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MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

Compulsory Part Paper 2

Question No. Key Question No. Key
1. D 31. A
2. B 32. D
3. D 33. C
4. A 34. D
5. B 35. A

6. B 36. C
7. A 37. B
8. C 38. C
9. A 39. D
10. C 40. B

11. C 41. D
12. D 42. C
13. C 43. D
14. B 44. A
15. C 45. A

16. A
17. D
18. C
19. D
20. B

21. A
22. B
23. B
24. C
25. B

26. B
27. A
28. B
29. D
30. A

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MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

Solutions to Paper 2
1. D 5. B
82n + 222  93n + 333 = (23)2n + 222  (32)3n + 333 32  3  p 12  p
f(3) = =
= 26n + 666  36n + 666 q q
= (2  3)6n + 666 ( 3) 2  ( 3)  p 6 p
f(3) = =
= 66n + 666 q q
f(3)  f(3) = 3
2. B 12  p 6 p
 = 3
4x2  12xy + 9y2  2x + 3y q q
= (2x  3y)2  (2x  3y) 6
= 3
= (2x  3y)(2x  3y  1) q
q = 2
3. D
b 2c 6. B
 =5
ac ca Maximum error
b 2c
 =5 1
ac ac =  2 cm
2
b  2c
=5 = 1 cm
ac
Least possible length of a side of the square
b  2c = 5a  5c
= (10  1) cm
3c = 5a  b
5a  b = 9 cm
c= Least possible area of the square
3
= 92 cm2
4. A = 81 cm2
Rewrite the given equation as:
7. A
3 p  q  1  2 p
 Solving 4x < 6  x:
1  2 p  p  q  6
3x < 6
5 p  q  1 .................................... (1)
i.e.  x > 2 ........................... (1)
3 p  q  7 .................................. (2) Solving 5(x + 1) > 17 + x:
(1)  3  (2)  5: 8q = 32
5x + 5 > 17 + x
q = 4
4x > 12
Alternative method:
x > 3 ....................... (2)
Rewrite the given equation as:
∵ x must satisfy (1) and (2).
3 p  q  p  q  6 ∴ The solutions are x > 3.

1  2 p  p  q  6
2 p  2q  6 .............................. (3)
i.e. 
3 p  q  7 .................................. (4)
(3)  3  (4)  2: 8q = 32
q = 4

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MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

8. C 8 
2 n   3n
2
mx  4x + m = 3 2m  3n 3
=  
2
mx  4x + m  3 = 0 4m  n 8 
4 n   n
∵ The equation has equal roots. 3 
∴ (4)2  4m(m  3) = 0 25
n
16  4m2 + 12m = 0 = 3
35
m2  3m  4 = 0 n
3
(m  4)(m + 1) = 0 5
m = 4 or 1 =
7
∴ (2m + 3n) : (4m + n) = 5 : 7
9. A
Let M and C be the marked price and the cost 12. D
respectively. Let T(n) be the number of dots in the nth
Selling price = M(1  20%) = 0.8M pattern.
Selling price = C(1  50%) = 0.5C T(1) = 3
∴ 0.8M = 0.5C T(2) = T(1) + 2(1) = 3 + 2 = 5
M = 0.625C T(3) = T(2) + 2(2) = 5 + 4 = 9
CM T(4) = T(3) + 2(3) = 9 + 6 = 15
r% =  100%
C T(5) = T(4) + 2(4) = 15 + 8 = 23
r C  0.625C
= T(6) = T(5) + 2(5) = 23 + 10 = 33
100 C
T(7) = T(6) + 2(6) = 33 + 12 = 45
r
= 0.375 ∴ The number of dots in the 7th pattern is
100
r = 37.5 45.

10. C 13. C
y = a(x + 2)2  b x2 kx 2
Since w  ,w= , where k  0.
= a(x2 + 4x + 4)  b y y
= ax2 + 4ax + 4a  b kx 2
w=
Coefficient of x2 = a y
∵ The graph opens upward. 2
x 1
∴ a>0 =
w y k
From the figure, 2
 x2  2
y-intercept of the graph < 0   =  1 
4a  b < 0 w y  k
 
b > 4a x4 1
2
= 2
w y k
11. C x4
m:n=8:3 ∴ must be constant.
w2 y
m 8
=
n 3
8
m= n
3

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MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

14. B 15. C
In △BCE, Each interior angle of ABCDEFGH
BE2 + CE2 = (52 + 122) cm2 = 169 cm2 (8  2)  180
=
BC2 = 132 cm2 = 169 cm2 8
∵ BE2 + CE2 = BC2 = 135
∴ BEC = 90 In △ABH,
Let AB = x cm. ∵ AB = AH
Area of ABCD = 156 cm2 ∴ ABH = AHB
( AB  DC )CE ABH + AHB + BAH = 180
= 156 cm2
2 ABH + ABH + 135 = 180
( x  6)(12) 2ABH = 45
= 156
2 ABH = 22.5
x + 6 = 26 ∵ △CDB  △ABH (SAS)
x = 20 ∴ CBD = AHB = ABH = 22.5
A F E 5 cm B CBD + DBH + ABH = ABC
22.5 + DBH + 22.5 = 135
12 cm 13 cm DBH = 90
∵ △ABH  △CDB
∴ BH = DB
D 6 cm C ∴ BHD = BDH = p
Construct DF  AB. In △BDH,
Then CDFE is a rectangle. BDH + BHD + DBH = 180
DF = CE = 12 cm p + p + 90 = 180
FE = DC = 6 cm 2p = 90
AF = AB  BE  FE p = 45
= (20  5  6) cm AHB + BHG = AHG
= 9 cm 22.5 + BHG = 135
In △ADF, BHG = 112.5
AD2 = AF2 + DF2 In △BGH,
AD = 9 2  12 2 cm BGH + GBH + BHG = 180
= 15 cm q + r + 112.5 = 180
q + r = 67.5
p + q + r = 45 + 67.5
= 112.5

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MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

16. A 18. C
Let r and h be the base radius and the height of AF : FB = 1 : 3
the right circular cone respectively. AF 1
=
Percentage change in the volume FB 3
1 1 FB = 3AF
π[r (1  10%)]2 h(1  15%)  πr 2 h
AB : DC = 1 : 2
= 3 3  100%
1 2 AB 1
πr h =
3 DC 2
= (1.028 5  1)  100% AF  FB 1
=
= 2.85% DC 2
∴ The volume is increased by 2.85%. AF  3 AF 1
=
DC 2
17. D 4 AF 1
=
∵ QR = UR DC 2
AF 1
∴ UQR = QUR = 30 =
DC 8
∵ PQ // UR
∵ △AEF ~ △CED (AAA)
∴ PQU = QUR = 30 AE AF 1
In △QRS, ∴ = =
CE CD 8
QRS + SQR = PSQ Area of ADE AE
=
QRS + 30 = 80 Area of CED CE
QRS = 50 Area of ADE 1
=
∵ RT is the angle bisector of PRQ. 64 cm 2 8
1 Area of △ADE = 8 cm2
∴ QRT = QRS
2 Area of ABC AB
=
1 Area of ADC DC
=  50
2 Area of ABC 1
=
= 25 (8  64) cm 2 2
In △QRT, Area of △ABC = 36 cm2
PTR = PQU + UQR + QRT Area of ABCD
= 30 + 30 + 25 = area of △CED + area of △ADE +
= 85 area of △ABC
= (64 + 8 + 36) cm2
= 108 cm2

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MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

19. D 21. A
I. ∵ BD is an axis of symmetry of ABCD. tan (90   ) cos (360   ) sin (180   )

∴ AB = BC and AD = CD. sin 210 cos120
∵ ABCD is a parallelogram. 1
  cos
tan  sin 
∴ AB = CD and AD = BC. = 
sin (180  30) cos (180  60)
i.e. AB = BC = CD = AD
1
∴ I is true.   cos
sin 
II. From I, ABCD is a rhombus. sin 
= cos 
∴ AC  BD  sin 30  cos 60
∴ II is true. cos  sin 
2
= 
III. ∵ ABCD is a rhombus and AC = BD. 1 1
sin 
∴ ABCD is a square. 2 2
In △BEM and △CEN, 2 cos 2
=  2 sin 
BM = CN sin 
ME = NE 2 cos 2   2 sin 2 
=
BE = CE sin 
∴ △BEM  △CEN (SSS) 2(cos   sin 2  )
2
=
∴ III is true. sin 
∴ I, II and III are true. 2
=
sin 

20. B
∵ △APQ ~ △DQC (AAA) 22. B
C
PQ AQ
∴ =
QC DC 103
D
AQ
=
AD
AQ A B
=
AQ  DQ Join AD and BD.
2 DQ In △BCD,
=
2 DQ  DQ CBD + BDC + BCD = 180
2 DQ CBD + BDC + 103 = 180
=
3DQ CBD + BDC = 77
2  
BDC : CBD = BC : CD = 4 : 3
=
3 4
In △CPQ, ∴ BDC = 77 
43
PQ = 44
tan PCQ =
QC ADB = 90
2 ABC + ADC = 180
=
3 ABC + ADB + BDC = 180
PCQ = 34, cor. to the nearest degree
ABC + 90 + 44 = 180
ABC = 46

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MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

Alternative method: 24. C

C I. L1: y + ax = b
103 y = ax + b
D
L2: y + cx = d
y = cx + d
A B From the figure,
Join AC and BD. slope of L1 < slope of L2 < 0
Let BAC = x. a < c < 0
BAC : CBD = BC : CD
  a>c>0
x 4 ∴ I is true.
= II. Substitute y = 0 into y + cx = d.
CBD 3
3 0 + cx = d
CBD = x
4 x=
d
BDC = BAC = x c
In △BCD, ∴ x-intercept of L2 =
d
BDC + BCD + CBD = 180 c
3 From the figure,
x + 103 + x = 180 x-intercept of L2 < 1
4
7 d
<1
x = 77
4 c
x = 44 d < c (∵ c > 0)
ACB = 90 ∴ II is not true.
In △ACB, III. Substitute x = 1 and y = 0 into y + ax = b.
ACB + BAC + ABC = 180 0 + a(1) = b
90 + 44 + ABC = 180 a=b
ABC = 46 ∵ d<c
∴ d > c
23. B ad>ac
y
b  d > a  c (∵ a = b )
P(x , 1) ∴ III is true.
150 ∴ Only I and III are true.
x
Q O

Refer to the figure.

PQ
tan POQ =
OQ
1
tan (180  150) =
OQ
1
tan 30 =
OQ
1 1
=
3 OQ
OQ = 3
x = OQ =  3

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MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

25. B 27. A
Substitute x = 1 and y = –2 into I. 4x2 + 4y2 – 4kx – 4ky + k2 = 0
6x + ky – 16 = 0. k2
x2 + y2 – kx – ky + =0
6(1) + k(–2) – 16 = 0 4
–2k = 10 Coordinates of the centre of C
k = –5  k k 
=  , 
6 6 6  2 2 
Slope of L2 = – =– =
k 5 5 k k
= , 
∵ L1  L2 2 2
1 1 5 ∴ I is not true.
∴ Slope of L1 = = = 
slope of L2 6 6 II. Radius of C
5 2 2 2
k k k
The equation of L1 is = 
    
5 2 2 4
y – (–2) =  (x – 1) 2
6 k k2 k2
=    
6y + 12 = 5x + 5 2 4 4
5 x + 6y + 7 = 0 2
k
=  
2
26. B
k
The locus of P is a circle with MN as a =
2
diameter.
Area of C
Coordinates of the centre of the locus of P 2
k
 8 4 =  
=  ,  2
 2 2
πk 2
= (4 , 2) =
4
The centre of the locus of P is the mid-point of
∴ II is true.
MN.
III. Distance between the origin and the centre
a2
∴ = 2 2 2
2 k  k 
=   0    0
a + 2 = 4 2  2 
a = 6 k
2
= 2 
2
2
k
>  
2
k
=
2
∴ The origin lies outside C.
∴ III is not true.
∴ Only II is true.

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MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

28. B When x  5, the data are arranged in

The required expected number of tokens ascending order as follows:
17 8 5 3, 4, 5, x, x
=0 + 60  + 120 
30 30 30 Median = 5
= 36 ∴ The median is not less than 3.
∴ II must be true.
29. D III. When 1  x  4,
Denote the 5 balls as A, B, C, D and E. mode  4
The 1st ball drawn 45
upper quartile = = 4.5
A B C D E 2
∴ The mode and the upper quartile are
The 2nd ball drawn

A     
B      not equal.
C      When x = 5,
     mode = 5
D
x5 55
E      upper quartile = = =5
2 2
( represents the two balls are of the same ∴ The mode and the upper quartile are
colour and  represents the two balls are of equal.
different colours) When x  6,
The required probability mode = x
20 xx
= upper quartile = =x
25 2
4 ∴ The mode and the upper quartile are
=
5 equal.
∴ III is not necessarily true.
30. A ∴ Only II must be true.
I. Mean
3 x  x  45 31. A
=
5
1
2 x  12 The graph of y = f(x) is reduced to of the
= 2
5
∵ 2x + 12 is an even number and is not original along the x-axis to become the graph
of y = f(2x). The y-intercepts of the graphs of
necessarily divisible by 5.
2 x  12 y = f(x) and y = f(2x) are the same. The
∴ is not necessarily an integer. x-intercepts of the graphs of y = f(2x) and
5
∴ I is not necessarily true. y = g(x) are the same.
II. When 1  x  3, the data are arranged in y-intercept of the graph of y  g ( x)
ascending order as follows: y-intercept of the graph of y  f (2 x)
x, x, 3, 4, 5 8
=
Median = 3 4
=2
When x = 4, the data are arranged in
The graph of y = f(2x) is enlarged to 2 times
ascending order as follows:
the original along the y-axis to become the
3, 4, x, x, 5
graph of y = g(x).
Median = x = 4
∴ g(x) = 2f(2x)

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MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

32. D 35. A
3 2
x  log3 y = 2x I. Let a and d be the first term and the
(log3 y)3  log3 y2 = 2 log3 y common difference of the sequence
(log3 y)3  2 log3 y = 2 log3 y respectively.
(log3 y)3  4 log3 y = 0 x200 + 300 = x100
(log3 y)[(log3 y)2  4] = 0 a + (200  1)d + 300 = a + (100  1)d
(log3 y)(log3 y + 2)(log3 y  2) = 0 100d = 300
log3 y = 0 or log3 y = 2 or log3 y = 2 d = 3
y = 1 or y = 32 or y = 32 x18 + x20 = 92
1 a + (18  1)d + a + (20  1)d = 92
y = 1 or y= or y=9
9 2a + 36d = 92
2a + 36(3) = 92
33. C 2a = 200
5  45 + 83 a = 100
= (22 + 1)(22)5 + (23)3 ∴ The first term of the sequence is 100.
= (22 + 1)210 + 29 ∴ I is true.
= 212 + 210 + 29 II. x1 + x2 + x3 + … + x2 018
= 1011000000000 2 2 018
= [2(100)  (2 018  1)(3)]
2
34. D = 5 903 659
Note that i2 = 1, i3 = i and i4 = 1. < 5 900 000
For any positive integer n, = 5.9  106
i4n = 1, i4n + 1 = i, i4n + 2 = 1 and i4n + 3 = i. ∴ II is true.
i + 2i2 + 4i3 + 8i4 + 16i5 + … + 65 536i17 III. Let xk be a positive term.
= i + 2i2 + 22i3 + 23i4 + 24i5 + … + 216i17 xk > 0
= i + 2(1) + 22(i) + 23(1) + 24i + 25(1) + 100 + (k  1)(3) > 0
26(i) + 27(1) + 28i + 29(1) + 210(i) + 100  3k + 3 > 0
211(1) + 212i + 213(1) + 214(i) + 215(1) + 3k > 103
216i 103
k< (= 34.33…)
= 2 + 23  25 + 27  29 + 211  213 + 215 + 3
∵ k is an integer.
(1  22 + 24  26 + 28  210 + 212  214 + 216)i
∴ The greatest value of k is 34.
 2[1  (2 2 )8 ] 1[1  (2 2 ) 9 ]
= + i ∵ Common difference < 0
1  (2 2 ) 1  (2 2 )
∴ x34 is the smallest positive term of the
= 26 214 + 52 429i
sequence.
∴ III is not true.
∴ Only I and II are true.

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MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

36. C 38. C
2
y = 4x + 24x + 36 Let AB = x.
y = 4(x2 + 6x + 9) In △ABE,
y = 4(x + 3)2 ∵ AE = BE
y = 2(x + 3) or y = 2(x + 3) (rejected) ∴ BAE = ABE = 35
y = 2x + 6 BAE + ABE + AEB = 180
∵ The slope and the intercept on the vertical 35 + 35 + AEB = 180
AEB = 110
axis of the graph representing the linear
By the sine formula,
relation between x and y are 2 and 6
AE AB
respectively. =
sin ABE sin AEB
∴ The answer is C. x sin 35
AE =
sin 110
37. B ∵ AD // BC
 x  y  4  0 .................................... (1) ∴ DAE = AEB = 110

 x  3 y  6  0 ................................. (2) In △ADE, by the cosine formula,
(2)  (1): 2y  2 = 0 DE2 = AE2 + AD2 – 2  AE  AD  cos DAE
2y = 2 DE
y=1 2
 x sin 35   x sin 35 
2
Substitute y = 1 into (1). =    x  2  x cos110
 sin 110   sin 110 
x+14=0
sin 2 35 2 sin 35 cos110
x=3 = x 1
2
sin 110 sin 110
∴ (3 , 1) is the point of intersection of
 1.337 946 99x
x + y  4 = 0 and x + 3y  6 = 0.
EC = BC  BE
y
x sin 35
=x
x+y4=0 sin 110
 sin 35 
R = x 1  
 sin 110 
(3 , 1)
x + 3y  6 = 0 In △CDE, by the cosine formula,
x cos CDE
O
CD 2  DE 2  EC 2
∵ mx + y + 1 attains its maximum value at =
2  CD  DE
(3 , 1) only. 2
2  
2 sin 35 
∴ Slope of mx + y + 1 = 0 x  (1.337 946 99 x)   x1  
  sin 110 
< slope of x + y – 4 = 0 
2 x(1.337 946 99 x)
m 1
 <   sin 35 
2
1 1 2
1  (1.337 946 99)  1  
m>1 =  sin 110 
2(1.337 946 99)
CDE = 10, cor. to the nearest degree

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OXFORD UNIVERSITY PRESS
MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

39. D ∵ △OTC ~ △ABC (AAA)

B C TC OC
Q ∴ =
BC AC
A D 8 cm (6  4) cm
=
BC (6  6  4) cm
BC = 12.8 cm
BT = BC  TC
H
R G = (12.8  8) cm
P = 4.8 cm
F E
Let R be the mid-point of FG.
41. D
Join PR and QR.
∵ y-coordinates of A and C are equal.
Then QRP = 90 and QPR is the angle
∴ AC is a horizontal line.
between PQ and the plane EFGH.
AB = (73  1) 2  (98  2) 2 = 120
Let x be the length of each side of the cube.
1 BC = (73  145) 2  (98  2) 2 = 120
Then PR = x and QR = x.
2 ∵ AB = BC and AC is a horizontal line.
In △PQR, ∴ H and Q lie on the vertical line x = 73.
QR Let (73 , s) and (73 , t) be the coordinates of H
tan QPR =
PR and Q respectively.
x
tan  = ∵ AH  BC
1
x ∴ Slope of AH  slope of BC = 1
2
=2 s2 98  2
 = 1
73  1 73  145
s  2 = 54
40. B
s = 56
Let D be the mid-point of BC.
4 cm C Coordinates of D
D
O  73  145 98  2 
8 cm = , 
 2 2 
A
T = (109 , 50)
∵ QD  BC
∴ Slope of QD  slope of BC = 1
B
t  50 98  2
Let O be the centre of the circle and r cm be  = 1
73  109 73  145
the radius of the circle.
t  50 = 27
Join OT.
t = 23
∵ BC is the tangent to the circle at T.
HQ = s – t
∴ OT  BC
= 56 – 23
In △OTC,
= 33
OT2 + TC2 = OC2
r2 + 82 = (r + 4)2
r2 + 64 = r2 + 8r + 16
48 = 8r
r=6
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OXFORD UNIVERSITY PRESS
MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

42. C 44. A
The required number of ways Note that the lowest score and the lower
= P35  P37 quartile of the distribution are 56 marks and
= 12 600 62 marks respectively.
∴ The scores of Leon and Mary are
43. D 56 marks and 62 marks respectively.
Number of male soldiers Let x marks be the mean of the distribution.
= 18  6 56  x
= 1.5
= 12 8
56  x = 12
The required probability
x = 68
= 1  P(more than 4 female soldiers)
The required standard score
C56  C312  C66  C212 62  x
=1 =
C818 8
7 62  68
=1 =
221 8
214 = 0.75
=
221
Alternative method: 45. A
Number of male soldiers abcd e f
I. m1 =
= 18  6 6
= 12 6m1 = a + b + c + d + e + f
The required probability 7 a  7b  7c  7 d  7e  7 f  7 m1
m2 =
C06  C812  C16  C712  C 26  C612  7
= a + b + c + d + e + f + m1
C36  C512  C46  C 412
= = 6m1 + m1
C818
= 7m1
214
= ∴ I must be true.
221
II. For the group of numbers {a, b, c, d, e, f},
minimum value  m1  maximum value.
Range of {a, b, c, d, e, f, m1}
= range of {a, b, c, d, e, f}
= r1
Range of {7a, 7b, 7c, 7d, 7e, 7f, 7m1}
= 7  range of {a, b, c, d, e, f, m1}
∴ r2 = 7r1
∴ II must be true.

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OXFORD UNIVERSITY PRESS
MOCK 17(I) COMPULSORY PART PAPER 2 SOLUTION

III. Suppose a, b, c, d, e and f are 6 different

numbers.
(a  m1 ) 2  (b  m1 ) 2  (c  m1 ) 2 
(d  m1 ) 2  (e  m1 ) 2  ( f  m1 ) 2
v1 =
6
6v1 = (a  m1) + (b  m1)2 + (c  m1)2 +
2

(d  m1)2 + (e  m1)2 + (f  m1)2

( 7 a  m 2 ) 2  ( 7b  m 2 ) 2 
(7c  m2 ) 2  (7 d  m2 ) 2 
( 7 e  m2 ) 2  ( 7 f  m2 ) 2 
(7 m1  m2 ) 2
v2 =
7
(7 a  7 m1 ) 2  (7b  7 m1 ) 2 
(7c  7 m1 ) 2  (7 d  7m1 ) 2 
(7e  7 m1 ) 2  (7 f  7m1 ) 2 
(7 m1  7m1 ) 2
=
7
49[(a  m1 )  (b  m1 ) 2  (c  m1 ) 2 
2

(d  m1 ) 2  (e  m1 ) 2  ( f  m1 ) 2 ]
=
7
= 7  6v1
= 42v1
< 49v1
∴ III is not true.
∴ Only I and II must be true.

© Oxford University Press 2017 P.14