Math 32....how was it? :/

[ffaadyy]

What an easy paper it was except for the last part of the Vector question. Still wont be losing more than 4 marks InshAllah.

 

[Saiyan]

my mechanics paper was so bad i lost 18 marks but p3 went well just gonna lose 5-6 marks...............got A in as can i get a* plzzzzzzzzzzzzzzz helpppppppppppppppppp....................................

You have to mention the percentage or else I cannot help you. Mechanics is a very sensitive paper in the sense that students can suffer a B because of their performance. S1 S2 M1 and M2 requires at least 40-38 marks to score an A and 42-45 to sore an A*. This is because of the fact that the whole marks of the paper is only 50 which makes the grading a bit more rigid.

 

[Saiyan]

What an easy paper it was except for the last part of the Vector question. Still wont be losing more than 4 marks InshAllah.

You must be kidding! I am an A* student but still found the paper difficult. Solved the whole vector question but got stuck in the differential equation.

 

[ffaadyy]

You must be kidding! I am an A* student but still found the paper difficult. Solved the whole vector question but got stuck in the differential equation.

You must be kidding! I am an A* student but still found the paper difficult. Solved the whole vector question but got stuck in the differential equation.

I am an A* student too and the mathematics coach at my school. This's how you had to do the differential question:


dy/dx=e(2x+y)
dy/dx=e^(2x) x e^( y )
[1/e^( y )] dy = e^(2x) dx
e^(-y) dy = e^(2x) dx

Integrate both the sides.

-e^(-y) = [e^(2x)/2] + c

Put x=0 and y=0 to find the value of 'c'.

-e^(-y) = [e^(2x)/2] + c
-e^(-0) = [e^(0)/2] + c
-1 = (1/2) + c
-1 - (1/2) = c
-3/2 = c

Put back this value of 'c' in the integrated solution.

-e^(-y) = [e^(2x)/2] + c
-e^(-y) = [e^(2x)/2] - (3/2)

Multiply the equation by a '-' sign.

- {-e^(-y) = [e^(2x)/2] - (3/2)}
e^(-y) = (3/2) - [e^(2x)/2]

Put 'ln' on both the sides.

ln e^(-y) = ln {(3/2) - [e^(2x)/2]}
-y = ln {(3/2) - [e^(2x)/2]}
y = - ln {(3/2) - [e^(2x)/2]}

Therefore, the answer was 'y = - ln {(3/2) - [e^(2x)/2]}'.

 

[princesskt]

You have to mention the percentage or else I cannot help you. Mechanics is a very sensitive paper in the sense that students can suffer a B because of their performance. S1 S2 M1 and M2 requires at least 40-38 marks to score an A and 42-45 to sore an A*. This is because of the fact that the whole marks of the paper is only 50 which makes the grading a bit more rigid.

last year there were no marks.......i got a in stats,and p1 gave p3 and mechanics this year......can p 3 overcome my deficiency in mechanics overall....

 

[ffaadyy]

Answer to the volume of revolution question was pie/4(e^2 - 1).

 

[ffaadyy]

I did by exactly the same method and probably came down to sucha an equation!!
But my friends say tht as it is nt proven as an Identity so u cannot multiply (cos X) on both sides to cut it!!!
Any idea about it!??

And gyz wat was the answer to volume of revo Q?
I got pi((e^2)/2 -(1/2)) !!!!
And my friends who integrated lnx whole squared are getting the same answers!! :/

y = x^(1/2) ln x

Volume = pie ( y )^2
Volume = pie [ x^(1/2) ln x ]^2
Volume = pie [x (ln x)^2]

Now use integration by parts to integrate 'x (ln x)^2'

x (ln x)^2
(x^2/2)(ln x)^2 - [x^2 /2 * 2 ln x * 1/x]
(x^2/2)(ln x)^2 - [x ln x]

Now integrate 'x ln x'

x ln x
(x^2/2) ln x - [x^2/2 * 1 /x]
(x^2/2) ln x - [x/2 ]
(x^2/2) ln x - x^2/4

(x^2/2)(ln x)^2 - [x ln x]
(x^2/2)(ln x)^2 - [ (x^2/2) ln x - x^2/4 ]
(x^2/2)(ln x)^2 - (x^2/2) ln x + x^2/4

Now put the limits 'e' (upper) and '1' (lower).

Volume = pie [(e^2/2)(ln e)^2 - (e^2/2) ln e + e^2/4] - [(1^2/2)(ln 1)^2 - (1^2/2) ln 1 + 1^2/4 ]
Volume = pie [(e^2/2) - (e^2/2) + e^2/4] - [ 1/4 ]
Volume = pie [e^2/4] - [ 1/4 ]
Volume = (pie/4)(e^2 - 1)

Therefore, the volume is '(pie/4)(e^2 - 1)'.

 

[Saiyan]

I am an A* student too and the mathematics coach at my school. This's how you had to do the differential question:


dy/dx=e(2x+y)
dy/dx=e^(2x) x e^( y )
[1/e^( y )] dy = e^(2x) dx
e^(-y) dy = e^(2x) dx

Integrate both the sides.

-e^(-y) = [e^(2x)/2] + c

Put x=0 and y=0 to find the value of 'c'.

-e^(-y) = [e^(2x)/2] + c
-e^(-0) = [e^(0)/2] + c
-1 = (1/2) + c
-1 - (1/2) = c
-3/2 = c

Put back this value of 'c' in the integrated solution.

-e^(-y) = [e^(2x)/2] + c
-e^(-y) = [e^(2x)/2] - (3/2)

Multiply the equation by a '-' sign.

- {-e^(-y) = [e^(2x)/2] - (3/2)}
e^(-y) = (3/2) - [e^(2x)/2]

Put 'ln' on both the sides.

ln e^(-y) = ln {(3/2) - [e^(2x)/2]}
-y = ln {(3/2) - [e^(2x)/2]}
y = - ln {(3/2) - [e^(2x)/2]}

Therefore, the answer was 'y = - ln {(3/2) - [e^(2x)/2]}'.

That was brilliant!

 

[Saiyan]

last year there were no marks.......i got a in stats,and p1 gave p3 and mechanics this year......can p 3 overcome my deficiency in mechanics overall....

Can't say it that way as mechanics was also tough. I happen to be an S2 student

 

[smzimran]

I am an A* student too and the mathematics coach at my school. This's how you had to do the differential question:


dy/dx=e(2x+y)
dy/dx=e^(2x) x e^( y )
[1/e^( y )] dy = e^(2x) dx
e^(-y) dy = e^(2x) dx

Integrate both the sides.

-e^(-y) = [e^(2x)/2] + c

Put x=0 and y=0 to find the value of 'c'.

-e^(-y) = [e^(2x)/2] + c
-e^(-0) = [e^(0)/2] + c
-1 = (1/2) + c
-1 - (1/2) = c
-3/2 = c

Put back this value of 'c' in the integrated solution.

-e^(-y) = [e^(2x)/2] + c
-e^(-y) = [e^(2x)/2] - (3/2)
----------------------------------- <-- Here I took the 2 in the denominator and sent it to the y side and later sent it back, guess that is also right?
Multiply the equation by a '-' sign.

- {-e^(-y) = [e^(2x)/2] - (3/2)}
e^(-y) = (3/2) - [e^(2x)/2]

Put 'ln' on both the sides.

ln e^(-y) = ln {(3/2) - [e^(2x)/2]}
-y = ln {(3/2) - [e^(2x)/2]}
y = - ln {(3/2) - [e^(2x)/2]}

Therefore, the answer was 'y = - ln {(3/2) - [e^(2x)/2]}'.

See the red colour addition i made ?

 

[ffaadyy]

Cosec2x=secx+cotx plz

cosec 2x = sec x + cot x
(1/sin 2x) = (1/cos x) + (cos x/sin x)
(1/2 sin x cos x) = (sin x + cos^2 x)/(sin x cos x)

'sin x cos x' present in both the denominators get cancelled.

(1/2 sin x cos x) = (sin x + cos^2 x)/(sin x cos x)
(1/2) = (sin x + cos^2 x)
(1/2) = sin x + (1 - sin^2 x)
1 = 2 sin x + 2 - 2 sin^2 x
2 sin^2 x - 2 sin x - 2 + 1 = 0
2 sin^2 x - 2 sin x - 1 = 0

a=2, b=-2, c=-1.

Find the value of 'sin x' using '[-b +- (b^2 - 4ac)^(1/2)]/2(a)]'.

sin x = 1.37 (calculator gives a math error therefore we reject this value) and sin x = -0.37.

sin x = - 0.37

As 'sin x' is negative, the angle would be either in the 3rd quadrant (180+x) or the 4th quadrant (360-x).

sin x = 0.37
x = 21.5

180 + 21.5 = 201.5
360 - 21.5 = 338.5

Therefore, the 2 angles are '201.5' and '338.5'.

 

[ffaadyy]

See the red colour addition i made ?

That's also correct, I presume. The final answer would've been 'y = (-1/2) ln (3 - e^2x) then.

 

[smzimran]

That's also correct, I presume. The final answer would've been 'y = (-1/2) ln (3 - e^2x) then.

Yes something like that i dont exactly remember, though!

I had an excellent paper but i also think that the paper was difficult, specially in terms of length!

 

[ffaadyy]

yes but..what about the binomial one?

Its answer was '1 - x + (x^2)/2'.

 

[ffaadyy]

Yes something like that i dont exactly remember, though!

I had an excellent paper but i also think that the paper was difficult, specially in terms of length!

Do you remember the question in which we had to find the stationary points first and then the nature of that point? It was question 6 or question 7, I guess

 

[smzimran]

Do you remember the question in which we had to find the stationary points first and then the nature of that point? It was question 6 or question 7, I guess

No
But i remember it was minimum point!

 

[Saiyan]

Its answer was '1 - x + (x^2)/2'.

That is correcto!

 

[Saiyan]

Do you remember the question in which we had to find the stationary points first and then the nature of that point? It was question 6 or question 7, I guess

That was question 6. The points were pie, pie/12 and 5pie/12 and the least value which is pie/12 is maximum

 

[ffaadyy]

That was question 6. The points were pie, pie/12 and 5pie/12 and the least value which is pie/12 is maximum

Do you remember that question? What was it?

 

[beststriker]

can anyone solve the volume ques plz??