Answer to ordinary level paper one 2, 3 ,6

armbruster

If anyone's got the time, I'd appreciate a look at the solutions. thanks!

philologos

wait for the marking scheme?

muggyog

when does that come out or where to you get t??

jessie11

comes out after results

fh041205

armbruster wrote: »

If anyone's got the time, I'd appreciate a look at the solutions. thanks!

Is it maths? If it is and you have the questions I'll might be able to help.

Carsinian Thau

2. (a)

3(4x+5)-2(6x+4)
= 12x + 15 – 12x – 8
= 7

(b)

(i) X^2 – 4x + 1 = 0

X=(-b±√b^2-4ac)/2a
X=(4±√16-4)/2
X=(4±√12)/2
X=(4±√3x4)/2
X=(4±2√3)/2
X=2±√3

(ii) (5^x)/3 = (5^6)/75
(5^x)/3 = (15625)/75
(5^x)/3 = 208.333333333
5^x=625
log5^x=log625
xlog5=log625
x=log625/log5
x=4

(c)

(i) X^2 + 4x + 4
=x^2 + 2x + 2x + 4
=x(x+2) + 2(x+2)
=(x+2)(x+2)

(ii) √(x2 + 4x + 4) + √(x^2 + 2x +1)
= √{(x+2)(x+2)} + √{( x^2 + x + x + 1)}
= √{(x+2)^2} + √{x(x+1)+1(x+1)}
= (x+2) + √{(x+1)(x+1)}
= (x+2) + √{(x+1)^2}
= (x+2) + (x+1)
= 2x + 3

(iii) √(x2 + 4x + 4) + √(x^2 + 2x +1)=x^2
2x + 3 = x^2
X^2 – 2x – 3 = 0
X= (-b ±√b^2 -4ac)/2a
X= (2 ±√2+12)/2
X= (2±√16)/2
X= (2±4)/2
X= (1±2)
But x>0, therefore x=1+2=3


3. (a)
a(x + 5) = 8
ax +5a =8
ax = 8 – 5a
x = (8 – 5a)/a

(b)

(i) X – y =1
X^2 + y^2 = 25

From i: X = 1 + y

From ii: (1+y)^2 + y^2 = 25
1 + 2y + y^2 + y^2 = 25
2 y^2 + 2y = 24
Y^2 + y = 12
Y^2 + y – 12 = 0
Y^2 + 4y – 3y - 12 = 0
Y(y+4)-3(y+4) = 0
(y+4)(Y-3)= 0
Therefore, y = -4 or y = 3

When y= -4; x=1+y=1+ -4=-3
When y= 3; x=1+y=1+3=4

(ii) X-y^2
When y=-4 and x=-3
x-y^2 = -3 – (-4^2)
= -3 – (16)
= -3 – 16
= -19
When y=3 and x=4
x-y^2=4 – (3^2)
=4 – (9)
=-5

(c)

(i) Intersects y axis
 X=0 and y=3
3=0+0+c
C=3

Intersects x axis
 Y=0 and x=-1
0=(-1)^2 – b + c
0=1 – b + c
B=1+c
B=1+3
B=4

(ii) Drop a perpendicular line from the top point of the triangle until it cuts the base in half. You end up with a right angled triangle with a hypotenuse of √(x^2 + 1) and a base of x.
Let y be the vertical height of the triangle
By Pythagoras’s theorem:

[√(x^2 + 1)]^2 = x^2 + y^2
X^2 + 1 = x^2 + y^2
1=y^2
Therefore, y (the vertical height of the triangle) = 1.

6. (a)
G(x)=2x-5
G(x)=19

2x-5=19
2x=24
X=12

(b)
f(x)=3x^2 + 5
f(x+h)=3(x+h)^2 + 5
f(x+h)=3(x^2 + h^2 + 2hx) + 5
f(x+h)=3x^2 + 3h^2 + 6hx + 5

Lim(h->0) [f(x+h)-f(x)]/h
Lim(h->0)[ 3x^2 + 3h^2 + 6hx + 5 – 3x^2 – 5]/h
Lim(h->0)[3h^2 + 6hx]/h
Lim(h->0)[3h + 6x]
=3(0) + 6x
= 6x

(c)

(i) Product rule dictates d/dx((u/v))=[vu’-uv’]/v^2
-> [(1-x^3)(2x-1)-(x^2-x)(-3x^2)]/(1-x^3)^2
=[2x – 1 – 2x^4 + 3x^4 – 3x^3]/(1 + x^6 – 2x^3)
=(x^4 – 3x^3 + 2x – 1)/(x^6 – 2x^3 + 1)

(ii) Slope at (0,0) means f’(0)

(0-3(0)+2(0)-1)/(0-2(0)+1)=-1 Therefore, slope = -1

But slope also equals tan of the angle the line makes with the x axis.
Tan(135)=-1. Therefore, both slopes are equal and are -1.
Therefore, tangent to the curve y = f (x) at the point (0, 0) makes an angle of
135° with the positive sense of the x-axis.