How was your math paper 33?

[Henry930821]

For vectors, the first part, p = 9 and part (ii) I forgot my answer.. sorry about that.. hehe.. Oh yeah, partial fraction I got A = 6, B = -2, C = 1.. did anyone get the same values as mine?

Yup! Values for partial fraction are correct

 

[Kandinsky]

If needed, the answers:

1) x = 5/(e-1)

2) (a) R = 25, alpha = 16.26 degrees, the answer to (b) is 59.10 degre

3) y = 1/2 (x^2 + 4)^3

4) dy/dx = (2t+3)/ 3. gradient = 2

5) f'(-1/2) = 6e

integral = -0.75

6) alpha = -2. beta = 1.67

7) A = 1/24. k = 10

8)p = 9
the point of intersection is (7,1,-2) and equation of plane is 10x-11y-7z = 73.

9) 6/(3-x) - (2x-1)/(x^2+1)
expansion: 3 - 4/3 * x - 7/9 *x^2 + 56/27*x^3

10) w = +- sqrt(8) i
the region R - the circle with center (4,4) and radius 2

3.66 <= modulus z <= 7.66

0.424<= argument z <= 1.15

 

[Henry930821]

If needed, the answers:

1) x = 5/(e-1)

2) (a) R = 25, alpha = 16.26 degrees, the answer to (b) is 59.10 degre

3) y = 1/2 (x^2 + 4)^3

4) dy/dx = (2t+3)/ 3. gradient = 2

5) f'(-1/2) = 6e

integral = -0.75

6) alpha = -2. beta = 1.67

7) A = 1/24. k = 10

8)p = 9
the point of intersection is (7,1,-2) and equation of plane is 10x-11y-7z = 73.

9) 6/(3-x) - (2x-1)/(x^2+1)
expansion: 3 - 4/3 * x - 7/9 *x^2 + 56/27*x^3

10) w = +- sqrt(8) i
the region R - the circle with center (4,4) and radius 2

3.66 <= modulus z <= 7.66

0.424<= argument z <= 1.15

Wow this is good stuff! how did you get the answers by the way? 100% correct?

 

[Kandinsky]

Checked the questions with all the useful staff like wolphram alpha and so on)
so, if any doubts/questions, ask)

 

[Meo]

If needed, the answers:

1) x = 5/(e-1)

2) (a) R = 25, alpha = 16.26 degrees, the answer to (b) is 59.10 degre

3) y = 1/2 (x^2 + 4)^3

4) dy/dx = (2t+3)/ 3. gradient = 2

5) f'(-1/2) = 6e

integral = -0.75

6) alpha = -2. beta = 1.67

7) A = 1/24. k = 10

8)p = 9
the point of intersection is (7,1,-2) and equation of plane is 10x-11y-7z = 73.

9) 6/(3-x) - (2x-1)/(x^2+1)
expansion: 3 - 4/3 * x - 7/9 *x^2 + 56/27*x^3

10) w = +- sqrt(8) i
the region R - the circle with center (4,4) and radius 2

3.66 <= modulus z <= 7.66

0.424<= argument z <= 1.15

Excuse me kind sir, can you post question 1? I do have the same answer but I just want to check.

 

[Kandinsky]

No problem.
1st question: ln(x+5) = ln(x) + 1

 

[issac0215]

Did you guys sit for mechanics paper? Which paper do you guys prefer p33 or p43?

Of course paper 43.
Maybe it is because paper 4 syllabus is almost same with physic AS..
So it is easier for me..

 

[issac0215]

(7,-1,-2) and -11x+10y-7z=-73?

yaya..i think so..wow u can remember the whole answer -.-

 

[Meo]

No problem.
1st question: ln(x+5) = ln(x) + 1

thank you. ^^ oh and, i think the point for vectors is wrong, I got my question paper back and redid that question and i got (7, -1, 2) , not sure if it's correct though.

 

[Kandinsky]

thank you. ^^ oh and, i think the point for vectors is wrong, I got my question paper back and redid that question and i got (7, -1, 2) , not sure if it's correct though.

can you then post the question? I'll recheck that.

 

[ninesone]

yes the number of equation of plane is weird..i remember it is ?x+?y+?z=-73..i have the same answer with my friend

yess! i got -73 too,

 

[Henry930821]

Anyone sat for stats paper 63? How was it?

 

[Kandinsky]

and who sat for mechanics paper 53 ? =)

 

[Meo]

can you then post the question? I'll recheck that.

sorry I took a while to reply, my teacher just returned my paper after attempting the paper herself.

8. Two lines have equations

r = (5i + j -4k) + s (i - j +3k) and r = (pi +4j - 2k) + t (2i + 5j - 4k)

where p is a constant. It is given that the lines intersect.
i. Find the value of p and determine the coordinates of the point of intersection.
ii. Find the equation of the plane containing the two lines, giving your answer in the form of ax + by + cz = d where a, b, c and d are integers.

Alhamdulillah, most of my answers matched yours, except for 7.ii (this is due to pure carelessness), 8.i and 8.ii

for 8.i I got (7, -1, 2)
and for 8.ii. I got -11x +10y +7z = -73

 

[Kandinsky]

Can you write the equations properly?

 

[Meo]

sorry I tried to post it in the vector form but i couldnt so i just did the i j k

 

[Kandinsky]

Yep. That's correct(your answer). Maybe I recalled that wrongly.

 

[Meo]

If needed, the answers:

1) x = 5/(e-1)

2) (a) R = 25, alpha = 16.26 degrees, the answer to (b) is 59.10 degre

3) y = 1/2 (x^2 + 4)^3

4) dy/dx = (2t+3)/ 3. gradient = 2

5) f'(-1/2) = 6e

integral = -0.75

6) alpha = -2. beta = 1.67

7) A = 1/24. k = 10

8)p = 9
the point of intersection is (7,1,-2) and equation of plane is 10x-11y-7z = 73.

9) 6/(3-x) - (2x-1)/(x^2+1)
expansion: 3 - 4/3 * x - 7/9 *x^2 + 56/27*x^3

10) w = +- sqrt(8) i
the region R - the circle with center (4,4) and radius 2

3.66 <= modulus z <= 7.66

0.424<= argument z <= 1.15

bro, I've checked the mark scheme, and for no. 3 the answer is just y = 1/2 (x^2 + 4)

why is (x^2 + 4) not cubed? I redid the question and still I have the cube.

 

[Kandinsky]

bro, I've checked the mark scheme, and for no. 3 the answer is just y = 1/2 (x^2 + 4)

why is (x^2 + 4) not cubed? I redid the question and still I have the cube.

OK, that's a mistake of CIE. Take a look at the line before the answer. It is:
ln y = 3 ln (x^2 +4) + ln 32 - 3 ln 4

Simplify this:
ln y = ln {(x^2+4)^3 * 32 : (64)}
ln y = ln {(x^2+4)^3 : 2}
hence
y = 1/2 (x^2+4)^3;
Alternatively, substitute the wrong answer y = 1/2 * (x^2+4) without cube into the equation, you should come to contradiction.