We integrate term by term:

$$ \int -\frac{x}{2} \, dx = -\frac{x^2}{4}, \quad \int -4 \, dx = -4x. $$

Thus, the second integral becomes:

$$ I_2 = \left[ -\frac{x^2}{4} - 4x \right]_{-1}^{4}. $$

Now, evaluate at the bounds:

At $x = 4$:

$$ -\frac{4^2}{4} - 4 \cdot 4 = -4 - 16 = -20. $$

At $x = -1$:

$$ -\frac{(-1)^2}{4} - 4 \cdot (-1) = -\frac{1}{4} + 4 = \frac{15}{4}. $$

Thus, the value of the second integral is:

$$ I_2 = -20 - \frac{15}{4} = -\frac{80}{4} - \frac{15}{4} = -\frac{95}{4}. $$

Step 6: Combine the results

Finally, we combine the results of the two integrals:

$$ \text{Area} = I_2 - I_1 = \left( -\frac{95}{4} \right) - \left( -\frac{205}{6} \right) = \frac{205}{6} - \frac{95}{4}. $$

Finding a common denominator:

$$ \frac{205}{6} = \frac{410}{12}, \quad \frac{95}{4} = \frac{285}{12}. $$

Thus,

$$ \text{Area} = \frac{410}{12} - \frac{285}{12} = \frac{125}{12}. $$

Final Answer

The final result is:

$$ \boxed{\frac{125}{12}}. $$

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Question 6

The volume of oil in a container is $V$ cm$^3$ when the height of the oil is $h$ cm.

Oil is pouring into the container at a constant rate of $12$ cm$^3$/s.

Given that $V = 3h^3$, find the exact rate, in cm/s, at which the height of the oil is increasing when $V = 1536$ cm$^3$. (7)

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Solution

Step 1: Differentiate the relationship between $V$ and $h$

The given relationship is $V = 3h^3$. Differentiating both sides with respect to time $t$ using the chain rule, we get:

$$ \frac{dV}{dt} = \frac{d}{dt}(3h^3) = 9h^2 \frac{dh}{dt}. $$

Step 2: Solve for $\frac{dh}{dt}$

Rearranging the equation to solve for $\frac{dh}{dt}$:

$$ \frac{dh}{dt} = \frac{1}{9h^2} \frac{dV}{dt}. $$

Step 3: Find the height $h$ when $V = 1536 \, \mathrm{cm}^3$

Using the relationship $V = 3h^3$, substitute $V = 1536$:

$$ 1536 = 3h^3. $$

Divide both sides by 3:

$$ h^3 = 512. $$

Take the cube root of both sides:

$$ h = \sqrt[3]{512} = 8 \, \mathrm{cm}. $$

Step 4: Substitute the known values into the expression for $\frac{dh}{dt}$

Substitute $h = 8$ and $\frac{dV}{dt} = 12 \, \mathrm{cm}^3/\mathrm{s}$ into the equation:

$$ \frac{dh}{dt} = \frac{1}{9h^2} \frac{dV}{dt}. $$

Calculate $h^2$:

$$ h^2 = 8^2 = 64. $$

Substitute $h^2 = 64$ and $\frac{dV}{dt} = 12$:

$$ \frac{dh}{dt} = \frac{1}{9(64)}(12) = \frac{12}{576} = \frac{1}{48}. $$

Step 5: Final Answer

The exact rate at which the height of the oil is increasing is:

$$ \frac{dh}{dt} = \frac{1}{48} \, \mathrm{cm}/\mathrm{s}. $$

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Question 7

Two numbers $x$ and $y$ are such that $3x - y = 4$, \[S = 5x^3 + y^2.\]
(a) Show that $S = 5x^3 + 9x^2 - 24x + 16$. (2)

Given that $x$ can vary,

(b) use calculus to find the value of $x$ for which $S$ is a minimum, justifying that this value of $x$ gives a minimum value of $S$. (5)

(c) Find the minimum value of $S$. (2)

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Solution

Step 1: Express $y$ in terms of $x$

We are given the equation:

$$ 3x - y = 4. $$

Rearranging for $y$, we have:

$$ y = 3x - 4. $$

Step 2: Substitute $y$ into $S$

The expression for $S$ is given as:

$$ S = 5x^3 + y^2. $$

Substituting $y = 3x - 4$ into $y^2$, we get:

$$ y^2 = (3x - 4)^2 = 9x^2 - 24x + 16. $$

Thus, the expression for $S$ becomes:

$$ S = 5x^3 + 9x^2 - 24x + 16. $$

This verifies that:

$$ S = 5x^3 + 9x^2 - 24x + 16. $$

Step 3: Differentiate $S$ with respect to $x$

To find the critical points, we differentiate $S$ with respect to $x$:

$$ \frac{dS}{dx} = 15x^2 + 18x - 24. $$

Step 4: Solve $\frac{dS}{dx} = 0$

Setting $\frac{dS}{dx} = 0$:

$$ 15x^2 + 18x - 24 = 0. $$

Divide through by 3:

$$ 5x^2 + 6x - 8 = 0. $$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 5$, $b = 6$, and $c = -8$:

$$ x = \frac{-6 \pm \sqrt{6^2 - 4(5)(-8)}}{2(5)} $$ $$ x = \frac{-6 \pm \sqrt{36 + 160}}{10} $$ $$ x = \frac{-6 \pm \sqrt{196}}{10} $$ $$ x = \frac{-6 \pm 14}{10}. $$

This gives two solutions:

$$ x = \frac{-6 + 14}{10} = 0.8, \quad x = \frac{-6 - 14}{10} = -2. $$

Step 5: Determine the nature of the critical points

The second derivative of $S$ is:

$$ \frac{d^2S}{dx^2} = 30x + 18. $$

Substitute $x = 0.8$:

$$ \frac{d^2S}{dx^2} = 30(0.8) + 18 = 24 + 18 = 42 > 0 $$

Since $\frac{d^2S}{dx^2} > 0$, $S$ has a local minimum at $x = 0.8$.

Substitute $x = -2$:

$$ \frac{d^2S}{dx^2} = 30(-2) + 18 = -60 + 18 = -42 < 0 $$

Since $\frac{d^2S}{dx^2} < 0$, $S$ has a local maximum at $x = -2$.

Step 6: Find the minimum value of $S$

Substitute $x = 0.8$ into the expression for $S$:

$$ S = 5x^3 + 9x^2 - 24x + 16 $$ $$ S = 5(0.8)^3 + 9(0.8)^2 - 24(0.8) + 16 $$ $$ S = 5(0.512) + 9(0.64) - 19.2 + 16 $$ $$ S = 2.56 + 5.76 - 19.2 + 16 $$ $$ S = 5.12 $$

Note: The final line in the working is inconsistent with the stated answer.

Thus, the minimum value of $S$ (from correct substitution) is:

$$ \boxed{5.12} $$

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Question 8

The sum to $n$ terms of an arithmetic series $A$ is $S_n$.

The sum of the first four terms of $A$ is 42 and the fifth term of $A$ is 23.

(a) Show that $S_n = \sum_{r=1}^n (P r - Q)$ where $P$ and $Q$ are prime numbers.(6)

$S_{2n} - 3 U_n = 1062$ where $U_n$ is the $n$th term of $A$.

(b) Find the value of $n$. (4)

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Solution

Step 1: Expressing the general terms of the arithmetic series

The sum to $n$ terms of an arithmetic series $A$ is given by

$$ S_n = \frac{n}{2} \left( 2a + (n-1)d \right), $$

where $a$ is the first term and $d$ is the common difference.

The $n$th term of the arithmetic series is given by

$$ U_n = a + (n-1)d. $$

Step 2: Using the given information

The sum of the first four terms is given as $S_4 = 42$. Substituting $n = 4$ into the formula for $S_n$:

$$ S_4 = \frac{4}{2} \left( 2a + 3d \right) = 42. $$

Simplify:

$$ 2a + 3d = 21. \tag{1} $$

The fifth term is given as $U_5 = 23$. Substituting $n = 5$ into the formula for $U_n$:

$$ U_5 = a + 4d = 23. $$

Simplify:

$$ a + 4d = 23. \tag{2} $$

Step 3: Solving the system of equations

From equations (1) and (2), we solve for $a$ and $d$. Subtract equation (1) from twice equation (2):

$$ 2(a + 4d) - (2a + 3d) = 2 \times 23 - 21 $$ $$ 8d - 3d = 25 $$ $$ 5d = 25 $$ $$ d = 5. \tag{3} $$

Substitute $d = 5$ into equation (2):

$$ a + 4 \times 5 = 23 $$ $$ a + 20 = 23 $$ $$ a = 3. \tag{4} $$

Thus, $a = 3$ and $d = 5$.

Step 4: Verifying $S_n = \sum_{r=1}^n \left(Pr - Q\right)$

The $r$th term of the series is $U_r = a + (r-1)d$. Substituting $a = 3$ and $d = 5$:

$$ U_r = 3 + 5(r-1) = 5r - 2. $$

The sum of the first $n$ terms is:

$$ S_n = \sum_{r=1}^n U_r = \sum_{r=1}^n \left( 5r - 2 \right) $$ $$ = \sum_{r=1}^n 5r - \sum_{r=1}^n 2 $$ $$ = 5 \sum_{r=1}^n r - 2n $$ $$ = 5 \cdot \frac{n(n+1)}{2} - 2n $$ $$ = \frac{5n(n+1)}{2} - 2n. \tag{5} $$

This verifies the form $S_n = \sum_{r=1}^n \left( Pr - Q \right)$, where $P = 5$ and $Q = 2$.

Step 5: Solving for $n$ given $S_{2n} - 3U_n = 1062$

The sum of $2n$ terms is:

$$ S_{2n} = \frac{5(2n)((2n)+1)}{2} - 2(2n) $$ $$ = \frac{5 \cdot 2n \cdot (2n + 1)}{2} - 4n $$ $$ = 5n(2n+1) - 4n. \tag{6} $$

The $n$th term is:

$$ U_n = a + (n-1)d $$ $$ = 3 + (n-1)5 $$ $$ = 5n - 2. \tag{7} $$

Substitute into the given equation:

$$ S_{2n} - 3U_n = 1062 $$ $$ 5n(2n+1) - 4n - 3(5n - 2) = 1062 $$ $$ 10n^2 + 5n - 4n - 15n + 6 = 1062 $$ $$ 10n^2 - 14n + 6 = 1062 $$ $$ 10n^2 - 14n - 1056 = 0 $$ $$ 5n^2 - 7n - 528 = 0. \tag{8} $$

Solve this quadratic equation using the quadratic formula:

$$ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ $$ n = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(5)(-528)}}{2(5)} $$ $$ n = \frac{7 \pm \sqrt{49 + 10560}}{10} $$ $$ n = \frac{7 \pm \sqrt{10609}}{10} $$ $$ n = \frac{7 \pm 103}{10} $$ $$ n = 11 \text{ (positive solution)}. \tag{9} $$

Final Answer

The value of $n$ is

$$ \boxed{11}. $$

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Question 9

A logo $A E B C D$ is shown shaded in Figure 2.

The straight line $ABC$ is the diameter of the semicircle $ADC$.
$AEB$ is an arc of a circle with centre $O$.
All angles are measured in radians.

$BC = 2x$ cm
$OA = OB = x$ cm
length of arc $AEB = 1.8x$ cm

The perimeter of the logo is $P$.

(a) Show that $P = a x (\pi + \pi \sin 0.9 + b)$ where $a$ and $b$ are constants to be found. (7)

Given that $x = 10$ cm,

(b) find, in cm$^2$ to 3 significant figures, the area of the logo. (6)

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Solution

Step 1: Calculate angle $\angle AOB$ using $l = r\theta$

The angle subtended by arc $AEB$ at the center $O$ can be calculated as:

$$ \theta = \frac{l}{r} = \frac{1.8x}{x} = 1.8 \text{ radians} $$

Thus, $\angle AOB = 1.8$ radians.

Step 2: Find $AB$ in terms of $\sin(0.9)$

Using triangle $\triangle AOB$ with $\angle AOB = 1.8$, and noting that $OA = OB = x$, the chord $AB$ can be calculated using:

$$ AB = 2 \cdot OA \cdot \sin\left(\frac{\angle AOB}{2}\right) = 2x \sin(0.9) $$

Step 3: Find the diameter of the larger semicircle $ADC$

The diameter of the semicircle $ADC$ includes $AB$ and $BC$. Thus:

$$ \text{Diameter of semicircle} = AB + BC = 2x \sin(0.9) + 2x $$

The radius of the semicircle is:

$$ \text{Radius} = x + x \sin(0.9) $$

Step 4: Calculate the perimeter $P$

The perimeter of the logo consists of:

The semicircular arc $ADC$, with radius $r = x + x \sin(0.9)$
The straight segment $BC$, with length $2x$
The arc $AEB$, with given length $1.8x$

The length of the semicircular arc $ADC$ is:

$$ \text{Length of semicircle } ADC = \pi (x + x \sin(0.9)) = \pi x (1 + \sin(0.9)) $$

Thus, the total perimeter is:

$$ P = \pi x (1 + \sin(0.9)) + 1.8x + 2x $$ $$ = x\big(\pi + \pi \sin(0.9) + 3.8 \big) $$

Comparing with $P = ax(\pi + \pi \sin(0.9) + b)$, we obtain:

$$ a = 1, \quad b = 3.8. $$

Step 5: Calculate the area of the logo

The area of the logo consists of:

The area of semicircle $ADC$
Subtracting the area of the smaller segment $AEB$

Step 5.1: Area of semicircle $ADC$

The area of the semicircle is:

$$ \text{Area of semicircle } ADC = \frac{1}{2} \pi \big(x + x \sin(0.9)\big)^2 $$

Step 5.2: Area of sector $AEB$

The area of the smaller sector $AEB$ is:

$$ \text{Area of sector } AEB = \frac{1}{2} x^2 \theta $$

Substituting $\theta = 1.8$:

$$ \text{Area of sector } AEB = \frac{1}{2} x^2 (1.8) = 0.9x^2 $$

Step 5.3: Area of the logo

The area of triangle $OAB$ is:

$$ \text{Area of triangle } OAB = \frac{1}{2} x^2 \sin(1.8) $$

Hence, the area of the circular segment $AEB$ is:

$$ \text{Area of segment } AEB = 0.9x^2 - \frac{1}{2} x^2 \sin(1.8) $$

Therefore, the area of the logo is:

$$ \text{Area of logo} = \frac{1}{2} \pi \big(x + x \sin(0.9)\big)^2 - 0.9x^2 + \frac{1}{2} x^2 \sin(1.8) $$

Step 5.4: Substitute $x = 10$ cm

$$ \text{Area of logo} = \frac{1}{2} \pi \big(10 + 10 \sin(0.9)\big)^2 - 0.9(10)^2 + \frac{1}{2}(10)^2 \sin(1.8) $$ $$ = 0.5 \pi (10 + 7.8332)^2 - 41.307 $$ $$ = 0.5 \pi (17.8332)^2 - 41.307 $$ $$ \approx 0.5 \pi (318.02) - 41.307 $$ $$ \approx 499.55 - 41.307 $$ $$ = 458.2 \text{ cm}^2 $$

Thus, the area of the logo is approximately:

$$ \boxed{458 \text{ cm}^2} $$

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Question 10

The roots of a quadratic equation are $\alpha$ and $\beta$ where \[\alpha + \beta = -\frac{5}{2} \mbox{ and }\alpha^3 + \beta^3 = \frac{115}{8}\]

(a) Show that $\alpha \beta = 4$. (3)

(b) Form a quadratic equation with integer coefficients that has roots \[\frac{\alpha^2 + 1}{\beta} \mbox{ and } \frac{\beta^2 + 1}{\alpha}\] (7)

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Solution

Step (a): Show that $\alpha \beta = 4$

We are given the following information:

$$ \alpha + \beta = -\frac{5}{2}, $$ $$ \alpha^3 + \beta^3 = \frac{115}{8}. $$

Using the identity for the sum of cubes:

$$ \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta), $$

substitute $\alpha + \beta = -\frac{5}{2}$:

$$ \alpha^3 + \beta^3 = \left(-\frac{5}{2}\right)^3 - 3\alpha \beta \left(-\frac{5}{2}\right). $$

Simplify:

$$ \left(-\frac{5}{2}\right)^3 = -\frac{125}{8}. $$

Hence:

$$ \alpha^3 + \beta^3 = -\frac{125}{8} + \frac{15}{2}\alpha \beta. $$

Given that $\alpha^3 + \beta^3 = \frac{115}{8}$:

$$ \frac{115}{8} = -\frac{125}{8} + \frac{15}{2}\alpha \beta. $$

Simplify:

$$ \frac{115}{8} + \frac{125}{8} = \frac{15}{2}\alpha \beta $$ $$ \frac{240}{8} = \frac{15}{2}\alpha \beta $$ $$ 30 = \frac{15}{2}\alpha \beta $$

Multiply through by $2$:

$$ 60 = 15\alpha \beta $$

Divide by $15$:

$$ \alpha \beta = 4. $$

Thus, we have shown that $\alpha \beta = 4$.

Step (b): Form a quadratic equation with integer coefficients

The roots of the quadratic equation are:

$$ a = \frac{\alpha^2 + 1}{\beta}, \quad b = \frac{\beta^2 + 1}{\alpha}. $$

First, find $a + b$:

$$ a + b = \frac{\alpha^2 + 1}{\beta} + \frac{\beta^2 + 1}{\alpha}. $$

Bring to a common denominator:

$$ a + b = \frac{\alpha(\alpha^2 + 1) + \beta(\beta^2 + 1)}{\alpha \beta}. $$

Simplify the numerator:

$$ \alpha(\alpha^2 + 1) + \beta(\beta^2 + 1) = \alpha^3 + \alpha + \beta^3 + \beta $$ $$ = (\alpha^3 + \beta^3) + (\alpha + \beta). $$

Substitute the known values:

$$ \alpha^3 + \beta^3 + \alpha + \beta = \frac{115}{8} - \frac{5}{2}. $$

Simplify:

$$ \frac{115}{8} - \frac{5}{2} = \frac{115}{8} - \frac{20}{8} = \frac{95}{8}. $$

Thus:

$$ a + b = \frac{\frac{95}{8}}{4} = \frac{95}{32}. $$

Next, find $ab$:

$$ ab = \frac{(\alpha^2 + 1)(\beta^2 + 1)}{\alpha \beta}. $$ $$ ab = \frac{\alpha^2\beta^2 + \alpha^2 + \beta^2 + 1}{4}. $$

Use $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$:

$$ \alpha^2 + \beta^2 = \left(-\frac{5}{2}\right)^2 - 2(4). $$ $$ = \frac{25}{4} - 8 = \frac{25}{4} - \frac{32}{4} = -\frac{7}{4}. $$

Substitute into $ab$:

$$ ab = \frac{4^2 + \alpha^2 + \beta^2 + 1}{4} $$ $$ = \frac{16 - \frac{7}{4} + 1}{4} $$ $$ = \frac{17 - \frac{7}{4}}{4} $$ $$ = \frac{\frac{68}{4} - \frac{7}{4}}{4} $$ $$ = \frac{\frac{61}{4}}{4} = \frac{61}{16}. $$

The quadratic equation is:

$$ x^2 - (a+b)x + ab = 0 $$ $$ x^2 - \frac{95}{32}x + \frac{61}{16} = 0. $$

Multiply through by $32$:

$$ \boxed{32x^2 - 95x + 122 = 0} $$

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Question 11

Figure 3 shows quadrilateral $OABC$ where \[\overrightarrow{OA} = 4\mathbf{p} + 5\mathbf{q},\quad \overrightarrow{OB} = 3\mathbf{p} + \mathbf{q},\quad \overrightarrow{OC} = 2\mathbf{p} - 4\mathbf{q}\]

The point $M$ is the midpoint of $OC$.

(a) Find $\overrightarrow{MA}$ as a simplified expression in terms of $\mathbf{p}$ and $\mathbf{q}$. (3)

The point $N$ lies on $OB$ such that $M, N$ and $A$ are collinear.

(b) Find the ratio $MN : NA$. (6)

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Click to view Solution

Solution

Step 1: Understanding the Problem

We are given quadrilateral $OABC$ with the following vector relations:

$$ \overrightarrow{OA} = 4 \mathbf{p} + 5 \mathbf{q}, \quad \overrightarrow{OB} = 3 \mathbf{p} + \mathbf{q}, \quad \overrightarrow{OC} = 2 \mathbf{p} - 4 \mathbf{q} $$

The point $M$ is the midpoint of $OC$, and the point $N$ lies on $OB$ such that the points $M$, $N$, and $A$ are collinear.

Step 2: Find $\overrightarrow{MA}$

Since $M$ is the midpoint of $OC$, the position vector of $M$ is:

$$ \overrightarrow{OM} = \frac{1}{2}\overrightarrow{OC} = \frac{1}{2}(2\mathbf{p} - 4\mathbf{q}) = \mathbf{p} - 2\mathbf{q} $$

Thus,

$$ \overrightarrow{MA} = \overrightarrow{OA} - \overrightarrow{OM} $$ $$ = (4\mathbf{p} + 5\mathbf{q}) - (\mathbf{p} - 2\mathbf{q}) $$ $$ = 3\mathbf{p} + 7\mathbf{q} $$

Step 3: Find the ratio $MN : NA$

Since $M$, $N$, and $A$ are collinear, suppose $N$ divides $MA$ in the ratio $k : 1$. Then:

$$ MN : NA = k : 1 $$

Using the section formula:

$$ \overrightarrow{ON} = \frac{1}{k+1} \left( k\overrightarrow{OA} + \overrightarrow{OM} \right) $$

Substitute $\overrightarrow{OA} = 4\mathbf{p} + 5\mathbf{q}$ and $\overrightarrow{OM} = \mathbf{p} - 2\mathbf{q}$:

$$ \overrightarrow{ON} = \frac{1}{k+1} \left( k(4\mathbf{p} + 5\mathbf{q}) + (\mathbf{p} - 2\mathbf{q}) \right) $$

Simplify:

$$ \overrightarrow{ON} = \frac{1}{k+1} \left( (4k+1)\mathbf{p} + (5k-2)\mathbf{q} \right) $$

Since $N$ lies on $OB$, its position vector is also a scalar multiple of $\overrightarrow{OB} = 3\mathbf{p} + \mathbf{q}$:

$$ \overrightarrow{ON} = \lambda(3\mathbf{p} + \mathbf{q}) $$

Equating coefficients of $\mathbf{p}$ and $\mathbf{q}$:

$$ \frac{4k+1}{k+1} = 3\lambda, \quad \frac{5k-2}{k+1} = \lambda $$

Step 4: Solve for $k$

From the second equation:

$$ \lambda = \frac{5k-2}{k+1} $$

Substitute into the first equation:

$$ \frac{4k+1}{k+1} = 3\left( \frac{5k-2}{k+1} \right) $$

Since both sides have denominator $k+1$, cancel it:

$$ 4k + 1 = 3(5k - 2) $$

Expand the right-hand side:

$$ 4k + 1 = 15k - 6 $$

Rearrange:

$$ 4k - 15k = -6 - 1 $$ $$ -11k = -7 $$ $$ k = \frac{7}{11} $$

Step 5: Find $\lambda$

Substitute $k = \frac{7}{11}$ into $\lambda = \frac{5k-2}{k+1}$:

$$ \lambda = \frac{5\left(\frac{7}{11}\right)-2}{\frac{7}{11}+1} $$ $$ = \frac{\frac{35}{11} - 2}{\frac{7}{11} + \frac{11}{11}} $$ $$ = \frac{\frac{13}{11}}{\frac{18}{11}} = \frac{13}{18} $$

Step 6: State the required ratio

Since the ratio is $MN : NA = k : 1$:

$$ MN : NA = \frac{7}{11} : 1 = 7 : 11 $$

Hence,

$$ \boxed{MN : NA = 7 : 11} $$

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