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Prove, by exhaustion, that if $n$ is an integer and $2 \leqslant n \leqslant 7$, then $A = n^2 + 2$ is not divisible by $4$. (4)
$\displaystyle \begin{aligned} \text{For } n = 2, \quad & A = 2^2 + 2 = 6. && \frac{6}{4} = 1.5 \text{ (not an integer)} \\ \text{For } n = 3, \quad & A = 3^2 + 2 = 11. && \frac{11}{4} = 2.75 \text{ (not an integer)} \\ \text{For } n = 4, \quad & A = 4^2 + 2 = 18. && \frac{18}{4} = 4.5 \text{ (not an integer)} \\ \text{For } n = 5, \quad & A = 5^2 + 2 = 27. && \frac{27}{4} = 6.75 \text{ (not an integer)} \\ \text{For } n = 6, \quad & A = 6^2 + 2 = 38. && \frac{38}{4} = 9.5 \text{ (not an integer)} \\ \text{For } n = 7, \quad & A = 7^2 + 2 = 51. && \frac{51}{4} = 12.75 \text{ (not an integer)} \end{aligned}$Since $A$ is not divisible by $4$ for all integers $n$ where $2 \leqslant n \leqslant 7$, the statement is proven by exhaustion.
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Given that $a$ and $b$ are positive constants, solve the simultaneous equations
$\displaystyle \begin{aligned} \log_6 a + \log_6 b &= 2 \\ \frac{a}{b} &= 144 \end{aligned}$Show each step of your working giving exact values for $a$ and $b$. (6)
$\displaystyle \begin{aligned} \log_6 a + \log_6 b &= 2 \\ \log_6(ab) &= 2 \quad \text{(using product law of logarithms)} \\ ab &= 6^2 \\ ab &= 36 \quad \text{--- (1)} \\ \frac{a}{b} &= 144 \implies a = 144b \quad \text{--- (2)} \\ \text{Substitute (2) into (1):} \quad & (144b)b = 36 \\ & 144b^2 = 36 \\ & b^2 = \frac{36}{144} = \frac{1}{4} \\ & b = \frac{1}{2} \quad \text{(since } b \text{ is a positive constant)} \\ \text{Substitute } b = \frac{1}{2} \text{ into (2):} \quad & a = 144\left(\frac{1}{2}\right) = 72 \\ & \therefore a = 72, \quad b = \frac{1}{2} \end{aligned}$ -
$\mathrm{f}(x) = 2x^3 - 3px^2 + x + 4p$
Given that $(x - 4)$ is a factor of $\mathrm{f}(x)$,
(a) show that the value of $p$ is $3$. (2)Using this value of $p$,(b) find the remainder when $\mathrm{f}(x)$ is divided by $(x + 2)$. (2)(c) factorise $\mathrm{f}(x)$ completely. (3)(a) By the Factor Theorem, if $(x - 4)$ is a factor, $\mathrm{f}(4) = 0$.
$\displaystyle \begin{aligned} 2(4)^3 - 3p(4)^2 + (4) + 4p &= 0 \\ 2(64) - 3p(16) + 4 + 4p &= 0 \\ 128 - 48p + 4 + 4p &= 0 \\ 132 - 44p &= 0 \\ 44p &= 132 \implies p = 3 \quad \text{(shown)} \end{aligned}$(b) Substitute $p = 3$ into $\mathrm{f}(x)$:
$\displaystyle \begin{aligned} \mathrm{f}(x) &= 2x^3 - 3(3)x^2 + x + 4(3) \\ &= 2x^3 - 9x^2 + x + 12 \end{aligned}$
By the Remainder Theorem, the remainder is $\mathrm{f}(-2)$.
$\displaystyle \begin{aligned} \mathrm{f}(-2) &= 2(-2)^3 - 9(-2)^2 + (-2) + 12 \\ &= 2(-8) - 9(4) - 2 + 12 \\ &= -16 - 36 - 2 + 12 = -42 \end{aligned}$
The remainder is $-42$.(c) Since $(x - 4)$ is a factor, we can write:
$2x^3 - 9x^2 + x + 12 = (x - 4)(ax^2 + bx + c)$
By inspection of the first and last terms: $a = 2 \text{ and } -4c = 12 \implies c = -3$.
$2x^3 - 9x^2 + x + 12 = (x - 4)(2x^2 + bx - 3)$
Comparing coefficients of $x^2$: $-9 = -8 + b \implies b = -1$.
$\displaystyle \begin{aligned} \text{The quadratic factor is } & 2x^2 - x - 3. \\ \text{Factorising the quadratic: } & 2x^2 - x - 3 = (2x - 3)(x + 1) \\ \therefore \mathrm{f}(x) &= (x - 4)(2x - 3)(x + 1) \end{aligned}$ -
Figure 1 shows a sketch of part of the graph of $y = (1 + x^2)^5, x \geqslant 0$.
Figure 1Complete the table below giving your values of $y$ rounded to 4 decimal places. (2)
$x$ $0$ $0.1$ $0.2$ $0.3$ $0.4$ $y$ $1.0000$ $1.0510$ $1.2167$ $1.5386$ $2.1003$ Use the trapezium rule with 4 strips to estimate the approximate value, to 3 decimal places, for
$\displaystyle \int_{0}^{0.4} (1 + x^2)^5 \, \mathrm{d}x $ (4)$\displaystyle \begin{aligned} \text{When } x = 0.1, \quad & y = (1 + 0.1^2)^5 = (1.01)^5 \approx 1.05101 \rightarrow 1.0510 \\ \text{When } x = 0.3, \quad & y = (1 + 0.3^2)^5 = (1.09)^5 \approx 1.53862 \rightarrow 1.5386 \end{aligned}$
Using the Trapezium Rule:
$\displaystyle \begin{aligned} h &= \frac{0.4 - 0}{4} = 0.1 \\ \int_{0}^{0.4} (1 + x^2)^5 \, \mathrm{d}x &\approx \frac{h}{2} \left[ y_0 + y_4 + 2(y_1 + y_2 + y_3) \right] \\ &\approx \frac{0.1}{2} \left[ 1.0000 + 2.1003 + 2(1.0510 + 1.2167 + 1.5386) \right] \\ &\approx 0.05 \left[ 3.1003 + 2(3.8063) \right] \\ &\approx 0.05 \left[ 3.1003 + 7.6126 \right] \\ &\approx 0.05 \left[ 10.7129 \right] \approx 0.535645 \\ &\approx 0.536 \text{ (to 3 d.p.)} \end{aligned}$ -
The $n$th term of a geometric series is $t_n$ and the common ratio is $r$.
Given that $t_3 + t_6 = \frac{28}{81}$ and $t_3 - t_6 = \frac{76}{405}$
(a)(i) show that $r = \frac{2}{3}$.(a)(ii) find the first term of the series. (5)(b) Find the sum to infinity of this geometric series. (2)(a) (i) $t_n = a r^{n-1} \implies t_3 = a r^2 \text{ and } t_6 = a r^5$
$\displaystyle \begin{aligned} a r^2 + a r^5 &= \frac{28}{81} \implies a r^2(1 + r^3) = \frac{28}{81} \quad \text{--- (1)} \\ a r^2 - a r^5 &= \frac{76}{405} \implies a r^2(1 - r^3) = \frac{76}{405} \quad \text{--- (2)} \end{aligned}$
Divide (1) by (2):
$\displaystyle \begin{aligned} \frac{a r^2(1 + r^3)}{a r^2(1 - r^3)} &= \frac{28/81}{76/405} \\ \frac{1 + r^3}{1 - r^3} &= \frac{28}{81} \times \frac{405}{76} = \frac{28 \times 5}{76} = \frac{140}{76} = \frac{35}{19} \\ 19(1 + r^3) &= 35(1 - r^3) \\ 19 + 19r^3 &= 35 - 35r^3 \\ 54r^3 &= 16 \implies r^3 = \frac{16}{54} = \frac{8}{27} \\ r &= \sqrt[3]{\frac{8}{27}} = \frac{2}{3} \quad \text{(shown)} \end{aligned}$
(ii) Substitute $r = \frac{2}{3}$ into (1):
$\displaystyle \begin{aligned} a \left(\frac{2}{3}\right)^2 \left(1 + \left(\frac{2}{3}\right)^3\right) &= \frac{28}{81} \\ a \left(\frac{4}{9}\right) \left(1 + \frac{8}{27}\right) &= \frac{28}{81} \\ a \left(\frac{4}{9}\right) \left(\frac{35}{27}\right) &= \frac{28}{81} \\ a \left(\frac{140}{243}\right) &= \frac{28}{81} \implies a = \frac{28}{81} \times \frac{243}{140} = \frac{1}{5} \times 3 = \frac{3}{5} \end{aligned}$
The first term $a = \frac{3}{5}$ (or $0.6$).(b) $\displaystyle \begin{aligned} S_\infty &= \frac{a}{1 - r} = \frac{3/5}{1 - 2/3} = \frac{3/5}{1/3} = \frac{3}{5} \times \frac{3}{1} = \frac{9}{5} \text{ (or } 1.8\text{)} \end{aligned}$ -
A circle with centre $O$ has equation $x^2 - 2x + y^2 + 10y - 19 = 0$.
(a)(i) Find the coordinates of $O$.(a)(ii) Find the radius of the circle. (4)Point $P$ has coordinates $(7, -2)$.(b) Verify that $P$ lies on the circle. (1)(c) Find the equation of the tangent to the circle at $P$. Give your answer in the form $ax + by + c = 0$ where $a, b$ and $c$ are integers. (4)(a) Complete the square for $x$ and $y$:
$\displaystyle \begin{aligned} (x^2 - 2x) + (y^2 + 10y) &= 19 \\ (x - 1)^2 - 1^2 + (y + 5)^2 - 5^2 &= 19 \\ (x - 1)^2 - 1 + (y + 5)^2 - 25 &= 19 \\ (x - 1)^2 + (y + 5)^2 &= 19 + 1 + 25 = 45 \end{aligned}$
(i) Coordinates of centre $O$ are $(1, -5)$.
(ii) $r^2 = 45 \implies \text{Radius } r = \sqrt{45} = 3\sqrt{5}$.(b) Substitute $P(7, -2)$ into the circle's equation $(x - 1)^2 + (y + 5)^2 = 45$:
$\displaystyle \begin{aligned} \text{LHS} &= (7 - 1)^2 + (-2 + 5)^2 \\ &= (6)^2 + (3)^2 = 36 + 9 = 45 \end{aligned}$
Since LHS = RHS (45 = 45), point $P$ lies on the circle.(c) Gradient of radius $OP$:
$\displaystyle \begin{aligned} m_{OP} &= \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - (-5)}{7 - 1} = \frac{3}{6} = \frac{1}{2} \end{aligned}$
Since the tangent is perpendicular to the radius at $P$:
$\displaystyle \begin{aligned} \text{Gradient of tangent } m &= - \frac{1}{m_{OP}} = -2 \end{aligned}$
Equation of the tangent at $P(7, -2)$:
$\displaystyle \begin{aligned} y - y_1 &= m(x - x_1) \\ y - (-2) &= -2(x - 7) \\ y + 2 &= -2x + 14 \implies 2x + y - 12 = 0 \end{aligned}$ -
Two numbers $x$ and $y$ are such that $3x + y = 15$.
The sum of the squares of $3x$ and $y$ is $S$.(a) Show that $S = 18x^2 - 90x + 225$. (3)Using calculus,(b) find the value of $x$ for which $S$ is a minimum, justifying that this value of $x$ gives a minimum value of $S$. (4)(c) Find the minimum value of $S$. (2)(a) $\displaystyle \begin{aligned} 3x + y &= 15 \implies y = 15 - 3x \\ S &= (3x)^2 + y^2 \\ S &= 9x^2 + (15 - 3x)^2 \\ S &= 9x^2 + 225 - 90x + 9x^2 \\ S &= 18x^2 - 90x + 225 \quad \text{(shown)} \end{aligned}$ (b) $\displaystyle \begin{aligned} \frac{\mathrm{d}S}{\mathrm{d}x} &= 36x - 90 \end{aligned}$
For a minimum, $\frac{\mathrm{d}S}{\mathrm{d}x} = 0$:
$\displaystyle \begin{aligned} 36x - 90 &= 0 \implies 36x = 90 \implies x = 2.5 \end{aligned}$
Justification: $\frac{\mathrm{d}^2S}{\mathrm{d}x^2} = 36$. Since $\frac{\mathrm{d}^2S}{\mathrm{d}x^2} > 0$, $S$ is a minimum when $x = 2.5$.(c) Substitute $x = 2.5$ into $S$:
$\displaystyle \begin{aligned} S_{\text{min}} &= 18(2.5)^2 - 90(2.5) + 225 \\ S_{\text{min}} &= 18(6.25) - 225 + 225 = 112.5 \end{aligned}$ -
(a) Find the first 4 terms of the binomial expansion, in ascending powers of $x$, of$\displaystyle \left(1 - \frac{x}{4}\right)^9 $giving each term in its simplest form. (3)(b) Use your expansion to estimate the value of $(0.975)^9$, giving your answer to 4 decimal places. (3)
(a) $\displaystyle \begin{aligned} \left(1 - \frac{x}{4}\right)^9 &= 1^9 + \binom{9}{1}(1)^8\left(-\frac{x}{4}\right) + \binom{9}{2}(1)^7\left(-\frac{x}{4}\right)^2 + \binom{9}{3}(1)^6\left(-\frac{x}{4}\right)^3 + \dots \\ &= 1 + 9\left(-\frac{x}{4}\right) + 36\left(\frac{x^2}{16}\right) + 84\left(-\frac{x^3}{64}\right) + \dots \\ &= 1 - \frac{9}{4}x + \frac{9}{4}x^2 - \frac{21}{16}x^3 \end{aligned}$ (b) Let $\left(1 - \frac{x}{4}\right) = 0.975 \implies \frac{x}{4} = 0.025 \implies x = 0.1$.
$\displaystyle \begin{aligned} (0.975)^9 &\approx 1 - \frac{9}{4}(0.1) + \frac{9}{4}(0.1)^2 - \frac{21}{16}(0.1)^3 \\ &\approx 1 - 0.225 + 0.0225 - 0.0013125 \\ &\approx 0.7961875 \approx 0.7962 \quad \text{(to 4 d.p.)} \end{aligned}$ -
The line with equation $y = 3x + 10$ intersects the curve with equation $y = -x^2 + x + 13$ at the points $P$ and $Q$ as shown in Figure 2.
Figure 2(a) Use algebra to find the coordinates of $P$ and the coordinates of $Q$. (4)The shaded region $S$ is bounded by the line and the curve as shown in Figure 2.(b) Use calculus to find the exact area of $S$. (7)(a) Set equations equal:
$\displaystyle \begin{aligned} 3x + 10 &= -x^2 + x + 13 \\ x^2 + 2x - 3 &= 0 \\ (x + 3)(x - 1) &= 0 \implies x = -3 \quad \text{or} \quad x = 1 \end{aligned}$
$\displaystyle \begin{aligned} \text{When } x &= -3, \quad y = 3(-3) + 10 = 1 \implies P(-3, 1) \\ \text{When } x &= 1, \quad y = 3(1) + 10 = 13 \implies Q(1, 13) \end{aligned}$
Coordinates are $P(-3, 1)$ and $Q(1, 13)$.(b) $\displaystyle \begin{aligned} \text{Area of } S &= \int_{-3}^{1} (\text{Curve} - \text{Line}) \, \mathrm{d}x \\ &= \int_{-3}^{1} ((-x^2 + x + 13) - (3x + 10)) \, \mathrm{d}x \\ &= \int_{-3}^{1} (-x^2 - 2x + 3) \, \mathrm{d}x \\ &= \left[ -\frac{x^3}{3} - x^2 + 3x \right]_{-3}^{1} \\ &= \left( -\frac{1^3}{3} - 1^2 + 3(1) \right) - \left( -\frac{(-3)^3}{3} - (-3)^2 + 3(-3) \right) \\ &= \left( -\frac{1}{3} - 1 + 3 \right) - \left( \frac{27}{3} - 9 - 9 \right) \\ &= \left( \frac{5}{3} \right) - (-9) = \frac{5}{3} + \frac{27}{3} = \frac{32}{3} \end{aligned}$ -
(a) Solve for $0 \leqslant x \leqslant 180^\circ$, giving your answers in degrees to 1 decimal place,$\displaystyle 2\tan(2x + 30^\circ) = 3 $(4)(b) Find, for $0 \leqslant x \leqslant \pi$, all the solutions of$\displaystyle 6\cos^2 x + \sin x - 4 = 0 $giving your answers in radians to 3 significant figures.
You must show clearly how you obtained your answers. (6)(a) $\displaystyle \begin{aligned} 2\tan(2x + 30^\circ) &= 3 \\ \tan(2x + 30^\circ) &= 1.5 \end{aligned}$
Let $\theta = 2x + 30^\circ$. For $0 \leqslant x \leqslant 180^\circ \implies 30^\circ \leqslant \theta \leqslant 390^\circ$.
$\displaystyle \begin{aligned} \text{Basic angle } &= \arctan(1.5) \approx 56.3099^\circ \\ \theta &= 56.3099^\circ, \quad 180^\circ + 56.3099^\circ = 236.3099^\circ \end{aligned}$
$\displaystyle \begin{aligned} 2x + 30^\circ &= 56.3099^\circ \implies 2x = 26.3099^\circ \implies x = 13.2^\circ \text{ (to 1 d.p.)} \\ 2x + 30^\circ &= 236.3099^\circ \implies 2x = 206.3099^\circ \implies x = 103.2^\circ \text{ (to 1 d.p.)} \end{aligned}$
$x = 13.2^\circ, \quad 103.2^\circ$(b) $\displaystyle \begin{aligned} 6\cos^2 x + \sin x - 4 &= 0 \\ 6(1 - \sin^2 x) + \sin x - 4 &= 0 \\ 6 - 6\sin^2 x + \sin x - 4 &= 0 \\ 6\sin^2 x - \sin x - 2 &= 0 \\ (3\sin x - 2)(2\sin x + 1) &= 0 \end{aligned}$
$\sin x = \frac{2}{3} \quad \text{or} \quad \sin x = -\frac{1}{2}$
Since $0 \leqslant x \leqslant \pi$, $\sin x$ must be positive. Therefore, $\sin x = -\frac{1}{2}$ has no solution.
$\displaystyle \begin{aligned} \sin x &= \frac{2}{3} \\ x &= \arcsin\left(\frac{2}{3}\right) \approx 0.7297 \text{ rad} \\ x &= \pi - 0.7297 \approx 2.4118 \text{ rad} \end{aligned}$
$x = 0.730, \quad 2.41 \text{ (to 3 s.f.)}$
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