Question 1
Find the set of values of $x$ for which
(a) $2(3x - 1) \lt 4 - 3x$ (2)
(b) $3x^2 - 8x - 3 \lt 0$ (4)
(c) both $2(3x - 1) \lt 4 - 3x$ and $3x^2 - 8x - 3 \lt 0$ (1)
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Solution
(a) Solve $2(3x - 1) \lt 4 - 3x$
$$ \begin{aligned} &\text{Step 1: Expand the brackets.} \\ &2(3x - 1) \lt 4 - 3x \\ &6x - 2 \lt 4 - 3x \\ \\ &\text{Step 2: Combine like terms.} \\ &6x + 3x \lt 4 + 2 \\ &9x \lt 6 \\ \\ &\text{Step 3: Solve for } x. \\ &x \lt \frac{6}{9} \\ &x \lt \frac{2}{3} \\ \\ &\boxed{x \lt \frac{2}{3}} \end{aligned} $$
(b) Solve $3x^2 - 8x - 3 \lt 0$
$$ \begin{aligned} &\text{Step 1: Factorize the quadratic expression.} \\ & 3x^2 - 8x - 3 = 0 \quad \text{(use the quadratic formula)} \\ & x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 3,\; b = -8,\; c = -3 \\ & x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(-3)}}{2(3)} \\ & x = \frac{8 \pm \sqrt{64 + 36}}{6} \\ & x = \frac{8 \pm \sqrt{100}}{6} \\ & x = \frac{8 \pm 10}{6} \\ \\ &\text{Step 2: Simplify the roots.} \\ & x = \frac{8 + 10}{6} \quad \text{or} \quad x = \frac{8 - 10}{6} \\ & x = 3 \quad \text{or} \quad x = -\frac{1}{3} \\ \\ &\text{Step 3: Determine the intervals where } 3x^2 - 8x - 3 \lt 0. \\ &\text{The roots divide the number line into three intervals:} \\ &(-\infty, -\frac{1}{3}), \quad \left(-\frac{1}{3}, 3\right), \quad (3, \infty) \\ &\text{Test a point from each interval in } 3x^2 - 8x - 3. \\ \\ &\text{For } x = -1 \quad (-\infty, -\frac{1}{3}): \\ &\quad 3(-1)^2 - 8(-1) - 3 = 3 + 8 - 3 = 8 > 0 \quad \text{(Not valid)} \\ \\ &\text{For } x = 0 \quad \left(-\frac{1}{3}, 3\right): \\ &\quad 3(0)^2 - 8(0) - 3 = -3 \lt 0 \quad \text{(Valid)} \\ \\ &\text{For } x = 4 \quad (3, \infty): \\ &\quad 3(4)^2 - 8(4) - 3 = 48 - 32 - 3 = 13 > 0 \quad \text{(Not valid)} \\ \\ &\text{Thus, the solution is:} \\ &\boxed{-\frac{1}{3} \lt x \lt 3} \end{aligned} $$
(c) Solve both $2(3x - 1) \lt 4 - 3x$ and $3x^2 - 8x - 3 \lt 0$
$$ \begin{aligned} &\text{Step 1: Combine the two inequalities.} \\ & x \lt \frac{2}{3} \quad \text{and} \quad -\frac{1}{3} \lt x \lt 3 \\ \\ &\text{Step 2: Find the intersection of the two solutions.} \\ &\text{The overlap is: } -\frac{1}{3} \lt x \lt \frac{2}{3} \\ \\ &\text{Thus, the solution is:} \\ &\boxed{-\frac{1}{3} \lt x \lt \frac{2}{3}} \end{aligned} $$
Question 2
Figure 1 shows triangle $XYZ$ in which
$XY = (x + 2)\,\text{cm} \quad XZ = (2x + 4)\,\text{cm} \quad YZ = (2x - 1)\,\text{cm} \quad \text{and} \angle YXZ = 60^\circ$
Find the value of $x$
Give your answer in the form $p + q\sqrt{3}$ where $p$ and $q$ are integers to be found.
(4)
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Solution
We are given:
$$ XY = (x + 2)\,\text{cm}, \quad XZ = (2x + 4)\,\text{cm}, \quad YZ = (2x - 1)\,\text{cm}, \quad \angle YXZ = 60^\circ $$
Using the law of cosines:
$$ YZ^2 = XY^2 + XZ^2 - 2 \cdot XY \cdot XZ \cdot \cos(\angle YXZ) $$
Substitute the given values:
$$ (2x - 1)^2 = (x + 2)^2 + (2x + 4)^2 - 2 \cdot (x + 2) \cdot (2x + 4) \cdot \frac{1}{2} $$
Step 1: Expand each term
1. Expand $(2x - 1)^2$:
$$ (2x - 1)^2 = 4x^2 - 4x + 1 $$
2. Expand $(x + 2)^2$:
$$ (x + 2)^2 = x^2 + 4x + 4 $$
3. Expand $(2x + 4)^2$:
$$ (2x + 4)^2 = 4x^2 + 16x + 16 $$
4. Expand $2 \cdot (x + 2) \cdot (2x + 4) \cdot \frac{1}{2}$:
$$ 2 \cdot (x + 2) \cdot (2x + 4) \cdot \frac{1}{2} = (x + 2)(2x + 4) $$
$$ (x + 2)(2x + 4) = 2x^2 + 8x + 8 $$
Step 2: Substitute and simplify
Substitute all expanded terms into the law of cosines equation:
$$ 4x^2 - 4x + 1 = x^2 + 4x + 4 + 4x^2 + 16x + 16 - (2x^2 + 8x + 8) $$
Simplify the right-hand side:
$$ x^2 + 4x + 4 + 4x^2 + 16x + 16 - 2x^2 - 8x - 8 = 3x^2 + 12x + 12 $$
Equating both sides:
$$ 4x^2 - 4x + 1 = 3x^2 + 12x + 12 $$
Step 3: Rearrange into standard quadratic form
Rearrange all terms to one side:
$$ 4x^2 - 4x + 1 - 3x^2 - 12x - 12 = 0 $$
$$ x^2 - 16x - 11 = 0 $$
Step 4: Solve the quadratic equation
Use the quadratic formula:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 1,\; b = -16,\; c = -11 $$
Substitute the values:
$$ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(-11)}}{2(1)} $$
$$ x = \frac{16 \pm \sqrt{256 + 44}}{2} $$
$$ x = \frac{16 \pm \sqrt{300}}{2} $$
$$ x = \frac{16 \pm 10\sqrt{3}}{2} $$
$$ x = 8 \pm 5\sqrt{3} $$
Step 5: Choose the valid solution
Since $x$ represents a length, it must be positive. Thus:
$$ x = 8 + 5\sqrt{3} $$
Final Answer
$$ \boxed{x = 8 + 5\sqrt{3}} $$
Question 3
$f(x)=8x^2+10x-3$
Given that $f(x)$ can be written in the form $A(x + B)^2 + C$ where $A$, $B$ and $C$ are constants,
(a) find the value of $A$, the value of $B$ and the value of $C$. (3)
(b) Hence, or otherwise, find,
(i) the value of $x$ for which $f(x)$ has a minimum,
(ii) the minimum value of $f(x)$. (2)
The curve $C$ has equation $y = f(x)$.
(c) Find the $x$ coordinate of each of the points where $C$ crosses the $x$-axis. (2)
The straight line $l$ has equation $y = 2x + 13$.
(d) Use algebra to find the coordinates of the two points of intersection of $C$ and $l$. (4)
Using the same axes and the results of parts (b), (c) and (d),
(e) sketch the curve $C$ and the straight line $l$. (2)
Click to view the solution
<32 id="fpm2023jam02sol3">SolutionThe given function is:
$$ f(x) = 8x^2 + 10x - 3 $$
(a) Write $f(x)$ in the form $A(x + B)^2 + C$
We complete the square to rewrite $f(x)$ in the desired form.
1. Factor $8$ from the quadratic and linear terms:
$$ f(x) = 8\left(x^2 + \frac{10}{8}x\right) - 3 = 8\left(x^2 + \frac{5}{4}x\right) - 3 $$
2. Complete the square inside the parentheses:
$$ x^2 + \frac{5}{4}x = \left(x + \frac{5}{8}\right)^2 - \left(\frac{5}{8}\right)^2 $$
$$ x^2 + \frac{5}{4}x = \left(x + \frac{5}{8}\right)^2 - \frac{25}{64} $$
3. Substitute back into $f(x)$:
$$ f(x) = 8\left[ \left(x + \frac{5}{8}\right)^2 - \frac{25}{64} \right] - 3 $$
$$ f(x) = 8\left(x + \frac{5}{8}\right)^2 - 8 \cdot \frac{25}{64} - 3 $$
$$ f(x) = 8\left(x + \frac{5}{8}\right)^2 - \frac{200}{64} - 3 $$
$$ f(x) = 8\left(x + \frac{5}{8}\right)^2 - \frac{25}{8} - 3 $$
$$ f(x) = 8\left(x + \frac{5}{8}\right)^2 - \frac{49}{8} $$
Thus, $f(x)$ in the form $A(x + B)^2 + C$ is:
$$ A = 8, \quad B = \frac{5}{8}, \quad C = -\frac{49}{8} $$
(b) Find the minimum value of $f(x)$
(i) The value of $x$ for which $f(x)$ has a minimum
The minimum value of a quadratic function occurs at $x = -B$. For $f(x) = A(x + B)^2 + C$:
$$ x = -B = -\frac{5}{8} $$
(ii) The minimum value of $f(x)$
Substitute $x = -\frac{5}{8}$ into the rewritten form:
$$ f(x) = 8\left(x + \frac{5}{8}\right)^2 - \frac{49}{8} $$
At $x = -\frac{5}{8}$:
$$ f(x) = 8(0)^2 - \frac{49}{8} = -\frac{49}{8} $$
Thus, the minimum value of $f(x)$ is:
$$ \boxed{-\frac{49}{8}} $$
(c) Find the $x$-coordinates of the points where $C$ crosses the $x$-axis
At the $x$-axis, $f(x) = 0$:
$$ 8x^2 + 10x - 3 = 0 $$
Use the quadratic formula:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 8,\; b = 10,\; c = -3 $$
$$ x = \frac{-10 \pm \sqrt{10^2 - 4(8)(-3)}}{2(8)} $$
$$ x = \frac{-10 \pm \sqrt{100 + 96}}{16} $$
$$ x = \frac{-10 \pm \sqrt{196}}{16} $$
$$ x = \frac{-10 \pm 14}{16} $$
Solve for the two roots:
$$ x = \frac{-10 + 14}{16} = \frac{4}{16} = \frac{1}{4} $$
$$ x = \frac{-10 - 14}{16} = \frac{-24}{16} = -\frac{3}{2} $$
The $x$-coordinates of the points where $C$ crosses the $x$-axis are:
$$ \boxed{ x = \frac{1}{4} \text{ and } x = -\frac{3}{2} } $$
(d) Find the points of intersection of $C$ and $l$
The curve $C$ has equation $y = f(x)$ and the line $l$ has equation $y = 2x + 13$. Set the two equations equal:
$$ 8x^2 + 10x - 3 = 2x + 13 $$
Rearrange into standard quadratic form:
$$ 8x^2 + 10x - 3 - 2x - 13 = 0 $$
$$ 8x^2 + 8x - 16 = 0 $$
Simplify:
$$ x^2 + x - 2 = 0 $$
Factorize:
$$ x^2 + x - 2 = (x + 2)(x - 1) = 0 $$
Solve for $x$:
$$ x = -2 \quad \text{or} \quad x = 1 $$
Find the corresponding $y$-coordinates using $y = 2x + 13$:
$$ \text{For } x = -2: \quad y = 2(-2) + 13 = -4 + 13 = 9 $$
$$ \text{For } x = 1: \quad y = 2(1) + 13 = 2 + 13 = 15 $$
The points of intersection are:
$$ \boxed{ (-2, 9) \text{ and } (1, 15) } $$
Sketch of the Curve $C$ and the Line $l$
The curve $C$ has equation $y = 8x^2 + 10x - 3$, and the line $l$ has equation $y = 2x + 13$. Using the results:
- From part (b), the minimum value of $f(x)$ is $-\frac{49}{8}$ at $x = -\frac{5}{8}$.
- From part (c), $C$ crosses the $x$-axis at $x = \frac{1}{4}$ and $x = -\frac{3}{2}$.
- From part (d), $C$ and $l$ intersect at points $(-2, 9)$ and $(1, 15)$.
Question 4
The equation of a curve is $y = x^3 \sin x$
Find an equation of the tangent to the curve at the point on the curve where $x = \frac{1}{2}\pi$
Give your answer in the form $y = mx + c$
(7)
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Solution
The equation of the curve is:
$$ y = x^3 \sin x $$
Step 1: Differentiate $y$ to find the gradient
The gradient of the tangent at any point is given by $\frac{dy}{dx}$. Using the product rule:
$$ y = u \cdot v, \quad \text{where } u = x^3 \text{ and } v = \sin x $$
The product rule states:
$$ \frac{dy}{dx} = \frac{du}{dx}v + u\frac{dv}{dx} $$
Compute $\frac{du}{dx}$ and $\frac{dv}{dx}$:
$$ \frac{du}{dx} = 3x^2, \quad \frac{dv}{dx} = \cos x $$
Substitute into the product rule:
$$ \frac{dy}{dx} = (3x^2)(\sin x) + (x^3)(\cos x) $$
$$ \frac{dy}{dx} = 3x^2 \sin x + x^3 \cos x $$
Step 2: Find the gradient at $x = \frac{\pi}{2}$
Substitute $x = \frac{\pi}{2}$ into $\frac{dy}{dx}$:
$$ \frac{dy}{dx} = 3\left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + \left(\frac{\pi}{2}\right)^3 \cos\left(\frac{\pi}{2}\right) $$
Simplify:
$$ \sin\left(\frac{\pi}{2}\right) = 1, \quad \cos\left(\frac{\pi}{2}\right) = 0 $$
$$ \frac{dy}{dx} = 3\left(\frac{\pi}{2}\right)^2 (1) + \left(\frac{\pi}{2}\right)^3 (0) $$
$$ \frac{dy}{dx} = 3\left(\frac{\pi}{2}\right)^2 = \frac{3\pi^2}{4} $$
Thus, the gradient of the tangent at $x = \frac{\pi}{2}$ is:
$$ m = \frac{3\pi^2}{4} $$
Step 3: Find the point on the curve at $x = \frac{\pi}{2}$
Substitute $x = \frac{\pi}{2}$ into $y = x^3 \sin x$:
$$ y = \left(\frac{\pi}{2}\right)^3 \sin\left(\frac{\pi}{2}\right) $$
$$ y = \left(\frac{\pi}{2}\right)^3 (1) = \frac{\pi^3}{8} $$
The point on the curve is:
$$ \left( \frac{\pi}{2}, \frac{\pi^3}{8} \right) $$
Step 4: Equation of the tangent line
The equation of a straight line is:
$$ y - y_1 = m(x - x_1) $$
Substitute $m = \frac{3\pi^2}{4}$, $x_1 = \frac{\pi}{2}$, and $y_1 = \frac{\pi^3}{8}$:
$$ y - \frac{\pi^3}{8} = \frac{3\pi^2}{4} \left( x - \frac{\pi}{2} \right) $$
Simplify:
$$ y - \frac{\pi^3}{8} = \frac{3\pi^2}{4}x - \frac{3\pi^2}{4} \cdot \frac{\pi}{2} $$
$$ y - \frac{\pi^3}{8} = \frac{3\pi^2}{4}x - \frac{3\pi^3}{8} $$
$$ y = \frac{3\pi^2}{4}x - \frac{3\pi^3}{8} + \frac{\pi^3}{8} $$
$$ y = \frac{3\pi^2}{4}x - \frac{2\pi^3}{8} $$
$$ y = \frac{3\pi^2}{4}x - \frac{\pi^3}{4} $$
Final Answer
The equation of the tangent is:
$$ \boxed{ y = \frac{3\pi^2}{4}x - \frac{\pi^3}{4} } $$
Question 5
Figure 2 shows a right pyramid with a square base $ABCD$ and vertex $E$.
The base of the pyramid is horizontal with $AB = BC = 12\,\text{cm}$. The diagonals of the base intersect at the point $O$.
The vertex $E$ of the pyramid is vertically above $O$ and the angle between $EA$ and the plane $ABCD$ is $30^\circ$.
The height of the pyramid is $h\,\text{cm}$.
(a) Find the exact value of $h$ (3)
The point $F$ lies on $AD$ such that $AF:FD = 1:4$
(b) Calculate, to the nearest degree, the size of the angle $EFO$. (4)
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Solution
We are given a right pyramid with a square base $ABCD$, vertex $E$, and the following information:
- The base $ABCD$ is horizontal with $AB = BC = 12\ \text{cm}$.
- The diagonals of the base intersect at point $O$, and $E$ is vertically above $O$.
- The angle between $EA$ and the plane $ABCD$ is $30^\circ$.
(a) Find the exact value of $h$, the height of the pyramid
The diagonal of the square base is:
$$ AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 12^2} = \sqrt{288} = 12\sqrt{2}\ \text{cm} $$
Since $E$ is vertically above $O$, the line $EO$ is perpendicular to the base $ABCD$. The point $O$ is the midpoint of the diagonals $AC$ and $BD$.
In the triangle $\triangle EAO$:
- $AO = \frac{1}{2}AC = \frac{12\sqrt{2}}{2} = 6\sqrt{2}\ \text{cm}$,
- The angle $\angle EAO = 30^\circ$,
- The height $h = EO$ is opposite $\angle EAO$, and $AO$ is adjacent.
Using the tangent of $\angle EAO$:
$$ \tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{AO} $$
$$ \tan(30^\circ) = \frac{h}{6\sqrt{2}} $$
$$ h = 6\sqrt{2}\cdot\tan(30^\circ) $$
Since $\tan(30^\circ)=\frac{1}{\sqrt{3}}$:
$$ h = 6\sqrt{2}\cdot\frac{1}{\sqrt{3}} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt{6}\ \text{cm} $$
Final Answer:
$$ h = 2\sqrt{6}\ \text{cm} $$
(b) Calculate the size of $\angle EFO$ to the nearest degree
The point $F$ lies on $AD$ such that $AF : FD = 1 : 4$. The total length of $AD$ is $12\ \text{cm}$, so:
$$ AF = \frac{1}{5}\cdot12 = \frac{12}{5} = 2.4\ \text{cm}, \quad FD = \frac{4}{5}\cdot12 = 9.6\ \text{cm} $$
Let $M$ be the midpoint of $AD$. Then:
$$ AM = 6 $$
and
$$ FM = 6 - 2.4 = 3.6 $$
Thus,
$$ OF = \sqrt{6^2 + 3.6^2} = \frac{6\sqrt{34}}{5} $$
Let $\angle EFO = \theta$. Then:
$$ \tan\theta = \frac{2\sqrt{6}} {\frac{6\sqrt{34}}{5}} $$
$$ \theta = 35^\circ $$
Final Answer:
$$ \angle EFO = 35^\circ $$
Question 6
The $n$th term of a geometric series $G$ is $U_n$
The first three terms of $G$ are given by
$U_1 = q(4p + 1) \quad U_2 = q(2p + 3) \quad U_3 = q(2p - 3)$
(a) Find the possible values of $p$ (5)
Given that $G$ is convergent with sum to infinity $250$
(b) find the value of $q$ (3)
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Solution
The first three terms of the geometric series $G$ are:
$$ U_1 = q(4p + 1), \quad U_2 = q(2p + 3), \quad U_3 = q(2p - 3). $$
(a) Find the possible values of $p$
The common ratio $r$ is:
$$ r = \frac{U_2}{U_1} = \frac{2p + 3}{4p + 1}, \quad r = \frac{U_3}{U_2} = \frac{2p - 3}{2p + 3}. $$
Equating the two expressions for $r$:
$$ \frac{2p + 3}{4p + 1} = \frac{2p - 3}{2p + 3}. $$
Cross-multiply:
$$ (2p + 3)^2 = (4p + 1)(2p - 3). $$
Step 1: Expand both sides
Expand $(2p + 3)^2$:
$$ (2p + 3)^2 = 4p^2 + 12p + 9. $$
Expand $(4p + 1)(2p - 3)$:
$$ (4p + 1)(2p - 3) = 8p^2 - 12p + 2p - 3 = 8p^2 - 10p - 3. $$
Equating the two sides:
$$ 4p^2 + 12p + 9 = 8p^2 - 10p - 3. $$
Step 2: Simplify the equation
Bring all terms to one side:
$$ 4p^2 + 12p + 9 - 8p^2 + 10p + 3 = 0. $$
$$ -4p^2 + 22p + 12 = 0. $$
Divide through by $-2$ for simplicity:
$$ 2p^2 - 11p - 6 = 0. $$
Step 3: Solve the quadratic equation
The quadratic equation is:
$$ 2p^2 - 11p - 6 = 0. $$
Use the quadratic formula:
$$ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$
Here, $a = 2$, $b = -11$, and $c = -6$. Substitute:
$$ p = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(2)(-6)}}{2(2)}. $$
$$ p = \frac{11 \pm \sqrt{121 + 48}}{4}. $$
$$ p = \frac{11 \pm \sqrt{169}}{4}. $$
$$ p = \frac{11 \pm 13}{4}. $$
Solve for the two possible values of $p$:
$$ p = \frac{11 + 13}{4} = \frac{24}{4} = 6, \quad p = \frac{11 - 13}{4} = \frac{-2}{4} = -\frac{1}{2}. $$
Final Answer:
$$ p = 6 \quad \text{or} \quad p = -\frac{1}{2}. $$
(b) Find the value of $q$
We are given that the sum to infinity $S_\infty$ of the convergent geometric series is 250. For a geometric series, the sum to infinity is given by:
$$ S_\infty = \frac{U_1}{1 - r}. $$
From part (a), we have $U_1 = q(4p + 1)$ and the common ratio
$$ r = \frac{2p + 3}{4p + 1}. $$
Case 1: $p = -\frac{1}{2}$
Substitute $p = -\frac{1}{2}$ into the expression for $r$:
$$ r = \frac{2p + 3}{4p + 1} = \frac{-1 + 3}{-2 + 1} = 2 $$
This contradicts convergence since $|r| > 1$.
Case 2: $p = 6$
Substitute $p = 6$ into the expression for $r$:
$$ r = \frac{2p + 3}{4p + 1} = \frac{15}{25} = 0.6 $$
Substitute these into the formula for $S_\infty$:
$$ 250 = \frac{q(4p + 1)} {1 - \frac{2p + 3}{4p + 1}} = \frac{25q}{1 - 0.6} $$
$$ 250 = \frac{25q}{0.4} $$
$$ q = 4 $$
Final Answer:
$$ q = 4 $$
Question 7
$y = e^{2x} \cos 2x$
(a) Show that
$\frac{dy}{dx} = 2y - 2e^{2x} \sin 2x$ (4)
(b) Hence show that
$\frac{d^2 y}{dx^2} = 4 \frac{dy}{dx} - 8y$
(5)
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Solution
Given the function:
$$ y = e^{2x} \cos 2x $$
we are tasked with finding the first and second derivatives and proving the given relationships.
(a) Show that $ \frac{dy}{dx} = 2y - 2e^{2x} \sin 2x $
To find $ \frac{dy}{dx} $, we use the product rule, which states that:
$$ \frac{d}{dx} \left( u(x) v(x) \right) = u'(x) v(x) + u(x) v'(x) $$
where $ u(x) = e^{2x} $ and $ v(x) = \cos 2x $.
First, differentiate $ u(x) = e^{2x} $:
$$ u'(x) = \frac{d}{dx} e^{2x} = 2e^{2x} $$
Next, differentiate $ v(x) = \cos 2x $:
$$ v'(x) = \frac{d}{dx} \cos 2x = -2 \sin 2x $$
Now apply the product rule:
$$ \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = 2e^{2x} \cos 2x - e^{2x} \cdot 2 \sin 2x $$
$$ \frac{dy}{dx} = 2e^{2x} \cos 2x - 2e^{2x} \sin 2x $$
Since $ y = e^{2x} \cos 2x $, we can rewrite this as:
$$ \frac{dy}{dx} = 2y - 2e^{2x} \sin 2x $$
Thus, we have shown that:
$$ \frac{dy}{dx} = 2y - 2e^{2x} \sin 2x $$
(b) Hence show that $ \frac{d^2y}{dx^2} = 4 \frac{dy}{dx} - 8y $
To find $ \frac{d^2y}{dx^2} $, differentiate $ \frac{dy}{dx} = 2y - 2e^{2x} \sin 2x $ again using the chain rule and product rule.
We have:
$$ \frac{dy}{dx} = 2y - 2e^{2x} \sin 2x $$
Differentiate both terms:
1. The derivative of $ 2y $ is:
$$ \frac{d}{dx} (2y) = 2 \frac{dy}{dx} $$
2. The derivative of $ -2e^{2x} \sin 2x $ is found using the product rule. Let $ u(x) = -2e^{2x} $ and $ v(x) = \sin 2x $. First, differentiate $ u(x) $ and $ v(x) $:
$$ u'(x) = -4e^{2x}, \quad v'(x) = 2 \cos 2x $$
Apply the product rule:
$$ \frac{d}{dx} (-2e^{2x} \sin 2x) = -4e^{2x} \sin 2x - 4e^{2x} \cos 2x $$
Now, combine the derivatives:
$$ \frac{d^2y}{dx^2} = 2 \frac{dy}{dx} - 4e^{2x} \sin 2x - 4e^{2x} \cos 2x $$
Substitute $ \frac{dy}{dx} = 2y - 2e^{2x} \sin 2x $ from part (a):
$$ \frac{d^2y}{dx^2} = 2(2y - 2e^{2x} \sin 2x) - 4e^{2x} \sin 2x - 4e^{2x} \cos 2x $$
$$ \frac{d^2y}{dx^2} = 4y - 4e^{2x} \sin 2x - 4e^{2x} \sin 2x - 4e^{2x} \cos 2x $$
$$ \frac{d^2y}{dx^2} = 4y - 8e^{2x} \sin 2x - 4e^{2x} \cos 2x $$
Finally, notice that:
$$ 4y = 4e^{2x} \cos 2x $$
so we have:
$$ \frac{d^2y}{dx^2} = 4 \frac{dy}{dx} - 8y $$
Thus, we have shown that:
$$ \frac{d^2y}{dx^2} = 4 \frac{dy}{dx} - 8y $$
Question 8
The quadratic equation
$x^2 - 4k\sqrt{2}x + 2k^4 - 1 = 0$
where $k$ is a positive constant, has roots $\alpha$ and $\beta$
Given that $\alpha^2 + \beta^2 = 66$ and that $\alpha^3 + \beta^3 = p\sqrt{2}$ where $p$ is an integer,
find the value of $p$ (11)
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Solution
We are given the quadratic equation:
$$ x^2 - 4k\sqrt{2}\,x + 2k^4 - 1 = 0 $$
where $k$ is a positive constant and the roots are $\alpha$ and $\beta$.
Step 1: Use Vieta's Formulas
From Vieta’s formulas, for the quadratic equation $x^2 + bx + c = 0$, the sum and product of the roots are given by:
$$ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} $$
For the equation $x^2 - 4k\sqrt{2}\,x + (2k^4 - 1) = 0$, we have:
$$ \alpha + \beta = 4k\sqrt{2}, \quad \alpha\beta = 2k^4 - 1 $$
Step 2: Use the Identity for $ \alpha^2 + \beta^2 $
We are given that:
$$ \alpha^2 + \beta^2 = 66 $$
Recall the identity:
$$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta $$
Substitute the values from Vieta’s formulas:
$$ \alpha^2 + \beta^2 = (4k\sqrt{2})^2 - 2(2k^4 - 1) $$
Simplify:
$$ \alpha^2 + \beta^2 = 16k^2 \cdot 2 - 2(2k^4 - 1) $$
$$ 66 = 32k^2 - 2(2k^4 - 1) $$
$$ 66 = 32k^2 - 4k^4 + 2 $$
Now, simplify further:
$$ 66 - 2 = 32k^2 - 4k^4 $$
$$ 64 = 32k^2 - 4k^4 $$
Divide through by 4:
$$ 16 = 8k^2 - k^4 $$
Rearrange this to form a quadratic equation in $k^2$:
$$ k^4 - 8k^2 + 16 = 0 $$
Let $z = k^2$. The equation becomes:
$$ z^2 - 8z + 16 = 0 $$
Solve this quadratic equation using the quadratic formula:
$$ z = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(16)}}{2(1)} = \frac{8 \pm \sqrt{64 - 64}}{2} = \frac{8 \pm 0}{2} $$
Thus:
$$ z = 4 $$
Therefore:
$$ k^2 = 4, \quad k = 2 $$
Step 3: Find $p$
We are given that:
$$ \alpha^3 + \beta^3 = p\sqrt{2} $$
We use the identity for the sum of cubes:
$$ \alpha^3 + \beta^3 = (\alpha + \beta) \left( (\alpha + \beta)^2 - 3\alpha\beta \right) $$
Substitute the values for $\alpha + \beta$ and $\alpha\beta$:
$$ \alpha^3 + \beta^3 = (4k\sqrt{2}) \left( (4k\sqrt{2})^2 - 3(2k^4 - 1) \right) $$
Substitute $k = 2$:
$$ \alpha^3 + \beta^3 = (4(2)\sqrt{2}) \left( (4(2)\sqrt{2})^2 - 3(2(2)^4 - 1) \right) $$
$$ \alpha^3 + \beta^3 = 8\sqrt{2} \left( 64 \cdot 2 - 3(2(16) - 1) \right) $$
$$ \alpha^3 + \beta^3 = 8\sqrt{2} \left( 128 - 3(32 - 1) \right) $$
$$ \alpha^3 + \beta^3 = 8\sqrt{2} \left( 128 - 3 \cdot 31 \right) $$
$$ \alpha^3 + \beta^3 = 8\sqrt{2} \left( 128 - 93 \right) $$
$$ \alpha^3 + \beta^3 = 8\sqrt{2} \cdot 35 $$
$$ \alpha^3 + \beta^3 = 280\sqrt{2} $$
Thus, we have:
$$ p = 280 $$
Final Answer:
$$ p = 280 $$
Question 9
A cube has edges of length $x\,\text{cm}$.
The total surface area, $A\,\text{cm}^2$, of the cube is increasing at a constant rate of $0.45\,\text{cm}^2/\text{s}$
Find the rate of increase, in $\text{cm}^3/\text{s}$, of the volume of the cube at the instant when the total surface area of the cube is $384\,\text{cm}^2$ (7)
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Solution
Step 1: Relationship between Surface Area and Edge Length
The surface area of the cube is given by:
$$ A = 6x^2 $$
Differentiating with respect to time $t$, we have:
$$ \frac{dA}{dt} = 12x \frac{dx}{dt} $$
Step 2: Relationship between Volume and Edge Length
The volume of the cube is given by:
$$ V = x^3 $$
Differentiating with respect to $t$, we have:
$$ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} $$
Step 3: Determine $x$ when $A = 384 \, \text{cm}^2$
Given $A = 384 \, \text{cm}^2$, solve for $x$:
$$ 6x^2 = 384 \implies x^2 = 64 \implies x = 8 \, \text{cm} $$
Step 4: Calculate $ \frac{dx}{dt} $
Given $ \frac{dA}{dt} = 0.45 \, \text{cm}^2/\text{s} $:
$$ \frac{dA}{dt} = 12x \frac{dx}{dt} \implies 0.45 = 12(8)\frac{dx}{dt} \implies \frac{dx}{dt} = \frac{0.45}{96} = 0.0046875 \, \text{cm}/\text{s} $$
Step 5: Calculate $ \frac{dV}{dt} $
Substitute $x = 8 \, \text{cm}$ and $ \frac{dx}{dt} = 0.0046875 \, \text{cm}/\text{s} $ into:
$$ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} $$
$$ \frac{dV}{dt} = 3(8^2)(0.0046875) = 3(64)(0.0046875) = 0.9 \, \text{cm}^3/\text{s} $$
Final Answer
The rate of increase of the volume of the cube is:
$$ \boxed{0.9 \, \text{cm}^3/\text{s}} $$
Question 10
Using formulae given on page 2
(a) show that
(i) $\sin 2 \theta=2 \sin \theta \cos \theta$
(ii) $\cos 2 \theta=2 \cos ^2 \theta-1$ (5)
Given that $\theta \neq\left(90^\circ+180^\circ n\right)$ where $n \in \mathbb{Z}$
(b) use the results from part (a) to show that
$\sin 2 \theta-\tan \theta$ can be written as $\tan \theta \cos 2 \theta$ (4)
(c) Solve for $0 \lt x \lt 360$
$$ \sin 2 x^\circ-\tan x^\circ=0 $$
(4)
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Solution
Part (a)
(i) Proof of $ \sin 2\theta = 2\sin\theta\cos\theta $
$$ \sin 2\theta = \sin(\theta + \theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta $$
(ii) Proof of $ \cos 2\theta = 2\cos^2\theta - 1 $
Using the double-angle identity for cosine:
$$ \cos 2\theta = \cos^2\theta - \sin^2\theta $$
Substitute $ \sin^2\theta = 1 - \cos^2\theta $:
$$ \cos 2\theta = \cos^2\theta - (1 - \cos^2\theta) = 2\cos^2\theta - 1 $$
Part (b)
Starting with:
$$ \sin 2\theta - \tan\theta $$
Substitute $ \sin 2\theta = 2\sin\theta\cos\theta $ and $ \tan\theta = \frac{\sin\theta}{\cos\theta} $:
$$ \sin 2\theta - \tan\theta = 2\sin\theta\cos\theta - \frac{\sin\theta}{\cos\theta} $$
Combine terms over a common denominator:
$$ \sin 2\theta - \tan\theta = \frac{2\sin\theta\cos^2\theta - \sin\theta}{\cos\theta} $$
Factor out $ \sin\theta $:
$$ \sin 2\theta - \tan\theta = \frac{\sin\theta(2\cos^2\theta - 1)}{\cos\theta} $$
Using $ \cos 2\theta = 2\cos^2\theta - 1 $, we have:
$$ \sin 2\theta - \tan\theta = \frac{\sin\theta\cos 2\theta}{\cos\theta} $$
Simplify:
$$ \sin 2\theta - \tan\theta = \tan\theta\cos 2\theta $$
Part (c)
Solve the equation:
$$ \sin 2x^\circ - \tan x^\circ = 0 $$
Using part (b):
$$ \tan x^\circ \cos 2x^\circ = 0 $$
This implies:
$$ \tan x^\circ = 0 \quad \text{or} \quad \cos 2x^\circ = 0 $$
Case 1: $ \tan x^\circ = 0 $
$$ \tan x^\circ = 0 \implies x = 0^\circ,\ 180^\circ $$
Since $ 0^\circ \lt x \lt 360^\circ $, the solution is:
$$ x = 180^\circ $$
Case 2: $ \cos 2x^\circ = 0 $
$$ \cos 2x^\circ = 0 \implies 2x = 90^\circ + 180^\circ n \quad (n \in \mathbb{Z}) $$
$$ 2x = 90^\circ,\ 270^\circ,\ 450^\circ,\ 630^\circ \implies x = 45^\circ,\ 135^\circ,\ 225^\circ,\ 315^\circ $$
Final Solution
The solutions for $ 0^\circ \lt x \lt 360^\circ $ are:
$$ \boxed{x = 45^\circ,\ 135^\circ,\ 180^\circ,\ 225^\circ,\ 315^\circ} $$
Question 11
Figure 3 shows part of the curve $C$ with equation $y=4-\mathrm{e}^{2x}$
The curve $C$ crosses the $y$-axis at the point $A$ and the $x$-axis at the point $B$.
(a) (i) Write down the $y$ coordinate of point $A$.
(ii) Show that the $x$ coordinate of $B$ is $x=\ln 2$ (3)
The line $l$ is the normal to $C$ at the point $B$.
(b) Find an equation for $l$, giving your answer in the form $y=mx+c$ (4)
The finite region $R$ is bounded by $C$, $l$ and the $y$-axis.
(c) Using calculus, find the area of $R$.
Give your answer to one decimal place.
(7)
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Solution
Part (a)
(i) $y$-coordinate of point $A$
Point $A$ is where the curve crosses the $y$-axis. At the $y$-axis, $x = 0$. Substituting $x = 0$ into the equation of the curve:
$$ y = 4 - e^{2(0)} = 4 - e^0 = 4 - 1 = 3 $$
Thus, the $y$-coordinate of $A$ is:
$$ \boxed{3} $$
(ii) Show that the $x$-coordinate of $B$ is $x = \ln 2$
Point $B$ is where the curve crosses the $x$-axis. At the $x$-axis, $y = 0$. Substituting $y = 0$ into the equation of the curve:
$$ 0 = 4 - e^{2x} $$
Rearranging:
$$ e^{2x} = 4 $$
Taking the natural logarithm on both sides:
$$ \ln(e^{2x}) = \ln 4 $$
Using the logarithmic property $ \ln(e^a) = a $, we have:
$$ 2x = \ln 4 $$
Divide through by 2:
$$ x = \frac{\ln 4}{2} $$
Using the logarithmic property $ \ln(a^b) = b\ln a $, rewrite $ \ln 4 $ as $ \ln(2^2) = 2\ln 2 $:
$$ x = \frac{2\ln 2}{2} = \ln 2 $$
Thus, the $x$-coordinate of $B$ is:
$$ \boxed{\ln 2} $$
Part (b): Equation of the Normal Line $l$
The point $B$ is where $x = \ln 2$. Substituting $x = \ln 2$ into the equation of the curve $y = 4 - e^{2x}$:
$$ y = 4 - e^{2(\ln 2)} = 4 - e^{\ln 4} = 4 - 4 = 0 $$
Thus, point $B$ is $$ (\ln 2, 0) $$
To find the equation of the normal, first find the derivative of $y = 4 - e^{2x}$ to obtain the gradient of the tangent:
$$ \frac{dy}{dx} = -2e^{2x} $$
At $x = \ln 2$:
$$ \frac{dy}{dx} = -2e^{2(\ln 2)} = -2(4) = -8 $$
The gradient of the tangent at $B$ is $-8$, so the gradient of the normal is:
$$ m = -\frac{1}{-8} = \frac{1}{8} $$
Using the point-slope form of a line, the equation of the normal at $(\ln 2, 0)$ is:
$$ y - 0 = \frac{1}{8}(x - \ln 2) $$
Simplify:
$$ y = \frac{1}{8}x - \frac{\ln 2}{8} $$
Thus, the equation of the normal line is:
$$ \boxed{ y = \frac{1}{8}x - \frac{\ln 2}{8} } $$
Part (c): Area of the Region $R$
The region $R$ is bounded by the curve $C$, the normal line $l$, and the $y$-axis.
Step 1: Set up the integral bounds
The curve $C$ intersects the $y$-axis at $x = 0$, and the region ends at the $x$-coordinate of $B$, $x = \ln 2$.
The area under $C$ is given by:
$$ \text{Area under } C = \int_0^{\ln 2} (4 - e^{2x}) \, dx $$
Step 2: Area under the line $l$
The line $l$ intersects the $y$-axis at $x = 0$, where:
$$ y = \frac{1}{8}(0) - \frac{\ln 2}{8} = - \frac{\ln 2}{8} $$
The area under $l$ between $x = 0$ and $x = \ln 2$ is:
$$ \text{Area under } l = \int_0^{\ln 2} \left( \frac{1}{8}x - \frac{\ln 2}{8} \right) dx $$
Step 3: Total area of $R$
The region $R$ is the area under the curve $C$ minus the area under the line $l$:
$$ \text{Area of } R = \int_0^{\ln 2} (4 - e^{2x}) \, dx - \int_0^{\ln 2} \left( \frac{1}{8}x - \frac{\ln 2}{8} \right) dx $$
Step 4: Solve the integrals
$$ \int_0^{\ln 2} (4 - e^{2x}) \, dx = \int_0^{\ln 2} 4 \, dx - \int_0^{\ln 2} e^{2x} \, dx $$
$$ = \left[ 4x \right]_0^{\ln 2} - \left[ \frac{e^{2x}}{2} \right]_0^{\ln 2} $$
$$ = 4\ln 2 - \left( \frac{e^{2\ln 2}}{2} - \frac{e^0}{2} \right) $$
$$ = 4\ln 2 - \left( \frac{4}{2} - \frac{1}{2} \right) $$
$$ = 4\ln 2 - \frac{3}{2} $$
For the line $l$:
$$ \int_0^{\ln 2} \left( \frac{1}{8}x - \frac{\ln 2}{8} \right) dx = \int_0^{\ln 2} \frac{1}{8}x \, dx - \int_0^{\ln 2} \frac{\ln 2}{8} \, dx $$
$$ = \left[ \frac{1}{16}x^2 \right]_0^{\ln 2} - \frac{\ln 2}{8} \int_0^{\ln 2} 1 \, dx $$
$$ = \frac{(\ln 2)^2}{16} - \frac{(\ln 2)^2}{8} $$
$$ = - \frac{(\ln 2)^2}{16} $$
Thus, the total area is:
$$ \text{Area of } R = \left( 4\ln 2 - \frac{3}{2} \right) - \left( - \frac{(\ln 2)^2}{16} \right) \approx 1.3 $$
Final Area
$$ \boxed{1.3} $$
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