Bansal Mock Test-1 (Answer Key + Solution) Specially for JEE Mains Examination- ConceptsMadeEasy.com

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DATE: 19-11-2019 JEE-MAIN MOCK TEST-1 XII

COURSE

NUCLEUS

TEST CODE

1 1 2 5 8

Q.No.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans

3

4

3

2

1

4

4

1

4

1

1

1

4

2

2

Q.No.

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Ans

2

1

2

3

2

4

4

3

3

4

1

3

2

4

2

Q.No.

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

Ans

3

1

1

1

3

3

3

2

1

4

4

1

2

3

2

Q.No.

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

Ans

2

2

1

4

1

2

2

2

1

1

2

4

2

3

3

PC

IOC

OC

PC

IOC

OC

PC

IOC

OC

PC

IOC

OC

PC

IOC

OC

Q.No.

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

Ans

2

2

4

4

2

4

4

3

2

3

2

2

4

2

2

PC

IOC

OC

PC

IOC

OC

PC

IOC

OC

PC

IOC

OC

PC

IOC

OC

Q.No.

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

Ans

1

2

2

3

2

3

1

3

2

1

2

3

1

4

4

HINTS & SOLUTIONS M AT H E M AT I C S Q.1

    r  a   b  c

  taking dot with b  c    [ r b c ] = [a b c ] where a  (0,1,1) ;   b  (1,1,1) and c  ( 2,1, 0)

0 1 1  [a b c ] = 1  1 1 = 0 – (0 – 2) + 1(–1 + 2) = 3 2 1 0 and

x y z  1 1 1 = = [ r b c] 2 1 0

x(0 + 1) – y(0 – 2) + z(–1 + 2) = x + 2y + z hence equation of plane is x + 2y + z = 3; 

p=

3 = 6

XII MT-1 [JEE Main]

3 Ans. ] 2

 

 

        Q.2 a  b  V  u  d  a [ a c d ] b               (a . V ) b  ( b . V ) a  ( u . a ) d  ( u . d ) a  [ a c d ] b           [ a c d ] b [ b c d ]a  [ b c a ]d [ b c d ]a [ a c d ] b      [ b c a ] d  0  a , b , c are coplanar ]

Q.3

Clearly minimum value of a2 + b2 + c2 O(0,0,0)

3x + 2y + z =7 P (a, b, c)

2

 | (3(0)  2(0)  (0)  7 |  49 7   = = = units. 2 2 2  2 (3)  (2)  (1)  14 

(This is possible when P(a, b, c) is foot of perpendicular from O(0, 0, 0) on the plane.) Page # 1


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