Vectors Question #4

Vectors Question #4

JC Mathematics

Another interesting vectors question.

The fixed point A has position vector a relative to a fixed point O. A variable point P has position vector r relative to O. Find the locus of P if r \bullet (ra) = 0.

Vectors Question #3

Vectors Question #3

JC Mathematics

This is a question a student sent me a few days back, and I shared with my class.

Find the Cartesian equation of the locus of all points (plane) that is equidistant of the xy plane and xz plane.

The following should aid students to visualise.

xy-, xz-, yz-planes

Sidenote: I think Vectors is a very important topic for 9758 as its applications are wide. Students should do their best to understand the topic. I will share a few more applied questions next week when I have time.

A little reminder to students doing Calculus now

A little reminder to students doing Calculus now

JC Mathematics

When \frac{dy}{dx} = 0, it implies we have a stationary point.

To determine the nature of the stationary point, we can do either the first derivative test or the second derivative.

The first derivative test:

First Derivative Test

Students should write the actual values of \alpha^-, \alpha, \alpha^+ and \frac{dy}{dx} in the table.

We use this under these two situations:
1. \frac{d^2y}{dx^2} is difficult to solve for, that is, \frac{dy}{dx} is tough to be differentiated
2. \frac{d^2y}{dx^2} = 0

The second derivative test:

Second Derivative Test

Other things students should take note is concavity and drawing of the derivative graph.

Vectors Question #2

Vectors Question #2

JC Mathematics

If c = |a| b + |b| a, where a , b and c are all non-zero vectors, show that c bisects the angle between a and b.

Thoughts on A’levels H2 Mathematics 2016 Paper 2

Thoughts on A’levels H2 Mathematics 2016 Paper 2

JC Mathematics

I’ll keep this short since we are all busy. One thing about paper 1 we saw, there were many unknowns.

So topics which I think will come out…

Differentiation – I think a min/max problem will come out, possibly with r and h both not given and asked to express r in terms of h. But students should revise a on the properties of curves with differentiation; given a curve equation with an unknown, for instance y=/frac{x^2+kx+1}{x-1}, find the range of k such that there is stationary points.

Complex Number – Loci will definitely come out. I’m saying they will combine with trigonometry.

Integration – Modulus integration hasn’t really been tested. Else a question on Area/ Volume could be tested, and I’ll say they need students to do some
Conics too.

For statistics, my students should have gotten the h1 stats this year. And if it’s an indicator, then it should not be a struggle.

I expect PnC and probability to be combined. Conditional Probability in a poisson question should be tested too, so do revise it well. For hypothesis testing, students should be careful of their formula and read really carefully about the alternative hypothesis. Also, :9 know that the formulas for poison PDF and binompdf are both given in mf15. Lastly, know when to use CLT.

All the best!

Random Questions from 2016 Prelims #12

Random Questions from 2016 Prelims #12

JC Mathematics

HCI/1/6

A group of boys want to set up a camping tent. They lay down a rectangular tarp OABC on the horizontal ground with OA = 3 m and AB = 1.5 m and secure the points D and E vertically above O and B respectively, such that .

HCI/1/6
HCI/1/6

Assume that the tent takes the shape as shown above with 6 triangular surfaces and a rectangular base. The point O is taken as the origin and the unit vectors i, j and k are taken to be in the direction of , and respectively.

(i) Show that the line DE can be expressed as r = 2k+\lambda(2i+j), \lambda \in \mathbb{R}.

(ii) Find the Cartesian equation of the plane ADE.

(iii) Determine the acute angle between the planes ADE and OABC. Hence, or otherwise, find the acute angle between the planes ADE and CDE.

Note: Question can be made harder and trickier should Origin, O be placed in the center of the base OACB.

2016 A-level H1 Mathematics (8864) Paper 1 Suggested Solutions

2016 A-level H1 Mathematics (8864) Paper 1 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. KS will hold no liability for any errors. Comments are entirely personal opinions.

As these workings and answers are rushed out asap, please pardon me for my mistakes and let me know if there is any typo. Many thanks.
I’ll try my best to attend to the questions as there is H2 Math Paper 2 coming up.

Question 1
(i)
\frac{d}{dx} [2 \text{ln}(3x^2 +4)]

= \frac{12x}{3x^2+4}

(ii)
\frac{d}{dx} [\frac{1}{2(1-3x)^2}]

= \frac{6}{2(1-3x)^3}

= \frac{3}{(1-3x)^3}


Question 2

2e^{2x} \ge 9 - 3e^x

2u^2 + 3u - 9 \ge 0

(2u-3)(u+3) \ge 0

\Rightarrow u \le -3 \text{~or~} u \ge \frac{3}{2}

e^x \le -3 (rejected since e^x > 0) or e^x \ge \frac{3}{2}

\therefore x \ge \text{ln} \frac{3}{2}


Question 3
(i)

Graph for 3i
Graph for 3i

(ii)
Using GC, required answer = -1.606531 \approx -1.61 (3SF)

(iii)
When x = 0.5, y = 0.35653066

y -  0.35653066 = \frac{-1}{-1.606531}(x-0.5)

y = 0.622459 x +0.04530106

y = 0.622 x + 0.0453 (3SF)

(iv)

\int_0^k e^{-x} -x^2 dx

= -e^{-x} - \frac{x^3}{3} \bigl|_0^k

= -e^{-k} - \frac{k^3}{3} + 1


Question 4
(i)
y = 1 + 6x - 3x^2 -4x^3

\frac{dy}{dx} = 6 - 6x - 12x^2

Let \frac{dy}{dx} = 0

6 - 6x - 12x^2 = 0

1 - x - 2x^2 = 0

2x^2 + x - 1 = 0

x = -1 or \frac{1}{2}

When x =-1, y = -4

When x = \frac{1}{2}, y = 2.75

Coordinates = (-1, -4) \text{~or~} (0.5, 2.75)

(ii)
\frac{dy}{dx} = 6 - 6x - 12x^2

\frac{d^2y}{dx^2} = - 6 - 24x

When x = -1, \frac{d^2y}{dx^2} = 18 > 0. So (-1, -4) is a minimum point.

When x = 0.5, \frac{d^2y}{dx^2} = -18 \textless 0. So (0.5, 2.75) is a maximum point.

(iii)

Graph for 4iii
Graph for 4iii

x-intercept = (-1.59, 0) \text{~and~} (1, 0) \text{~and~} (-0.157, 0)

(iv)
Using GC, \int_0.5^1 y dx = 0.9375


Question 5
(i)
Area of ABEDFCA = \frac{1}{2}(2x)(2x)\text{sin}60^{\circ} -  \frac{1}{2}(y)(y)\text{sin}60^{\circ}

2\sqrt{3} = \sqrt{3}x^2 - \frac{\sqrt{3}}{4}y^2

2 = x^2 - \frac{y^2}{4}

4x^2 - y^2 =8

(ii)

Perimeter = 10

4x+2y + (2x-y) = 10

6x + y = 10

y = 10 - 6x

4x^2 - (10-6x)^2 = 8

4x^2 - 100 +120x -36x^2 = 8

32x^2 -120x+108=0

x=2.25 \text{~or~} 1.5

When x=2.25, y = -3.5 (rejected since y >0)

When x=1.5, y = 1


Question 6
(i)
The store manager has to survey \frac{1260}{2400} \times 80 = 42 male students and \frac{1140}{2400} \times 80 = 38 female students in the college. He will do random sampling to obtain the required sample.

(ii)
Stratified sampling will give a more representative results of the students expenditure on music annually, compared to simple random sampling.

(iii)
Unbiased estimate of population mean, \bar{x} = \frac{\sum x}{n} = \frac{312}{80} = 3.9

Unbiased estimate of population variance, s^2 = \frac{1}{79}[1328 - \frac{312^2}{80}] = 1.40759 \approx 1.41


Question 7
(i)
[Venn diagram to be inserted]

(ii)
(a)
\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B) = 0.8
(b)
\text{P}(A \cup B) -  \text{P}(A \cap B) = 0.75

(iii)
\text{P}(A | B')

= \frac{\text{P}(A \cap B')}{\text{P}(B')}

= \frac{\text{P}(A) - \text{P}(A \cap B)}{1 - \text{P}(B)}

= \frac{0.6 - 0.05}{1-0.25}

= \frac{11}{15}


Question 8
(i)
Required Probability = \frac{1}{2} \times \frac{3}{10} \times \frac{2}{9} = \frac{1}{30}

(ii)
Find the probability that we get same color. then consider the complement.

Required Probability = 1 - \frac{1}{30} - \frac{1}{2} \times \frac{5}{10} \times \frac{4}{9} - \frac{1}{2} \times \frac{2}{10} \times \frac{1}{9} - \frac{1}{2} \times \frac{4}{6} \times \frac{3}{5} - \frac{1}{2} \times \frac{2}{6} \times \frac{1}{5} = \frac{11}{18}

(iii)
\text{P}(\text{Both~balls~are~red} | \text{same~color})

= \frac{\text{P}(\text{Both~balls~are~red~and~same~color})}{\text{P}(\text{same~color})}

= \frac{\text{P}(\text{Both~balls~are~red})}{\text{P}(\text{same~color})}

= \frac{1/30}{7/18}

= \frac{3}{35}


Question 9
(i)
Let X denote the number of batteries in a pack of 8 that has a life time of less than two years.

X \sim B(8, 0.6)
(a)
Required Probability =\text{P}(X=8) = 0.01679616 \approx 0.0168

(b)
Required Probability =\text{P}(X \ge 4) = 1 - \text{P}(X \le 3) = 0.8263296 \approx 0.826

(ii)
Let Y denote the number of packs of batteries, out of 4 packs, that has at least half of the batteries having a lifetime of less than two years.

Y \sim B(4, 0.8263296)

Required Probability =\text{P}(Y \le 2) = 0.1417924285 \approx 0.142

(iii)
Let W denote the number of batteries out of 80 that has a life time of less than two years.

W \sim B(80, 0.6)

Since n is large, np = 48 > 5, n(1-p)=32 >5

W \sim N(48, 19.2) approximately

Required Probability

= \text{P}(w \ge 40)

= \text{P}(w > 39.5) by continuity correction

= 0.9738011524
\approx 0.974


Question 10
(i)
Let X be the top of speed of cheetahs.
Let \mu be the population mean top speed of cheetahs.

H_0: \mu = 95

H_1: \mu \neq 95

Under H_0, \bar{X} \sim N(95, \frac{4.1^2}{40})

Test Statistic, Z = \frac{\bar{X}-\mu}{\frac{4.1}{\sqrt{40}}} \sim N(0,1)

Using GC, p=0.0449258443 \textless 0.05 \Rightarrow H_0 is rejected.

(ii)
H_0: \mu = 95

H_1: \mu > 95

For H_0 to be not rejected,

\frac{\bar{x}-95}{\frac{4.1}{\sqrt{40}}} \textless 1.644853626

\bar{x} \textless 96.06 \approx 96.0 (round down to satisfy the inequality)

\therefore \{ \bar{x} \in \mathbb{R}^+: \bar{x} \textless 96.0 \}


Question 11
(i)
[Sketch to be inserted]

(ii)
Using GC, r = 0.9030227 \approx 0.903 (3SF)

(iii)
Using GC, y = 0.2936681223 x - 1.88739083

y = 0.294 x - 1.89 (3SF)

(iv)
When x = 16.9, y = 3.0756 \approx 3.08(3SF)

Time taken = 3.08 minutes

Estimate is reliable since x = 16.9 is within the given data range and |r|=0.903 is close to 1.

(v)
Using GC, r = 0.5682278 \approx 0.568 (3SF)

(vi)
The answers in (ii) is more likely to represent since |r|=0.903 is close to 1. This shows a strong positive linear correlation between x and y.


Question 12
Let X, Y denotes the mass of the individual biscuits and its empty boxes respectively.

X \sim N(20, 1.1^2)

Y \sim N(5, 0.8^2)

(i)
\text{P}(X \textless 19) = 0.181651 \approx 0.182

(ii)
X_1 + ... + X_12 + Y \sim N(245, 15.16)

\text{P}(X_1 + ... + X_12 + Y > 248) = 0.2205021 \approx 0.221

(iii)
0.6X \sim N(12, 0.66^2)

0.2Y \sim N(1, 0.16^2)

Let A=0.6X and B = 0.2Y

A_1+...+A_12+B \sim N(145, 5.2528)

\text{P}(142 \textless A_1+...+A_12+B \textless 149) = 0.864257 \approx 0.864