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Mock p3 Solved

A level pure mathematic hard paper solved

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0% found this document useful (0 votes)

26 views19 pages

Mock p3 Solved

A level pure mathematic hard paper solved

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warriormahim

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PURE MATHEMATICS

Mensuration
Volume of sphere = 4
3
πr 3

Surface area of sphere = 4πr 2

Volume of cone or pyramid = 13 × base area × height

Area of curved surface of cone = πr × slant height

Arc length of circle = rθ ( θ in radians)

Area of sector of circle = 12 r 2θ ( θ in radians)

Algebra
For the quadratic equation ax 2 + bx + c = 0 :

−b ± b 2 − 4ac
x=
2a
For an arithmetic series:
un = a + (n − 1)d , S n = 12 n( a + l ) = 12 n{2a + (n − 1) d }

For a geometric series:

a(1 − r n ) a
un = ar n −1 , Sn =
1− r
(r ≠ 1) , S∞ =
1− r
( r <1 )

Binomial series:
n  n  n
(a + b) n = a n +   a n −1b +   a n − 2b 2 +   a n −3b3 + K + b n , where n is a positive integer
1  2  3
n n!
and   =
 r  r!(n − r )!
n(n − 1) 2 n(n − 1)(n − 2) 3
(1 + x) n = 1 + nx + x + x + K , where n is rational and x < 1
2! 3!

2
Trigonometry
sin θ
tan θ ≡
cos θ
cos 2 θ + sin 2 θ ≡ 1 , 1 + tan 2 θ ≡ sec 2 θ , cot 2 θ + 1 ≡ cosec 2 θ
sin( A ± B) ≡ sin A cos B ± cos A sin B
cos( A ± B) ≡ cos A cos B m sin A sin B

tan A ± tan B
tan( A ± B ) ≡
1 m tan A tan B
sin 2 A ≡ 2sin A cos A
cos 2 A ≡ cos 2 A − sin 2 A ≡ 2cos 2 A − 1 ≡ 1 − 2sin 2 A
2 tan A
tan 2 A ≡
1 − tan 2 A
Principal values:
− 12 π ⩽ sin −1 x ⩽ 12 π , 0 ⩽ cos −1 x ⩽ π , − 12 π < tan −1 x < 12 π

Differentiation
f( x ) f ′( x )

xn nx n −1
1
ln x
x
ex ex
sin x cos x
cos x − sin x
tan x sec 2 x
sec x sec x tan x
cosec x − cosec x cot x
cot x − cosec 2 x
1
tan −1 x
1 + x2
du dv
uv v +u
dx dx
du dv
v −u
u dx dx
v v2
dy dy dx
If x = f(t ) and y = g(t ) then = ÷
dx dt dt

3
Integration
(Arbitrary constants are omitted; a denotes a positive constant.)

f( x ) ∫ f( x ) dx
x n +1
xn (n ≠ −1)
n +1
1
ln x
x
ex ex
sin x − cos x
cos x sin x
sec 2 x tan x
1 1 x
tan −1  
x + a2
2
a a
1 1 x−a
ln ( x > a)
x − a2
2
2a x + a

1 1 a+x
a − x2
2
ln
2a a − x
( x < a)

dv du
∫ u dx dx = uv −∫ v dx dx
f ′( x)
∫ f ( x) dx = ln f ( x)

Vectors
If a = a1i + a2 j + a3k and b = b1i + b2 j + b3k then

a.b = a1b1 + a2b2 + a3b3 = a b cos θ

4
PMT

Solve the equation x − 2 =

1
1 3x . [3]

2 The sequence of values given by the iterative formula

xn xn3 + 100
xn+1 =
2 xn3 + 25
,

with initial value x1 = 3.5, converges to !.

(i) Use this formula to calculate ! correct to 4 decimal places, showing the result of each iteration
to 6 decimal places. [3]

(ii) State an equation satisfied by ! and hence find the exact value of !. [2]

3
ln y

(0.64, 0.76)

(1.69, 0.32)

x2
O

The variables x and y satisfy the equation y = Ae−kx , where A and k are constants. The graph of ln y
2

against x2 is a straight line passing through the points 0.64, 0.76 and 1.69, 0.32, as shown in the
diagram. Find the values of A and k correct to 2 decimal places. [5]

4 The polynomial ax3 − 20x2 + x + 3, where a is a constant, is denoted by p x. It is given that 3x + 1
is a factor of p x.

(i) Find the value of a. [3]

(ii) When a has this value, factorise p x completely. [3]

5
y

x
–a 3a

The diagram shows the curve with equation

x3 + xy2 + ay2 − 3ax2 = 0,
where a is a positive constant. The maximum point on the curve is M . Find the x-coordinate of M in
terms of a. [6]

© UCLES 2013 9709/32/M/J/13

PMT

1
6 (i) By differentiating , show that the derivative of sec x is sec x tan x. Hence show that if
cos x
y = ln sec x + tan x then = sec x.
dy
[4]
dx

(ii) Using the substitution x = ï3 tan 1, find the exact value of
3
Ô
1
3 + x2
dx,
1

expressing your answer as a single logarithm. [4]

7 (i) By first expanding cos x + 45Å, express cos x + 45Å − ï2 sin x in the form R cos x + !,
where R > 0 and 0Å < ! < 90Å. Give the value of R correct to 4 significant figures and the value
of ! correct to 2 decimal places. [5]

(ii) Hence solve the equation

cos x + 45Å − ï2 sin x = 2,
for 0Å < x < 360Å. [4]

in the form 2 + +
1 A B C
x 2x + 1 x 2x + 1
8 (i) Express 2
. [4]
x

(ii) The variables x and y satisfy the differential equation

y = x2 2x + 1
dy
,
dx
and y = 1 when x = 1. Solve the differential equation and find the exact value of y when x = 2.
Give your value of y in a form not involving logarithms. [7]

9 (a) The complex number w is such that Re w > 0 and w + 3w* = iw2 , where w* denotes the complex
conjugate of w. Find w, giving your answer in the form x + iy, where x and y are real. [5]

(b) On a sketch of an Argand diagram, shade the region whose points represent complex numbers
Ï which satisfy both the inequalities Ï − 2i ≤ 2 and 0 ≤ arg Ï + 2 ≤ 14 0. Calculate the greatest
value of Ï for points in this region, giving your answer correct to 2 decimal places. [6]

10 The points A and B have position vectors 2i − 3j + 2k and 5i − 2j + k respectively. The plane p has
equation x + y = 5.

(i) Find the position vector of the point of intersection of the line through A and B and the plane p.
[4]

(ii) A second plane q has an equation of the form x + by + cÏ = d, where b, c and d are constants.
The plane q contains the line AB, and the acute angle between the planes p and q is 60Å. Find
the equation of q. [7]

© UCLES 2013 9709/32/M/J/13

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12
, ,

10 - x

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A container in the shape of a cuboid has a square base of side x and a height of (10 - x) . It is given
that x varies with time, t , where t 2 0 . The container decreases in volume at a rate which is inversely
proportional to t .
1 1 20
When t = 10 , x= 2 and the rate of decrease of x is 37 .

(a) Show that x and t satisfy the differential equation

dx -1
= . [5]
dt 2t `20x - 3x 2j

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14
, ,

10 The equations of two straight lines are

r = i + j + 2ak + m (3i + 4j + ak) and r =- 3i - j + 4k + n (- i + 2j + 2k) ,

where a is a constant.

(a) Given that the acute angle between the directions of these lines is 14 r , find the possible values
of a. [6]

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15
, ,

(b) Given instead that the lines intersect, find the value of a and the position vector of the point of
intersection. [5]