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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.

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.

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#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

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...

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)^2bi) 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...

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#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

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

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#4

(Original post by

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

**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

(Original post by

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

**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.

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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