O-Level Additional Math

[jieheng]

lost boy:

Hi, ps help to solve sec one math


2. Find the sum of the even numbers from 50 to 100 inclusive.

a = first term , d = common difference

Tn (nth term) = a + (n-1)*d

Sn (sum of first n terms) = (n/2)*[ 2a + (n-1)*d ]

a = 50 , d = 2

Tn = a + (n-1)*d

100 = 50 + (n-1)*2

n = 26

Sn = (n/2)*[ 2a + (n-1)*d ]

S26 = (26/2)*[2*50 + (26-1)*2] = 1950

[jieheng]

lost boy:

Hi, ps help to solve sec one math


4. In the arithmetic progression whose first term is -27,the tenth term is equal to the sum of the first nine terms. Calculate the common difference.

4)

a = -27 , d = common difference

Tn (nth term) = a + (n-1)*d

T10 = -27 + (10-1)*d = -27 + 9d

Sn = (n/2)*[ 2a + (n-1)*d ]

S9 = (9/2)*[2(-27) + (9-1)*d] = (9/2)[-54 + 8d] = -243 + 36d

T10 = S9

-27 + 9d = -243 + 36d

27d = 216

d = 6

[jieheng]

lost boy:

Hi, ps help to solve sec one math



5. The sum of the first 6 terms of an arithmetical progression is 55.5 and the sum of the next 6 terms is 145.5. Find the common difference and first term.

5)

a = first term , d = common difference

Sn (sum of first n terms) = (n/2)*[ 2a + (n-1)*d ]

S6 = (6/2)*[ 2a + (6-1)*d ] = 55.5
6a + 15d = 55.5 ------ (1)

Sum of next 6 terms = Sum of first 12 terms - Sum of first 6 terms = S12 - S6

S12 - S6 = 145.5
(12/2)*[ 2a + (12-1)d ] - 55.5 = 145.5
12a + 66d = 201 ----- (2)

Solving equations (1) and (2) ,

a = 3 , d = 2.5

[lost boy]

Thanks JieHeng! Your help is much appreciated.

[S-H]

mathtuition88:

积少成多: How can doing at least one Maths question per day help you improve!


We all know the saying “an apple a day keeps the doctor away“. Many essential activities, like eating, exercising, sleeping, needs to be done on a daily basis.

Mathematics is no different!

Here is a surprising fact of how much students can achieve if they do at least one Maths question per day. (the question must be substantial and worth at least 5 marks)

This study plan is based on the concept of 积少成多, or “Many little things add up“. Also, this method prevents students from getting rusty in older topics, or totally forgetting the earlier topics. Also, this method makes use of the fact that the human brain learns during sleep, so if you do mathematics everyday, you are letting your brain learn during sleep everyday.

Let’s take the example of Additional Mathematics.

Exam is on 24/25 October 2013.
Let’s say the student starts the “One Question per day” Strategy on 20 May 2013

Days till exam: 157 days (22 weeks or 5 months, 4 days)

So, 157 days = 157 questions (or more!)
Read the full article at:
http://mathtuition88.com/2013/05/18/how-can-doing-at-least-one-maths-question-per-day-help-you-improve-maths-tuition-stud/

Well said!! Thank you for the advice. Will follow and hope to see improvement in my maths for my '0' level exam!

[mathtuition88]

S-H:

Well said!! Thank you for the advice. Will follow and hope to see improvement in my maths for my '0' level exam!

I am glad that you find the advice helpful!
All the best for your O level exams!

Perseverance can grind an iron rod into a needle, not to mention acing your O level exams!

(This legend is about Li Bai (李白), a great poet in Tang Dynasty. Li Bai was naughty and disliked study when he was a child. One day he saw an old woman grinding an iron rod on a big stone when he was playing by a river. Driven by curiosity, Li Bai came up and asked,
\"What are you doing, granny?\"
\"Grinding an iron rod,\" said the old women without stopping grinding.
\"Then what for?\" he asked again.
\"To make a sewing needle,\" was the answer.
\"What?!\" little Li Bai was puzzled, \"you want to grind so big a rod into a needle? It will take many years.\"
\"This doesn't matter. As long as I persevere in doing so, there is nothing you cannot achieve in the world. Certainly I can make a needle from the rod.\" Deeply moved by what the old woman said, Li Bai took effort to study since then and finally became one of the greatest poets in China.)

[S-H]

Hi can someone help me with this question, thanks!


Given that the line y+2nx-3m=0 is a tangent to the curve y=mx^2+nx+7m, where m and n are positive integers. Find the ratio of m:n and thus find the x-coordinate of the point where the line is tangent to the curve.

[jieheng]

S-H:

Hi can someone help me with this question, thanks!


Given that the line y+2nx-3m=0 is a tangent to the curve y=mx^2+nx+7m, where m and n are positive integers. Find the ratio of m:n and thus find the x-coordinate of the point where the line is tangent to the curve.

y+2nx-3m=0 => y= -2nx+3m

mx^2+nx+7m = = -2nx+3m

mx^2+3nx+4m = 0

Since the line is the tangent to the curve ,

D = b^2 - 4ac = 0

(3n)^2 - 4(m)(4m) = 0

9n^2 - 16m^2 = 0

(m/n)^2 = 9/16

m/n = 3/4 or m/n = - 3/4 (rej as m,n are positive integers)

m/n = 3/4 => 3n = 4m

mx^2+3nx+4m = 0

mx^2+4mx+4m = 0

m(x^2+4x+4) = 0

x^2+4x+4 = 0

(x+2)^2 = 0

x = -2

[S-H]

jieheng:

S-H:

Hi can someone help me with this question, thanks!


Given that the line y+2nx-3m=0 is a tangent to the curve y=mx^2+nx+7m, where m and n are positive integers. Find the ratio of m:n and thus find the x-coordinate of the point where the line is tangent to the curve.

y+2nx-3m=0 => y= -2nx+3m

mx^2+nx+7m = = -2nx+3m

mx^2+3nx+4m = 0

Since the line is the tangent to the curve ,

D = b^2 - 4ac = 0

(3n)^2 - 4(m)(4m) = 0

9n^2 - 16m^2 = 0

(m/n)^2 = 9/16

m/n = 3/4 or m/n = - 3/4 (rej as m,n are positive integers)

m/n = 3/4 => 3n = 4m

mx^2+3nx+4m = 0

mx^2+4mx+4m = 0

m(x^2+4x+4) = 0

x^2+4x+4 = 0

(x+2)^2 = 0

x = -2

Thank you very much jieheng!

[Vivian22]

Hi, can someone please help me with these 2 questions?


1. In the expansion of (2x^2 - x^-4)^n, in descending powers of x, if the coefficient of the third term is seven times that of the first term. Find the value of n.

2. A graph, y=5x(e^-x), x>0, has a stationary point at S. FInd the x-coordinate of S and the value of d^2y/dx^2 (second derivative) at S.

Thanks in advance.