O-Level Additional Math

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Re: O-Level Additional Math

Postby KSP2013777 » Sat May 21, 2016 2:37 pm

Thank you, Kenneth Tutor.
Can you help me with another question, Please?
Image
Thank you.

KSP2013777
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Re: O-Level Additional Math

Postby questionable_i » Sat May 21, 2016 5:36 pm

Can anyone help me? :thankyou: in advance.

Q1) A function f is defined by f(x)=|x^2-3x|. Find the values of x for which f(x)=x-3.

Q2) Find the x-coordinates of the points of intersection of the curve y=|16-x^2| and the line y=33.

Q3) Find the x-coordinates of all the points at which the graph of y=|x^2-9|-5 meets the x-axis.

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Re: O-Level Additional Math

Postby kennethTutor » Sun May 22, 2016 6:51 pm

KSP2013777 wrote:Thank you, Kenneth Tutor.
Can you help me with another question, Please?
Image
Thank you.


The fraction is an improper fraction, since the degree of the numerator is 4 while the degree of the denominator is 3.

Perform long division first. You obtain x + (9 - 3x^2)/(x^3 + 3x)

Finally perform partial fraction decomposition on (9 - 3x^2)/(x^3 + 3x)

Final ans should be: x + 3/x - 6x/(x^2 + 3)

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Re: O-Level Additional Math

Postby kennethTutor » Sun May 22, 2016 7:34 pm

questionable_i wrote:Can anyone help me? :thankyou: in advance.

Q1) A function f is defined by f(x)=|x^2-3x|. Find the values of x for which f(x)=x-3.

Q2) Find the x-coordinates of the points of intersection of the curve y=|16-x^2| and the line y=33.

Q3) Find the x-coordinates of all the points at which the graph of y=|x^2-9|-5 meets the x-axis.


For Q1, first we sub one equation into another.

|x^2 - 3x| = x - 3

For modulus, there is the positive and negative side.

For the positive (and removing the modulus),

x^2 - 3x = x - 3
x^2 - 4x + 3 = 0
(x - 3)(x - 1) = 0

x = 3 or x = 1

For the negative,

x^2 - 3x = -(x - 3)
x^2 - 3x = -x + 3
x^2 - 2x - 3 = 0
(x - 3)(x + 1) = 0

x = 3 or x = -1

So the values of x are: -1, 1 or 3


You can use the same approach for Q2 and Q3. For Q3, the equation of x-axis is y = 0.

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Re: O-Level Additional Math

Postby KSP2013777 » Mon May 23, 2016 8:30 pm

Please, can somebody help me solve this question?
Image
Is it tangent to curve means Discriminant =0? But how to use Discriminant if there is ln x? Please help. Thank you

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Re: O-Level Additional Math

Postby PiggyLalala » Mon May 23, 2016 8:39 pm

KSP2013777 wrote:Please, can somebody help me solve this question?
Image
Is it tangent to curve means Discriminant =0? But how to use Discriminant if there is ln x? Please help. Thank you



this question is not about discriminant = 0. Always remember, discriminant is only for quadratic equation. This is a differentiation question.

So differentiatate and set dy/dx = gradient of the tangent. Find x. Then the y coordinate. Lastly the value of k. Give it a try. 加油。。 :rahrah: :rahrah:

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Re: O-Level Additional Math

Postby KSP2013777 » Tue May 24, 2016 9:29 am

Thank you, piggy lalala. Can get my answer now. :smile:

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Re: O-Level Additional Math

Postby KSP2013777 » Thu Jun 02, 2016 9:25 pm

I need help for this question. Can somebody please help me.
Image
Thank you.

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Re: O-Level Additional Math

Postby questionable_i » Sat Jun 04, 2016 7:50 pm

Can anyone help me with this question? Thank you.

1) By expressing each of the following equations in the form a^x=b where a and b are real numbers, take logarithms to solve for x.

(a) 2^(x+1)=3^x

(b) Image

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Re: O-Level Additional Math

Postby PiggyLalala » Sun Jun 05, 2016 10:02 am

KSP2013777 wrote:I need help for this question. Can somebody please help me.
Image
Thank you.


Image

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