# As & A Level 9709 Mathematics Guess Paper 1

### Question

1 Find the term independent of $x$ in the expansion of $\left(x-\frac{3}{2 x}\right)^6$. 

Ans:

$\left(x-\frac{3}{2 x}\right)^6$
Term is ${ }^6 \mathrm{C}_3 \times x^3 \times\left(\frac{-3}{2 x}\right)^3$
$\rightarrow-67.5 \mathrm{oe}$

### Question

2 (i) Express the equation $\sin 2 x+3 \cos 2 x=3(\sin 2 x-\cos 2 x)$ in the form $\tan 2 x=k$, where $k$ is a constant. 
(ii) Hence solve the equation for $-90^{\circ} \leqslant x \leqslant 90^{\circ}$. Answer/Explanation

Ans:
\text { (i) } \quad \begin{aligned} & 2 \sin 2 x=6 \cos 2 x \\ & \tan 2 x=k \\ & \rightarrow \tan 2 x=3 \text { or } k=3 \end{aligned}
(ii)
\begin{aligned} x= & \left(\tan ^{-1}(\text { their } k)\right) \div 2 \\ & \left(71.6^{\circ} \text { or }-108.4^{\circ}\right) \div 2 \\ & x=35.8^{\circ},-54.2^{\circ} \\ x= & 0.624^{\mathrm{c}},-0.946^{\mathrm{c}} \\ x= & 0.198 \pi^{\mathrm{c}},-0.301 \pi^{\mathrm{c}} \end{aligned} ### Question 3 (i) Find the term independent ofx$in the expansion of$\left(\frac{2}{x}-3 x\right)^6$. (ii) Find the value of$a$for which there is no term independent of$x$in the expansion of$
\left(1+a x^2\right)\left(\frac{2}{x}-3 x\right)^6
$▶️Answer/Explanation Ans: 3(i)$\quad 6 \mathrm{C} 3\left(\frac{2}{x}\right)^3(-3 x)^3 \mathrm{SOI}$also allowed if seen in an expansion -4320 Identified as answer 3(ii)$\quad 6 \mathrm{C} 2\left(\frac{2}{x}\right)^4[(-) 3 x]^2 \quad$SOI clearly identified as critical term$15 a \times 16 \times 9-$their$4320(=0)a=2$### Question 4(i) Prove the identity$(\sin \theta+\cos \theta)(1-\sin \theta \cos \theta) \equiv \sin ^3 \theta+\cos ^3 \theta$.  (ii) Hence solve the equation$(\sin \theta+\cos \theta)(1-\sin \theta \cos \theta)=3 \cos ^3 \theta$for$0^{\circ} \leqslant \theta \leqslant 360^{\circ}$▶️Answer/Explanation Ans: 4(1).$(\sin \theta+\cos \theta)(1-\sin \theta \cos \theta)=\sin \theta+\cos \theta$. LHS$=\sin \theta+\cos \theta-\sin ^2 \theta \cos \theta-\sin \theta \cos ^2 \theta
=\sin \theta\left(1-\cos ^2 \theta\right)+\cos \theta\left(1-\sin ^2 \theta\right) \text { or }\left(s+c-c\left(1-c^2\right)-\mathrm{s}\left(1-\mathrm{s}^2\right)\right)
$Uses$\sin ^2 \theta+\cos ^2 \theta=1 \rightarrow \sin ^3 \theta+\cos ^3 \theta$(RHS)$\mathrm{Or}$or$\begin{aligned} & \text { LHS }=(\sin \theta+\cos \theta)\left(\sin ^2 \theta+\cos ^2 \theta-\sin \theta \cos \theta\right) \\ & =\sin ^3 \theta+\sin \theta \cos ^2 \theta-\sin ^2 \theta \cos \theta+\cos \theta \sin ^2 \theta+\cos ^3 \theta-\sin \theta \cos ^2 \theta=\sin ^3 \theta+\cos ^3 \theta\end{aligned}$4(ii)$\quad(\sin \theta+\cos \theta)(1-\sin \theta \cos \theta)=3 \cos ^3 \theta \rightarrow \sin ^3 \theta=2 \cos ^3 \theta\rightarrow \tan ^3 \theta=2 \rightarrow \theta=51.6^{\circ}$or$231.6^{\circ}$(only) ### Question 5 The first three terms of an arithmetic progression are$4, x$and$y$respectively. The first three terms of a geometric progression are$x, y$and 18 respectively. It is given that both$x$and$y$are positive. (i) Find the value of$x$and the value of$y$.  (ii) Find the fourth term of each progression.  ▶️Answer/Explanation Ans: 5(i) From the AP:$x-4=y-x$From the GP:$\frac{y}{x}=\frac{18}{y}$Simultaneous equations:$y^2-9 y-36=0$or$2 x^2-17 x+8=0$OR$4+d=x, 4+2 d=y \rightarrow \frac{4+2 d}{4+d}=r$oе$(4+d)\left(\frac{4+2 d}{4+d}\right)^2=18 \rightarrow 2 d^2-d-28=0$d=4 5(ii)$\quad$AP 4 th term$=16$GP 4 th term$=8 \times\left(\frac{12}{8}\right)^3$=27 ### Question The diagram shows a solid figure$A B C D E F$in which the horizontal base$A B C$is a triangle right-angled at$A$. The lengths of$A B$and$A C$are 8 units and 4 units respectively and$M$is the mid-point of$A B$. The point$D$is 7 units vertically above$A$. Triangle$D E F$lies in a horizontal plane with$D E, D F$and$F E$parallel to$A B, A C$and$C B$respectively and$N$is the mid-point of$F E$. The lengths of$D E$and$D F$are 4 units and 2 units respectively. Unit vectors$\mathbf{i}, \mathbf{j}$and$\mathbf{k}$are parallel to$\overrightarrow{A B}, \overrightarrow{A C}$and$\overrightarrow{A D}$respectively. (i) Find$\overrightarrow{M F}$in terms of$\mathbf{i}, \mathbf{j}$and$\mathbf{k}$.  (ii) Find$\overrightarrow{F N}$in terms of$\mathbf{i}$and$\mathbf{j}$.  (iii) Find$\overrightarrow{M N}$in terms of$\mathbf{i}, \mathbf{j}$and$\mathbf{k}$.  (iv) Use a scalar product to find angle$F M N$.  ▶️Answer/Explanation Ans: 6 (i)$\quad \mathbf{M F}=-4 \mathbf{i}+2 \mathbf{j}+7 \mathbf{k}$6(ii)$\quad \mathbf{F N}=2 \mathbf{i}-\mathbf{j}$6(iii)$\quad \mathbf{M N}=-2 \mathbf{i}+\mathbf{j}+7 \mathbf{k}$(iv) MF.MN$=8+2+49=59|\mathbf{M F}| \times|\mathbf{M N}|=\sqrt{4^2+2^2+7^2} \times \sqrt{2^2+1^2+7^2}\cos F M N=\frac{+/-59}{\sqrt{69} \times \sqrt{54}}F M N=14.9^{\circ}$or 0.259 ### Question 7 (a) Prove the identity$\frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta} \equiv \frac{2}{\cos \theta}$.  (b) Hence solve the equation$\frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}=\frac{3}{\sin \theta}$, for$0 \leqslant \theta \leqslant 2 \pi$.  ▶️Answer/Explanation Ans:$7($a)$
\frac{(1+\sin \theta)^2+\cos ^2 \theta}{\cos \theta(1+\sin \theta)}
$Use of$\sin ^2 \theta+\cos ^2 \theta=1 \rightarrow \frac{2+2 \sin \theta}{\cos \theta(1+\sin \theta)} \rightarrow \frac{2}{\cos \theta}$. 7 (b)$
\frac{2}{\cos \theta}=\frac{3}{\sin \theta} \rightarrow \tan \theta=1.5
\theta=0.983$or 4.12 (FT on second value for 1 st value$+\pi$) ### Question The diagram shows a symmetrical metal plate. The plate is made by removing two identical pieces from a circular disc with centre$C$. The boundary of the plate consists of two arcs$P S$and$Q R$of the original circle and two semicircles with$P Q$and$R S$as diameters. The radius of the circle with centre$C$is$4 \mathrm{~cm}$, and$P Q=R S=4 \mathrm{~cm}$also. (a) Show that angle$P C S=\frac{2}{3} \pi$radians.$$\ (b) Find the exact perimeter of the plate.  (c) Show that the area of the plate is$\left(\frac{20}{3} \pi+8 \sqrt{3}\right) \mathrm{cm}^2$.  ▶️Answer/Explanation Ans: 8(a) Either Let midpoint of$P Q$be$H$:$\sin H C P=\frac{2}{4} \Rightarrow$Angle$H C P=\frac{\pi}{6}$Or$\sin P S Q=\frac{4}{8} \Rightarrow$Angle$P S Q=\frac{\pi}{6}$Or using cosine rule: angle$P C Q=\frac{\pi}{3}$Or by inspection: triangle$P C Q$or$P C T$is equilateral so angle$P C Q=\frac{\pi}{3}$Angle$P C S=\pi-\frac{\pi}{6}-\frac{\pi}{6}=\frac{2}{3} \pi$8(b)$\quad$Perimeter$=2 \times 4 \times \frac{2 \pi}{3}$or$8 \pi-\frac{8 \pi}{3}+2 \pi \times 2\frac{28 \pi}{3}$Area sector$C P Q=\frac{1}{2} \times 4^2 \times \frac{\pi}{3}=\frac{8 \pi}{3}$Area of segment of large circle beyond$C P Q
=\frac{8 \pi}{3}-\frac{1}{2} \times 4^2 \times \sin \left(\frac{\pi}{3}\right)=\frac{8 \pi}{3}-4 \sqrt{3}
$Area of small semicircle$=\pi \times 2 \quad$or area of small circle$=\pi \times 2^2$Area of plate$=$Large circle$-[2 \times]$small semicircle$-[2 \times]$segment area$\pi \times 4^2-\pi \times 2^2-2 \times\left(\frac{8 \pi}{3}-4 \sqrt{3}\right)=\frac{20 \pi}{3}+8 \sqrt{3}$Alternative method for Question 8(c) Area of sector$P C S=\frac{1}{2} \times 4^2 \times \frac{2 \pi}{3}=\frac{16 \pi}{3}$Area of triangle$P C Q=\frac{1}{2} \times 4^2 \times \sin \frac{\pi}{3}=4 \sqrt{3}$Area of small semicircle$=\pi \times 2 \quad$or area of circle$=\pi \times 2^2$Area of plate$=[2 \times]$large sector$+[2 \times]$triangle$-[2 \times]$small semicircle$
2\left(\frac{16 \pi}{3}\right)+2(4 \sqrt{3})-\pi \times 2^2=\frac{20 \pi}{3}+8 \sqrt{3}
$### Question 9 The volume$V \mathrm{~m}^3$of a large circular mound of iron ore of radius$r \mathrm{~m}$is modelled by the equation$V=\frac{3}{2}\left(r-\frac{1}{2}\right)^3-1$for$r \geqslant 2$. Iron ore is added to the mound at a constant rate of$1.5 \mathrm{~m}^3$per second. (a) Find the rate at which the radius of the mound is increasing at the instant when the radius is$5.5 \mathrm{~m}$.  (b) Find the volume of the mound at the instant when the radius is increasing at$0.1 \mathrm{~m}$per second.  ▶️Answer/Explanation Ans: 9(a)$\left[\frac{\mathrm{d} V}{\mathrm{~d} r}=\right] \frac{9}{2}\left(r-\frac{1}{2}\right)^2$9(b)$\frac{\mathrm{d} r}{\mathrm{~d} t}=\frac{\mathrm{d} r}{\mathrm{~d} V} \times \frac{\mathrm{d} V}{\mathrm{~d} t}=\frac{1.5}{t h e i r \frac{\mathrm{d} V}{\mathrm{~d} r}}\left[=\frac{1.5}{\frac{9}{2}\left(5.5-\frac{1}{2}\right)^2}=\frac{1.5}{112.5}\right]$0.0133 or$\frac{3}{225}$or$\frac{1}{75}$[ metres per second]$\frac{\mathrm{d} V}{\mathrm{~d} r}$or their$\frac{\mathrm{d} V}{\mathrm{~d} r}=\frac{1.5}{0.1}$or 15 OR$0.1=\frac{1.5}{\text { their } \frac{\mathrm{d} V}{\mathrm{~d} r}}\left\lfloor=\frac{2 \times 1.5}{9\left(r-\frac{1}{2}\right)^2} \mathrm{OE}\right\rfloor\left[\frac{9}{2}\left(r-\frac{1}{2}\right)^2=15 \Rightarrow\right] r=\frac{1}{2}+\sqrt{\frac{10}{3}}$[Volume$=$] 8.13 AWRT ### Question 10 The equation of a curve is such that$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=6 x^2-\frac{4}{x^3}$. The curve has a stationary point at$\left(-1, \frac{9}{2}\right)$. (a) Determine the nature of the stationary point at$\left(-1, \frac{9}{2}\right)$.  (b) Find the equation of the curve.  (c) Show that the curve has no other stationary points.  (d) A point$A$is moving along the curve and the$y$-coordinate of$A$is increasing at a rate of 5 units per second. Find the rate of increase of the$x$-coordinate of$A$at the point where$x=1$.  ▶️Answer/Explanation Ans:$
\text { 10(b) } \begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x^3+\frac{2}{x^2}[+c] \\
& 0=-2+2+c \text { leading to } c= \\
& y=\frac{1}{2} x^4-\frac{2}{x}+(\text { their } c) x+k \\
& \frac{9}{2}=\frac{1}{2}+2+k \text { leading to } k= \\
& y=\frac{1}{2} x^4-\frac{2}{x}+2
\end{aligned}
$10(c)$\quad \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x^3+\frac{2}{x^2}=0$Leading to$x^5=-1$So only stationary point is when$x=-110(\mathrm{~d})
\begin{aligned}
& \text { At } x=1, \frac{\mathrm{d} y}{\mathrm{~d} x}= \\
& \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{\mathrm{d} x}{\mathrm{~d} y} \times \frac{\mathrm{d} y}{\mathrm{~d} t}=\frac{1}{4} \times 5 \\
& \frac{5}{4}
\end{aligned}
$### Question 11 The coordinates of points$A, B$and$C$are$A(5,-2), B(10,3)$and$C(2 p, p)$, where$p$is a constant. (a) Given that$A C$and$B C$are equal in length, find the value of the fraction$p$.  (b) It is now given instead that$A C$is perpendicular to$B C$and that$p$is an integer. (i) Find the value of$p$.  (ii) Find the equation of the circle which passes through$A, B$and$C$, giving your answer in the form$x^2+y^2+a x+b y+c=0$, where$a, b$and$c$are constants. ▶️Answer/Explanation Ans: 11(a)$
\begin{aligned}
& (5-2 p)^2+(p+2)^2=(10-2 p)^2+(3-p)^2 \\
& 25-20 p+4 p^2+p^2+4 p+4=100-40 p+4 p^2+9-6 p+p^2 \\
& 30 p=80 \rightarrow p=\frac{8}{3} \text { oe }
\end{aligned}
$11 (b)(i)$
\begin{aligned}
& m_{A C}=\frac{p+2}{2 p-5} \quad m_{B C}=\frac{p-3}{2 p-10} \\
& \frac{p+2}{2 p-5} \times \frac{p-3}{2 p-10}=-1 \\
& – \\
& p^2-p-6=-\left(4 p^2-30 p+50\right) \rightarrow 5 p^2-31 p+44 \quad(=0) \\
& \left.p=4 \text { (Ignore } p=\frac{11}{5}\right)
\end{aligned}
$11(b)(ii) Mid-point of$A B=(71 / 2,1 / 2)
r^2=2 \frac{1 / 2}{2}+2 \frac{1 / 2^2}{}\left[=\frac{50}{4}\right] \text { or } r=\sqrt{ }\left(2 \frac{1 / 2^2}{}+2 \frac{1 / 2^2}{}\right) \quad\left[=\frac{5 \sqrt{ } 2}{2}\right]
$Equation of circle is$(x-\text { their } 71 / 2)^2+(y-\text { their } 1 / 2)^2=$their$\frac{50}{4}
x^2+y^2-15 x-y+44=0
$### Question Find the term independent of$x$in each of the following expansions. (a)$\left(3 x+\frac{2}{x^{2}}\right)^6$ (b)$\left(3 x+\frac{2}{x^2}\right)^6\left(1-x^3\right)$ ▶️Answer/Explanation Ans: (a) The term independent of$x$will arise when the power of$x$is zero. We need to find the term in the expansion of$\left(3 x+\frac{2}{x^2}\right)^6$that has no$x$term. The general term in the binomial expansion of$(3x + \frac{2}{x^2})^6$is given by: $$\binom{6}{r} (3x)^{6-r}\left(\frac{2}{x^2}\right)^r$$ We want to find the value of$r$that makes the$x$term zero, which means that: $$(3x)^{6-r} \left(\frac{2}{x^2}\right)^r$$ coefficient of$x^{6-3r}$will be equivalent to zero .$6-3r=0\Rightarrow r=2$So, 3rd term will be independent term of the expansion$\left(3 x+\frac{2}{x^2}\right)^6$$$\binom{6}{2} (3x)^{6-2}\left(\frac{2}{x^2}\right)^2$$ We can simplify the expression using the binomial coefficient: $$\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15$$ Next, we can simplify the powers of$3x$and$2/x^2$: $$(3x)^{6-2} = (3x)^4 = 81x^4$$ $$\left(\frac{2}{x^2}\right)^2 = \frac{2^2}{x^4} = \frac{4}{x^4}$$ Putting it all together, we get: $$\binom{6}{2} (3x)^{6-2}\left(\frac{2}{x^2}\right)^2 = 15 \times 81x^4 \times \frac{4}{x^4} = 4860$$ Therefore, the value of the expression is$4860$. (b) To find the term independent of$x$in this expression, we need to look for the term in the expansion of$\left(3 x+\frac{2}{x^2}\right)^6\left(1-x^3\right)$that has no$x$term. To get this we will multiply the expression by$\left(1-x^3\right)\underbrace{\left(3 x+\frac{2}{x^2}\right)^6 x \times (1)}_{x^0} -\underbrace{\left(3 x+\frac{2}{x^2}\right)^6 x \times (1)}_{x^{-3}}$coefficient of$x^0$we already calculated in (a) which is equal to$4860$, Now we will calculate$x^{-3}6-3r=-3\Rightarrow r=3$So, 4rd term will be independent term of the expansion$\left(3 x+\frac{2}{x^2}\right)^6\left(1-x^3\right)$$$\binom{6}{3} (3x)^{6-3}\left(\frac{2}{x^2}\right)^3[\times (-x^3)]$$ On simplifying ,$-4320$Value of term independent will be$\underbrace{\left(3 x+\frac{2}{x^2}\right)^6 x \times (1)}_{4860} -\underbrace{\left(3 x+\frac{2}{x^2}\right)^6 x \times (1)}_{4320}=540$### Question Functions$f$and$g$are defined by$
\begin{array}{ll}
\mathrm{f}(x)=4 x-2, & \text { for } x \in \mathbb{R}, \\
\mathrm{g}(x)=\frac{4}{x+1}, & \text { for } x \in \mathbb{R}, x \neq-1 .
\end{array}
$(a) Find the value of$f g(7)$.  (b) Find the values of$x$for which$\mathrm{f}^{-1}(x)=\mathrm{g}^{-1}(x)$.  ▶️Answer/Explanation Ans: (a) To evaluate$fg(7)$, we first need to find$g(7)$: $$g(7) = \frac{4}{7+1} = \frac{1}{2}$$ Next, we substitute this value into$f$to get: $$f\left(g(7)\right) = f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right) – 2 = 0$$ Therefore,$fg(7) = 0$. (b) To find the values of$x$for which$\mathrm{f}^{-1}(x)=\mathrm{g}^{-1}(x)$, we need to find$x$such that$\mathrm{f}(\mathrm{f}^{-1}(x))=\mathrm{g}(\mathrm{g}^{-1}(x))=x$. Let$y=\mathrm{f}^{-1}(x)$. Then, we have$\mathrm{f}(y)=x$, which implies$4y-2=x$. Solving for$y$, we get$y=\frac{x+2}{4}$. Similarly, let$z=\mathrm{g}^{-1}(x)$. Then, we have$\mathrm{g}(z)=x$, which implies$\frac{4}{z+1}=x$. Solving for$z$, we get$z=\frac{4}{x}-1$. Now, we have$\mathrm{f}^{-1}(x)=y=\frac{x+2}{4}$and$\mathrm{g}^{-1}(x)=z=\frac{4}{x}-1$. Setting them equal to each other, we get$\frac{x+2}{4}=\frac{4}{x}-1$. Simplifying this equation, we get$x^2-2x-8=0$, which can be factored as$(x-4)(x+2)=0$. Therefore, the solutions are$x=4$and$x=-2$. To check that these values of$x$satisfy the condition$\mathrm{f}(\mathrm{f}^{-1}(x))=\mathrm{g}(\mathrm{g}^{-1}(x))=x$, we substitute them into the expressions we obtained earlier for$\mathrm{f}^{-1}(x)$and$\mathrm{g}^{-1}(x)$. For$x=4$, we have$y=\frac{x+2}{4}=\frac{6}{4}=\frac{3}{2}$and$z=\frac{4}{x}-1=0$, so$\mathrm{f}(\mathrm{f}^{-1}(x))=\mathrm{f}(y)=4$and$\mathrm{g}(\mathrm{g}^{-1}(x))=\mathrm{g}(z)=4$, which satisfies the condition. For$x=-2$, we have$y=\frac{x+2}{4}=0$and$z=\frac{4}{x}-1=-2$, so$\mathrm{f}(\mathrm{f}^{-1}(x))=\mathrm{f}(y)=-2$and$\mathrm{g}(\mathrm{g}^{-1}(x))=\mathrm{g}(z)=-2$, which also satisfies the condition. Therefore, the solutions are$x=4$and$x=-2$. ### Question A woman’s basic salary for her first year with a particular company is$\$30000$ and at the end of the year she also gets a bonus of $\$ 600$. (a) For her first year, express her bonus as a percentage of her basic salary.$$At the end of each complete year, the woman’s basic salary will increase by$3 \%$and her bonus will increase by$\$100$.
(b) Express the bonus she will be paid at the end of her 24th year as a percentage of the basic salary paid during that year. 

Ans:

(a) The bonus of $\$ 600$is a fraction of the woman’s basic salary of$\$30000$. To find out what that fraction is, we can calculate:

$$\frac{\ 600}{\ 30000} = 0.02$$

So the bonus is $0.02$ times the basic salary. To express this as a percentage, we multiply by $100$:

$$0.02 \times 100 = 2\%$$

Therefore, the bonus of $\$ 600$is$2\%$of the woman’s basic salary of$\$30000$ for her first year.

(b)

For the basic salary, we can use the formula for the compound interest:

$$T_{24}=a(1+r)^{24}$$

where $a=30000$ is the initial amount, $r=0.03$ is the annual interest rate, and $T_{24}$ is the amount after 24 years. Plugging in the values, we get:

$$T_{24}=30000(1+0.03)^{24} \approx 59,207.60$$

So the woman’s basic salary at the end of her 24th year is approximately $\$ 59,207.60$. For the bonus, we know that the woman receives$\$600$ in the first year and an additional $\$ 100$each year thereafter. So in the 24th year, she will receive: $$\text{bonus} = \ 600 + (\text{number of years} – 1) \times \ 100 = \ 2900$$ Finally, we can express the bonus as a percentage of the basic salary in the 24th year: $$\text{percentage} = \frac{\text{bonus}}{\text{basic salary}} \times 100\% = \frac{2900}{59,207.60} \times 100\% \approx 4.89\%$$ Therefore, the bonus of$\$2900$ she will be paid at the end of her 24th year is approximately $4.89\%$ of the basic salary paid during that year, which is approximately $\$ 59,207.60\$.

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