MATSEC

Pure Mathematics

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September 2019 Paper 1 Question 1

We would like to solve:
(1+\cos(2x)\frac{dy}{dx}=2(2y+1)\sin(2x)
Separating y and x on the opposite sides of the equation:

\begin{aligned} \frac{1}{2y+1}dy&=\frac{2\sin(2x)}{1+\cos(2x)}dx\\ \int\frac{1}{2y+1}dy&=\int\frac{2\sin(2x)}{1+\cos(2x)}dx\ \ \ \ ...(1) \end{aligned}
Integrating the right side of Equation (1) by substitution:

\text{Let }I=\int\frac{2\sin(2x)}{1+\cos(2x)}dx
\begin{aligned} \text{Let }u&=1+\cos(2x)\\ \frac{du}{dx}&=-2\sin(2x)\\ du&=-2\sin(2x)dx \end{aligned}
\begin{aligned} I &=-\int\frac{1}{u} du\\ &=-\ln\lvert u\rvert\\ &=-\ln\lvert 1+\cos(2x)\rvert \end{aligned}
Performing integration on both sides of Equation (1):

\frac{1}{2}\ln\lvert 2y+1\rvert=-\ln\lvert 1+\cos(2x)\rvert+C
Using the trigonometric identity \cos2A=2\cos^2 A-1:

\begin{aligned} \frac{1}{2}\ln\lvert 2y+1\rvert&=-\ln\lvert 2\cos^2 x\rvert+C\\ \ln\lvert 2y+1\rvert^{\frac{1}{2}}&=\ln\lvert 2\cos^2 x\rvert^{-1}+C \end{aligned}
Since the input of a square root function is positive and 2\cos^2 x is positive:

\begin{aligned} \ln( 2y+1)^{\frac{1}{2}}&=\ln( 2\cos^2 x)^{-1}+C\\ e^{\ln(2y+1)^{\frac{1}{2}}}&=e^{\ln( 2\cos^2 x)^{-1}+C}\\ e^{\ln(2y+1)^{\frac{1}{2}}}&=e^{\ln( 2\cos^2 x)^{-1}}e^{C} \end{aligned}
Letting A=e^C:
(2y+1)^{\frac{1}{2}}=A( 2\cos^2 x)^{-1}
The General Solution is:
( 2y+1)^{\frac{1}{2}}= \frac{1}{2}A\sec^2 x\ \ \ \ ...(2)
Let x=0 and y=0 in Equation (2):
\begin{aligned} ( 2(0)+1)^{\frac{1}{2}}&= \frac{1}{2}A\sec^2 (0)\\ A&=2 \end{aligned}
Substituting A=2 in Equation (2), we get the Particular Solution:

( 2y+1)^{\frac{1}{2}}= \sec^2 x

September 2019 Paper 1 Question 2

(a) Consider the curve:
x^2-xy+y^2=12\ \ \ \ ...(1)
Differentiating all terms by x:

\begin{aligned} \frac{d(x^2)}{dx}-\frac{d(xy)}{dx}+\frac{d(y^2)}{dx}&=\frac{d(12)}{dx}\\ 2x-(x\frac{dy}{dx}+y)+2y\frac{dy}{dx}&=0\\ 2x-x\frac{dy}{dx}-y+2y\frac{dy}{dx}&=0\\ (2y-x)\frac{dy}{dx}&=y-2x\\ \frac{dy}{dx}&=\frac{y-2x}{2y-x}\ \ \ \ ...(2) \end{aligned}
To find the stationary points, let \frac{dy}{dx}=0 in Equation (2). Therefore:

\begin{aligned} \frac{y-2x}{2y-x}&=0\\ y-2x&=0\\ y&=2x \end{aligned}
To find the x-values of the stationary points, substitute y=2x in Equation (1):

\begin{aligned} x^2-xy+y^2&=12\\ x^2-x(2x)+(2x)^2&=12\\ x^2-2x^2+4x^2&=12\\ 3x^2&=12\\ x^2&=4\\ x&=\pm 2. \end{aligned}
To find the y-values of the stationary points, substitute x=\pm 2 in y=2x:

When x=-2,y=2(-2)=-4.

When x=2,y=2(2)=4.

Hence the coordinates of the stationary points are (-2,-4) and (2,4).

(b) Let us find \frac{d^2y}{dx^2} by differentiating Equation (2) (found in Part (a)) using the quotient rule.

\begin{aligned} u&=y-2x\\ u'&=\frac{dy}{dx}-2 \end{aligned} \begin{aligned} \qquad \end{aligned} \begin{aligned} v&=2y-x\\ v'&=2\frac{dy}{dx}-1 \end{aligned}
By the quotient rule:

\begin{aligned} \frac{d^2y}{dx^2}&=\frac{vu'-uv'}{v^2}\\ &=\frac{(2y-x)(\frac{dy}{dx}-2)-(y-2x)(2\frac{dy}{dx}-1)}{(2y-x)^2}\\ &=\frac{2y\frac{dy}{dx}-x\frac{dy}{dx}-4y+2x-2y\frac{dy}{dx}+4x\frac{dy}{dx}+y-2x}{(2y-x)^2}\\ &=\frac{3x\frac{dy}{dx}-3y}{(2y-x)^2} \end{aligned}
At the stationary points, \frac{dy}{dx}=0. Hence:

\begin{aligned} \frac{d^2y}{dx^2}&=-\frac{3y}{(2y-x)^2}\\ &=-\frac{3y}{(x-2y)^2}\\ (x-2y)\frac{d^2y}{dx^2}&=-\frac{3y}{(x-2y)} \end{aligned}
Comparing this formula with (x-2y)\frac{dy^2}{dx^2}=k, we get:

k=-\frac{3y}{x-2y}
At the stationary point (-2,-4): k=-\frac{3(-4)}{-2-2(-4)}=2

At the stationary point (2,4): k=-\frac{3(4)}{2-2(4)}=2

Hence, k=2.

(c) Consider the second order derivative at the stationary points \frac{d^2y}{dx^2}=-\frac{3y}{(2y-x)^2} as found in Part (b).

At (-2,-4):

\begin{aligned} \frac{d^2y}{dx^2}&=-\frac{3(-4)}{(2(-4)-(-2))^2}\\ &=\frac{12}{36}\\ &>0 \end{aligned}
Hence (-2,-4) is a minimum point.

At (2,4):

\begin{aligned} \frac{d^2y}{dx^2}&=-\frac{3(4)}{(2(4)-2)^2}\\ &=-3\\ &\lt 0 \end{aligned}

Hence (2,4) is a maximum point.

September 2019 Paper 1 Question 3

(a) The equations of the two lines l_1 and l_2 can be written as:

\begin{aligned} \bf{r}&=\begin{pmatrix}-5\\1\\10\end{pmatrix}+\lambda\begin{pmatrix}-3\\0\\4\end{pmatrix}\\ &=\begin{pmatrix}-5-3\lambda\\1\\10+4\lambda\end{pmatrix} \end{aligned} \begin{aligned} \qquad\qquad \end{aligned} \begin{aligned} \bf{r}&=\begin{pmatrix}3\\0\\-9\end{pmatrix}+\mu\begin{pmatrix}1\\1\\7\end{pmatrix}\\ &=\begin{pmatrix}3+\mu\\\mu\\-9+7\mu\end{pmatrix} \end{aligned}

To find the point of intersection P we equate the two lines:

\begin{pmatrix}-5-3\lambda\\1\\10+4\lambda\end{pmatrix}=\begin{pmatrix}3+\mu\\\mu\\-9+7\mu\end{pmatrix}

The 1^st, 2^nd and 3^rd row give the following three equations:

\begin{aligned} -5-3\lambda=3+\mu \end{aligned} \begin{aligned} \qquad \end{aligned} \begin{aligned} 1=\mu \end{aligned} \begin{aligned} \qquad \end{aligned} \begin{aligned} 10+4\lambda=-9+7\mu \end{aligned}
The 2^nd equation gives us immediately that \mu=1. Let \mu=1 in the equation of line l_2 to obtain the coordinates of P:
\begin{aligned} \bf{r}&=\begin{pmatrix}3+1\\1\\-9+7(1)\end{pmatrix}\\ &=\begin{pmatrix}4\\1\\2\end{pmatrix} \end{aligned} Hence: P\begin{pmatrix}4\\1\\2\end{pmatrix}.

(b) We need to show that there exists a real number \mu that satisfies the equation:

\begin{pmatrix}5\\2\\5\end{pmatrix}=\begin{pmatrix}3\\0\\-9\end{pmatrix}+\mu\begin{pmatrix}1\\1\\7\end{pmatrix}
Solving for \mu:
\begin{aligned} \begin{pmatrix}2\\2\\14\end{pmatrix}&=\mu\begin{pmatrix}1\\1\\7\end{pmatrix}\\ \therefore \mu&=2 \end{aligned}
Since \mu exists, Q\begin{pmatrix}5\\2\\5\end{pmatrix} lies on the line l_2. \square

(c) Let M \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}.

Consider the vector \vec{MQ}, by the triangle law:

\begin{aligned} \vec{MQ}&=\vec{OQ}-\vec{OM}\\ &=\begin{pmatrix}5-x_1\\2-y_1\\5-z_1\end{pmatrix} \end{aligned}
\vec{MQ} is perpendicular to the line l_1. Hence \vec{MQ} is perpendicular to \begin{pmatrix}-3\\0\\4\end{pmatrix} (the directional vector of l_1). Hence:
\begin{aligned} \begin{pmatrix}-3\\0\\4\end{pmatrix}\cdot\begin{pmatrix}5-x_1\\2-y_1\\5-z_1\end{pmatrix}&=0\\ -3(5-x_1)+4(5-z_1)&=0\\ 3x_1-4z_1&=-5\ \ \ \ ...(1) \end{aligned}
On the other hand, since M lies on the line l_1:

\begin{aligned} \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}&=\begin{pmatrix}-5\\1\\10\end{pmatrix}+\lambda\begin{pmatrix}-3\\0\\4\end{pmatrix}\\ &=\begin{pmatrix}-5-3\lambda\\1\\10+4\lambda\end{pmatrix} \end{aligned}
This gives us the following 3 equations:

\begin{aligned} x_1=-5-3\lambda\ \ \ \ ...(2) \end{aligned} \begin{aligned} \qquad \end{aligned} \begin{aligned} y_1=1\ \ \ \ ...(3) \end{aligned} \begin{aligned} \qquad \end{aligned} \begin{aligned} z_1=10+4\lambda\ \ \ \ ...(4) \end{aligned}
Equation (2) gives us immediately that y_1=1. Now let us solve Equations (2) and (4) simulataneously to get rid of \lambda. Make \lambda subject of the formula in Equation (4) and insert it in Equation (2):
\begin{aligned} z_1&=10+4\lambda\\ \therefore\lambda&=\frac{z_1-10}{4} \end{aligned} \begin{aligned} \qquad \end{aligned} \begin{aligned} x_1&=-5-3\lambda\\ x_1&=-5-3(\frac{z_1-10}{4})\\ \therefore x_1&=-\frac{3}{4}z_1+\frac{10}{4}\ \ \ \ ...(5) \end{aligned}
Let us solve Equations (1) and (5) simulataneously to obtain values of x_1 and z_1. Inserting Equation (5) in Equation (1) gives:

\begin{aligned} 3x_1-4z_1&=-5\\ 3(-\frac{3}{4}z_1+\frac{10}{4})-4z_1&=-5\\ \therefore z_1&=2\\ \end{aligned}
Inserting z_1=2 in Equation (5) gives:
\begin{aligned} x_1&=-\frac{3}{4}(2)+\frac{10}{4}\\ \therefore x_1&=1 \end{aligned}
Hence M=\begin{pmatrix}1\\1\\2\end{pmatrix}.

(d) Consider the right-angled triangle \triangle PQM.

\begin{aligned} \lvert\vec{MP}\rvert&=\lvert \vec{OP}-\vec{OM}\rvert \text{ (by the triangle law)}\\ &=\begin{vmatrix} \begin{pmatrix}4\\1\\-2\end{pmatrix}-\begin{pmatrix}1\\1\\2\end{pmatrix}\end{vmatrix}\\ &=\begin{vmatrix}\begin{pmatrix}3\\0\\-4\end{pmatrix}\end{vmatrix}\\ &=\sqrt{9+0+16}\\ &=5 \end{aligned}
\begin{aligned} \lvert\vec{MQ}\rvert&=\lvert \vec{OQ}-\vec{OM}\rvert \text{ (by the triangle law)}\\ &=\begin{vmatrix} \begin{pmatrix}5\\2\\5\end{pmatrix}-\begin{pmatrix}1\\1\\2\end{pmatrix}\end{vmatrix}\\ &=\begin{vmatrix}\begin{pmatrix}4\\1\\3\end{pmatrix}\end{vmatrix}\\ &=\sqrt{16+1+9}\\ &=\sqrt{26} \end{aligned}
\begin{aligned} \therefore\text{Area of }\triangle PMQ&=\frac{\text{length}\times \text{breadth}}{2}\\ &=\frac{5\sqrt{26}}{2}\\ &=\frac{5}{2}\sqrt{26} \end{aligned}

September 2019 Paper 1 Question 4

* This solution involves 3 topics: Remainder & Factor Theorem, Partial Fractions, and the Binomial Theorem

(a) Consider the function f(x)=4x^3+4x^2-7x+2.

Let x=-2:
\begin{aligned} f(-2)&=4(-2)^3+4(-2)^2-7(-2)+2\\ &=4(-8)+4(4)+14+2\\ &=0 \end{aligned}
\therefore\ x=-2 is a root of 4x^3+4x^2-7x+2.
\therefore\ (x+2) is a factor of 4x^3+4x^2-7x+2.

Now let us factorise f(x) completely. Let:

4x^3+4x^2-7x+2\equiv (x+2)(ax^2+bx+c)\ \ \ \ ...(1)
Expanding the RHS of the identity:

4x^3+4x^2-7x+2\equiv ax^3+(2a+b)x^2+(c+2b)x+2c
Equating coefficients of x^3:
4=a Equating coefficients of x^2:
\begin{aligned} 4&=2a+b\\ 4&=2(4)+b\text{ since }a=4\\ b&=-4 \end{aligned}
Equating coefficients of x:
\begin{aligned} -7&=c+2b\\ -7&=c+2(-4)\text{ since }b=-4\\ c&=1 \end{aligned}
Hence Equation (1) becomes:

\begin{aligned} 4x^3+4x^2-7x+2&\equiv (x+2)(4x^2-4x+1)\\ &\equiv (x+2)(2x-1)(2x-1)\\ &\equiv (x+2)(2x-1)^2 \end{aligned}

(b) (i) \begin{aligned} f(x)=&\frac{10-17x+14x^2}{4x^3+4x^2-7x+2}\\ =&\frac{10-17x+14x^2}{(x+2)(2x-1)^2}\text{ (by Part (a))} \end{aligned}
This is the case where we have repeated linear factors in the denominator. Hence the function is split in partial fractions as follows:

\frac{10-17x+14x^2}{(x+2)(2x-1)^2}\equiv \frac{A}{x+2}+\frac{B}{(2x-1)}+\frac{C}{(2x-1)^2}
Finding a common denominator for the LHS:

\begin{aligned} \frac{10-17x+14x^2}{(x+2)(2x-1)^2}&\equiv \frac{A(2x-1)^2+B(x+2)(2x-1)+C(x+2)}{(x+2)(2x-1)^2}\\ \therefore 10-17x+14x^2&\equiv A(2x-1)^2+B(x+2)(2x-1)+C(x+2)\ \ \ \ ...(2) \end{aligned}
Let x=\frac{1}{2} in Equation (2):
\begin{aligned} 10-17(\frac{1}{2})+14(\frac{1}{2})^2&=C(\frac{1}{2}+2)\\ \therefore C&=2 \end{aligned}
Let x=-2 in Equation (2):

\begin{aligned} 10-17(-2)+14(-2)^2&=A(-4-1)^2\\ \therefore A&=4 \end{aligned} Let x=0 in Equation (2):
\begin{aligned} 10&=A(1)+B(2)(-1)+C(2)\\ 10&=A-2B+2C\\ 10&=4-2B+2(2)\text{ (since }A=4\text{ and }C=2)\\ \therefore B&=-1 \end{aligned}
Therefore:
f(x)=\frac{4}{x+2}-\frac{1}{(2x-1)}+\frac{2}{(2x-1)^2}

(b) (ii) Let us obtain each fraction of f(x) in a form that could be expanded by the Binomial Theorem.

\begin{aligned} f(x)&=\frac{4}{x+2}-\frac{1}{(2x-1)}+\frac{2}{(2x-1)^2}\\ &=4(x+2)^{-1}-(2x-1)^{-1}+2(2x-1)^{-2}\\ &=4(2+x)^{-1}-(-1+2x)^{-1}+2(-1+2x)^{-2}\\ &=4[2(1+\frac{x}{2})]^{-1}-[-1(1-2x)]^{-1}+2[-1(1-2x)]^{-2}\\ &=4(\frac{1}{2})(1+\frac{x}{2})^{-1}-\frac{1}{(-1)}(1-2x)^{-1}+2\frac{1}{(-1)^2}(1-2x)^{-2}\\ &=2(1+\frac{x}{2})^{-1}+(1-2x)^{-1}+2(1-2x)^{-2} \end{aligned}
Now expanding each of these three terms using the Binomial Theorem gives:

\begin{aligned} f(x)&=2[1+(-1)(\frac{x}{2})+\frac{(-1)(-2)}{1\cdot2}(\frac{x}{2})^2+\frac{(-1)(-2)(-3)}{1\cdot 2 \cdot 3}(\frac{x}{2})^3+\cdots]\\ &\ \ \ +[1+(-1)(-2x)+\frac{(-1)(-2)}{1\cdot 2}(-2x)^2+\frac{(-1)(-2)(-3)}{1\cdot 2 \cdot 3}(-2x)^3+\cdots]\\ &\ \ \ +2[1+(-2)(-2x)+\frac{(-2)(-3)}{1\cdot 2}(-2x)^2+\frac{(-2)(-3)(-4)}{1\cdot 2 \cdot 3}(-2x)^3+\cdots]\\ &=2[1-\frac{1}{2}x+\frac{x^2}{4}-\frac{x^3}{8}+\cdots]\\ &\ \ \ +[1+2x+4x^2+8x^3+\cdots]\\ &\ \ \ +2[1+4x+12x^2+32x^3+\cdots]\\ &=5+9x+28\frac{1}{2}x^2+71\frac{3}{4}x^3+\cdots \end{aligned}
Let us find the range of values for which this expansion is valid. The following two inequalities must be satisfied.
\begin{aligned} -1\lt\frac{x}{2}\lt 1\\ -2\lt x \lt 2 \end{aligned} \begin{aligned} \qquad\qquad \end{aligned} \begin{aligned} -1\lt -2x \lt 1\\ -\frac{1}{2}\lt x \lt \frac{1}{2} \end{aligned}
The intersection of these two intervals is: -\frac{1}{2}\lt x \lt \frac{1}{2}

September 2019 Paper 1 Question 5

Solution Matrix Trans

September 2019 Paper 1 Question 6

Xi curve sketching

September 2019 Paper 1 Question 7

(a) Let I=\int(3x^2-x+1)\sin{x}dx
\begin{aligned} u&=3x^2-x+1\\ u'&=6x-1 \end{aligned} \begin{aligned} \qquad \end{aligned} \begin{aligned} v&=\int \sin{x} dx\\ v&=-\cos{x} \end{aligned}
Integrating by parts:
\begin{aligned} I&=uv-\int vu'dx\\ &=(3x^2-x+1)(-\cos{x})-\int(-\cos{x})(6x-1)dx\\ &=(-3x^2+x-1)\cos{x}+\int(6x-1)\cos{x}dx \end{aligned}
Let J=\int(6x-1)\cos{x}dx
\begin{aligned} u&=6x-1\\ u'&=6 \end{aligned} \begin{aligned} \qquad \end{aligned} \begin{aligned} v&=\int \cos{x} dx\\ v&=\sin{x} \end{aligned}
Integrating by parts:
\begin{aligned} J&=uv-\int vu'dx\\ &=(6x-1)\sin{x}-\int 6\sin{x}dx\\ &=(6x-1)\sin{x}+6\cos{x}+C \end{aligned}
\begin{aligned} \therefore I&=(-3x^2+x-1)\cos{x}+J\\ &=(-3x^2+x-1)\cos{x}+(6x-1)\sin{x}+6\cos{x}+C\\ &=(-3x^2+x+5)\cos{x}+(6x-1)\sin{x}+C \end{aligned}

(b) (i) Let y=a^x.
\begin{aligned} y&=a^x\\ &=(e^{\ln{a}})^x\\ &=e^{(\ln{a})x}\text{ (using the Laws of Indices)} \end{aligned}
\begin{aligned} \therefore\frac{dy}{dx}&=e^{(\ln{a})x}\ln{a}\text{ (by the Chain Rule)}\\ &=(e^{\ln{a}})^x\ln{a}\text{ (using the Laws of Indices)}\\ &=a^x\ln{a}\ \square \end{aligned}

(b) (ii) Let I=\int 2^x\sqrt{2^x+1}dx and let us use the substitution u=2^x:
\begin{aligned} u&=2^x\ \ \ \ ...(1)\\ \frac{du}{dx}&=2^x\ln{2}\\ \frac{du}{\ln{2}}&=2^x dx\ \ \ \ ...(2) \end{aligned}
\begin{aligned} I&=\int 2^x\sqrt{2^x+1}dx\\ &=\int \frac{\sqrt{u+1}du}{\ln{2}}\text{ (substituting Equations (1) and (2))}\\ &=\frac{1}{\ln{2}}\int (u+1)^{\frac{1}{2}}du\\ &=\frac{2(u+1)^{\frac{3}{2}}}{3\ln{2}}+C\\ &=\frac{2(2^x+1)^{\frac{3}{2}}}{3\ln{2}}+C \end{aligned}
\begin{aligned} &\therefore\int_0^3 2^x\sqrt{2^x+1}dx\\ &=\begin{bmatrix}\frac{2(2^x+1)^{\frac{3}{2}}}{3\ln{2}}\end{bmatrix}_0^3\\ &=\frac{2(2^3+1)^{\frac{3}{2}}}{3\ln{2}}-\frac{2(2^0+1)^{\frac{3}{2}}}{3\ln{2}}\\ &=\frac{2(27-\sqrt{8})}{3\ln{2}} \end{aligned}

September 2019 Paper 1 Question 9

(a) Let us count the number of 6-digit positive integers.


The 1^st digit can be chosen in 9 ways. Any value from 1,2,3,4,5,6,7,8,9 is possible. Note that 0 is not included because otherwise we would obtain an integer having 5 or less digits.

Each of the 2^nd, 3^rd, \cdots , 6^th digits can be chosen in 10 ways. Any value from 0,1,2,3,4,5,6,7,8,9 is possible.

By the multiplication principle, the number of 6-digit positive integers is:

9\times10\times10\times10\times10\times10=900000
Now let us count the number of 6-digit positive integers that begin and end with an even digit.


The 1^st digit can be chosen in 4 ways. Any value from 2,4,6,8 is possible. Again 0 is not included because otherwise we would obtain an integer having 5 or less digits.

The 6^th digit can be chosen in 5 ways. Any value from 0,2,4,6,8 is possible.

Each of the 2^nd, 3^rd, 4^th, 5^th digits can be chosen in 10 ways. Any value from 0,1,2,3,4,5,6,7,8,9 is possible.

By the multiplication principle, the number of 6-digit positive integers beginning and ending in an even digit is:

4\times10\times10\times10\times10\times5=200000
\begin{aligned}\therefore\text{Probability}&=\frac{200000}{900000}\\&=\frac{2}{9}\end{aligned}
(b) (i) The set of 6-digit positive integers containing the digit 7 once can be partitioned into the following sets:

➀ The set of 6-digit positive integers containing the digit 7 only in the 1^st place
➁ The set of 6-digit positive integers containing the digit 7 only in the 2^nd place
➂ The set of 6-digit positive integers containing the digit 7 only in the 3^rd place
\qquad\qquad\vdots
➅ The set of 6-digit positive integers containing the digit 7 only in the 6^th place

➀ Let us count the number of 6-digit positive integers containing the digit 7 only in the 1^st place.


The 1^st digit can be chosen in just 1 way, by choosing it to be 7.

The remaining digits can be chosen in 9 ways, that is, any digit except 7.

By the multiplication principle, the number of 6-digit positive integers containing the digit 7 only in the 1^st place is:

1\times9\times9\times9\times9\times9=59049
➁ Let us count the number of 6-digit positive integers containing the digit 7 only in the 2^nd place.


The 1^st digit can be chosen in 8 ways, by choosing any digit from 1,2,3,4,5,6,8,9

The 2^nd digit can be chosen in just 1 way, by choosing it to be 7.

The remaining digits can be chosen in 9 ways, that is, any digit except 7.

By the multiplication principle, the number of 6-digit positive integers containing the digit 7 only in the 2^nd place is:

8\times1\times9\times9\times9\times9=52488
➂ - ➅ The cases for the 6-digit positive integers containing 7 only in the 3^rd, 4^th, 5^th and 6^th place are similar to previous case.

Therefore the number of 6-digit positive integers containing the digit 7 once is:

59049+5(52488)=321489

(b) (ii) The set of 6-digit positive integers containing the digit 7 and 8 only once can be partitioned into the following sets:

➀ The set of 6-digit positive integers containing the digit 7 and 8 only once and either 7 or 8 is in the 1^st place.
➁ The set of 6-digit positive integers containing the digit 7 and 8 only once and neither 7 nor 8 is in the 1^st place.

➀ Let us count the number of 6-digit positive integers containing the digit 7 and 8 only once and either 7 or 8 is in the 1^st place.

Let us assume that the digit 7 and 8 occupy the first two places.


The 1^st digit can be chosen in 2 ways, by choosing either 7 or 8.

The 2^nd digit can be chosen in just 1 way. If 7 was chosen for the 1^st place, then 8 is chosen and vice-versa.

The remaining digits can be chosen in 8 ways, that is, any digit except 7 or 8.

Moreover there are 5 ways how to choose the place where to put 7 or 8 apart from the 1^st place.

Therefore by the multiplication principle, the number of 6-digit positive integers containing the digit 7 and 8 only once and either 7 or 8 is in the 1^st place: is:

5\times2\times 1\times8\times8\times8\times8=40960
➁ Let us count the number of 6-digit positive integers containing the digit 7 and 8 only once and neither 7 nor 8 is in the 1^st place.

Let us assume that the digit 7 and 8 occupy the 2^nd and 3^th places.


The 1^st digit can be chosen in 7 ways, by choosing any digit from 1,2,3,4,5,6,9.

The 2^nd digit can be chosen in 2 ways, by choosing either 7 or 8. The 3^rd digit can be chosen in just 1 way. If 7 was chosen for the 2^nd place, then 8 is chosen and vice-versa.

The remaining digits can be chosen in 8 ways, that is, any digit except 7 or 8.

Moreover there are \begin{pmatrix}5\\2\end{pmatrix} ways how to choose the two places where to put the 7 and 8, excluding the 1^st place.

Therefore by the multiplication principle, the number of 6-digit positive integers containing the digit 7 and 8 only once and neither 7 nor 8 is in the 1^st place is:

\begin{pmatrix}5\\2\end{pmatrix}\times 7\times2\times1\times8\times8\times8\times8=71680
Therefore, the number of 6-digit positive integers containing the digit 7 and 8 only once is:

40960+71680=112640

September 2019 Paper 1 Question 10

(a) Let the given function be expressed as the square of a quadratic expression:

\begin{aligned} x^4+12x^3+46x^2+px+q&\equiv (ax^2+bx+c)^2\\ &\equiv (ax^2+bx+c)(ax^2+bx+c) \end{aligned}
Equating coefficients of x^4:

\begin{aligned} 1&=a^2\\ \therefore a&=\pm 1 \end{aligned}
Let us assume that a=1.

Note that we could have chosen a to be -1 and then we would get values of \ b and c with opposite signs than the values obtained in this case. Ultimately both cases will give us the same values for p and q.

Equating coefficients of x^3:

\begin{aligned} 12&=ab+ba\\ 12&=2ab\\ 12&=2(1)b\text{ (since }a=1)\\ \therefore b&=6 \end{aligned}
Equating coefficients of x^2:

\begin{aligned} 46&=ac+b^2+ca\\ 46&=2ac+b^2\\ 46&=2(1)c+6^2\text{ (since }a=1\text{ and }b=6)\\ \therefore c&=5 \end{aligned}
Equating coefficients of x:

\begin{aligned} p&=bc+cb\\ p&=2bc\\ p&=2(6)(5)\text{ (since }b=6\text{ and }c=5)\\ \therefore p&=60 \end{aligned}
Equating constants:

\begin{aligned} q&=c^2\\ q&=5^2 \text{ (since }c=5)\\ \therefore q&=25 \end{aligned}
Hence: p=60 and q=25.

(b) Let us simplify the given equation to one in quadratic form:

\frac{x}{x-a}+\frac{x}{x-b}=1+c
Making a common denominator for the L.H.S.:

\frac{x(x-b)+x(x-a)}{(x-a)(x-b)}=1+c
Taking the denominator on the L.H.S. as a factor on the R.H.S.:

x(x-b)+x(x-a)=(1+c)(x-a)(x-b)
Expanding:

x^2-bx+x^2-ax=(1+c)x^2-(1+c)(a+b)x+(1+c)ab
Putting all the terms on the L.H.S.:

(1-c)x^2+c(a+b)x-ab(1+c)=0\ \ \ \ ...(1)
Recall that in general, the quadrating equation a'x^2+b'x+c'=0 has real and equal roots if and only if b'^2-4a'c'=0. Hence considering Equation (1):

\begin{aligned} c^2(a+b)^2+4(1-c)(ab)(1+c)&=0\\ c^2(a+b)^2+4(1-c^2)(ab)&=0\\ c^2(a+b)^2+4ab-4c^2(ab)&=0\\ c^2[(a+b)^2-4ab]+4ab&=0\\ c^2(a-b)^2&=-4ab\text{ (note that }(a+b)^2-4ab=(a-b)^2)\\ \therefore c^2&=-\frac{4ab}{(a-b)^2}\ \square \end{aligned}
\begin{aligned} c^2&=-\frac{4ab}{(a-b)^2}\\ &=1-1-\frac{4ab}{(a-b)^2}\\ &=1-\frac{(a-b)^2+4ab}{(a-b)^2}\\ &=1-\frac{(a+b)^2}{(a-b)^2}\text{ (since }(a+b)^2-4ab=(a-b)^2)\\ &=1-\begin{pmatrix}\frac{a+b}{a-b}\end{pmatrix}^2\ \square \end{aligned}
Note that c is a real number, \therefore c^2\geq 0. Suppose that c^2=0:

\begin{aligned} \therefore 1-\begin{pmatrix}\frac{a+b}{a-b}\end{pmatrix}^2&=0\\ (a+b)^2&=(a-b)^2\\ ab&=0 \end{aligned}
\therefore a=0\text{ or }b=0
This is not possible since both a and b should be non-zero. Hence c^2\neq 0. Hence c^2 >0.

On the other hand, \begin{pmatrix}\frac{a+b}{a-b}\end{pmatrix}^2 is a non-negative number. Therefore the maximum of c^2=1-\begin{pmatrix}\frac{a+b}{a-b}\end{pmatrix}^2 is 1 and this is obtained when a=-b. Hence c^2\leq 1.

\therefore 0\lt c^2\leq 1\ \square

September 2019 Paper 2 Question 1

(a) Let the Integrating Factor be:

\begin{aligned} \text{I.F.}&=e^{\int (2x+1)dx}\\ &=e^{x^2+x} \end{aligned}
We multiply the original equation by the I.F. to get an exact differential equation:

\begin{aligned}e^{x^2+x}\frac{dy}{dx}+(2x+1)ye^{x^2+x}-e^{-x^2}e^{x^2+x}&=0\\e^{x^2+x}\frac{dy}{dx}+(2x+1)ye^{x^2+x}-e^x&=0\end{aligned}
Using the product rule in reverse:

\frac{d}{dx}(e^{x^2+x}y)-e^x=0
Therefore:
\begin{aligned} \frac{d}{dx}(e^{x^2+x}y)&=e^x\\ \int\frac{d}{dx}(e^{x^2+x}y)dx&=\int e^x dx\\ e^{x^2+x}y&=e^x+C \end{aligned}

(b) Consider \frac{d^2 y}{dx^2}+4\frac{dy}{dx}+4y=0

The auxiliary equation is:

\begin{aligned} m^2+4m+4&=0\\ (m+2)(m+2)&=0\\ m&=-2 \end{aligned}
We have real and repeat roots. Therefore the Complimentary Function (C.F.) is:

\begin{aligned} y=(A+Bx)e^{-2x} \end{aligned}
Let the Particular Integral be:

\begin{aligned} y&=ax^2+bx+c&\ \ \ \ ...(1)\\ \therefore\frac{dy}{dx}&=2ax+b &\ \ \ \ ...(2)\\ \text{and }\frac{d^2 y}{dx^2}&=2a &\ \ \ \ ...(3) \end{aligned}
Substituting Equations (1), (2) and (3) in the original 2^nd Order Differential Equation gives us:

2a+4(2ax+b)+4(ax^2+bx+c)\equiv 8x^2
Let us find the values of a, b and c.

Equating the coefficients of x^2:

\begin{aligned} 4a&=8\\ a&=2 \end{aligned}
Equating the coefficients of x:

\begin{aligned} 8a+4b&=0\\ 8(2)+4b&=0\text{ (since }a=2)\\ b&=-4 \end{aligned}
Equating constants:

\begin{aligned} 2a+4b+4c&=0\\ 2(2)+4(-4)+4c&=0\text{ (since }a=2\text{ and }b=-4)\\ c&=3 \end{aligned}
Therefore the Particular Integral (P.I.) is:

y=2x^2-4x+3
The General Solution (G.S.) is:

\begin{aligned} y&=\text{C.F.}+\text{P.I.}\\ y&=(A+Bx)e^{-2x}+2x^2-4x+3\ \ \ \ ...(4) \end{aligned}
We will use the initial conditions to find A and B in order to find the Particular Solution (P.S.).

Substituting x=0 and y=0 in Equation (4):

\begin{aligned} 0&=Ae^{-2(0)}+2(0)^2-4(0)+3\\ \therefore A&=-3 \end{aligned}
By differentiating Equation (4) we get:

\frac{dy}{dx}=-2(A+Bx)e^{-2x}+Be^{-2x}+4x-4\ \ \ \ (5)
Substituting x=0 and \frac{dy}{dx}=0 in Equation (5):

\begin{aligned} 0&=-2(A+0)e^{-2(0)}+Be^{-2(0)}+4(0)-4\\ &B=-2 \end{aligned}
By substituting A=-3 and B=-2 in the G.S. (Equation (4)), we get the Particular Solution (P.S.):

\begin{aligned} y&=(-3-2x)e^{-2x}+2x^2-4x+3\\ y&=-(3+2x)e^{-2x}+2x^2-4x+3 \end{aligned}

September 2019 Paper 2 Question 2

*****Includes Newton Rhapson Trapezium Rule and Simpson's Rule *****

2 (a) Consider the function f(x)=\sin(2x)+\ln(x)

\begin{aligned} f(0.2)&=\sin(2(0.2))+\ln(0.2)\\ &=\sin(0.4)+\ln(0.2)\\ &=-1.22\\ &\lt 0 \end{aligned}
\begin{aligned} f(1)&=\sin(2(1))+\ln(1)\\ &=\sin(2)+\ln(1)\\ &=0.91\\ &\gt 0 \end{aligned}
Since the function f changes its sign in the interval [0.2,1], it has a root in such an interval.

Now let us use Newton-Rhapson's Method to find an approximation of this root.

\begin{aligned} f(x)&=\sin(2x)+\ln(x)\\ f'(x)&=2\cos(2x)+\frac{1}{x} \end{aligned}
1^st iteration:

x_1=0.6

\begin{aligned} f(0.6)&=\sin(2(0.6))+\ln(0.6)=0.4212\\ f'(0.6)&=2\cos(2(0.6))+\frac{1}{0.6}=2.3914 \end{aligned}
By Newton-Rhapson's Approximation:

\begin{aligned} x_2&=x_1-\frac{f(x_1)}{f'(x_1)}\\ &=0.6-\frac{f(0.6)}{f'(0.6)}\\ &=0.6-\frac{0.4212}{2.3914}\\ &=0.4239 \end{aligned}
2^nd iteration:

x_2=0.4239

\begin{aligned} f(0.4239)&=\sin(2(0.4239))+\ln(0.4239)=-0.1086\\ f'(0.4239)&=2\cos(2(0.4239))+\frac{1}{0.4239}=3.6826 \end{aligned}
By Newton-Rhapson's Approximation:

\begin{aligned} x_3&=x_2-\frac{f(x_2)}{f'(x_2)}\\ &=0.4239-\frac{f(0.4239)}{f'(0.4239)}\\ &=0.4239-\frac{0.1086}{3.6826}\\ &=0.4533 \end{aligned}
(b) Since the interval width h=1 we are going to consider the value of the function y=\frac{1}{\ln (x+4)} at the following x-values:

x	-1	0	1	2	3
y=\frac{1}{\ln (x+4)}	y_0=\frac{1}{\ln 3}	y_1=\frac{1}{\ln 4}	y_2=\frac{1}{\ln 5}	y_3=\frac{1}{\ln 6}	y_4=\frac{1}{\ln 7}

(i) Using the Trapezoidal Rule with n=4:

\begin{aligned} \int_{-1}^{3}\frac{1}{\ln(x+4)}dx&\simeq \frac{1}{2}h \{(y_0+y_4)+2(y_1+y_2+y_3)\}\\ &=\frac{1}{2}(1) \begin{Bmatrix}(\frac{1}{\ln 3}+\frac{1}{\ln 7})+2(\frac{1}{\ln 4}+\frac{1}{\ln 5}+\frac{1}{\ln 6})\end{Bmatrix}\text{ (since }h=1)\\ &=2.613 \end{aligned} (ii) Using Simpson's Rule with n=4:

\begin{aligned} \int_{-1}^{3}\frac{1}{\ln(x+4)}dx&\simeq \frac{1}{3}h \{(y_0+y_4)+4(y_1+y_3)+2(y_2)\}\\ &=\frac{1}{3}(1) \begin{Bmatrix}(\frac{1}{\ln 3}+\frac{1}{\ln 7})+4(\frac{1}{\ln 4}+\frac{1}{\ln 6})+2(\frac{1}{\ln 5})\end{Bmatrix}\text{ (since }h=1)\\ &=2.595 \end{aligned}
The error for the Trapezoidal Rule estimation is:

\text{Exact}-\text{Estimate}=2.593-2.613=-0.02
The error for the Simpson's Rule estimation is:

\text{Exact}-\text{Estimate}=2.593-2.595=-0.002
The Simpson's Rule Estimation gives a lower absolute error, therefore it offers a better estimate.

September 2019 Paper 2 Question 5

(a) Proof by Induction.

Induction Basis: Let n=8.

\begin{aligned} \text{L.H.S.}&=n^2=8^2=64\\ \text{R.H.S.}&=7k+1=7(8)+1=57 \end{aligned}
\therefore\text{L.H.S.}\gt\text{R.H.S.}
\therefore Result holds for n=8.

Induction Hypothesis: Assume that the result holds for n=k, that is, k^2 \gt 7k+1.

Induction Step: Let us show that the result holds for n=k+1.

\begin{aligned} \text{L.H.S.}&=n^2\\ &=(k+1)^2\\ &=k^2+2k+1\\ &> (7k+1) +2k+1\text{ (by the Induction Hypothesis)}\\ &=9k+2\\ &=7k+7+2k-5\\ &=7(k+1)+2k-5\\ &> 7(k+1)+1 \text{ (note that since }k\geq 8\text{, then it follows that }2k-5>1)\\ &=\text{R.H.S.} \end{aligned}
\therefore\text{L.H.S.}\gt\text{R.H.S.}
\therefore Result holds for n=k+1.

\therefore Result holds by Induction for all n\geq 8. \square

(b) (i) Let n=1 in the recurrence relation u_{n+1}=\frac{2 u_n-1}{3}:

\begin{aligned} u_2&=\frac{2u_1-1}{3}\\ &=\frac{2(1)-1}{3}\text{ (since }u_1=1)\\ &=\frac{1}{3} \end{aligned}
Let n=2 in the recurrence relation:

\begin{aligned} u_3&=\frac{2u_2-1}{3}\\ &=\frac{2(\frac{1}{3})-1}{3}\text{ (since }u_2=\frac{1}{3})\\ &=-\frac{1}{9} \end{aligned}
Let n=3 in the recurrence relation:

\begin{aligned} u_4&=\frac{2u_3-1}{3}\\ &=\frac{2(-\frac{1}{9})-1}{3}\text{ (since }u_3=-\frac{1}{9})\\ &=-\frac{11}{27} \end{aligned}
Let n=4 in the recurrence relation:

\begin{aligned} u_5&=\frac{2u_4-1}{3}\\ &=\frac{2(-\frac{11}{27})-1}{3}\text{ (since }u_4=-\frac{11}{27})\\ &=-\frac{49}{81} \end{aligned}
Hence the first 5 terms of the sequence are: 1,\frac{1}{3},-\frac{1}{9},-\frac{11}{27},-\frac{49}{81}.

(ii) Consider the equation u_n=3(\frac{2}{3})^n-1.

Induction Basis: Let n=1.

\begin{aligned} \text{L.H.S.}&=u_1\\ &=1\text{ (from Part (b) (i))} \end{aligned}
\begin{aligned}\text{R.H.S.}&=3\begin{pmatrix}\frac{2}{3}\end{pmatrix}^1-1\\&=1 \end{aligned}
\therefore\text{L.H.S.}=\text{R.H.S.}
\therefore Result holds for n=1.

Induction Hypothesis: Assume that the result holds for n=k, that is, u_k=3(\frac{2}{3})^k-1.

Induction Step: Let us show that the result holds for n=k+1.

\begin{aligned} \text{L.H.S.}&=u_{k+1}\\ &=\frac{2u_k-1}{3}\text{ using the recurrence relation given in Part (b) (i))}\\ &=\frac{2(3(\frac{2}{3})^k-1)-1}{3}\text{ (by the Induction Hypothesis)}\\ &=\frac{2(3)(\frac{2}{3})^k-2-1}{3}\\ &=\frac{2(3)(\frac{2}{3})^k-3}{3}\\ &=2\begin{pmatrix}\frac{3}{3}\end{pmatrix}\begin{pmatrix}\frac{2}{3}\end{pmatrix}^k-1\\ &=3\begin{pmatrix}\frac{2}{3}\end{pmatrix}^{k+1}-1\\ &=\text{R.H.S.} \end{aligned}
\therefore \text{L.H.S.}= \text{R.H.S.}
\therefore Result holds for n=k+1.

\therefore Result holds by Induction for all n\geq 1. \square

September 2020 Paper 2 Question 7

**** NOTE this question is a combination of Maclaurin's Series and Polar Coordinates **** (a) (i) Maclaurin's Theorem states that:

f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+\frac{x^4}{4!}f^{iv}(0)+\cdots\ \ \ \ ...(1)

\begin{aligned} f(x)&=(1+e^{-x})^2\\ f'(x)&=2(1+e^{-x})(-e^{-x})\\ &=-2e^{-x}(1+e^{-x})\\ &=-2e^{-x}-2e^{-2x}\\ f''(x)&=2e^{-x}+4e^{-2x}\\ f'''(x)&=-2e^{-x}-8e^{-2x}\\ f^{iv}(x)&=2e^{-x}+16e^{-2x} \end{aligned} \begin{aligned} \qquad \end{aligned} \begin{aligned} f(0)&=(1+e^{-0})^2=4\\ \\ \\ f'(0)&=-2-2=-4\\ f''(0)&=2+4=6\\ f'''(0)&=-2-8=-10\\ f^{iv}(0)&=2+16=18 \end{aligned}
Substituting the values for f'(0), f''(0), f'''(0) and f^{iv}(0) in the Maclaurin's expansion, given by Equation (1) results in:

\begin{aligned} (1+e^{-x})^2&=4+x(-4)+\frac{x^2}{2!}(6)+\frac{x^3}{3!}(-10)+\frac{x^4}{4!}(18)+\cdots\\ &=4-4x+3x^2-\frac{5}{3}x^3+\frac{3}{4}x^4+\cdots \end{aligned}
(ii) Consider the sequence: f'(0), f''(0), f'''(0),f^{iv}(0),\cdots

In Part (a) (i) it was shown that this sequence is given by:

-2-2, 2+4,-2-8,2+16,\cdots
The general term of such as sequence is:

\begin{aligned} f^{(n)}(0)&=2(-1)^n+2^n(-1)^n\\ &=(2+2^n)(-1)^n \end{aligned}
Therefore, for n\geq 1:

\begin{aligned} \text{The coefficient of }x^n&=\frac{f^{(n)}(0)}{n!}\\ &=\frac{(2+2^n)(-1)^n}{n!} \end{aligned}
(b) (i) To convert an equation from Cartesian to Polar form, we use the equations:

\begin{aligned} x^2+y^2&=r^2&\ \ \ \ ...(2)\\ x&=r\cos\theta&\ \ \ \ ...(3)\\ \text{and }y&=r\sin\theta&\ \ \ \ ...(4)\end{aligned}
\begin{aligned} (x^2+y^2)^{\frac{3}{2}}&=x^2+5y^2\\ (r^2)^{\frac{3}{2}}&=r^2\cos^2\theta+5r^2\sin^2\theta\text{ (using Equations (2), (3) and (4))}\\ r^3&=r^2(cos^2\theta+5\sin^2\theta)\\ r&=\cos^2\theta+5\sin^2\theta\\ r&=\cos^2\theta+5(1-\cos^2\theta)\text{ (since }\sin^2 A +\cos^2 A=1)\\ r&=5-4\cos^2\theta\\ r&=5-4\begin{pmatrix}\frac{\cos 2\theta+1}{2}\end{pmatrix}\text{ (since }\cos 2A =2\cos^2A-1)\\ r&=5-2\cos 2\theta-2\\ r&=3-2\cos 2\theta\ \square \end{aligned}
(ii) Let us find the points where the curve touches the pole. Such points have their r-value equal to 0. Hence let r=0.

\begin{aligned} r&=0\\ 3-2\cos 2\theta&=0\\ 2\cos 2\theta&=3\\ \cos 2\theta&=1.5 \end{aligned}
This equation has no solution for \theta since the range of the cosine function is [1,1] and 1.5 is outside this range.

Hence there are no points with r=0.

Hence the curve does not touch the pole. (Note: You will visualise this in part (iii) below)

Therefore there are no tangents to the curve at the pole. \square

(iii)

\theta	0	\frac{\pi}{6}	\frac{\pi}{3}	\frac{\pi}{2}	\frac{2\pi}{3}	\frac{5\pi}{6}	\pi	\frac{7\pi}{6}	\frac{4\pi}{3}	\frac{3\pi}{2}	\frac{5\pi}{3}	\frac{11\pi}{6}	2\pi
r=3-2\cos2\theta	1	2	4	5	4	2	1	2	4	5	4	2	1

(iv) Let A be the area enclosed by the curve between 0 and \frac{\pi}{2}.

Then A=\frac{11\pi}{4}, since it is one quarter of the whole area enclosed by the curve which is given to be 11\pi.

By using the formula for the area enclosed by a polar curve:

\begin{aligned} A&=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}r^2 d\theta\\ A&=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(3-2\cos 2\theta)^2 d\theta\\ \frac{11\pi}{4}&=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(3-2\cos 2\theta)^2 d\theta\\ \therefore\int_{0}^{\frac{\pi}{2}}(3-2\cos 2\theta)^2 d\theta &=\frac{11\pi}{2} \end{aligned}

September 2019 Paper 2 Question 10

(a) \begin{aligned} &\mathbb{P}[\text{\textquotedblleft Alarm 1 fails\textquotedblright}\cap\text{\textquotedblleft Alarm 2 fails\textquotedblright}\cap\text{\textquotedblleft Alarm 3 fails\textquotedblright}]\\ =&\mathbb{P}[\text{\textquotedblleft Alarm 1 fails\textquotedblright}]\mathbb{P}[\text{\textquotedblleft Alarm 2 fails\textquotedblright}]\mathbb{P}[\text{\textquotedblleft Alarm 3 fails\textquotedblright}]\\ &\ \ \ \ \ \ \ \ \text{ (by independence: }\mathbb{P}[A\cap B]=\mathbb{P}[A]\mathbb{P}[B])\\ =&(0.01)(0.01)(0.01) \end{aligned}
(b) \begin{aligned} \mathbb{P}[\text{\textquotedblleft 1 Black ball\textquotedblright}]&=\mathbb{P}[1^{st}\text{ selection is White}\cap 2^{nd}\text{ selection is White}]\\ &=\mathbb{P}[1^{st}\text{ sel. is White}]\mathbb{P}[2^{nd}\text{ sel. is White}|1^{st}\text{ sel. is White}]\\ &\ \ \ \ \ \ \ \ \text{ (since: }\mathbb{P}[A|B]=\frac{\mathbb{P}[A\cap B]}{\mathbb{P}[B]})\\ &=\begin{pmatrix}\frac{1}{2}\end{pmatrix}\begin{pmatrix}\frac{2}{3}\end{pmatrix}\\ &=\frac{1}{3} \end{aligned}
\begin{aligned} \mathbb{P}[\text{\textquotedblleft 2 Black balls\textquotedblright}]&=\mathbb{P}[1^{st}\text{ selection is Black}\cap 2^{nd}\text{ selection is White}]\\ &\ \ +\mathbb{P}[1^{st}\text{ selection is White}\cap 2^{nd}\text{ selection is Black}]\\ &=\mathbb{P}[1^{st}\text{ sel. is Black}]\mathbb{P}[2^{nd}\text{ sel. is White}|1^{st}\text{ sel. is Black}]\\ &\ \ +\mathbb{P}[1^{st}\text{ sel. is White}]\mathbb{P}[2^{nd}\text{ sel. is Black}|1^{st}\text{ sel. is White}]\\ &=\begin{pmatrix}\frac{1}{2}\end{pmatrix}\begin{pmatrix}\frac{1}{3}\end{pmatrix}+\begin{pmatrix}\frac{1}{2}\end{pmatrix}\begin{pmatrix}\frac{1}{3}\end{pmatrix}\\ &=\frac{1}{3} \end{aligned}
\begin{aligned} \mathbb{P}[\text{\textquotedblleft 3 Black balls\textquotedblright}]&=\mathbb{P}[1^{st}\text{ selection is Black}\cap 2^{nd}\text{ selection is Black}]\\ &=\mathbb{P}[1^{st}\text{ sel. is Black}]\mathbb{P}[2^{nd}\text{ sel. is Black}|1^{st}\text{ sel. is Black}]\\ &=\begin{pmatrix}\frac{1}{2}\end{pmatrix}\begin{pmatrix}\frac{2}{3}\end{pmatrix}\\ &=\frac{1}{3} \end{aligned}
(c) \begin{aligned} \mathbb{P}[\text{Biased Coin}|H]=\frac{wwww}{www} \end{aligned}
To Continue

September 2020 Paper 1 Question 1a

This is the solution of the question mentioned above. This is the solution of the question mentioned above. This is the solution of the question mentioned above.

\(x^2/2=4\)

September 2020 Paper 2 Question 4

Here it is.

May 2019 Paper 2 Question 10