class: sz1-a(revision) jee-main model date: 02-01-21 time: … · 2021. 1. 4. · narayana co...
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Narayana CO Schools
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Class: SZ1-A(Revision) JEE-MAIN MODEL Date: 02-01-21
Time: 3hrs RTM-04 Max. Marks: 300
Jee-Main_Model
SZ1-A Physics Initial Key Dt. 02-01-2021
Q.No. 01 02 03 04 05 06 07 08 09 10
Ans. 1 3 3 1 4 2 4 2 4 1
Q.No. 11 12 13 14 15 16 17 18 19 20
Ans. 3 2 1 2 1 3 2 2 1 1
Q.No. 21 22 23 24 25
Ans. 6.00 10.55 to 10.60
34.28 to 34.30
10.00 40.00
SZ1-A Chemistry Initial Key Dt. 02-01-2021
Q.No. 26 27 28 29 30 31 32 33 34 35
Ans. 3 3 4 1 4 4 2 1 4 1
Q.No. 36 37 38 39 40 41 42 43 44 45
Ans. 4 1 3 4 1 3 1 1 3 2
Q.No. 46 47 48 49 50
Ans. 6.46 21.37 1.20 to
1.25 124.50 -0.90
SZ1-A Mathematics Initial Key Dt. 02-01-2021
Q.No. 51 52 53 54 55 56 57 58 59 60
Ans. 3 2 4 1 3 4 3 4 1 4
Q.No. 61 62 63 64 65 66 67 68 69 70
Ans. 2 2 1 4 4 1 2 2 4 4
Q.No. 71 72 73 74 75
Ans. 0.00 3.00 0.00 1.00 1.00
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SZ1-A_JEE-MAIN_RTM-04_Key&Solutions_Exam.Dt.02-01-21
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01. 1 2p p p= +
Powers added up individual powers will be 2D
02. Increases
03. Power of the lenses does not depend on its size
04. The ciliary muscles do not contract and the eye is least strained
05. size of object directly depends
08. 0 10 , 5f cm Fe cm= =
0 0 0
1 1 1
v u f− =
0
1 11 1
0.5v= − = −
Object distance of eyepiece ( )1x cm= +
1 1 1
30 1 5x+ =
+
6x =
1 6x+ =
5x cm=
13.
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14.
15.
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16.
19.
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23.
25.
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30.
31. q and w represents path function
32.
42.
43.
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44.
45.
46.
47.
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48.
49.
50.
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51. System has unique solution if det. of coefficinet 0=
1 1 1
2 1 1 0 0
3 2
−
52. AX B=
2 3 6 4
5 7 14 , , 1
3 2 4 0
x
A X y B
z
−
= − = = −
now 0A = .
The equations either have no solution or an infinite number of
solutions to decide about this we proceed to find ( ).adj A B
0 0 0
22 26 58
11 13 29
adjA
= − − − −
( )
0
. 114 0
57
adj A B
= − −
no solution
53. AX B=
1 1 1 6
0 1 2 8
0 3 6 24
A B
=
2 2 1R R R⎯⎯→ −
3 3 1R R R⎯⎯→ −
1 1 1 6
0 1 2 8
0 0 0 0
= =
Rank A = rank [A][B] = 2
System has infinitely many solution.
System is equivalent to 6, 2 8x y z y z+ + = + =
8 2 , 2z y x= = − = − +
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54. 0A =
Now
4 6 2 4 6 2
6 9 3 6 9 3
2 3 1 2 3 1
T
adj A
− −
= − − = − − − −
( )
2
2
2 2
4 6 2 1 4 6 2
6 9 3 6 9 3 0
2 3 1 2 3
adj A B
− − +
= − − = − + − = − − +
Since system is consistent
22 6 4 0 1,2 − + = =
55. Augumented martrix
1 1 1 5 1 1 1 5
1 2 0 1 1 5
1 2 3 9 0 0 3 9
A B
= = − − − −
2 2 1
3 3 1
R R R
R R R
⎯⎯→ −
⎯⎯→ −
Rank A = Rank [AB] if 3
The system has unique solution if 3
If 3, 9 = = , rank A = rank [AB] = 2 < no. of unknowns
The system has infinitely many solution if 3, 9 = =
If 3, 9 = , rank A = 2, rank [AB] = 3 no solution
56. System has non-trival solution 0
a b c
b c a
c a b
=
3 3 3 3 0a b c abc + + − =
( )( )( )2 2 0a b c c bw aw c bw aw + + + + + + =
Since , ,a b c are positive 0a b c+ + , either 2 0aw bw c+ + = or
4 2 0aw bw c+ + =
Hence roots of the equation 2 0at bt c+ + = are 2,w w
57.
1 1 1
1 1 1 0
1 1 1
a
b
c
−
− =
−
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ab bc ca abc + + =
( )1
2 2 2 3
3
ab bc caa b c
+ +
( )1
3 3abc
( )min 27abc =
59. The system has infinite solutions if
1 0
0 1 0 1,1
0 1
a
a a
a
= = −
For 1a = we have the system 0, 0, 0x y y z z x+ = + = + = which have
only one solution 0x y z= = =
1a = −
60. Non-trival solution if
1 1
1 1 0
1 1
−
− − =
−
( )2 1 0 − =
0,1, 1 = −
61. 1 2A A A I− = + +
3 2I A A A= + +
33 6 19
35 4 30
30 13 22
A
− −
= − −
3 26 11A A A I− + =
multiplying both the sides by 1A−
1 2 6 11A A A I− = − +
6, 11 = − =
5+ =
62. ( ) ( )( ) ( ) 1 1T
TTBB I A I A I A I A− −
= − + + −
( ) ( )( )1 TI A I A I A−
= − + +
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( ) ( )( ) ( )1
1 TI A I A I A I A
−− = − + − −
A is skew-symmetric TA A= −
( ) ( ) ( ) ( )11 TI A I A I A I A−−
− − + −
( ) ( )( )( )1 1
I A I A I A I A− −
= − − + +
.TBB I I I = =
63. ( ) ( )1
*A I A I A−
= + −
( ) ( )1
** * *A I A I A−
= + −
( ) ( ) ( ) ( ) ( )1 1 1
* 2I A I A I A I A I A I A I− − −
+ = + + + + − = + ( )1
2 I A−
= +
( ) ( )1 1
*2
I A I A−
+ = +
( ) ( ) ( ) ( ) ( )1 1 1
* 2I A I A I A I A I A I A A− − −
− = + + − + − = +
( ) ( )11
** 22
A I A I A A A−
= + + =
64. A has rank 3 0A
( )
1
1 0
1
b c
y a b c y b c
b y c
+ + + +
+
( )
1
0 0 0
0 0
b c
y a b c y
y
+ + + ( )2 0y y a b c + + +
65. matrix A has all integer entries
cofactor of each entry is integer
adjoine A has integers entries
( ) ( )11
A adj A adj AA
− = =
66. T TPP I P P= =
TQ PAP=
T T T TP Q P PAP AP = =
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2005 2004 2004 2 2003 ...T T T TP Q P P Q Q P AP Q P A P Q P= = =
( ) ( )2004 2004 2005T TA P QP A P PA A= = =
21 2 1
...0 1 0 1
nn
A A
= =
20051 2005
0 1A
=
20051 2005
0 1
TP Q P
=
67. TA A I=
2 2 2 1a b c + + =
0ab bc ca+ + =
Since , , 0a b c
0ab bc ca+ +
3 3 3a b c + + is exists
( )( )3 3 3 2 2 2 4a b c a b c a b c ab bc ca+ + = + + + + − − − =
68. characteristic equation of A is
0A I− =
3 26 11 6 0 − + − =
A satisfies characteristic equation 3 26 11 6 0A A A I− + − =
1 26 6 11A A A I− = − +
6, 11C d= − =
69. 3 3ij
P a
=
2i jij ijb a+=
3 3ij
Q b
=
11 12 13
21 22 23
31 32 33
, 2
a a a
P a a a P
a a a
= =
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11 12 13 11 12 13
21 22 23 21 22 23
31 32 33 31 32 33
4 8 16
8 16 32
16 32 64
b b b a a a
Q b b b a a a
c c c a a a
= =
Determinant of
11 12 13 11 12 13
21 22 33 21 22 23
31 32 33 31 32 33
4 8 16
8 16 32 4 8 16 2 2 2
16 32 64 4 4 4
a a a a a a
Q a a a a a a
a a a a a a
= =
11 12 13
2 3 4 1 2 1 13
21 22 23
31 32 33
4 8 16 2 4 2 2 2 2 2 2 2
a a a
a a a
a a a
= = =
70. 2TP P I= +
2 TP P I= +
2 2P P I I= + +
4 3P P I= +
3 3P I= −
PX X= −
71. 3 3 2 2,P Q P Q Q P= =
P Q
Subtracting 3 2 3 2P P Q Q Q P− = −
( ) ( )2 2 0P P Q Q P Q− + − =
( )( )2 2 0P Q P Q+ − =
If ( )2 2 0P Q P Q+ − =
If 2 2 0P Q+ , then 2 2P Q+ is invertible
0P Q − =
Controduction
Hence 2 2 0P Q+ =
72. for no solution
1 8 4
3 3 1
k k
k k k
+=
+ −
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2 4 3 8k k k+ + =
2 4 3 0, 1,3k k k− + = =
If 8 4 1
1 false1 3 2
k
= +
If 8 4 3
3 true6 9 1
k
= =−
3k =