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secondary 4 | A Maths
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Date Posted: 3 years ago
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done
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Not sure if this approach is accepted by your teacher
Date Posted:
3 years ago
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For part b, as requested
Date Posted:
3 years ago
Sie help to solve 2sin4x+cos4x=2 for 0 lessthan or equal to x less than or equal to 1
2 sin 4x + cos 4x = 2 for 0 <= x <= 1 (rad)
Expressing the LHS using R formula,
R = sqrt (2^2 + 1^2)
= sqrt 5
Phase shift alpha = taninv (1/2)
LHS becomes sqrt 5 times sin (4x + taninv 0.5)
sqrt 5 times sin (4x + taninv 0.5) = 2
sin (4x + taninv 0.5) = 2 / sqrt 5
Basic angle, gamma
= sininv (2 / sqrt 5)
We are solving for angles of 4x within 0 and 4 rad (4 rad is around 1.3 pi which includes first quad, 2nd quad and a portion of the 3rd quad)
4x + taninv 0.5 = 1st quad, 2nd quad, 1st quad next cycle (reject, it’s too far away to be accepted as a solution)
Expressing the LHS using R formula,
R = sqrt (2^2 + 1^2)
= sqrt 5
Phase shift alpha = taninv (1/2)
LHS becomes sqrt 5 times sin (4x + taninv 0.5)
sqrt 5 times sin (4x + taninv 0.5) = 2
sin (4x + taninv 0.5) = 2 / sqrt 5
Basic angle, gamma
= sininv (2 / sqrt 5)
We are solving for angles of 4x within 0 and 4 rad (4 rad is around 1.3 pi which includes first quad, 2nd quad and a portion of the 3rd quad)
4x + taninv 0.5 = 1st quad, 2nd quad, 1st quad next cycle (reject, it’s too far away to be accepted as a solution)
gamma = sininv (2 / sqrt 5)
4x + taninv 0.5 = gamma, pi - gamma
4x = gamma - taninv 0.5 or pi - gamma - taninv 0.5
We then proceed to solve for x from there
I have no calculator with me now, but judging from the numbers, both solutions should be accepted
4x + taninv 0.5 = gamma, pi - gamma
4x = gamma - taninv 0.5 or pi - gamma - taninv 0.5
We then proceed to solve for x from there
I have no calculator with me now, but judging from the numbers, both solutions should be accepted
Ok sir thank you
One more question sir
Find all the exact angles between 0 and pi, which satisfy the equation
Sin(x - pi/5) - cos pi/10 =10
Find all the exact angles between 0 and pi, which satisfy the equation
Sin(x - pi/5) - cos pi/10 =10
sin (x - pi/5) - cos (pi/10) = 10
This one seems to have no solution at all, since
10 + cos (pi/10)
exceeds 1 and sin (x - ...) > 1 has no solution at all
This one seems to have no solution at all, since
10 + cos (pi/10)
exceeds 1 and sin (x - ...) > 1 has no solution at all
Answer is 3pi/5 , 4pi/5
If RHS = 0 then there are solutions
But RHS = 10 and I doubt there are any solutions
Unless you meant
sin (x - pi/5) - cos (pi/10) = 0
But RHS = 10 and I doubt there are any solutions
Unless you meant
sin (x - pi/5) - cos (pi/10) = 0
Sorry sir RHS equal to 0
I look at it later, but you have to do something like this
sin (x - pi/5) - cos (pi/10) = 0
sin (x - pi/5) = cos (pi/10)
But we know that sin (pi/2 - theta) = cos theta
Using this, we can say that cos (pi/10) = sin (2pi/5)
(The equivalent statement for this in degrees is cos 18 = sin 72)
sin (x - pi/5) = sin 2pi/5
Basic angle = sininv (sin 2pi/5), which is 2pi/5 itself
x - pi/5 = 1st/2nd
x - pi/5 = 2pi/5 or pi - 2pi/5
x = 3pi/5 or 4pi/5
sin (x - pi/5) - cos (pi/10) = 0
sin (x - pi/5) = cos (pi/10)
But we know that sin (pi/2 - theta) = cos theta
Using this, we can say that cos (pi/10) = sin (2pi/5)
(The equivalent statement for this in degrees is cos 18 = sin 72)
sin (x - pi/5) = sin 2pi/5
Basic angle = sininv (sin 2pi/5), which is 2pi/5 itself
x - pi/5 = 1st/2nd
x - pi/5 = 2pi/5 or pi - 2pi/5
x = 3pi/5 or 4pi/5
Thank you sir
Sir Help to solve |2x-3|+6x = |9-6x|+4
|2x - 3| - 6x = |9 - 6x| + 4
|2x - 3| - |9 - 6x| = 4 + 6x
Since the positions within the modulus brackets are interchangeable,
|2x - 3| - |6x - 9| = 4 + 6x
|2x - 3| - 3 |2x - 3| = 4 + 6x
Factorising,
(1 - 3) |2x - 3| = 4 + 6x
-2 |2x - 3| = 4 + 6x
|2x - 3| = -2 - 3x (*)
So, we have
2x - 3 = -2 - 3x or 2x - 3 = - (-2 - 3x)
5x = 1 or 2x - 3 = 2 + 3x
x = 1/5 or x = -5
The solution of x = 1/5 will be rejected due to failure in the (*) equation. I have yet to inspect in the original equation whether x = -5 is valid, but it is valid in the (*) question.
|2x - 3| - |9 - 6x| = 4 + 6x
Since the positions within the modulus brackets are interchangeable,
|2x - 3| - |6x - 9| = 4 + 6x
|2x - 3| - 3 |2x - 3| = 4 + 6x
Factorising,
(1 - 3) |2x - 3| = 4 + 6x
-2 |2x - 3| = 4 + 6x
|2x - 3| = -2 - 3x (*)
So, we have
2x - 3 = -2 - 3x or 2x - 3 = - (-2 - 3x)
5x = 1 or 2x - 3 = 2 + 3x
x = 1/5 or x = -5
The solution of x = 1/5 will be rejected due to failure in the (*) equation. I have yet to inspect in the original equation whether x = -5 is valid, but it is valid in the (*) question.
Sir but answer is x=1
Alamak
I miswrote the very first line
Wait
I miswrote the very first line
Wait
Yes sir 4-6x
|2x - 3| + 6x = |9 - 6x| + 4
I wrote wrongly as |2x - 3| - 6x the previous time
|2x - 3| - |9 - 6x| = 4 - 6x
Since the positions within the modulus brackets are interchangeable,
|2x - 3| - |6x - 9| = 4 - 6x
|2x - 3| - 3 |2x - 3| = 4 - 6x
Factorising,
(1 - 3) |2x - 3| = 4 - 6x
-2 |2x - 3| = 4 - 6x
|2x - 3| = -2 + 3x (*)
So, we have
2x - 3 = -2 + 3x or 2x - 3 = - (-2 + 3x)
x = -1 or 2x - 3 = 2 - 3x
x = -1 or 5x = 5
x = -1 or x = 1
We reject x = -1 due to failure in the (*) equation, but allow x = 1
But you must still sub back into the very first equation to check that the solution x = 1 is valid
I wrote wrongly as |2x - 3| - 6x the previous time
|2x - 3| - |9 - 6x| = 4 - 6x
Since the positions within the modulus brackets are interchangeable,
|2x - 3| - |6x - 9| = 4 - 6x
|2x - 3| - 3 |2x - 3| = 4 - 6x
Factorising,
(1 - 3) |2x - 3| = 4 - 6x
-2 |2x - 3| = 4 - 6x
|2x - 3| = -2 + 3x (*)
So, we have
2x - 3 = -2 + 3x or 2x - 3 = - (-2 + 3x)
x = -1 or 2x - 3 = 2 - 3x
x = -1 or 5x = 5
x = -1 or x = 1
We reject x = -1 due to failure in the (*) equation, but allow x = 1
But you must still sub back into the very first equation to check that the solution x = 1 is valid
My answer x=1 or x=-1
But in answer sheet x=1 only
Why sir x=-1 rejected
But in answer sheet x=1 only
Why sir x=-1 rejected
So every time we need to substitute?
Or any other condition?
Yes, substitution back into the original equation is a must
|2x - 3| = -2 + 3x
Supposing we sub back x = -1 into this equation,
LHS = |-5|
RHS = -5
These are not equal.
This is because the outputs of modulus functions such as |2x - 3| must be non negative, ie
|2x - 3| >= 0
Since the RHS is negative when x = -1, we know that x = -1 is invalid.
If you do the same for x = 1, the validity holds, so x = 1 is a solution (and in fact the only one for the question).
Supposing we sub back x = -1 into this equation,
LHS = |-5|
RHS = -5
These are not equal.
This is because the outputs of modulus functions such as |2x - 3| must be non negative, ie
|2x - 3| >= 0
Since the RHS is negative when x = -1, we know that x = -1 is invalid.
If you do the same for x = 1, the validity holds, so x = 1 is a solution (and in fact the only one for the question).
X^3-2x^2-x+3 / (x^2-2x+1) = x - 2/(x-1) + 1/ (x-1)^2.
Hence find integral of x^3-2x^2-x-4/(x^2-2x+1) dx.
How to solve
Hence find integral of x^3-2x^2-x-4/(x^2-2x+1) dx.
How to solve
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