Additional Mathematics 4037/22

[Saad Mughal]

but how? wat was plotted against wat?

It wasn't about the graph only. We had to replace 'x' not '3^x'.

 

[Moiz Awan]

You have to put points in equation and then solve two obtained equations simultaneously.

 

[Shadow]

It wasn't about the graph only. We had to replace 'x' not '3^x'.

You have to put points in equation and then solve two obtained equations simultaneously.

shit happens x_x it was worth 4 marks?

 

[job wilson]

1. y=25x^4 - 20x^2 + 4
2. (i) A, E
(ii) C,D
5. 4+3root5
The vector question: OC = 10i -24j and unit vector OC = 5/13i - 12/13j
Circular measure question: r=12.7 cm, A=14.6 cm^2
Two possible coordinates of C (0,10) and (4,2)
In that natural logarithm question, e^x = 3 and x = ln 3
Coordinates of k and c were 9 and 5.
Last question: Area was 16.9 units.

the answer to the nuture of turning point was one was max and another minimum?????

 

[Shadow]

yea

 

[Snowysangel]

How'd u find the coordinates of C? :S and how many marks was the last question for?

 

[Shadow]

How'd u find the coordinates of C? :S and how many marks was the last question for?

If u r talking about C in the triangle wala ques, v had to simultaneously solve it ... I did it wrong

 

[Saad Mughal]

How'd u find the coordinates of C? :S and how many marks was the last question for?

Simultaneously. One was a quadratic equation, one was a linear equation.

 

[Iishrak]

how did u guys solve that trig one, like the one u had with cos =x , 270<x<360 and then u had to find cosec or something

 

[redd]

how did u guys solve that trig one, like the one u had with cos =x , 270<x<360 and then u had to find cosec or something

sin ^2=1 - cos^2
sin = underroot (1- cos^2)
sin= underroot (1-x^2) (substituting cos with x)
cosec= 1 divided by sin
hence cosec = 1 divided by underoot(1-x^2)

 

[Iishrak]

sin ^2=1 - cos^2
sin = underroot (1- cos^2)
sin= underroot (1-x^2) (substituting cos with x)
cosec= 1 divided by sin
hence cosec = 1 divided by underoot(1-x^2)

o my godd, that's the same answer i gave,but i did in other way i kinda draw that 4 quadrant, and made the p x cordinate and denominator was one so yeah , i thot it was wrongg cz i literally never did any mahts like that DDD , haha im so happy now, inshallah an a star now no doubt, now i want allah to give me minimun 6 a's i did so bad in physics amd computing

 

[Iishrak]

whats the answer of the functions? talking abt both number 1 and 2

 

[Iishrak]

If u r talking about C in the triangle wala ques, v had to simultaneously solve it ... I did it wrong

just one we had to do it with simultaneous, the other one like where u were given the length was easy, as the x-cordinate was already given b , and c were smae

 

[Haris Bin Zahid]

sin ^2=1 - cos^2
sin = underroot (1- cos^2)
sin= underroot (1-x^2) (substituting cos with x)
cosec= 1 divided by sin
hence cosec = 1 divided by underoot(1-x^2)

When you took the underroot, you had to discard the positive answer (sin x is negative in the fourth quadrant). Hence, the final answer was
-1/√1-p^2 .

 

[Snowysangel]

just one we had to do it with simultaneous, the other one like where u were given the length was easy, as the x-cordinate was already given b , and c were smae

How were b and c the same? I formed two equations, one through pythagorus and the other through the distance formula...and then I dont know what the hell I did how many marks was the area question for, the very last one?

 

[UzairNavid]

The question in which y=k(3^x) + c was there.

I don't remember this question. :/ Could you please remind me what they were asking, and which question it was a part of?

 

[Shadow]

sin ^2=1 - cos^2
sin = underroot (1- cos^2)
sin= underroot (1-x^2) (substituting cos with x)
cosec= 1 divided by sin
hence cosec = 1 divided by underoot(1-x^2)

thats correct just a 'p' in place of 'x' as the ans had to b given in terms of p

 

[redd]

thats correct just a 'p' in place of 'x' as the ans had to b given in terms of p

Yeh, i replied to the question quoted. It was P there...

 

[redd]

When you took the underroot, you had to discard the positive answer (sin x is negative in the fourth quadrant). Hence, the final answer was
-1/√1-p^2 .

Yeh ofcourse...

 

[Shadow]

When you took the underroot, you had to discard the positive answer (sin x is negative in the fourth quadrant). Hence, the final answer was
-1/√1-p^2 .

Damn! I wrote the positive one ... m losing 1 mark