2014 H2 Maths Prelim Papers - RJC P1 solution.pdf
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RAFFLES INSTITUTION 2014 YEAR 6 PRELIMINARY EXAMINATION Qn. [Marks] 1
Solution f ( x)
[4]
x 2 ax b b xa x x
f '( x) 1
b 0 x b x2
x-intercepts
x
b ,0 and b ,0 y
b, 0
0
b, 0
x=0 2(i) [2]
Using G.C. n6
u6 1.0817 1
n7
u7 0.71334 1
Least n 7 (ii)
Since un 20 for all values of n , then
[2]
1 5 20 20 r(20) 1 r 20 4 100 1 100
(iii)
4 l As n , un l and un 1 l , l l 1 3 100
[2]
l 0 (N.A.) or
4 l 1 1 3 100 l 25
1
MATHEMATICS PAPER 1 (9740 / 1) Higher 2
Qn. [Marks] 3(i) [3]
Solution
f ( x) 1 x 1 x 2
2
2 2 2 3 4 1 x 1 x x 1! 2! 1 x 1 2 x 2 3 x 4
1 x 2 x 2 2 x3 3x 4 (ii)
Observe that
[3]
1 x
2 2
1
2 x 2 2 3 x 4 1!
2!
2 3 r!
r
2
,
the coefficient of x 2 r 1 is 1 (r 1) 1 r
4
(r 1)
1 (r 1) x 2 r
1 2 x 2 3x 4 Since f ( x) 1 x 1 x 2
r 1
(r 1).
Let Pn be the statement
[4]
n
sin(2rx)sin x sin nx sin(n 1) x for n
.
r 1
1
L.H.S. = sin(2rx)sin x sin 2 x sin x
When n = 1,
r 1
R.H.S. = sin x sin 2 x L.H.S. = R.H.S.
Hence P1
is
true.
Assume Pk is true for some k
,
k
i.e.
sin(2rx)sin x sin kx sin(k 1) x r 1
To prove that Pk 1 is true, k 1
i.e.
sin(2rx)sin x sin k 1 x sin k 2 x r 1
k 1
L.H.S.
= sin(2rx)sin x r 1 k
sin(2rx )sin x sin 2(k 1) x sin x r 1
sin kx sin(k 1) x sin 2(k 1) x sin x
sin kx sin(k 1) x 2sin(k 1) x cos(k 1) x sin x 2
x2r
Qn. [Marks]
Solution
sin(k 1) x sin kx 2cos(k 1) x sin x sin(k 1) x sin kx sin(k 2) x sin kx
sin k 2 x sin k 1 x R.H.S. OR k 1
L.H.S. sin(2rx )sin x r 1 k
sin(2rx)sin x sin 2(k 1) x sin x r 1
sin kx sin(k 1) x sin 2(k 1) x sin x
1 1 cos(2k 1) x cos x cos(2k 3) x cos(2k 1) x 2 2
1 cos(2k 3) x cos x 2
1 2k 3 1 2k 3 1 2sin x sin x 2 2 2
sin k 2 x sin k 1 x = R.H.S. Hence Pk is true Pk 1 is true, and since P1 is true, by mathematical induction , Pn is true for all n
[3]
.
sin 3x sin 4 x dx sin x 2
4
=
sin(2rx) dx.
4
=
3
2
r 1
2
sin 2 x sin 4 x sin 6 x dx.
4
1 1 1 2 = cos 2 x cos 4 x cos6 x 4 6 2 4
=
1 1 1 1 1 0 1 1 1 0 = 6 2 4 6
3
Qn. [Marks] 5
Solution Equation of the new curve is y 3
[4]
1 . 2x 1
(0, 4)
x
[4] y
y (0, 4)
(0, 4)
y=3
y=3
x
From the graph, for f ( x) f x , x 6(i) [1] (ii) [2]
a
x
1 1 2 or x . 2 2 3
b represents the length of projection of OA onto OB. b 2
3a b 102 100 2 2 9 a b 6a b 100 6a b 9(3)2 (5) 2 100 6
Therefore a b 1. (iii) [2]
Let N be the foot of the perpendicular from A to the line OB. ON a
b 1 . b 5
Using Pythagoras Theorem,
4
Qn. [Marks]
Solution 2
1 224 AN OA ON 3 . 25 5 2
AN
2
2
2
224 4 14 2.9933 2.99 (3sf). 25 5
1 Area of triangle OAB OB AN 2 14 7.48 (3sf). 2
(iv)
(a 2b) a k (2a 3b) a for some constant k.
[4]
Now, a 0, b 0 and a b 1 a b , so a and b are non-zero and non-parallel vectors.
( 1)a 2b ka 3kb ( 1 k )a (3k 2)b Hence 3k 2 0 k 7(a)
2 5 and k 1 . 3 3
C1 is a circle centred at (0,0) with radius 5.
[2]
C2 :
x2 y2 1 is an ellipse centred at (0,0) with length of the 100 100 a b 10 10 ) and vertical axis 2( ) . horizontal axis 2( a b
Note : a b length of the horizontal axis > length of vertical axis To get 4 points of intersection, we need :
10 10 5 0 a 4 and 5b 4 a b OR
x2 y 2 x2 y2 1. 1 with C2 : Compare C1 : x y 25 100 100 25 25 a b 2
2
For them to intersects at 4 points, 100 100 25 and 25 b a
b 4 and 0 a 4 since a 0 is given.
5
Qn. [Marks] (b) [3]
Solution
x 2 y 2 25
C1 :
x2 y2 C2 : 2 10 10 3
2
1
y 25 x 2 100 x 2 y 3
C1 and C2 intersect at x 3.9528 (5 s.f.) (from GC) Thus area of the required region
3.9528 100 x 2 3.9528 2 dx 25 x 2 dx 10 5 3 22.3 (3 s.f.) OR x 2 y 2 25
C1 :
C2 : x 2 9 y 2 100
x 25 y 2
x 100 9 y 2
C1 and C2 intersect at y 3.0619 (5 s.f.) (from GC) Thus area of the required region 2 0
3.0619
100 9 y 2 25 y 2 dy
22.3 (3 s.f.)
(c) [4]
x 2 y 2 25
C1 : C2 :
x2 y2 102 10 2
2
1
Required Volume 5 4 3 x 2 dy 5 5 3
5
5
100 4 y dy 500 3 2
x 2 25 y 2 x 2 100 4 y 2
Note :
4 3 5 is the volume of sphere 3 formed when rotating the circle
5
4 500 100 y y 3 3 5 3
about the y axis.
500 500 500 500 500 3 3 3 500
6
Qn. [Marks] 8(a)
Solution
z4 8 3 i 0
[6]
z4 8 3 i 5
16 e 6
i
5 2k i 6 16 e
5 1 k i 24 2
z 2e
Im
, k 0 , 1 , 2
2
–2
2
–2
2 e 24 i 7
w2 aw * b 0
(b) [6]
Re
w aw * b * 0* w * aw * * b* 0 2
2
w *2 a w * * b 0 , a* a and b* b since a real. 2 Hence, w* is a root of z az * b 0 .
z 2 6 z * 9 0
x iy 2 6 x iy 9 0 x 2 y 2 2ixy 6 x 6iy 9 0 x 2 y 2 6 x 9 2 y x 3 i 0 Compare imaginary parts, y 0 or x 3 .
Consider real parts: 2 When y 0 , x 6 x 9 0 which gives x 3
When x 3 , 32 y 2 18 9 0 giving y 6 Hence z 3 , 3 6i , 3 6i
7
and b are
Qn. [Marks] 9 (i) [3]
Solution
5 0 5 OA 0 , OC 3 , OR 3 , so 0 0 2 0 5 6 n AR CR 3 0 10 . 2 2 15 6 5 6 Therefore 1 : r 10 0 10 30. 15 0 15
(ii) [2]
The angle between
1 is cos
1
0 and the horizontal base, which has normal 0 , 1
6 0 10 . 0 15 1 15 cos 1 142.136 . 36 100 225 0 0 1 19
The acute angle is 180 142.136 37.9 . (1 dec. pl.) (iii)
By symmetry, the acute angle between 2 and the horizontal base is
[1]
also 37.864 Hence the angle between 1 and 2 is 2(37.864 ) =
75.7 (1 dec. pl.). (iv) [3]
0 0 0 OP 0 , OS 3 , SP 3 , therefore 2 2 0 0 0 0 OX OS SX 3 3 3(1 ) . 2 0 2
l XY :
l XY :
5 r OY XY , , and with OY 2 , 1 5 5 r 2 3 1 , . 1 1
8
Qn. [Marks]
Solution 5 r 3 , . 2
Clearly lOR : (v) [4]
When the lines XY and OR intersect, 5 5
2 1
1
5
3 2 3 1 .
Therefore,
1 3 2 (3 1) 2 1 Solving gives
1 2 1 , and . 3 3 3
OW OR, so OW : OR 2 : 3. 10(i) [3]
Since x tan , y sec and tan2 1 sec2
x 2 1 y 2 for x 0 and y 1 y C
x (ii)
y
[7]
Q cot ,0
x tan , y sec ;
x
dy sec tan tan sin dx sec2 sec
Equation of tangent at P is y sec sin x tan
9
Qn. [Marks]
Solution At Q , 0 sec sin x tan .
sin 1 sin 2 1 cos 2 x cos sin cos sin cos sin cos x cot Tangent at P intersects x-axis at Q cot ,0
Equation of normal at P is y sec At R , 0 sec
x
1 x tan sin
1 x tan sin
sin tan 2 tan cos
Normal at P intersects x-axis at R 2 tan ,0 Triangle PQR is a right-angled triangle in circle, so QR is a diameter. 2 tan cot QR A 2 2 2
1/ tan tan 2
2
2
2
1 A tan [shown] 2 tan (iii) [2]
2
1 1 1 2 2 t 2 t 1 2 2 t 1 2 4t 4t 2t 2
1 t 0 2t
[1]
For 0
2
, t tan > 0
1 So minimizing A tan over 0 is equivalent to 2 2 tan 2
2
1 minimizing A t for t 0 2t 2
1 A tan 2 from above result. 2 tan Hence the minimum value of A = 2
10
Qn. [Marks]
Solution
11(i) When x = 0, [1] (ii)
d2 y d2 y 1 y and y 1 when x 0 ) . (since dx 2 dx 2
Differentiating (1) with respect to x twice gives
[2]
d3 y dy 3 dx dx
and
d4 y d2 y . dx 4 dx 2
d2 y dy 1 [from (i)] When x 0, 1 (given) and dx dx 2 Hence, when x 0,
d3 y d4 y 1 and 4 1. dx3 dx
Maclaurin’s theorem gives
1 1 1 2 1 2 1 4 x x x x 1! 2! 3! 4! 1 2 1 3 1 4 y 1 x x x x 2 6 24 y 1
(iii)
Differentiating
[4]
dy u implicitly with respect to x gives dx
d 2 y 1 1 du dx 2 2 u dx
y
1 dx du 2 dy dx
Hence,
(since
1 dx dy d2 y = ) y and u 2 dx dx u dy
du 2 y. (shown) dy
Alternative Method 2
dy dy u u . dx dx 2
dy Differentiating u implicitly with respect to x gives dx 2 du du dx du dy d y 2 y. (shown) 2( y ) 2 2 dy dx dy dx dx dx
Integrating
du 2 y with respect to y gives dy
u 2 y dy y 2 A, where A is an arbitrary constant.
11
Qn. [Marks]
Solution Using u 1 when y 1, we have A 2, Hence, u 2 y 2 .
(iv)
Substituting u 2 y 2 into
[4]
dy u gives dx
dy 2 y2 , dx
which we can integrate via
1
dy 1 2 y 2 dx 1 dy 1dx 2 y2 y sin 1 xB 2 y 2 sin x B , where B is an arbitrary constant Using y 1 when x 0, we have B sin 1
1 . 2 4
Hence y 2 sin x , i.e. P 2 and Q . 4 4 (v) [2]
y 2 sin x 4 2 sin x cos cos x sin 4 4 sin x cos x (since cos
= sin
4 x2 x4 x3 x 1 3! 2! 4! 2 3 4 x x x 1 x 2 6 24
=
1 ) 2
4 (from MF15)
12
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