2019 IJMB mathematics paper 2

SECTION A (20%)

1. Given the points $P(2,-3)$ and $Q(-2,5)$, find the coordinates of the mid-point $A$ of $PQ$ and its distance from $B(-1,3)$. [05 marks]

2. Calculate the value of $M$ of the distribution with probability density function $f(x) = M\sin x$, where $0\le x\le\pi$. [04 marks]

3. Calculate $P(A)$ given that $P(B)=0.72$, $P(A\cup B)=0.98$, if $A$ and $B$ are mutually exclusive events. [03 marks]

4. Find the equation of the circle whose centre is $(-2,1)$ and radius $\sqrt5$. [04 marks]

5. Find the eccentricity and foci of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{4}=1$. [04 marks]

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SECTION B: CO-ORDINATE GEOMETRY

6. (a) Show that the equation $y^2-2y-4x+5=0$ represents a parabola. Find the coordinates of its vertex, focus, and the equations of its directrix and axis. [10 marks]

(b) Find the equation of the circle which passes through the points $(1,2)$, $(1,0)$, and $(0,2)$. [10 marks]

7. (a) Show that the equation $x^2+4y^2+2x+8y+1=0$ represents an ellipse. Hence, find the value of its eccentricity, centre, and the equations of its directrices. [10 marks]

(b) Show that the locus of points which move so that its distance from the origin is twice its distance from the point $(3,0)$ is a circle. [10 marks]

8. (a) Find the equations of the tangents to the hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{4}=1$ from the point $(-2,1)$. [10 marks]

(b) Sketch the curve $r = 2(1-\sin\theta)$ in polar form, for $0\le\theta\le360°$. [10 marks]

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SECTION C: STATISTICS

9. (a) The number of deaths in a small town follows a Poisson law with an average of 2 deaths a week. Find the probability that in a certain week:

(i) at least two deaths occur; (ii) at most two deaths occur; (iii) there are exactly two deaths. [12 marks]

(b) If a card is randomly selected from a deck of cards, find the probability that it is:

(i) A queen or a Diamond; (ii) A king of Diamonds. [08 marks]

10. (a) If 20% of the bolts produced by a machine are defective, determine, using the binomial distribution, the probability that out of 5 bolts selected at random:

(i) 1; (ii) 0; and (iii) less than 2 will be defective. [12 marks]

(b) The age of students in a certain class is normally distributed with mean 18 years and standard deviation of 4 years. Determine the probability of a student with age between 15 years and 25 years. [08 marks]

11. The following are marks obtained by eight students out of a maximum of 10 marks for each subject in a test:

Statistics $X$	6	8	7	6	9	5	8	7
Economics $Y$	7	6	8	8	5	9	7	6

(i) Verify that for the above data $\bar{x}=\bar{y}=7$

(ii) Calculate the correlation coefficient, $r$.

(iii) Interpret the value of this coefficient.

(iv) Calculate Spearman’s rank correlation coefficient.

(v) Interpret the value of the coefficient.

(vi) Do the values in (ii) and (iv) above show any correspondence?

(vii) Plot the points $(x,y)$ on the scatter diagram.

(viii) Does the scatter diagram indicate the same type of result as in (ii) and (iv)?

(ix) Calculate the equation of the line of $x$ on $y$. [20 marks]

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COMPLETE SOLUTIONS

SECTION A

Q1. Midpoint $A$ of $PQ$ and Distance from $B(-1,3)$

Given: $P(2,-3)$, $Q(-2,5)$

Midpoint $A$:

$$A = \left(\frac{2+(-2)}{2}, \frac{-3+5}{2}\right) = \left(\frac{0}{2}, \frac{2}{2}\right)$$

$$\boxed{A = (0,1)}$$

Distance $AB$ where $B(-1,3)$:

$$AB = \sqrt{(0-(-1))^2 + (1-3)^2} = \sqrt{1+4}$$

$$\boxed{AB = \sqrt5\ \text{units}}$$

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Q2. Find $M$ for $f(x) = M\sin x$, $0\le x\le\pi$

For a valid probability density function:

$$\int_0^{\pi} f(x)\,dx = 1$$

$$M\int_0^{\pi}\sin x\,dx = 1$$

$$M[-\cos x]_0^{\pi} = 1$$

$$M[-\cos\pi + \cos0] = 1$$

$$M[1+1] = 1$$

$$2M = 1$$

$$\boxed{M = \frac12}$$

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Q3. Find $P(A)$ given Mutually Exclusive Events

Given:

• $P(B) = 0.72$
• $P(A\cup B) = 0.98$
• $A$ and $B$ are mutually exclusive $\Rightarrow P(A\cap B)=0$

Formula: $P(A\cup B) = P(A)+P(B)-P(A\cap B)$

$$0.98 = P(A) + 0.72 - 0$$

$$P(A) = 0.98 - 0.72$$

$$\boxed{P(A) = 0.26}$$

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Q4. Equation of Circle — Centre $(-2,1)$, Radius $\sqrt5$

Standard form: $(x-h)^2+(y-k)^2=r^2$

$$(x-(-2))^2 + (y-1)^2 = (\sqrt5)^2$$

$$(x+2)^2 + (y-1)^2 = 5$$

Expanding:

$$x^2+4x+4+y^2-2y+1 = 5$$

$$\boxed{x^2+y^2+4x-2y=0}$$

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Q5. Eccentricity and Foci of $\dfrac{x^2}{25}+\dfrac{y^2}{4}=1$

Comparing with $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$:

• $a^2=25 \Rightarrow a=5$
• $b^2=4 \Rightarrow b=2$
• Since $a>b$, major axis is along the $x$-axis

$$c^2 = a^2-b^2 = 25-4 = 21 \Rightarrow c=\sqrt{21}$$

Eccentricity:

$$e = \frac{c}{a} = \frac{\sqrt{21}}{5}$$

$$\boxed{e = \frac{\sqrt{21}}{5} \approx 0.917}$$

Foci: $(\pm c, 0)$

$$\boxed{\text{Foci} = (\sqrt{21},0)\ \text{and}\ (-\sqrt{21},0)}$$

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SECTION B: CO-ORDINATE GEOMETRY

Q6(a). Show $y^2-2y-4x+5=0$ is a Parabola

Complete the square in $y$:

$$y^2-2y = 4x-5$$

$$(y-1)^2 - 1 = 4x-5$$

$$(y-1)^2 = 4x-4$$

$$(y-1)^2 = 4(x-1)$$

This is of the form $(Y)^2 = 4pX$ where $Y=y-1$, $X=x-1$, $4p=4 \Rightarrow p=1$

This is a parabola opening to the right. ✓

Vertex: $X=0,Y=0 \Rightarrow x=1,y=1$

$$\boxed{\text{Vertex} = (1,1)}$$

Focus: $X=p=1 \Rightarrow x-1=1 \Rightarrow x=2,y=1$

$$\boxed{\text{Focus} = (2,1)}$$

Directrix: $X=-p \Rightarrow x-1=-1$

$$\boxed{x=0}$$

Axis of symmetry: $Y=0 \Rightarrow y-1=0$

$$\boxed{y=1}$$

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Q6(b). Circle Through $(1,2)$, $(1,0)$, and $(0,2)$

General equation: $x^2+y^2+Dx+Ey+F=0$

Substituting $(1,2)$:

$$1+4+D+2E+F=0 \Rightarrow D+2E+F=-5\quad (1)$$

Substituting $(1,0)$:

$$1+0+D+0+F=0 \Rightarrow D+F=-1\quad (2)$$

Substituting $(0,2)$:

$$0+4+0+2E+F=0 \Rightarrow 2E+F=-4\quad (3)$$

From $(1)-(2)$:

$$2E=-4 \Rightarrow E=-2$$

From $(3)$: $F = -4-2(-2) = -4+4 = 0$

From $(2)$: $D = -1-F = -1-0 = -1$

Equation:

$$\boxed{x^2+y^2-x-2y=0}$$

Verification:

• $(1,2)$: $1+4-1-4=0$ ✓
• $(1,0)$: $1+0-1-0=0$ ✓
• $(0,2)$: $0+4-0-4=0$ ✓

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Q7(a). Show $x^2+4y^2+2x+8y+1=0$ is an Ellipse

Complete the square:

$$(x^2+2x) + 4(y^2+2y) + 1 = 0$$

$$(x+1)^2-1 + 4[(y+1)^2-1] + 1 = 0$$

$$(x+1)^2-1+4(y+1)^2-4+1=0$$

$$(x+1)^2+4(y+1)^2=4$$

Dividing by 4:

$$\frac{(x+1)^2}{4} + \frac{(y+1)^2}{1} = 1$$

This is of the form $\dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}=1$ with $a^2=4$, $b^2=1$ $\Rightarrow$ ellipse ✓

Centre: $(-1,-1)$

$$\boxed{\text{Centre} = (-1,-1)}$$

$a=2$, $b=1$

$$c^2=a^2-b^2=4-1=3 \Rightarrow c=\sqrt3$$

Eccentricity:

$$\boxed{e = \frac{\sqrt3}{2} \approx 0.866}$$

Directrices (along major axis, parallel to $y$-axis):

$$x+1 = \pm\frac{a}{e} = \pm\frac{2}{\sqrt3/2} = \pm\frac{4}{\sqrt3}$$

$$\boxed{x = -1 \pm \frac{4}{\sqrt3} = -1 \pm \frac{4\sqrt3}{3}}$$

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Q7(b). Locus — Distance from Origin is Twice Distance from $(3,0)$

Let $P(x,y)$ be any point on the locus.

Distance from origin: $r_1 = \sqrt{x^2+y^2}$

Distance from $(3,0)$: $r_2 = \sqrt{(x-3)^2+y^2}$

Condition: $r_1 = 2r_2$

$$\sqrt{x^2+y^2} = 2\sqrt{(x-3)^2+y^2}$$

Squaring:

$$x^2+y^2 = 4[(x-3)^2+y^2]$$

$$x^2+y^2 = 4[x^2-6x+9+y^2]$$

$$x^2+y^2 = 4x^2-24x+36+4y^2$$

$$0 = 3x^2-24x+36+3y^2$$

$$3x^2+3y^2-24x+36=0$$

$$x^2+y^2-8x+12=0$$

Completing the square:

$$(x-4)^2-16+y^2+12=0$$

$$(x-4)^2+y^2=4$$

This is a circle with centre $(4,0)$ and radius $2$. ✓ $\blacksquare$

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Q8(a). Tangents to Hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{4}=1$ from $(-2,1)$

Equation of tangent with slope $m$ to $\dfrac{x^2}{16}-\dfrac{y^2}{4}=1$:

$$y = mx \pm\sqrt{16m^2-4}$$

Passing through $(-2,1)$:

$$1 = m(-2) \pm\sqrt{16m^2-4}$$

$$1+2m = \pm\sqrt{16m^2-4}$$

Squaring:

$$(1+2m)^2 = 16m^2-4$$

$$1+4m+4m^2 = 16m^2-4$$

$$0 = 12m^2-4m-5$$

Quadratic formula:

$$m = \frac{4\pm\sqrt{16+240}}{24} = \frac{4\pm\sqrt{256}}{24} = \frac{4\pm16}{24}$$

$$m_1 = \frac{20}{24} = \frac56,\qquad m_2 = \frac{-12}{24} = -\frac12$$

Tangent 1 ($m=5/6$):

$$y-1 = \frac56(x+2)$$

$$6y-6 = 5x+10$$

$$\boxed{5x-6y+16=0}$$

Tangent 2 ($m=-1/2$):

$$y-1 = -\frac12(x+2)$$

$$2y-2 = -x-2$$

$$\boxed{x+2y=0}$$

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Q8(b). Sketch $r=2(1-\sin\theta)$, $0\le\theta\le360°$

Key values table:

$\theta$	$\sin\theta$	$r=2(1-\sin\theta)$
$0°$	$0$	$2$
$30°$	$0.5$	$1$
$90°$	$1$	$0$
$150°$	$0.5$	$1$
$180°$	$0$	$2$
$210°$	$-0.5$	$3$
$270°$	$-1$	$4$
$330°$	$-0.5$	$3$
$360°$	$0$	$2$

This is a cardioid pointing downward (minimum $r=0$ at $\theta=90°$, maximum $r=4$ at $\theta=270°$).

         θ=90° (r=0)
              *
         *         *
    *                   *
  *    θ=0°(r=2)          *
    *                   *
         *         *
              *
         θ=270° (r=4) ← maximum

(Plot in polar coordinates showing a downward-pointing cardioid)

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SECTION C: STATISTICS

Q9(a). Poisson Distribution — Deaths

Given: $\lambda = 2$ deaths/week, $X\sim\text{Poisson}(2)$

Poisson formula: $P(X=k) = \dfrac{e^{-\lambda}\lambda^k}{k!}$

$$e^{-2} = 0.1353$$

$$P(X=0) = \frac{e^{-2}\cdot2^0}{0!} = 0.1353$$

$$P(X=1) = \frac{e^{-2}\cdot2^1}{1!} = 0.2707$$

$$P(X=2) = \frac{e^{-2}\cdot2^2}{2!} = 0.2707$$

(i) $P(X\ge2)$ — at least two deaths:

$$P(X\ge2) = 1-P(X=0)-P(X=1) = 1-0.1353-0.2707$$

$$\boxed{P(X\ge2) = 0.5940}$$

(ii) $P(X\le2)$ — at most two deaths:

$$P(X\le2) = P(X=0)+P(X=1)+P(X=2) = 0.1353+0.2707+0.2707$$

$$\boxed{P(X\le2) = 0.6767}$$

(iii) $P(X=2)$ — exactly two deaths:

$$\boxed{P(X=2) = \frac{e^{-2}\cdot4}{2} = 0.2707}$$

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Q9(b). Card Probability

Deck: 52 cards, 4 suits, 13 cards each

(i) $P(\text{Queen or Diamond})$:

Using the addition rule:

$$P(Q\cup D) = P(Q)+P(D)-P(Q\cap D) = \frac{4}{52}+\frac{13}{52}-\frac{1}{52} = \frac{16}{52}$$

$$\boxed{P(\text{Queen or Diamond}) = \frac{4}{13} \approx 0.308}$$

(ii) $P(\text{King of Diamonds})$:

Only 1 king of diamonds in the deck:

$$\boxed{P(\text{King of Diamonds}) = \frac{1}{52} \approx 0.019}$$

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Q10(a). Binomial Distribution — Defective Bolts

Given: $n=5$, $p=0.20$ (defective), $q=0.80$

$X\sim B(5,0.2)$

Formula: $P(X=r) = \binom{n}{r}p^r q^{n-r}$

(i) $P(X=1)$ — exactly 1 defective:

$$P(X=1) = \binom{5}{1}(0.2)^1(0.8)^4 = 5\times0.2\times0.4096$$

$$\boxed{P(X=1) = 0.4096}$$

(ii) $P(X=0)$ — none defective:

$$P(X=0) = \binom{5}{0}(0.2)^0(0.8)^5 = 1\times1\times0.32768$$

$$\boxed{P(X=0) = 0.3277}$$

(iii) $P(X<2)$ — less than 2 defective:

$$P(X<2) = P(X=0)+P(X=1) = 0.3277+0.4096$$

$$\boxed{P(X<2) = 0.7373}$$

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Q10(b). Normal Distribution — Student Ages

Given: $\mu=18$, $\sigma=4$

Find $P(15\le X\le25)$

Standardizing:

$$Z_1 = \frac{15-18}{4} = -\frac34 = -0.75$$

$$Z_2 = \frac{25-18}{4} = \frac74 = +1.75$$

$P(-0.75\le Z\le1.75)$:

$$= P(Z\le1.75) - P(Z\le-0.75) = 0.9599-0.2266$$

$$\boxed{P(15\le X\le25) = 0.7333 \approx 73.33\%}$$

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Q11. Correlation and Regression

Data:

Student	$X$	$Y$	$X^2$	$Y^2$	$XY$
1	6	7	36	49	42
2	8	6	64	36	48
3	7	8	49	64	56
4	6	8	36	64	48
5	9	5	81	25	45
6	5	9	25	81	45
7	8	7	64	49	56
8	7	6	49	36	42
$\Sigma$	56	56	404	404	382

$n=8$

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(i) Verify $\bar{x}=\bar{y}=7$

$$\bar{x} = \frac{\Sigma X}{n} = \frac{56}{8} = 7\ \checkmark$$

$$\bar{y} = \frac{\Sigma Y}{n} = \frac{56}{8} = 7\ \checkmark$$

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(ii) Correlation Coefficient $r$

Formula:

$$r = \frac{n\Sigma XY - \Sigma X\Sigma Y}{\sqrt{[n\Sigma X^2-(\Sigma X)^2][n\Sigma Y^2-(\Sigma Y)^2]}}$$

Numerator:

$$n\Sigma XY - \Sigma X\Sigma Y = 8(382)-(56)(56) = 3056-3136 = -80$$

Denominator:

$$n\Sigma X^2-(\Sigma X)^2 = 8(404)-56^2 = 3232-3136 = 96$$

$$n\Sigma Y^2-(\Sigma Y)^2 = 8(404)-56^2 = 3232-3136 = 96$$

$$\sqrt{96\times96} = 96$$

$$\boxed{r = \frac{-80}{96} = -0.833}$$

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(iii) Interpretation of $r$

$r=-0.833$ indicates a strong negative correlation between Statistics and Economics marks. Students who score high in Statistics tend to score low in Economics and vice versa.

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(iv) Spearman’s Rank Correlation Coefficient

Ranking (1 = highest):

Student	$X$	Rank $X$ ($R_x$)	$Y$	Rank $Y$ ($R_y$)	$d=R_x-R_y$	$d^2$
1	6	6.5	7	4.5	2	4
2	8	2.5	6	7	−4.5	20.25
3	7	4.5	8	1.5	3	9
4	6	6.5	8	1.5	5	25
5	9	1	5	8	−7	49
6	5	8	9	1	7	49
7	8	2.5	7	4.5	−2	4
8	7	4.5	6	7	−2.5	6.25
					$\Sigma d^2$	166.5

(Tied ranks are averaged)

Formula:

$$r_s = 1 - \frac{6\Sigma d^2}{n(n^2-1)}$$

$$r_s = 1 - \frac{6\times166.5}{8(64-1)}$$

$$r_s = 1 - \frac{999}{504}$$

$$r_s = 1 - 1.982 = -0.982$$

$$\boxed{r_s \approx -0.982}$$

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(v) Interpretation of $r_s$

$r_s = -0.982$ indicates a very strong negative rank correlation — there is almost perfect inverse ranking between Statistics and Economics performance.

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(vi) Correspondence Between $r$ and $r_s$

Yes, both $r=-0.833$ and $r_s=-0.982$ show strong negative correlation, indicating correspondence in their conclusions. Both confirm that students performing well in Statistics tend to perform poorly in Economics. The Spearman coefficient is slightly stronger due to its rank-based nature.

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(vii) Scatter Diagram

Plot points $(X,Y)$:

Y
9 |          *
8 |       *     *
7 |  *             *
6 |     *             *
5 |          *
  |___________________
     5  6  7  8  9   X

Points: $(6,7),(8,6),(7,8),(6,8),(9,5),(5,9),(8,7),(7,6)$

The points show a downward trend from left to right.

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(viii) Scatter Diagram vs Calculated Coefficients

Yes, the scatter diagram confirms the same result as (ii) and (iv). The downward slope of the points from top-left to bottom-right is consistent with the negative correlation found in both $r=-0.833$ and $r_s=-0.982$.

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(ix) Equation of Line of $x$ on $y$

Formula: $x-\bar{x} = r\left(\dfrac{\sigma_x}{\sigma_y}\right)(y-\bar{y})$

Computing $\sigma_x$ and $\sigma_y$:

$$\sigma_x^2 = \frac{\Sigma X^2}{n}-\bar{x}^2 = \frac{404}{8}-49 = 50.5-49 = 1.5$$

$$\sigma_y^2 = \frac{\Sigma Y^2}{n}-\bar{y}^2 = \frac{404}{8}-49 = 1.5$$

$$\sigma_x = \sigma_y = \sqrt{1.5}$$

Regression coefficient of $x$ on $y$:

$$b_{xy} = r\cdot\frac{\sigma_x}{\sigma_y} = -0.833\times\frac{\sqrt{1.5}}{\sqrt{1.5}} = -0.833$$

Line of $x$ on $y$:

$$x-7 = -0.833(y-7)$$

$$x-7 = -0.833y+5.833$$

$$\boxed{x = -0.833y+12.833}$$

or equivalently: $x + 0.833y = 12.833$