[Ineedhelp998]

How would you work out part 8 : IV

and for question 9 part iii) why is it MT^2 = radius^2 - 2^2

and not MT^2 = radius^2 + 2^2

because doesn't MT = the hypotenuse so it would be a^2 + b^2 = c^2 ?

Thank you

[raheem94]

(Original post by Ineedhelp998)


How would you work out part 8 : IV

and for question 9 part iii) why is it MT^2 = radius^2 - 2^2

and not MT^2 = radius^2 + 2^2

because doesn't MT = the hypotenuse so it would be a^2 + b^2 = c^2 ?

Thank you

Question 8 (iv),

Differentiate,

It is an increasing function when,

[notnek]

8iv. If a function is increasing, what does that tell you about it's gradient while it's increasing?

9iii. MT is not the hypotenuse but it's hard for me to show you via text. Try to notice that CM is perpendicular to MT. So CT must be the hypotenuse.

[Ineedhelp998]

(Original post by notnek)
8iv. If a function is increasing, what does that tell you about it's gradient while it's increasing?

9iii. MT is not the hypotenuse but it's hard for me to show you via text. Try to notice that CM is perpendicular to MT. So CT must be the hypotenuse.

Doesn't it look like this?

[notnek]

(Original post by Ineedhelp998)
Doesn't it look like this?

ST is a chord i.e. a straight line connecting S and T. Your ST is not a chord.

And what makes you think angle MCT is a right-angle?

[Ineedhelp998]

(Original post by notnek)
ST is a chord i.e. a straight line connecting S and T. Your ST is not a chord.

Yes it is supposed to be a straight line, but hard to draw

[Ineedhelp998]

(Original post by notnek)
ST is a chord i.e. a straight line connecting S and T. Your ST is not a chord.

And what makes you think angle MCT is a right-angle?

hmm come to think of it I don't know

actually because mc and tc are perpendicular

[dongonaeatu]

(Original post by Ineedhelp998)


How would you work out part 8 : IV

and for question 9 part iii) why is it MT^2 = radius^2 - 2^2

and not MT^2 = radius^2 + 2^2

because doesn't MT = the hypotenuse so it would be a^2 + b^2 = c^2 ?

Thank you

for 8:iv

differentiate y=7+6x-x^2
so dy/dx= 6-2x
increasing function is where dy/dx>0
so 6-2x>0 just treat this like an equals sign and factorise
we have 2 which goes into both 6 and 2x so we put this outside the bracket
2(3-x)>0 so x=0 and x=3
because it is a cubic, you draw the graph of it and it crosses the x-axis at 0 and 3. Then shade in where the line is above 0 and write down this inequality

you should get 2 answers

[notnek]

(Original post by Ineedhelp998)
hmm come to think of it I don't know

actually because mc and tc are perpendicular

Why do you think that? Even in your drawing, angle MCT doesn't look like a right-angle.

Consider the triangle CST. This triangle is isosceles since it's made up of two radii. And by joining a line from the midpoint of the chord to the centre of the circle, you're splitting the triangle into two.

Can you see why angle CMT is a right-angle now?

[raheem94]

(Original post by notnek)
Why do you think that? Even in your drawing, angle MCT doesn't look like a right-angle.

Consider the triangle CST. This triangle is isosceles since it's made up of two radii. And by joining a line from the midpoint of the chord to the centre of the circle, you're splitting the triangle into two.

Can you see why angle CMT is a right-angle now?

May be my diagram will help you explaining him,

[Ineedhelp998]

(Original post by notnek)
Why do you think that? Even in your drawing, angle MCT doesn't look like a right-angle.

Consider the triangle CST. This triangle is isosceles since it's made up of two radii. And by joining a line from the midpoint of the chord to the centre of the circle, you're splitting the triangle into two.

Can you see why angle CMT is a right-angle now?

thanks now I get it, don't know where I got my diagram from :P

[notnek]

(Original post by dongonaeatu)
for 8:iv

differentiate y=7+6x-x^2
so dy/dx= 6-2x
increasing function is where dy/dx>0
so 6-2x>0 just treat this like an equals sign and factorise
we have 2 which goes into both 6 and 2x so we put this outside the bracket
2(3-x)>0 so x=0 and x=3
because it is a cubic, you draw the graph of it and it crosses the x-axis at 0 and 3. Then shade in where the line is above 0 and write down this inequality

you should get 2 answers

Firstly, full solutions are not permitted in this forum.

Secondly, the line in bold above is wrong. Why do you have two solutions for x in a linear equation? I think you're getting your inequality confused with

[Ineedhelp998]

(Original post by raheem94)
May be my diagram will help you explaining him,

Ohh thats how it looks

[dongonaeatu]

(Original post by notnek)
Firstly, full solutions are not permitted in this forum.

Secondly, the line in bold above is wrong. Why do you have two solutions for x in a linear equation? I think you're getting your inequality confused with

it is not wrong, and i didnt post full solutions. He has to draw the graph to get the answer. And i meant 2 solutions as in x> than a number or x< than a number is where its an increasing function.

[raheem94]

(Original post by notnek)
Firstly, full solutions are not permitted in this forum.

Secondly, the line in bold above is wrong. Why do you have two solutions for x in a linear equation? I think you're getting your inequality confused with

I have said him multiple times to not answer questions on the maths forum, i have still not seen a single correctly answered question from him.

This is the response he gave me a few minutes ago:

(Original post by dongonaeatu)
ok, sorry. I wont answer any questions from now on.

Now why did you answered again?

I think the OP had done this question, he didn't needed any more help. You should leave the questions for the more experienced mathematicians to answer.

[raheem94]

(Original post by dongonaeatu)
it is not wrong, and i didnt post full solutions. He has to draw the graph to get the answer. And i meant 2 solutions as in x> than a number or x< than a number is where its an increasing function.

You aren't clear with this concept yourself mate.

[dongonaeatu]

(Original post by raheem94)
You aren't clear with this concept yourself mate.

What did i do wrong? increasing function is where dy/dx>0
so you get the 2 x values from factorising and draw the graph of it and shade in where the graph is above 0 and this is when its an increasing function

[notnek]

(Original post by dongonaeatu)
it is not wrong, and i didnt post full solutions. He has to draw the graph to get the answer. And i meant 2 solutions as in x> than a number or x< than a number is where its an increasing function.

I was too quick to say that you'd posted a full solution when you hadn't actually given the solution. Sorry.

But your working is still incorrect. What would your answer be?

[raheem94]

(Original post by dongonaeatu)
What did i do wrong? increasing function is where dy/dx>0
so you get the 2 x values from factorising and draw the graph of it and shade in where the graph is above 0 and this is when its an increasing function

Done

This is what we needed to do.

[dongonaeatu]

(Original post by notnek)
I was too quick to say that you'd posted a full solution when you hadn't actually given the solution. Sorry.

But your working is still incorrect. What would your answer be?

okay, the answer would be
increasing function when x<0 and x>3