Need question solving.. (1 Viewer)

[Bokky]

I did my half yearly for Maths 2U and 3U and got pretty poor results.
But i have another assessment task, a test, where 50% of the exact same questions from my half yearly will appear again in this next paper (for both 2U and 3U), so thats why i need help with the ones i got wrong. THANX FOR ALL THOSE THAT HELP

here are a few :

1) (i) Find the point of intersection of the curves y=x^2+3x+2 and y=x^2-9x+14
(ii) Find the area of the region bounded by these 2 curves and lying above the x axis.

2) Find the exact area bounded by the curve y=ln x, the y axis, and the lines y=2 and y=6.

3) Show that the volume obtained by rotating about the x-axis the area Beneath the curve y=e^2x, from x=0 to x=3 has a magnitude pie/4(e^12-1).

4) If 2 ln x = ln (3x+4) fine, correct to 3 d.p the value of ln 2x.

5) A (2,6) B (5,-1) C (1,-3) and D(-2,-2) are the vertices of a quadrilateral.
(i) Find the perpendicular distance from A to BD.
(ii) Find the area of triangle ABD
(iii) Fine the area of the quadrilateral.

6) For what value of k will the lines 3x-ky-7=0 and x+2y-5=0 be
(i) parallel
(ii) perpendicular

7) Solve the equation 2^(x+1) = 9, giving the answer correct to 5 d.p

i know its a lot but thanx a lot for those who help, i really appreciate it

 

[FinalFantasy]

1) (i) Find the point of intersection of the curves y=x^2+3x+2 and y=x^2-9x+14
(ii) Find the area of the region bounded by these 2 curves and lying above the x axis.
answer:
i)to get intersection x²+3x+2=x²-9x+14
12x=12
x=1, now sub x=1 back into one of the equations, e.g y=(1)²+3(1)+2=6
.: point of intersection of those curves is (1,6)
ii)to find area bounded by those curves:

area=int. (x^2+3x+2) dx from x=-2 to x=1 +int. (x^2-9x+14) dx from x=1 to x=2
=[x³\3+3x²\2+2x] from -2 to 1 +[x³\3-9x²\2+14x] from 1 to 2

2) Find the exact area bounded by the curve y=ln x, the y axis, and the lines y=2 and y=6.
y=lnx
e^y=e^lnx
.: x=e^y
area=int. e^y dy from y=2 to y=6
area=[e^y] from 2 to 6

 

[Jago]

1.
(i) make the two equations equal each other
ie x^2+3x+2 = x^2 - 9x + 14

2. make y = f(x) into x = f(y), then integrate using the definites 2 and 6

3. basic volume question. make sure you square f(x) and put pi outside the integral

 

[Trev]

1) (i) Find the point of intersection of the curves y=x^2+3x+2 and y=x^2-9x+14
(ii) Find the area of the region bounded by these 2 curves and lying above the x axis.

i) y = x² + 3x + 2 (1)
y = x² - 9x + 14 (2)
Simultaneously:
x² + 3x + 2 = x² - 9x + 14
12x = 12
x = 1; therefore y = (1)x² - 9x + 14 + 3(1) + 2 = 6
Point of intersection is (1,6)

ii)
Area is given by area under curve between -1 and 1 [with equation (1)] and under curve between 1 and 2 [with equation (2)].

∫[x² + 3x + 2]dx (-1 to 1) + ∫[x² - 9x + 14]dx (from 1 to 2)
[x³/3 + (3x²)/2 + 2x] (from -1 to 1) + [x³/3 - (9x²)/2 + 14x] (from 1 to 2)
Area = 14/3 + 17/6 = 15/2 units²

 

[Trev]

6) For what value of k will the lines 3x-ky-7=0 and x+2y-5=0 be
(i) parallel
(ii) perpendicular

3x-ky-7=0
ky = 3x - 7
y = (3/k)x - 7/k (1)

x+2y-5=0
2y = -x + 5
y = (-1/2)x - 5/2 (2)

(i) Parallel, gradients equal eachother, so:
(3/k) = (-1/2); k = -6.
(ii) Perpendicular; k = 1/6. (m₁m₂ = -1)

 

[FinalFantasy]

3) Show that the volume obtained by rotating about the x-axis the area Beneath the curve y=e^2x, from x=0 to x=3 has a magnitude pie/4(e^12-1).

v=pi int. e^4x dx from 0 to 3
=pi [(1\4) e^4x] from 0 to 3
=pi\4 (e^12-1) units ³

4) If 2 ln x = ln (3x+4) fine, correct to 3 d.p the value of ln 2x.
2lnx=ln(3x+4)
ln x²=ln(3x+4)
x²=3x+4
x²-3x-4=0
(x-4)(x+1)=0
.: x=4 or x=-1
.: ln 2x=ln8=2.046
the log can't be negative so ignore -1

5) A (2,6) B (5,-1) C (1,-3) and D(-2,-2) are the vertices of a quadrilateral.
(i) Find the perpendicular distance from A to BD.
(ii) Find the area of triangle ABD
(iii) Fine the area of the quadrilateral.

i)to get perp. distance from A to BD, find equation of line BD and den use perpendicular distance formula:
equation of BD: (y+1)\(x-5)=(-2+1)\(-2-5)=1\7
7y+7=x-5
x-7y-12=0 is the equation BD
now use da formula, d=|Ax1+By1+C|\sqrt(A²+B²) with the point A(2,6)

(ii) Find the area of triangle ABD
after u get perp. distance from A to BD, just use area=1\2 times BD times that perpendicular distance

(iii) Fine the area of the quadrilateral.
juz get perp. distance from C to BD den do da same as above den add the areas

 

[Trev]

7) Solve the equation 2^(x+1) = 9, giving the answer correct to 5 d.p

2^(x+1) = 9
(x+1)ln2 = ln9
x+1 = (ln9)/(ln2)
x = (ln9)/(ln2) - 1
= (to 5 decimal places) 2.16993

 

[azza_3761]

6) gradient = -a/b
-1/2 = -3/k
k = 6

7) log(a)b = c b = a^c
9 = 2^(x+1)
log(2)9 = x + 1
x = log(2)9 - 1
= 2.16993 (5dp)

 

[Trev]

2) Find the exact area bounded by the curve y=ln x, the y axis, and the lines y=2 and y=6.

y = lnx
x = e^y
Area given by:
∫ e^ydy (from 2 to 6)
= e²(e⁴ - 1)

 

[FinalFantasy]

hahaha, so many ppl rush to answer dis one
i refresh den many posts already lol

 

[Bokky]

WOW that was quick and a lot of posts lol
u guys rock! ta

 

[Riddles]

if u have any more problems, don't hesitate to ask

 

[Bokky]

yeh, ive got a few more that im still stuck on...

1) If int. (2x-3)dx=10 from (a+2) and (a), fine the value of a.

2) Box A holds 12 red apples and 9 green apples.
Box B holds 8 red and 7 green apples.
Emily chooses a box at random and takes 2 apples from the box. By drawing a tree diagram, or otherwise, find the probability that the apples are the SAME colour.

3) ABCDEFGH is a regular octagon, with the centre O. If AE = 2 units, find the area of triangle AOB and hence the area of the octagon in simplest surd form (sorry no diagram).

4) Given that f(x)=4x^3+18x^3+27x-15 find k such that f ' (k) = f ' ' (k).

ta guys

 

[shafqat]

3) As AE = 2, AO = 1
Also BO = 1 (isos triangle)
Now angle AOB = 360/8 (revolution divided by 8)
= 45
So are of AOB = 1/2 ab sinC
= 1/2.1.1.1/sqrt2
= 1/2sqrt2
So are of octagon = 8/2sqrt2
= 2sqrt2

 

[shafqat]

For 4), f'(x) = 12x^2 + 36x + 27
f''(x) = 24x + 36
Putting them equal to one another, and simplifying, we get
4x^2 + 4x - 9 = 0
(2x - 1)(2x + 3) = 0
x = 1/2, -3/2

 

[shafqat]

Theres 1/2 of getting either box
For box A, its 12/21.11/20 for RR, and 9/21.8/20 for GG
For box B, its 8/15.7/14 for RR, and 7/15.6/14 for GG
So prob = 1/2 (12/21.11/20 + 9/21.8/20) + 1/2 (8/15.7/14 + 7/15.6/14)
= 10/21

 

[shafqat]

1) int. (2x-3)dx= from (a+2) and (a) = (a+2)^2 - 3(a+3) - a^2 + 3a
= 2a - 5 = 10
so a = 15/2