Currently Further Mathematics Answers

[15/10/2020, 10:10:20 AM] Ceo Oliveloaded: (1) find the value of y for which 5^2x + 1 – 26(5^x + 3)

(2i) if x=x² and y=5/1, Find dy/dx

(2ii) find the value of y if

| y – 1 2|
| | = 4
| 4 3|

(3) A matrix M has the ability of transforming the vertices A(4,0), B(4,4) and C(0,4) of a triangle ABC to their image A(12,0), B(12,8) and C(0,8). Find:

(i) M
(ii) |M|
(iii) The area of triangle A‘B’ C, If the area of triangle ABC is 8 square units

(4a) Show that the point (–2, –1) lies on the circle x² + y² + 2x + 6y + 5 = 0.
Hence, Determine the equation of the tangent to the circle at the point (–2, –1)

(4b) Find the length of the tangent to a circle x² + y² + 6x – 2y – 4 = 0. From the point (4,3)

(5) Two lorries ofasses 50kg and 62kg are traveling on the same rate with speeds 6m/s and 4m/s respectively in the same direction. If there was a collision and the two hook, with one what speed do they move together? Correct your answer to 1 decimal place

(6a) The vertices P, Q, R and S of a quadrilateral have position vectors p, q, r and s respectively relative to the origin O where
p=2i + 3j, q=5i + 3j, r=5i + 6j and s=2i + 6j.
i and j being unit vectors along OX and OY axes respectively.

(i) Find the magnitude of the vectors:
(I) PS and QR
(II) PQ and SR

(ii) Find the area of the quadrilateral PQRS

H(6b) Given that vectors a=2, b=3 and ab=3. Find the angle between a and b

2020 FURTHER MATHEMATICS ANSWER VERIFIED EXPO

OBJ
01-10: EAAEDDBADB
11-20: DBCDDCBBAA
21-30: CEABCCCBDC
31-40: ACBCBEDCAB

41-50: CEDACBADEC

THEORY
(2i)
F(x) = x³ – 6x² + 9x
FD/dx (fx) = 3x² – 12x + 9
Using standard deviation

(2ii) Gradient of f(x) at point A (2,2)
d/dx f(x) = 3x² – 12x + 9

At point A , x=2
= 3(2)² – 12(2) + 9
= 3(4) – 12(2) + 9
= 12 -24 + 9
= -3

(2iii)
Equation of Tangent at point A

y-y¹= m ( x-x¹)
but m= -3
at point A, y¹= 2¹ x¹= 2
y-2=-3(x-2)
y-2 =-3x+6

y=-3x +6 + 2=> y= 8-3x

(5)
Mass,m=150g, g=9.8m/s²
When the lift moves with a constant velocity acceleration
a=o
(i) Reaction,R=w=mg
R=mg
=150×9.8
=1470N

(ii) When the lift moves up word with acceleration 4.5m/s²
F=ma=R-mg
: . R=ma+mg
R=m(a+g)
R=150(4.5+9.8)
=150×14.3

=2145N

(9a)
By formula area of the sector is given by :
A = 1/2rθ , where θ is in radians
Therefore,
θr²/2 = 147

θr² = 294
θ = 294/r² ——-(1)
Also,
Perimeter of the sector will be 56cm
So,
P = θr + 2r
Therefore,
θr + 2r = 56
θr = 56-2r
θ = 56-2r/r

(9b)
Set the RHS of both equations equal since the LHS are equal
:. 294/r² = 56-2r/r
:. 294/r = 56-2r
294 = 56r – 2r²
= 2r² – 56r + 294 = 0
= r² – 28r + 147 = 0
= (r-7)(r-21) = 0
Either:
r-7 = 0
r=7
OR
r-21 = 0
r=21
And,
When r=21
θ=294/21² = 0.7rad
From equ(1) and when r=7

θ=294/7² =6rad

(10a)
2x² — 5x — 3 = 0
Following general quadratic formula
@ + ẞ = — b/a
@ẞ= c/a
x² — 5/2x — 3/2 = 0
— 5/2 = —(@ + ẞ) : (@ + ẞ) = 5/2
— 3/2 = @ẞ
Find 1/@ + 1/ẞ
:. (@ + ẞ)/@ẞ = 5/2 ÷ (—3/2)
5/2 × (— 2/3) = — 5/3
Thus , 1/@ + 1/ẞ = — 5/3

@² + ẞ² = (@ + ẞ)² — 2@ẞ
= (5/2)² — 2(—3/2)
= 25/4 + 3
= (25 + 12)/4 : 37/4
Hence, @² + ẞ² = 37/4

(10b)
Since they have equal roots
D = 0
b² = 4ac
(q + 2)² = 4(q)²
q² + 4q + 4 = 4q²
3q² — 4q — 4 = 0
3q² — 6q + 2q — 4 = 0
3q(q — 2) + 2(q — 2) = 0

thus,

q = — 2/3 and 2

(13ai)
Given: mass ,m =10kg
Force,F = 40N
Time, t = 0.5secs
Impulse, I = Ft = 40×0.5 = 20Ns

(13aii) Ft = m(v-u) where u= 0 (at rest)
20 = 10(v-0)
20 = 10v
V = 20/10 = 2m/s
Final speed = 2m/s

(13aiii)
Given: u=0 ; v=2m/s ; t=0.5secs

S= 1/2(u+v)t
S= 1/2(0+2)×0.5
S= 0.5 metres
Distance = 1/2 metre or 50cm

(13b)
Range R , = Time of flight × Horizontal component of speed

75 = T×35×cos38°
T = 75/35cos38° = 2.719secs

Vertical displacement= vertical component × Time of flight of speed
= Usinθ × T
= 35sin38 × 75/35cos38
= 75Tan38°
= 58.596 metres

~ 58.6 metres

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