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June 2006 - C1 and C2 Solutions

JUNE 2006 - C1


1.
∫(6x2+2+x−12)dx\huge \int(6x^2+2+x^{-\frac{1}{2}}) dx
Unparseable latex formula:

\huge = \frac{6x^3}{3} + 2x + 2x^{\frac{1}{2}+c


Unparseable latex formula:

\huge = \frac {2x}^3 + 2x + {2x}^{\frac{1}{2}}+ c



2.
x2−7x−18>0\huge x^2 - 7x - 18 > 0
(x−9)(x+2)>0\huge (x-9)(x+2) > 0
x<−2,x>9\huge x<-2, x>9

3.
(a) Parabola resting on x=-3, crosses y axis at (0,9)
(b) Parabola with no roots, crosses y at (0,9+k)

4.
(a) a1=3\huge a_1 = 3
an+1=3an−5\huge a_{n+1} = 3a_n-5
a2=4\huge a_2 = 4
a3=7\huge a_3 = 7

(b) a4=16\huge a_4 = 16
a5=43\huge a_5=43
Σa1+a2+a3+a4+a5=73\huge \Sigma{a_1+a_2+a_3+a_4+a_5} = 73

5.
(a) ddx(x4+6x)\huge \frac{d}{dx} (x^4 + 6\sqrt{x})
=ddx(x4+6x12)\huge = \frac{d}{dx} (x^4 + 6x^{\frac{1}{2}})
Unparseable latex formula:

\huge = {4x}^3 + {3x}^{-\frac{1}{2}



(b) ddx((x+4)2x)\huge \frac{d}{dx}(\frac{(x+4)^2}{x})
=ddx((x2+8x+16x)\huge = \frac{d}{dx}(\frac{(x^2+8x+16}{x})
=ddx(x+8+16x)\huge = \frac{d}{dx}(x+8+\frac{16}{x})
=ddx(x+8+16x−1)\huge = \frac{d}{dx}(x+8+16x^{-1})
=1−16x−2\huge = 1-16x^{-2}

6.
(a) (4+3)(4−3)\huge (4+\sqrt3)(4-\sqrt3)
Unparseable latex formula:

\huge = 16 +4{\sqrt{3}-4{\sqrt3} - 3


=13\huge = 13

(b) 264+3\huge \frac{26}{4+\sqrt3}
=(26)(4−3)(4+3)(4−3\huge = \frac{(26)(4-\sqrt3)}{(4+\sqrt3)( 4-\sqrt3}
=(26)(4−3)13\huge = \frac{(26)(4-\sqrt3)}{13}
=2(4−3)\huge = 2(4-\sqrt3)
=8−23\huge = 8-2{\sqrt3}

7. U11=9\huge U_11 = 9
S11=77\huge S_11 = 77
77=112(2a+10d)\huge 77 = \frac{11}{2}(2a+10d)
9=a+10d\huge 9 = a + 10d
a=5\huge a = 5
d=25\huge d = \frac{2}{5}

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Reply 1
Avatar for Kolya
Kolya
OP
17
8.
(a)
x2+2px+(3p+4)=0\huge x^2+2px+(3p+4)=0
b2−4ac=0\huge b^2-4ac =0
(2p)2−4(3p+4)=0\huge (2p)^2-4(3p+4)=0
4p2−12p+16=0\huge {4p}^2-12p+16 = 0
p2−3p+4=0\huge p^2-3p+4=0
(p−4)(p+1)\huge (p-4)(p+1)
p=4\huge p = 4

(b)
x2+8x+16=0\huge x^2+8x+16 = 0
(x+4)(x+4)\huge (x+4)(x+4)
x=−4\huge x= -4

9.
(a)
(x2−6x)(x−2)+3x\huge (x^2-6x)(x-2)+3x
=x3−8x2+15x\huge =x^3-8x^2+15x
=x(x2−8x+15)\huge =x(x^2-8x+15)

(b)
x(x−5)(x−3)\huge x(x-5)(x-3)

(c) Cubic graph crossing x axis at 0, 3 and 5.

10.
(a)
ddx=2x+3x2\huge \frac{d}{dx} = 2x + \frac{3}{x^2}
f(x)=∫(2x+3x2)dx\huge f(x) = \int(2x + \frac{3}{x^2}) dx
f(x)=x2−3x−1+c\huge f(x) = x^2-{3x}^{-1} + c
(3,152)\huge (3, \frac{15}{2})
152=8+c\huge {\frac{15}{2}}= 8 + c
c=−12\huge c = -\frac{1}{2}
f(x)=x2−3x−1−12\huge f(x) = x^2-{3x}^{-1}-{\frac{1}{2}}

(b)
f(−2)=4+32−12=5\huge f(-2) = 4 + \frac{3}{2} -\frac{1}{2} = 5

(c)
ddx=2x+3x2\huge \frac{d}{dx} = 2x + \frac{3}{x^2}
x=−2\huge x = -2
ddx=−4+34\huge \frac{d}{dx} = -4 +\frac{3}{4}
ddx=−134\huge \frac{d}{dx} = -\frac{13}{4}
y=−134x+c\huge y = -\frac{13}{4}x + c
(−2,5)\huge (-2, 5)
5=132+c\huge 5 = \frac{13}{2}+c
c=−32\huge c = -\frac{3}{2}
y=−134x−32\huge y = -\frac{13}{4}x - \frac{3}{2}
4y=−13x−6\huge 4y= -13x-6
13x+4y+6=0\huge 13x+4y+6=0

11.
(a)
P(−1,2)Q(11,8)\huge P(-1,2) Q(11,8)
Δy=6Δx=12\huge \Delta{y}=6 \Delta{x}=12
y=12x+c\huge y=\frac{1}{2}x+c
2=−12+c\huge 2 = -\frac{1}{2}+c
c=52\huge c= \frac{5}{2}
y=12x+52\huge y=\frac{1}{2}x+\frac{5}{2}

(b)
l2:y=−2x+c\huge l_2 : y=-2x+c
R(10,0)\huge R(10,0)
0=−20+c\huge 0 = -20+c
c=20\huge c = 20
y=−2x+20\huge y=-2x+20

l1y=12x+52\huge l_1 y=\frac{1}{2}x+\frac{5}{2}
l2y=−2x+20\huge l_2 y=-2x+20
−2x+20=12x+52\huge -2x+20=\frac{1}{2}x+\frac{5}{2}
52x=−352\huge \frac{5}{2}x=-\frac{35}{2}
5x=35\huge 5x=35
x=7\huge x=7
S(7,6)\huge S(7,6)

(c)
R(10,0)\huge R(10,0)
S(7,6)\huge S(7,6)
Δx−3Δy=6\huge \Delta{x}-3 \Delta{y}=6
Length=32+62\huge Length = \sqrt{3^2+6^2}
=45\huge = \sqrt{45}
=35\huge = {3}{\sqrt5}

(d)
A suitable diagram will give Area{PQR} = 45 square units.

End of C1 Solutions.
Reply 2
Avatar for N0_1_Indian
N0_1_Indian
C2 Solutions???????????? June 06.
Reply 3
Avatar for Kolya
Kolya
OP
17
Patience is a virtue :smile: C2 will be up soon (thanks to Insparato)
Reply 4
Avatar for Lykabelle
Lykabelle
12
Thanks for that:smile:
good job on the solutions, we appreciate it
Reply 6
Avatar for SuBB
SuBB
1
BEST thing ive seen all day :smile:

i should be getting above 95%

only way i could of lost marks is through unclear method or my messy handwriting :smile:
Reply 7
Avatar for N0_1_Indian
N0_1_Indian
LOL sorry. Appreciate ure time and effort. Peace
Reply 8
Avatar for Kolya
Kolya
OP
17
C2 Solutions (Thanks to Insparato)

1.
(2+x)6=6C0(2)6+6C1(x)(2)5+6C2(2)4(x)2\huge (2 + x)^6 = ^6C_0 (2)^6 + ^6C_1 (x)(2)^5 + ^6C_2 (2)^4(x)^2
=64+192x+240x2\huge = 64 + 192x + 240x^2

2.
∫12(3x2+5+4x2)dx\huge \int^2_1{({3x}^2+5+\frac{4}{x^2})}dx
=(x3+5x−4x−1)12\huge ={(x^3 + 5x - 4x^{-1})}^{2}_{1}
=(23+(5)(2)−(4)(2−1)−(13+(5)(1)−(4)(1−1)\huge =({2}^3 + (5)(2)-(4)(2^{-1})-({1}^3 + (5)(1)-(4)(1^{-1})
=(8+10−2)−(1+5−4)\huge =(8 + 10 -2)-(1+5-4)
=16−2=14\huge = 16-2 =14

3.
(i)
Log636=Log1036Log106=2\huge Log_{6}36 = \frac{Log_{10}36}{Log_{10}6} = 2

(ii)
2loga3+loga11=loga32+loga11=loga9+loga11=loga99\huge 2log_a3 + log_a11 = log_a32 + log_a11 = log_a9 + log_a11 = log_a99

4.
(a)
f(x)=2x3+3x2−29x−60\huge f(x) = 2x^3+{3x}^2-29x-60
f(−2)=2(−2)3+3(−2)2−29(−2)−60\huge f(-2) = 2{(-2)}^3+3{(-2)}^2-29(-2)-60
f(−2)=2(−8)+3(4)+58−60\huge f(-2) = 2(-8)+3(4) + 58-60
f(−2)=−16+12+58−60\huge f(-2) = -16+12+58-60
f(−2)=−6\huge f(-2) = -6
Remainder=−6\huge Remainder = -6

(b)
if (x+3)\huge (x+3) is a factorf(−3)=0\huge f(-3) = 0
f(−3)=(2)(−3)3+(3)(−3)2−(29)(−3)−60\huge f(-3) = (2){(-3)}^3+(3){(-3)}^2-(29)(-3)-60
f(−3)=(2)(−27)+27+87−60\huge f(-3) = (2)(-27)+27+87-60
f(−3)=−54+27+87−60\huge f(-3) = -54+27+87-60
f(−3)=−27+87−60=0\huge f(-3) = -27+87-60 = 0
f(−3)=0\huge f(-3) = 0
thus (x+3) must be a factor of f(x)

5.
(a)
Exponential curve crossing the y-axis at (0,1)

(b)
x=0f(x)=1.000\huge x = 0 f(x) = 1.000
x=0.6f(x)=1.933\huge x = 0.6 f(x) = 1.933
x=0.8f(x)=2.408\huge x = 0.8 f(x) = 2.408

(c)
Area=12h(y0+2(y1+y2+...+yn−1)+yn)\huge Area= \frac{1}{2}h(y_0 + 2(y_1 + y_2 +...+y_{n-1}) + y_n)
H=b−an=(1−0)5=0.2\huge H = b-\frac{a}{n} = \frac{(1-0)}{5} = 0.2
Area=12(0.2)(1+2(1.246+1.933+2.408)+3)\huge Area= \frac{1}{2}(0.2)(1+2(1.246+1.933+2.408)+3)
=0.1(1+14.278+3)\huge = 0.1(1+14.278+3)
=0.1(18.278)\huge = 0.1(18.278)
=1.8278\huge = 1.8278
=1.83units2\huge =1.83 {units}^2

6.
(a)
sinθ=5cosθ\huge sin\theta = 5cos\theta
sinθcosθ=5cosθcosθ\huge\frac{sin\theta}{cos\theta}=\frac{5cos\theta}{cos\theta}
=tanθ=5\huge =tan\theta = 5

(b)
tanθ=5\huge tan\theta=5
θ=78.7,258.7\huge \theta = 78.7, 258.7

7.
(a)
tangent gradient:
Unparseable latex formula:

\large y=3x&#8211;4=3


PQ;grad=−13\huge PQ; grad = -\frac{1}{3}
y=−13x+c\huge y= -\frac{1}{3}x+c
P(2,2)\huge P(2,2)
c=83\huge c = \frac{8}{3}
y=−13x+83\huge y= -\frac{1}{3}x+\frac{8}{3}

(b)
Q(x,1)\huge Q(x,1)
1=−13x+83\huge 1 = -\frac{1}{3}x+\frac{8}{3}
x=5\huge x=5

(c)
Centre(5,1)\huge Centre(5,1)
r=(5−2)2+(1−2)2\huge r = \sqrt{(5-2)^2+(1-2)^2}
r=10\huge r=\sqrt{10}
(x−a)2+(y−b)2=r2\huge (x-a)^2+(y-b)^2=r^2
(x−5)2+(y−1)2=10\huge (x-5)^2+(y-1)^2=10

8.
(a)
l=rθ\huge l=r\theta
L=(2.12)(0.65)=1.378m=1.38m(2dp)\huge L=(2.12)(0.65)=1.378m=1.38m (2dp)

(b)
A=12r2θ\huge A = \frac{1}{2r^2}\theta
A=12(2.12)2(0.65)\huge A = \frac{1}{2}(2.12)^2(0.65)
A=1.46m2\huge A =1.46m^2

(d)
A=12(a)(b)(sinC)\huge A = \frac{1}{2}(a)(b)(sinC)
=12(1.86)(2.12)(sin0.9208)\huge = \frac{1}{2}(1.86)(2.12)(sin0.9208)
=1.9716xSin0.9208=1.5696\huge =1.9716 x Sin 0.9208 = 1.5696
ABCDArea=1.4607+1.5696=3.03m2(2dp)\huge {ABCD} Area = 1.4607 + 1.5696 = 3.03m^2 (2dp)
Thats a kool markscheme except i think youve made an error on qu 10c

-4 + 3/4 = -13/4

y - 5 = -13/4(x+2)

4y = -13x -6

-13x - 4y -6 = 0

im proably wrong but what the hell lol
Reply 10
Avatar for Kolya
Kolya
OP
17
Corrected Dominic :smile:
Reply 11
Avatar for WaW™
WaW™
3
[email protected]
Thats a kool markscheme except i think youve made an error on qu 10c

-4 + 3/4 = -13/4

y - 5 = -13/4(x+2)

4y = -13x -6

-13x - 4y -6 = 0

im proably wrong but what the hell lol


is that just not the same just move everything to the left side and u get

13x + 4y + 6 = 0 :biggrin:

so neither of u were wrong!
thanks a lot for solutions
i just have some probs with printing this
Reply 13
Avatar for Kolya
Kolya
OP
17
Maybe take a screenshot and copy that into Word.
tnx ill try these
Baal_k
5.
(a)
Exponential curve crossing the y-axis at (1,0)

(0,1)?
Reply 16
Avatar for Kolya
Kolya
OP
17
C2 SOLUTIONS (Continued)

9.
(a)
S∞=25\huge S_\infty = 25
(a)(r)=4\huge (a)(r) = 4
a1−r=25\huge \frac{a}{1-r} = 25
a=4R\huge a= \frac{4}{R}
4/r1−r=25\huge \frac{4/r}{1-r} = 25
4r=25(1−r)\huge \frac{4}{r} = 25(1-r)
4r=25−25r\huge \frac{4}{r} = 25 - 25r
4=25r−25r2\huge 4 = 25r-25r^2
25r2−25r+4=0\huge 25r^2 -25r + 4 = 0

(b)
(5r−1)(5r−4)=0\huge (5r-1)(5r-4) = 0
r=15r=45\huge r = \frac{1}{5} r = \frac{4}{5}

(c)
a=20,a=5\huge a = 20, a = 5

(d)
Sn=a(1−r)1−rn\huge S_n = \frac{a(1-r)}{1-r^n}
a1−r=25\huge \frac{a}{1-r}=25
Sn=25(1−rn)\huge S_n = 25(1-r^n)

(e)
r=45\huge r = \frac{4}{5}
Unparseable latex formula:

\huge 25(1-(\frac{4}{5}}^n) > 24


1−(45)n=2425\huge 1-(\frac{4}{5})^n = \frac{24}{25}
(45)n=1−2425\huge (\frac{4}{5})^n = 1-\frac{24}{25}
(45)n=4100\huge (\frac{4}{5})^n = \frac{4}{100}
nlog1045=log104100\huge n log_{10} \frac{4}{5} = log_{10} \frac{4}{100}
n=log100.04/log100.8\huge n = log_{10} 0.04 / log_{10} 0.8
n=14.425\huge n = 14.425
n=15\huge n = 15

10.
(a)
y=x3−8x2+20x\huge y = x^3-{8x}^2 + 20x
dydx=3x2−16x+20\huge \frac {dy}{dx} = 3x^2 - 16x+20
dy/dx=0\huge dy/dx = 0
3x2−16x+20=0\huge 3x^2-16x+20 = 0
(3x−10)(x−2)\huge (3x-10)(x-2)
x=2,x=103\huge x = 2, x = \frac{10}{3}
a:x=2\huge a: x= 2
b:x=103\huge b: x= \frac{10}{3}

(b)
dydx=3x2−16x+20\huge \frac{dy}{dx}= 3x^2-16x+20
d2ydx2=6x−16\huge \frac{d^2y}{dx^2} = 6x-16
subx=2intod2ydx2\huge sub x = 2 into \frac{d^2y}{dx^2}
d2ydx2=6(2)−16\huge \frac{d^2y}{dx^2} = 6(2) -16
=12−16=−4\huge= 12-16 = -4
d2ydx2=−4\huge \frac{d^2y}{dx^2} = -4
Sod2ydx2<0\huge So \frac{d^2y}{dx^2} < 0
a=maximum\huge a = maximum

(c)
∫(x3−8x2+20x)dx\huge \int (x^3-8x^2+20x) dx
x44−8x33+20x22+c\huge \frac{x^4}{4}-\frac{{8x}^3}{3}+\frac{{20}x^2}{2}+c
=x44−8x33+10x2+c\huge = \frac{x^4}{4}-\frac{{8x}^3}{3}+{10x}^2+c

{d)
∫02(x3−8x2+20x)\huge \int^2_0(x^3-8x^2+20x)
[14x4−83x3+10x2]02\huge [\frac{1}{4}x^4-\frac{8}{3}x^3+{10x}^2]^2_0
=[(14)(24)−(643)+(10)(22)]−[0]\huge = [(\frac{1}{4})(2^4)-(\frac{64}{3})+(10)(2^2)]-[0]
=[4−643+40]=683\huge = [4-\frac{64}{3}+40] = \frac{68}{3}
Triangle Area from A directly downwards to point X and to point N
A y-coordinate
Sub x=2\large x = 2 into y=x3−8x2+20x\large y = x^3-{8x}^2+20x
y=23−(8)(22)+(20)(2)\huge y = 2^3-(8)(2^2) +(20)(2)
y=8−32+40\huge y = 8-32+40
y=16\huge y = 16
Height=16\huge Height = 16
Area of AXN = 12(bh)\large \frac{1}{2}(bh)
Area of AXN = 12(43)(16)\large \frac{1}{2}(\frac{4}{3})(16)
Area of AXN = 323\large \frac{32}{3}

Total Area of Region R
Area of Triangle AXN+ ∫02(x3−8x2+20x)dx\large \int^2_0 (x^3-{8x}^2+20x) dx

Total area of region R=323+683\large R = \frac{32}{3}+\frac{68}{3}
Total area of region R=1003units2\large R = \frac{100}{3}units^2
Reply 17
Avatar for Kolya
Kolya
OP
17
Please post any corrections. Thank you :smile:
5
(c)
Area=12h(y0+2(y1+y2+etc)+yn)\huge Area= \frac{1}{2}h(y_0 + 2(y_1 + y_2 +etc ) + y_n)
H=b−an=(1−0)5=0.2\huge H = \frac {b-a}{n} = \frac{(1-0)}{5} = 0.2
Area=12(0.2)(1+2(1.246+1.552+1.933+2.408)+3)\huge Area= \frac{1}{2}(0.2)(1+2(1.246+1.552 + 1.933+2.408)+3)
=0.1(1+14.278+3)\huge = 0.1(1+14.278+3)
=0.1(18.278)\huge = 0.1(18.278)
=1.8278\huge = 1.8278
=1.83units2\huge =1.83 {units}^2

7.
(a)
gradoftangent(y=3x−4)=3\huge grad of tangent (y=3x-4) = 3
PQ;grad=−13\huge PQ; grad = -\frac{1}{3}
y=−13x+c\huge y= -\frac{1}{3}x+c
P(2,2)\huge P(2,2)
c=83\huge c = \frac{8}{3}
y=−13x+83\huge y= -\frac{1}{3}x+\frac{8}{3}

Slight changes in the actual text to make it clearer.

Just copy the tex'd ive used into your original post by the quote button and ill delete these once the changes have been made.
9.

(a)
S∞=25\huge S_\infty = 25
(a)(r)=4\huge (a)(r) = 4
a1−r=25\huge \frac{a}{1-r} = 25
a=4R\huge a= \frac{4}{R}
4/r1−r=25\huge \frac{4/r}{1-r} = 25
4r=25(1−r)\huge \frac{4}{r} = 25(1-r)
4r=25−25r\huge \frac{4}{r} = 25 - 25r
4=25r−25r2\huge 4 = 25r-25r^2
25r2−25r+4=0\huge 25r^2 -25r + 4 = 0

(e)

r=45\huge r = \frac{4}{5}
Unparseable latex formula:

\huge 25(1-(\frac{4}{5}}^n) > 24


1−(45)n=2425\huge 1-(\frac{4}{5})^n = \frac{24}{25}
(45)n=1−2425\huge (\frac{4}{5})^n = 1-\frac{24}{25}
(45)n=4100\huge (\frac{4}{5})^n = \frac{4}{100}
nlog1045=log104100\huge n log_{10} \frac{4}{5} = log_{10} \frac{4}{100}
n=log100.04/log100.8\huge n = log_{10} 0.04 / log_{10} 0.8
n=14.425\huge n = 14.425
n=15\huge n = 15

10.
(a)
y=x3−8x2+20x\huge y = x^3-{8x}^2 + 20x
dydx=3x2−16x+20\huge \frac {dy}{dx} = 3x^2 - 16x+20
dy/dx=0\huge dy/dx = 0
3x2−16x+20=0\huge 3x^2-16x+20 = 0
(3x−10)(x−2)\huge (3x-10)(x-2)
x=2,x=103\huge x = 2, x = \frac{10}{3}
a:x=2\huge a: x= 2
b:x=103\huge b: x= \frac{10}{3}

(b)
dydx=3x2−16x+20\huge \frac{dy}{dx}= 3x^2-16x+20
d2ydx2=6x−16\huge \frac{d^2y}{dx^2} = 6x-16
subx=2intod2ydx2\huge sub x = 2 into \frac{d^2y}{dx^2}
d2ydx2=6(2)−16\huge \frac{d^2y}{dx^2} = 6(2) -16
=12−16=−4\huge= 12-16 = -4
d2ydx2=−4\huge \frac{d^2y}{dx^2} = -4
Sod2ydx2<0\huge So \frac{d^2y}{dx^2} < 0
a=maximum\huge a = maximum


Again mostly just tidying up and a few things that needed alil redoing.

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