Mathematics: Post your doubts here!

[Yousif Mukkhtar]

minato112 Can you explain why there is no turning point?

 

[Minato112]

minato112 Can you explain why there is no turning point?

Sorry for l8 rep. Well I think I might have made a mistake about that but now its all clear.

Look at the function g(x). Its turning point is (1,8). But when you look at its domain, you will notice that it starts right only after the turning point, i.e x > 1. Therefore it is a one-one function. You get it or you want me to sketch the curve and show you?

 

[Yousif Mukkhtar]

Sorry for l8 rep. Well I think I might have made a mistake about that but now its all clear.

Look at the function g(x). Its turning point is (1,8). But when you look at its domain, you will notice that it starts right only after the turning point, i.e x > 1. Therefore it is a one-one function. You get it or you want me to sketch the curve and show you?

Thanks, as always.

 

[Minato112]

Thanks, as always.

You understood it?

 

[Yousif Mukkhtar]

You understood it?

Yeah I did.

 

[Ammirul Shafiq]

salam..
anyone, how to solve this question?
tan 3x = 4 tan x
180<x<0

 

[Binyamine]

Firs

salam..
anyone, how to solve this question?
tan 3x = 4 tan x
180<x<0

First consider tan 3x = tan ( 2x + x ) = ( tan 2x + tan x ) / ( 1 - tan 2x tanx )

What is tan 2x = 2 tanx / ( 1 - ( tanx ) ^ 2 )

Try it...

 

[Ammirul Shafiq]

Firs


First consider tan 3x = tan ( 2x + x ) = ( tan 2x + tan x ) / ( 1 - tan 2x tanx )

What is tan 2x = 2 tanx / ( 1 - ( tanx ) ^ 2 )

Try it...

got it.thanks!

 

[Ammirul Shafiq]

(i) Express 8 cos θ + 15 sin θ in the form Rcos(θ − α), where R > 0 and 0< α < 90
Give the value of α correct to 2 decimal places.
(ii) Hence solve the equation 8 cos θ + 15 sin θ = 12, giving all solutions in the interval 0< θ < 360
How to find the angle for question ii) ? I answered question 1 already, but still confuse how to get the angle for question ii)

 

[Binyamine]

(i) Express 8 cos θ + 15 sin θ in the form Rcos(θ − α), where R > 0 and 0< α < 90
Give the value of α correct to 2 decimal places.
(ii) Hence solve the equation 8 cos θ + 15 sin θ = 12, giving all solutions in the interval 0< θ < 360
How to find the angle for question ii) ? I answered question 1 already, but still confuse how to get the angle for question ii)

We have to convert to R Cos...so a is going to be the coefficient of cos and b is the coefficient of sin
Recall; R = Square root of a^2 + b ^2
α = tan inverse of ( b/a)

R = ( 8^2 + 15^2 ) ^ (1/2)
= 17
α = tan inverse of ( 15/8 )
= 61.9 ( 1 d.p)
Now for second part

8 cos θ + 15 sin θ = 12

we already know that 8 cos θ + 15 sin θ is infact 17 Cos ( θ − 61.9 )

17 Cos ( θ − 61.9 ) = 12
Cos ( θ − 61.9 ) = 12/17
Find key angle = cos inverse of ( 12/17 )

Use it in first and fourth quadrant.

θ − 61.9 = 45.1 , 314.9
hence θ = 107, 376.8

 

[M00uon]

We have to convert to R Cos...so a is going to be the coefficient of cos and b is the coefficient of sin
Recall; R = Square root of a^2 + b ^2
α = tan inverse of ( b/a)

R = ( 8^2 + 15^2 ) ^ (1/2)
= 17
α = tan inverse of ( 15/8 )
= 61.9 ( 1 d.p)
Now for second part

8 cos θ + 15 sin θ = 12

we already know that 8 cos θ + 15 sin θ is infact 17 Cos ( θ − 61.9 )

17 Cos ( θ − 61.9 ) = 12
Cos ( θ − 61.9 ) = 12/17
Find key angle = cos inverse of ( 12/17 )

Use it in first and fourth quadrant.

θ − 61.9 = 45.1 , 314.9
hence θ = 107, 376.8

If the same type of question is asked but instead of "8 cos θ + 15 sin θ in the form Rcos(θ − α)" its 8 sin θ + 15 cos θ ? will α be still 61.9

Looking forward for your prompt response

 

[Binyamine]

If the same type of question is asked but instead of "8 cos θ + 15 sin θ in the form Rcos(θ − α)" its 8 sin θ + 15 cos θ ? will α be still 61.9

Looking forward for your prompt response

Obviously the answer is going to be diferent. The point is not what is being converted, instead it is TO WHAT IT IS BEING CONVERTED.

Let me explain to you.

R^2 = a ^2 + b^2
i.e R = SquareRoot ( a ^2 + b^2 )

and α = tan inverse of ( b/a)

Now if we are converting to R COS, then a is going to be the coefficient of cos and b the coefficient of sin

If we were converting to R SIN, then a is going to be the coefficient of sin and b the coefficient of cos

And the best part, is that in examination, it already tells you to what form to convert it to.

 

[M00uon]

Obviously the answer is going to be diferent. The point is not what is being converted, instead it is TO WHAT IT IS BEING CONVERTED.

Let me explain to you.

R^2 = a ^2 + b^2
i.e R = SquareRoot ( a ^2 + b^2 )

and α = tan inverse of ( b/a)

Now if we are converting to R COS, then a is going to be the coefficient of cos and b the coefficient of sin

If we were converting to R SIN, then a is going to be the coefficient of sin and b the coefficient of cos

And the best part, is that in examination, it already tells you to what form to convert it to.

Great thanks alot!
there is one question bothering me , its related to chapter FUNCTION
I don't know how to find domain & range and also stuck in the last part
I have uploaded the question and awating for your reply.(This question is from my school test paper)

 

[Yousif Mukkhtar]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_11.pdf
Can someone explain
q2) I got the differentiation part, but how can we use the first derivative to show that all gradients are positive?

and q7)i)
I used this method:
x+3y+4x+y+3x+2y=48
8x+6y=48
6y=48-8x
y=(48-8x)/6
y=6-4x/3

 

[Binyamine]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_11.pdf
Can someone explain
q2) I got the differentiation part, but how can we use the first derivative to show that all gradients are positive?

and q7)i)
I used this method:
x+3y+4x+y+3x+2y=48
8x+6y=48
6y=48-8x
y=(48-8x)/6
y=6-4x/3

Question 2.

When you differentiated you got dy/dx = 9 x^2 - 12 x + 4

What you should next do is to complete the square.

= 9 ( x^2 - 4/3 x ) + 4
= 9 [ ( x - 2/3 ) ^ 2 - 4/9 ] + 4
= 9 ( x - 2/3 ) ^ 2 - 4 + 4
= 9 ( x - 2/3 ) ^ 2

which will be positive for any value of x since when we square any number it become positive. And a positive number multiplied by 9 is still positive.

Question 7 (i)
You were on the right track until you made a division mistake.
x+3y+4x+y+3x+2y=48
8x+6y=48
6y=48-8x
y=(48-8x)/6

Divide every term by 2

y = ( 24 - 4 x ) / 3
or y = 8 - (4/3) x

7 (ii) Area = 3xy + 3 xy
= 6 xy
= 6 x ( (48-8x)/6 )
= x ( 48-8x)
= 48 x - 8 x^2

7(iii) THis you can do, you find dA/dx, and the value of x at stationary point. Find an expression for d A^2/ dx^2 and substitute the value of x for which we have a stationary point in it. You will see that d A^2/ dx^2 is negative, hence a maximum.

 

[Binyamine]

Great thanks alot!
there is one question bothering me , its related to chapter FUNCTION
I don't know how to find domain & range and also stuck in the last part
I have uploaded the question and awating for your reply.(This question is from my school test paper)View attachment 19827

Have a look on the following videos.

Question 8 June 2012 Paper 1 9709/11/M/J/12
Question 8 :

Question 10 JUNE 2012 Paper 1 9709/12/M/J/12

Question 10 :

Question 11 JUNE 2012 Paper 1 9709/11/M/J/12

Question 11 :

Question 11 BES, 2013, Paper 1, specimen paper, cie, 9820


Question 11 :

 

[Yousif Mukkhtar]

Question 2.

When you differentiated you got dy/dx = 9 x^2 - 12 x + 4

What you should next do is to complete the square.

= 9 ( x^2 - 4/3 x ) + 4
= 9 [ ( x - 2/3 ) ^ 2 - 4/9 ] + 4
= 9 ( x - 2/3 ) ^ 2 - 4 + 4
= 9 ( x - 2/3 ) ^ 2

which will be positive for any value of x since when we square any number it become positive. And a positive number multiplied by 9 is still positive.

Question 7 (i)
You were on the right track until you made a division mistake.
x+3y+4x+y+3x+2y=48
8x+6y=48
6y=48-8x
y=(48-8x)/6

Divide every term by 2

y = ( 24 - 4 x ) / 3
or y = 8 - (4/3) x

7 (ii) Area = 3xy + 3 xy
= 6 xy
= 6 x ( (48-8x)/6 )
= x ( 48-8x)
= 48 x - 8 x^2

7(iii) THis you can do, you find dA/dx, and the value of x at stationary point. Find an expression for d A^2/ dx^2 and substitute the value of x for which we have a stationary point in it. You will see that d A^2/ dx^2 is negative, hence a maximum.

So y=8-(4/3)x is accepted even though it's not written in the marking scheme?

Anyways thanks man!

 

[Binyamine]

So y=8-(4/3)x is accepted even though it's not written in the marking scheme?

Anyways thanks man!

it should be accepted.
Thanks Allah for indeed he is our Creator and we need to be grateful to him.

 

[kronix6]

Th

6 sin^2Q + 6 sinQ cosQ = sinQ cosQ - cos^2Q
6 sin^2Q + cos^2Q = -5 sinQ cos Q

(Divide by cos^2Q everywhere) : 6 tan^2Q + 1 = -5 tan Q

6 tan^2Q + 5 tanQ + 1 = 0
(3 tanQ + 1) (2 tanQ + 1) = 0
tan Q = (-1/3) or tan Q = (-1/2)

Hope It Helps

anks man . i appreciate it

 

[Minato112]

Th

anks man . i appreciate it

No Problem!