Question
Solve the equation \(4 \sin \theta+\tan \theta=0\) for \(0^{\circ}<\theta<180^{\circ}\).
▶️Answer/Explanation
Ans:
\(\begin{aligned} & 4 \sin \theta+\tan \theta=0 \Rightarrow 4 \sin \theta+\frac{\sin \theta}{\cos \theta}[=0] \\ & \Rightarrow \sin \theta(4 \cos \theta+1)[=0 \Rightarrow \sin \theta=0 \text { or }] \cos \theta=-\frac{1}{4} \\ & \theta=104.5^{\circ}\end{aligned}\)
Question
(a) Find the first three terms in the expansion, in ascending powers of \(x\), of \((2+3 x)^4\). [2]
(b) Find the first three terms in the expansion, in ascending powers of \(x\), of \((1-2 x)^5\). [2]
(c) Hence find the coefficient of \(x^2\) in the expansion of \((2+3 x)^4(1-2 x)^5\). [2]
▶️Answer/Explanation
Ans:
(a) \(16+96 x+216 x^2\)
(b) \(1-10 x+40 x^2\)
(c) \((16 \times 40)-(10 \times 96)+(1 \times 216)\)
\(-104\)
Question
The diagram shows graphs with equations \(y=\mathrm{f}(x)\) and \(y=\mathrm{g}(x)\).
Describe fully a sequence of two transformations which transforms the graph of \(y=\mathrm{f}(x)\) to \(y=\mathrm{g}(x)\). [4]
▶️Answer/Explanation
Ans:
\(\{\) Stretch \(\}\{\) factor 2\(\}\) in \(y\)-direction \(\}\)
\(\{\) Translation \(\}\left(\begin{array}{c}\{-6\} \\ \{0\}\end{array}\right)\)
Question
The diagram shows a sector \(A B C\) of a circle with centre \(A\) and radius \(8 \mathrm{~cm}\). The area of the sector is \(\frac{16}{3} \pi \mathrm{cm}^2\). The point \(D\) lies on the \(\operatorname{arc} B C\).
Find the perimeter of the segment \(B C D\).
▶️Answer/Explanation
Ans:
$
\frac{1}{2} \times 8^2 \times \theta=\frac{16 \pi}{3} \Rightarrow \theta=\frac{\pi}{6}
$
Arc length \(=8 \times\) their \(\frac{\pi}{6}[=4.1887 \ldots]\)
$
[B C=] 2 \times 8 \sin \left(\frac{1}{2} \times \text { their } \frac{\pi}{6}\right)[=4.1411 \ldots]
$
$
\text { Perimeter }=8.33
$
Alternative methods for Question 4: 2nd M1 mark (use normal scheme for the other marks)
ALT \(1 B C^2=8^2+8^2-2 \times 8 \times 8 \cos \left(\right.\) their \(\left.\frac{\pi}{6}\right)[\Rightarrow B C=4.14 \ldots]\)
ALT \(2 B C^2=(8-4 \sqrt{3})^2+4^2[\Rightarrow B C=4.14 \ldots]\)
ALT \(3 \frac{B C}{\sin \left(\frac{\pi}{6}\right)}=\frac{8}{\sin \left(\frac{5 \pi}{12}\right)}[\Rightarrow B C=4.14 \ldots]\)
Question
The line with equation \(y=k x-k\), where \(k\) is a positive constant, is a tangent to the curve with equation \(y=-\frac{1}{2 x}\).
Find, in either order, the value of \(k\) and the coordinates of the point where the tangent meets the curve. [5]
▶️Answer/Explanation
Ans:
$
k x-k=-\frac{1}{2 x} \Rightarrow 2 k x^2-2 k x+1[=0]
$
OR quadratic in \(y: \quad x=\frac{y+k}{k} \Rightarrow y=-\frac{1}{2\left(\frac{y+k}{k}\right)} \Rightarrow 2 y^2+2 k y+k=0\)
$
\begin{aligned}
& b^2-4 a c[=0] \Rightarrow([-] 2 k)^2-4(2 k)(1)[=0] \\
& \text { or } 4 k^2-8 k[=0 \Rightarrow 4 k(k-2)=0]
\end{aligned}
$
OR using equation in \(y: \quad(2 k)^2-4(2)(k)=0\)
$
k=2 \text { only }
$
\(4 x^2-4 x+1=0\left[\Rightarrow(2 x-1)^2=0\right] \Rightarrow x=\frac{1}{2}\)
\(y=2 \times \frac{1}{2}-2=-1\)
Alternative method for Q5
\(\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2 x^2} \text { or } \frac{1}{2} x^{-2} \\ & {[y=] \frac{1}{2 x^2} x-\frac{1}{2 x^2}=-\frac{1}{2 x} \text { or } \frac{1}{x}=\frac{1}{2 x^2}\left[\Rightarrow 2 x^2-x=0\right]}\end{aligned}\)
\(\begin{aligned} & x=\frac{1}{2} \text { only } \\ & y=\left[2 \times \frac{1}{2}-2\right]=-1\end{aligned}\)
\(y=\left[2 \times \frac{1}{2}-2\right]=-1\)
\(k=2\)
Question
The first three terms of an arithmetic progression are \(\frac{p^2}{6}, 2 p-6\) and \(p\).
(a) Given that the common difference of the progression is not zero, find the value of \(p\). [3]
(b) Using this value, find the sum to infinity of the geometric progression with first two terms \(\frac{p^2}{6}\) and \(2 p-6\).
▶️Answer/Explanation
Ans:
\(\begin{aligned} & 2(2 p-6)=p+\frac{p^2}{6} \Rightarrow \frac{p^2}{6}-3 p+12[=0] \\ & \text { OR }(2 p-6)-\frac{p^2}{6}=p-(2 p-6) \Rightarrow \frac{p^2}{6}-3 p+12[=0] \\ & \text { OR } \frac{1}{6} d^2+d[=0]\end{aligned}\)
\(\begin{aligned} & p^2-18 p+72[=0] \Rightarrow(p-6)(p-12)[=0] \text { or } \frac{18 \pm \sqrt{(-18)^2-4(1)(72)}}{2} \\ & \text { OR } d\left(\frac{1}{6} d+1\right)[=0] \Rightarrow d=-6\end{aligned}\)
\(p=12\) only
b.For GP \(r=\left[\frac{2 p-6}{\frac{p^2}{6}}\right]=\frac{18}{24}\left[=\frac{3}{4}\right]\)
Sum to infinity \(=\frac{24}{1-\frac{3}{4}}=96\)
Question
A curve has equation \(y=2+3 \sin \frac{1}{2} x\) for \(0 \leqslant x \leqslant 4 \pi\).
(a) State greatest and least values of \(y\). [2]
(b) Sketch the curve.[2]
(c) State the number of solutions of the equation
$
2+3 \sin \frac{1}{2} x=5-2 x
$
for \(0 \leqslant x \leqslant 4 \pi\).[1]
▶️Answer/Explanation
Ans:
a.[Greatest =] 5
[Least =] ‒1
b.
(c) 1
Question
The functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined as follows, where \(a\) and \(b\) are constants.
$
\begin{aligned}
& \mathrm{f}(x)=1+\frac{2 a}{x-a} \text { for } x>a \\
& \mathrm{~g}(x)=b x-2 \text { for } x \in \mathbb{R}
\end{aligned}
$
(a) Given that \(\mathrm{f}(7)=\frac{5}{2}\) and \(\operatorname{gf}(5)=4\), find the values of \(a\) and \(b\). [4]
For the rest of this question, you should use the value of \(a\) which you found in (a).
(b) Find the domain of \(\mathrm{f}^{-1}\).[1]
(c) Find an expression for \(\mathrm{f}^{-1}(x)\).[3]
▶️Answer/Explanation
Ans:
a.\(\begin{aligned} & 1+\frac{2 a}{7-a}=\frac{5}{2}\left[\Rightarrow \frac{2 a}{7-a}=\frac{3}{2} \Rightarrow 7 a=21\right] \Rightarrow a=\ldots \\ & \text { OR } 1+\frac{2 a}{7-a}=\frac{5}{2}\left[\Rightarrow(7-a)+2 a=\frac{5}{2}(7-a)[\Rightarrow 7 a=21] \Rightarrow a=\ldots\right. \\ & a=3\end{aligned}\)
\(\begin{aligned} & \mathrm{f}(5)=1+\frac{2(\text { their } 3)}{5-\text { their } 3}=4[\Rightarrow 4 b-2=4] \Rightarrow b=\ldots \\ & \text { OR } \operatorname{gf}(5)=b\left(1+\frac{2(\text { their } 3)}{5-\text { their } 3}\right)-2[\Rightarrow 4 b-2=4] \Rightarrow b=\ldots\end{aligned}\)
\(b=\frac{3}{2}\)
(b) x > 1
c.EITHER \(x-1=\frac{6}{y-3}[\Rightarrow(y-3)(x-1)=6]\)
OR \(x=1+\frac{6}{y-3} \Rightarrow x(y-3)=(y-3)+6\)
\(y-3=\frac{6}{x-1}\) or \(y(x-1)=3 x+3\)
$
\left[\mathrm{f}^{-1}(x)\right]=3+\frac{6}{x-1}
$
Question
Water is poured into a tank at a constant rate of \(500 \mathrm{~cm}^3\) per second. The depth of water in the tank, \(t\) seconds after filling starts, is \(h \mathrm{~cm}\). When the depth of water in the tank is \(h \mathrm{~cm}\), the volume, \(V \mathrm{~cm}^3\), of water in the tank is given by the formula \(V=\frac{4}{3}(25+h)^3-\frac{62500}{3}\).
(a) Find the rate at which \(h\) is increasing at the instant when \(h=10 \mathrm{~cm}\). [3]
(b) At another instant, the rate at which h is increasing is 0.075 cm per second.
Find the value of V at this instant. [3]
▶️Answer/Explanation
Ans:
\(\frac{\mathrm{d} V}{\mathrm{~d} h}=\frac{4}{3} \times 3(25+h)^2[=4900\) when \(h=10]\)
\(\frac{\mathrm{d} V}{\mathrm{~d} h} \times \frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{\mathrm{d} V}{\mathrm{~d} t} \Rightarrow\) their \(” 4(25+10)^2 ” \times \frac{\mathrm{d} h}{\mathrm{~d} t}=500 \Rightarrow \frac{\mathrm{d} h}{\mathrm{~d} t}=\left[\frac{500}{4900}\right]\)
\(\frac{\mathrm{d} h}{\mathrm{~d} t}=0.102\left[\mathrm{cms}^{-1}\right]\)
b.\(\frac{\mathrm{d} V}{\mathrm{~d} t}=\frac{\mathrm{d} V}{\mathrm{~d} h} \times \frac{\mathrm{d} h}{\mathrm{~d} t} \Rightarrow 500=\) their \(” 4(25+h)^2 ” \times 0.075\)
\(\left[(25+h)^2=\frac{5000}{3}\right] \Rightarrow h=[15.8248 \ldots]\)
\(V=69900 \mathrm{~cm}^3\)
Question
The diagram shows part of the curve with equation \(y=\frac{4}{(2 x-1)^2}\) and parts of the lines \(x=1\) and \(y=1\). The curve passes through the points \(A(1,4)\) and \(B,\left(\frac{3}{2}, 1\right)\).
(a) Find the exact volume generated when the shaded region is rotated through \(360^{\circ}\) about the \(x\)-axis. [5]
(b) A triangle is formed from the tangent to the curve at \(B\), the normal to the curve at \(B\) and the \(x\)-axis.
Find the area of this triangle. [6]
▶️Answer/Explanation
Ans:
\([\pi] \int \frac{16}{(2 x-1)^4}[\mathrm{~d} x]=[\pi] \int 16(2 x-1)^{-4}[\mathrm{~d} x]=[\pi]\left(-\frac{16}{3 \times 2 \times(2 x-1)^3}\right)\)
\([\pi]\left(-\frac{16}{3 \times 2 \times(2 x-1)^3}\right)\)
\([\pi]\left(-\frac{16}{6 \times 8}+\frac{16}{6 \times 1}\right)\left[=[\pi] \frac{112}{48}=[\pi] \frac{7}{3}\right]\)
Volume of cylinder \(\left[=\pi \times 1^2 \times \frac{1}{2}\right]=\frac{1}{2} \pi\) OR \([\pi] \int_1^{1.5} 1[\mathrm{~d} x]=\frac{1}{2} \pi\)
Volume of revolution \(\left[=\frac{7}{3} \pi-\frac{1}{2} \pi\right]=\frac{11}{6} \pi\)
b.$
\left[\frac{\mathrm{d} y}{\mathrm{~d} x}=\right]\left\{-8(2 x-1)^{-3}\right\}\{\times 2\}
$
At \(B\) gradient \(=-2\)
Eqn of tangent \(y-1=\) their \(“-2 “\left(x-\frac{3}{2}\right)\) OR Eqn of normal \(y-1=\) their \(” \frac{1}{2} “\left(x-\frac{3}{2}\right)\)
Tangent crosses \(x\)-axis at 2 or normal crosses \(x\)-axis at \(-\frac{1}{2}\)
$
\text { Area }=\frac{5}{4}
$
Question
The equation of a curve is such that \(\frac{\mathrm{d} y}{\mathrm{~d} x}=6 x^2-30 x+6 a\), where \(a\) is a positive constant. The curve has a stationary point at \((a,-15)\).
(a) Find the value of \(a\). [2]
(b) Determine the nature of this stationary point. [2]
(c) Find the equation of the curve. [3]
(d) Find the coordinates of any other stationary points on the curve. [2]
▶️Answer/Explanation
Ans:
\(6 a^2-30 a+6 a=0[\Rightarrow 6 a(a-4)=0]\)
\(a=4\) only
b.\(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=12 x-30\) or correct values of \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) either side of \(x=4\)
At \(x=4, \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}>0 \therefore\) minimum or \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=18 \therefore\) minimum or concludes minimum from \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) values
c.\(\begin{aligned} & {[y=] \frac{6}{3} x^3-\frac{30}{2} x^2+6(\text { their } a) x[+c]} \\ & -15=2(\text { their } ” 4 “)^3-15(\text { their } ” 4 “)^2+6(\text { their } ” 4 “)^2+c \\ & y=2 x^3-15 x^2+24 x+1\end{aligned}\)
d.$
\frac{\mathrm{d} y}{\mathrm{~d} x}=6 x^2-30 x+6(\text { their } ” 4 “)[=0]
$
If correct, \([6](x-1)(x-4)[=0]\) or \(\frac{30 \pm \sqrt{(-30)^2-4(6)(24)}}{12}\)
Coordinates \((1,12)\)
Question
The diagram shows a circle \(P\) with centre \((0,2)\) and radius 10 and the tangent to the circle at the point \(A\) with coordinates \((6,10)\). It also shows a second circle \(Q\) with centre at the point where this tangent meets the \(y\)-axis and with radius \(\frac{5}{2} \sqrt{5}\).
(a) Write down the equation of circle \(P\). [1]
(b) Find the equation of the tangent to the circle P at A. [2]
(c) Find the equation of circle Q and hence verify that the y-coordinates of both of the points of intersection of the two circles are 11. [3]
(d) Find the coordinates of the points of intersection of the tangent and circle Q, giving the answers in surd form. [3]
▶️Answer/Explanation
Ans:
a. \(x^2+(y-2)^2=100\)
b.Gradient of radius \(=\left[\frac{10-2}{6-0}=\right] \frac{4}{3}\) or gradient of tangent \(=\frac{-3}{4}\)
Equation of tangent is \(y-10=-\frac{3}{4}(x-6) \quad\left[\Rightarrow y=-\frac{3}{4} x+\frac{29}{2}\right]\)
Equation of tangent is \(y-10=-\frac{3}{4}(x-6) \quad\left[\Rightarrow y=-\frac{3}{4} x+\frac{29}{2}\right]\)
c.Coordinates of centre of circle \(Q\) are \(\left(0\right.\), their \(\left.\frac{29}{2}\right)\)
Equation of circle \(Q\) is \(x^2+\left(y \text {-their } \frac{29}{2}\right)^2=\left(\frac{5 \sqrt{5}}{2}\right)^2\left[=\frac{125}{4}\right]\)
\(x^2+(11-2)^2=100 \Rightarrow x^2=19\) and \(x^2+\left(11-\frac{29}{2}\right)^2=\frac{125}{4} \Rightarrow x^2=19\)
OR e.g. \(\frac{125}{4}-\left(y-\frac{29}{2}\right)^2+(y-2)^2=100 \Rightarrow 25 y=275 \Rightarrow y=11\)
d.$
x^2+\left(-\frac{3}{4} x+\frac{29}{2}-\frac{29}{2}\right)^2=\frac{125}{4}\left[\Rightarrow \frac{25}{16} x^2=\frac{125}{4} \Rightarrow x^2=20\right]
$
or \(y^2-29 y+199[=0]\)
\(x= \pm 2 \sqrt{5}\) or \(y=\frac{29 \mp 3 \sqrt{5}}{2}\)
\(y\left[=\left(-\frac{3}{4} \times \pm \sqrt{20}\right)+\frac{29}{2}\right]=\frac{29 \mp 3 \sqrt{5}}{2}\)