O-Level Additional Math

[k1ndan]

fromnuaa:

answer: 1


3x3=9
9x9=81
1x1=1

k1ndan:

What is the last digit for 3^2010?

^: to the power of

Thanks

Hi,

Thanks for the answer, but can you elaborate more on how you get the answer. Look complicated to me.

Thanks.

[fromnuaa]

sorry answer should be 9


2010 mod 4 =2

3x3=9

k1ndan:

fromnuaa:

answer: 1

3x3=9
9x9=81
1x1=1

[quote=\"k1ndan\"]What is the last digit for 3^2010?

^: to the power of

Thanks

Hi,

Thanks for the answer, but can you elaborate more on how you get the answer. Look complicated to me.

Thanks.[/quote]

[k1ndan]

Hi all,


Please help to solve the followin Sec 1 Maths Questions:

Q1) 3,12,25,42 are the 1st four terms of a number sequence.

(a) What is the nth term of the sequence? Express in terms of n.

(b) What is 40th term?

(c) Which term is the number 1537?



Q2) Line 1: 2 + 6 = 8 = 2 x 2^2
Line 2: 2 + 6 + 10 = 18 = 2 x 3^2
Line 3: 2 + 6 + 10 +14 = 8 = 2 x 4^2

(a) Express the sum of S in terms of n in the nth line sequence.

Thanks.

[k1ndan]

Hi all,


Please help to solve the followin Sec 1 Maths Questions:

Q1) 3,12,25,42 are the 1st four terms of a number sequence.

(a) What is the nth term of the sequence? Express in terms of n.

(b) What is 40th term?

Which term is the number 1537?



Q2) Line 1: 2 + 6 = 8 = 2 x 2^2
Line 2: 2 + 6 + 10 = 18 = 2 x 3^2
Line 3: 2 + 6 + 10 +14 = 8 = 2 x 4^2

(a) Express the sum of S in terms of n in the nth line sequence.

Thanks.

[Dharma]

k1ndan:

Hi all,


Please help to solve the followin Sec 1 Maths Questions:

Q1) 3,12,25,42 are the 1st four terms of a number sequence.

(a) What is the nth term of the sequence? Express in terms of n.

(b) What is 40th term?

(c) Which term is the number 1537?



Q2) Line 1: 2 + 6 = 8 = 2 x 2^2
Line 2: 2 + 6 + 10 = 18 = 2 x 3^2
Line 3: 2 + 6 + 10 +14 = 8 = 2 x 4^2

(a) Express the sum of S in terms of n in the nth line sequence.

Thanks.

Q1) 3,12,25,42 are the 1st four terms of a number sequence.

(a) What is the nth term of the sequence? Express in terms of n.
T1 = 3
T2 = 3 + (5 + 4) = 3 + 5(1) + 4(1)
T3 = 3 + (5 + 4) + (5 + 4 + 4) = 3 + 5(2) + 4(1 + 2)
T4 = 3 + (5 + 4) + (5 + 4 + 4) + (5 + 4 + 4 +4) = 3 + 5(3) + 4(1 + 2 + 3)

Tn = 3 + 5(n-1) + 4(n-1)(n)/2 = 3 + 5n – 5 + 2n^2 – 2n = 2n^2 + 3n - 2


(b) What is 40th term?

T40 = 2(40)^2 + 3(40) – 2 = 4918

(c) Which term is the number 1537?

2n^2 + 3n – 2 = 1537
2n^2 + 3n – 1539 = 0
(2n + 57)(n – 27) = 0
2n + 57 = 0 or n – 27 = 0
n = -57/2 (rejected) or n = 27
27th term => 1537


Q2)
Line 1: 2 + 6 = 8 = 2 x 2^2
Line 2: 2 + 6 + 10 = 18 = 2 x 3^2
Line 3: 2 + 6 + 10 +14 = 32 = 2 x 4^2

(a) Express the sum of S in terms of n in the nth line sequence.

Sn = 2(n + 1)^2

[Guthix]

James Ang:

A Maths is easy to study in school but not easy to do it privately due to the need for indepth practice and learning. High level maths techniques and treatments are taught in Add Maths such as logarithm, further trigo, binomial equations, differentiation and integration, kinematics etc which E maths only student will not understand without proper coaching and guidance.

Actually should consider yourself lucky as compared to the Chinese...

[trytry]

Anyone can help my son in this question (sec 4 prelim):


The gradient of a curve at the point (x,y) is 3-4x. Given that the maximum value of the curve is -2, find the equation of the curve?

TIA.

[iFruit]

Let's say equation of the curve is y=f(x)


Then gradient f'(x) = 3-4x

Then f(x) = 3x - 2x^2 + C ...(taking integral)

f(x) is max when f'(x) = 0 so 3-4x=0 --> x=3/4

substituting the max value and x=3/4, we get

-2 = 3x3/4 -2(3/4)^2 + C --->C = -25/8

so f(x) = y = 3x -2x^2 -25/8

or

8y = 24x -16x^2 -25.

trytry:

Anyone can help my son in this question (sec 4 prelim):

The gradient of a curve at the point (x,y) is 3-4x. Given that the maximum value of the curve is -2, find the equation of the curve?

TIA.

[SKT]

Please help:


It is given that α and β are roots of the following equation:

x² + 2x - 107 = √(x² + 2x + 3)

Form an equation with roots α² and β².

[iFruit]

Let's say √(x² + 2x +3) = y


Then we can write the original equation as

x² + 2x +3 - 110 = √(x² + 2x + 3)

which is

y² - 110 = y ----> y² - 110 -y =0 ---> (y+10)(y-11)=0

so, y=11 or -10 -----> √(x² + 2x +3) = 11 or -10

Taking the postive value y = 11,

(x² + 2x + 3) = 121 ---> x² + 2x -118 = 0

considering the roots α and β for the equation,

α + β = -2, α.β = -118 ---> (α² + β²) = 4+236 = 240, α².β² = 118²

Now We need an equation with roots α² and β² or of the form

(x-α²)(x - β²) = x² -(α² + β²)x + α².β² = 0

or

x² -240x+ 118² = 0


You can do the same exercise for the negative value y = -10

SKT:

Please help:

It is given that α and β are roots of the following equation:

x² + 2x - 107 = √(x² + 2x + 3)

Form an equation with roots α² and β².