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Thread starter 14 years ago
#1
Hey

Could someone please take a few moments to work through some of these questions with me? the questions are ones i wasnt too sure about. they are from a mock i did on friday. it would be great if i had an idea of how i did. thanks.

4B) pi/12 and 5pi/12

5b ii) 2 roots?

6a) 1 + 12/(3x+2) + 36/(3x+2)^2 ??
bi) 1/3 ln |3x+2| ?
bii) simple, but i messed it up.

c) could somebody tell me where the 5 marks would be available here? how many marks would i get if i used my answer for part bii) for the third term? but did everything else correctly?

8b i) dx/dy = 3sec^2(3y) ?
ii) 1/12?

thanks again for anybody that can help. that would be great.
0
14 years ago
#2
4)b theta = 0 or pi/12
Notice the identity. Thie gives sin2O=2(sin2O)^2 given sin2O = x
this gives x = 2x^2 (notice quadratic equation...hence to roots)
solve for x gives 1/4 or 0 arc sin to find theta
5)b Two roots
6)a 1 - 12/(3x+2) + 36/(3x+2)^2
bi) 1/3 ln |3x+2|
ii) Use substitution
x = 2/3(tanu)^2
dx = 4/3tanu(secu)^2
this gives
f (2tanu^2 +2)^-2 (secu)^2tanu du
f tanu(secu)^2/(2secu^2)^2du
f 1/4 tanu (cosu)^2du
f 1/4 sinu cosu du
1/8 (sinu)^2 + c
c) In theory at ALevel should you mess up on cii) but still spot that you need to manipulate all three terms for your answer to c you can still get full marks provided you give the answer in terms of the term you messed up (of couse provided this is possible).

8) b) i)dx/dy = 3sec^2(3y)
ii) 1/12

DAMN I'm bored

Hmm a prospective Imperial medic....makes so much sense...
0
14 years ago
#3
4) a.) i.) sin2theta = 2sinthetacostheta
ii.) cos2theta = cos^2theta - sin^2theta = cos^2theta - (1 - cos^2theta) = 2cos^2theta - 1
iii.) tantheta(1 + cos2theta) = tantheta(2cos^2theta) = 2tanthetacosthetacostheta = 2sinthetacostheta = sin2theta

b.) tantheta(1 + cos2theta) = 2sin^2(2theta)
---> sin2theta = 2sin^2(2theta)
---> sin2theta = 1/2
For interval 0 < 2theta < 2Pi:
---> 2theta = Pi/6
---> 2theta = 5Pi/6

Solutions For Interval 0 < theta < Pi:
theta = Pi/12, 5Pi/12

5.) a.) 3^x = 7
---> xln3 = ln7
---> x = ln7/ln3
---> x = 1.77 (3.S.F)

b.) i.) To sketch the graph of y = 7 - x^2:
When y = 0: x = (+/-) Sq.rt(7)
---> Graph Crosses (Sq.rt7, 0) and (-Sq.rt7, 0).
When x = 0: y = 7
---> Graph Crosses (0, 7)
* Graph of y = 7 - x^2 is a parabola passing through these 3 points.

To sketch the graph of y = 3^x:
Graph lies completely above and never crosses the x-axis.
When x = 0: y = 1
---> Graph Crosses (0, 1)
Graph has the same general shape of a graph of the form y = a^x. i.e.) Exponential growth.

ii.) No. roots to 3^x = 7 - x^2 = No. of intersections between the graphs of y = 7 - x^2 and y = 3^x.

6.) a.) y^2 = [1 + 6/(3x + 2)][1 + 6/(3x + 2) = 1 + 12/(3x + 2) + 36/(3x + 2)^2

b.) i.) Int. 1/(3x + 2) dx. = Int. 3/[3(3x + 2)] dx. = 1/3 Int. 3/(3x + 2) dx. = 1/3 ln|3x + 2| + k

b.) ii.) Int. 1/(3x + 2)^2 dx. = Int. (3x + 2)^(-2) dx. = 1/[3(-1)] . (3x + 2)^(-1) + k = -1/[3(3x + 2)] + k

c.) Volume (Solid) = Pi . Int (Between Limits 0 and 2) y^2 dx. = Pi[x + 4ln|3x + 2| - 12/(3x + 2)] (Between Limits 0 and 2) = Pi[(2 + 4ln8 - 3/2) - (4ln2 - 6)] = Pi(4ln8 - 4ln2 + 13/2) = Pi(4ln4 + 13/2) = 37.8 Cube Units (3.S.F)

b.) i.) x = tan3y
---> dx/dy = 3sec^2(3y)

ii.) dy/dx = 1/[3sec^2(3y)] = [cos^2(3y)]/3
When y = Pi/9: dy/dx = [(cos Pi/3)^2]/3 = (1/4)/3 = 1/12
0
14 years ago
#4
(Original post by Nima)
b.) tantheta(1 + cos2theta) = 2sin^2(2theta)
---> sin2theta = 2sin^2(2theta)
---> sin2theta = 1/2
For interval 0 < 2theta < 2Pi:
---> 2theta = Pi/6
---> 2theta = 5Pi/6
sin2theta = 2sin^2(2theta)---> sin2theta = 1/2 does not follow as there are clearly two roots to this. Admittedly I did not give the second value for 2theta=1/2 as I used arcsin.
(Original post by Nima)

ii.) No. roots to 3^x = 7 - x^2 = No. of intersections between the graphs of y = 7 - x^2 and y = 3^x.
Not sure how you came to that conclusion. You said in your post that the graph of f=3^x exists not below the positive number line for f given we stay within the real plane. Also the graph g=7 - x^2 starts above f given a value of x =0. However if we take the values of g for both x=-3 and x=3 then we find g exists below the x axis. Going out on a limb here because I can't prove it. But both functions f and g are continous then there must exist at least two points where they intercept (ie points where one is larger than the other then the other is larger, as g clearly must be less than f where |x|>3and g is greater than f where x=0.

Good luck on getting into medicine here (as far as I can tell you really don't need any luck/brains).
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