JEE Main 2021 Mathematics February 26 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. If 
? ?
a
 and 
? ?
b
 are perpendicular, then
? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ? ? ?
a a a a b
is equal to :
(1) ?
? ? ? ?
a b (2)
?
0
(3)
? ? ? ?
4
1
a b
2
(4)
? ? ? ?
4
a b
Answer (4)
Sol. Let ˆ c be a unit vector in the direction of  ?
?
?
a b .
? ? ? ? ? ? ?
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ a b c, b c a & c a b
? ? ? ?
? ?
? ?
ˆ a b a b c
? ?
2
ˆ
a a b a b b ? ? ? ?
? ?
? ? ?
? ? ? ?
3
ˆ a a a b a b c ? ? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ? ?
? ?
? ? ? ? ?
4
ˆ
a a a a b a b b
?
?
?
4
a b
2. The value of 
?
?
?
?
?
2 2
x
2
cos x
dx
1 3
 is :
(1)
?
4
(2)
? 2
(3)
?
2
(4)
? 4
Answer (1)
Sol. ? ?
a a
a 0
f(x)dx f(x) f(a x) dx
?
? ? ?
? ?
2 2 2 2 2
x x x
0
2
cos x cos x cos ( x)
dx dx
1 3 1 3 1 3
? ?
?
? ?
?
? ?
? ? ?
? ?
PART–C : MATHEMATICS
x 2 2
2 2
x x
0 0
1 3
cos x dx cos x dx
1 3 1 3
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
? ?
? ?
2
0
1
1 cos2x dx
2 4
?
?
? ? ?
?
3. The value of 
? ? ?
? ? ?
? ? ?
(a 1 ) (a 2) a 2 1
(a 2) (a 3) a 3 1
(a 3) (a 4) a 4 1
 is :
(1) 0 (2) ? ? ? (a 2)(a 3)(a 4)
(3) –2 (4) ? ? ? (a 1 )(a 2)(a 3)
Answer (3)
Sol. Given determinant is
2
2
a 3a 2 a 2 1
D a 5a 6 a 3 1
a 7a 12 a 4 1
? ? ?
? ? ? ?
? ? ?
R
3
 ? R
3
 – R
2
; R
2
 ? R
2
 – R
1
2
a 3a 2 a 2 1
2a 4 1 0
2a 6 1 0
? ? ?
? ?
?
Expanding by C
3
D = (2a + 4) – (2a + 6) = –2
4. The maximum slope of the curve
? ? ? ?
4 3 2
1
y x 5x 18x 19x
2
 occurs at the point :
(1) (2, 2) (2) (0, 0)
(3)
? ?
? ?
? ?
21
3,
2
(4) (2, 9)
Answer (1*)
Sol.
4 3 2
1
y x 5x 18x 19x
2
? ? ? ?
Slope = y ? = 2x
3
 – 15x
2
 + 36x – 19 = g(x) say
g ?(x) = 6x
2
 – 30x + 36 = 6(x – 2)(x – 3)
g ?(x) = 0 ? x = 2, 3
Slope g(x) has local maximum at x = 2
x = 2 ? y = 2
Local maximum at (2, 2)
[Note : Overall maximum (Absolute maximum)
value of slope is far greater than that at (2, 2)].
Page 2


JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. If 
? ?
a
 and 
? ?
b
 are perpendicular, then
? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ? ? ?
a a a a b
is equal to :
(1) ?
? ? ? ?
a b (2)
?
0
(3)
? ? ? ?
4
1
a b
2
(4)
? ? ? ?
4
a b
Answer (4)
Sol. Let ˆ c be a unit vector in the direction of  ?
?
?
a b .
? ? ? ? ? ? ?
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ a b c, b c a & c a b
? ? ? ?
? ?
? ?
ˆ a b a b c
? ?
2
ˆ
a a b a b b ? ? ? ?
? ?
? ? ?
? ? ? ?
3
ˆ a a a b a b c ? ? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ? ?
? ?
? ? ? ? ?
4
ˆ
a a a a b a b b
?
?
?
4
a b
2. The value of 
?
?
?
?
?
2 2
x
2
cos x
dx
1 3
 is :
(1)
?
4
(2)
? 2
(3)
?
2
(4)
? 4
Answer (1)
Sol. ? ?
a a
a 0
f(x)dx f(x) f(a x) dx
?
? ? ?
? ?
2 2 2 2 2
x x x
0
2
cos x cos x cos ( x)
dx dx
1 3 1 3 1 3
? ?
?
? ?
?
? ?
? ? ?
? ?
PART–C : MATHEMATICS
x 2 2
2 2
x x
0 0
1 3
cos x dx cos x dx
1 3 1 3
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
? ?
? ?
2
0
1
1 cos2x dx
2 4
?
?
? ? ?
?
3. The value of 
? ? ?
? ? ?
? ? ?
(a 1 ) (a 2) a 2 1
(a 2) (a 3) a 3 1
(a 3) (a 4) a 4 1
 is :
(1) 0 (2) ? ? ? (a 2)(a 3)(a 4)
(3) –2 (4) ? ? ? (a 1 )(a 2)(a 3)
Answer (3)
Sol. Given determinant is
2
2
a 3a 2 a 2 1
D a 5a 6 a 3 1
a 7a 12 a 4 1
? ? ?
? ? ? ?
? ? ?
R
3
 ? R
3
 – R
2
; R
2
 ? R
2
 – R
1
2
a 3a 2 a 2 1
2a 4 1 0
2a 6 1 0
? ? ?
? ?
?
Expanding by C
3
D = (2a + 4) – (2a + 6) = –2
4. The maximum slope of the curve
? ? ? ?
4 3 2
1
y x 5x 18x 19x
2
 occurs at the point :
(1) (2, 2) (2) (0, 0)
(3)
? ?
? ?
? ?
21
3,
2
(4) (2, 9)
Answer (1*)
Sol.
4 3 2
1
y x 5x 18x 19x
2
? ? ? ?
Slope = y ? = 2x
3
 – 15x
2
 + 36x – 19 = g(x) say
g ?(x) = 6x
2
 – 30x + 36 = 6(x – 2)(x – 3)
g ?(x) = 0 ? x = 2, 3
Slope g(x) has local maximum at x = 2
x = 2 ? y = 2
Local maximum at (2, 2)
[Note : Overall maximum (Absolute maximum)
value of slope is far greater than that at (2, 2)].
JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
5. In an increasing geometric series, the sum of
the second and the sixth term is 
25
2
and the
product of the third and fifth term is 25. Then,
the sum of 
th th
4 ,6
and 
th
8 term is equal to :
(1) 26 (2) 35
(3) 30 (4) 32
Answer (2)
Sol.
2 6
25
a a
2
? ?
2
3 5 2 6 4
a a 25 a a a ? ? ? ? ?
2
4 4
a 25 a 5 ? ? ?
a
2
 & a
6
 are roots of 
2
25
x x 25 0
2
? ? ?
5
x , 10
2
?
2 6
5
a , a 10
2
? ? ( ? GP is increasing)
a
4
 = 5
a
4
 =a
2
r
2 2 2
5
5 r r 2
2
? ? ? ?
a
8
 = a
6
r
2
 = 10 × 2 = 20
a
4
 + a
6
 + a
8
 = 5 + 10 + 20 = 35
6. The number of seven digit integers with sum of
the digits equal to 10 and formed by using the
digits 1, 2 and 3 only is :
(1) 77 (2) 42
(3) 82 (4) 35
Answer (1)
Sol. Combination of digits
3, 2, 1, 1, 1, 1, 1 ?
7!
42
5!
?
2, 2, 2, 1, 1, 1, 1 ?
7!
35
4! 3!
?
Total = 42 + 35 = 77
7. The sum of infinite series
? ? ? ? ?
2 3 4 5
2 7 12 17 22
1
3
3 3 3 3
+......is equal to :
(1)
13
4
(2)
9
4
(3)
11
4
(4)
15
4
Answer (1)
Sol.
2 3 4 5
2 3 4 5
2 3 4 5
2 7 12 17 22
S 1 ............
3
3 3 3 3
1 1 2 7 12 17
S ...........
3 3
3 3 3 3
2 1 5 5 5 5
S 1 .......
3 3
3 3 3 3
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ?
? ? ? ? ? ? ?
? ?
? ?
5 5
4 5
9 9
1 2
3 3
1
3 3
? ? ? ?
?
?
2 4 5 13
S
3 3 6 6
? ? ?
13
S
4
?
8. Consider the three planes
? ? ?
1
P : 3x 15y 21z 9 ,
? ? ?
2
P : x 3y z 5, and
? ? ?
3
P : 2x 10y 14z 5
Then, which one of the following is true ?
(1)
2 3
P and P are parallel
(2)
1 3
P and P are parallel
(3)
1 2
P and P are parallel
(4)
1 2
P ,P and 
3
P all are parallel
Answer (2)
Sol. Ratios of DRs of normals of P
1
 & P
3
 are
3 15 21
2 10 14
? ?
? Normals are parallel
? P
1
 || P
3
9. Let A be a symmetric matrix of order 2 with
integer entries. If the sum of the diagonal
elements of A
2
 is 1, then the possible number
of such matrices is :
(1) 6 (2) 1
(3) 4 (4) 12
Answer (3)
Sol. Let 
a c
A
c b
? ?
?
? ?
? ?
Page 3


JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. If 
? ?
a
 and 
? ?
b
 are perpendicular, then
? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ? ? ?
a a a a b
is equal to :
(1) ?
? ? ? ?
a b (2)
?
0
(3)
? ? ? ?
4
1
a b
2
(4)
? ? ? ?
4
a b
Answer (4)
Sol. Let ˆ c be a unit vector in the direction of  ?
?
?
a b .
? ? ? ? ? ? ?
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ a b c, b c a & c a b
? ? ? ?
? ?
? ?
ˆ a b a b c
? ?
2
ˆ
a a b a b b ? ? ? ?
? ?
? ? ?
? ? ? ?
3
ˆ a a a b a b c ? ? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ? ?
? ?
? ? ? ? ?
4
ˆ
a a a a b a b b
?
?
?
4
a b
2. The value of 
?
?
?
?
?
2 2
x
2
cos x
dx
1 3
 is :
(1)
?
4
(2)
? 2
(3)
?
2
(4)
? 4
Answer (1)
Sol. ? ?
a a
a 0
f(x)dx f(x) f(a x) dx
?
? ? ?
? ?
2 2 2 2 2
x x x
0
2
cos x cos x cos ( x)
dx dx
1 3 1 3 1 3
? ?
?
? ?
?
? ?
? ? ?
? ?
PART–C : MATHEMATICS
x 2 2
2 2
x x
0 0
1 3
cos x dx cos x dx
1 3 1 3
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
? ?
? ?
2
0
1
1 cos2x dx
2 4
?
?
? ? ?
?
3. The value of 
? ? ?
? ? ?
? ? ?
(a 1 ) (a 2) a 2 1
(a 2) (a 3) a 3 1
(a 3) (a 4) a 4 1
 is :
(1) 0 (2) ? ? ? (a 2)(a 3)(a 4)
(3) –2 (4) ? ? ? (a 1 )(a 2)(a 3)
Answer (3)
Sol. Given determinant is
2
2
a 3a 2 a 2 1
D a 5a 6 a 3 1
a 7a 12 a 4 1
? ? ?
? ? ? ?
? ? ?
R
3
 ? R
3
 – R
2
; R
2
 ? R
2
 – R
1
2
a 3a 2 a 2 1
2a 4 1 0
2a 6 1 0
? ? ?
? ?
?
Expanding by C
3
D = (2a + 4) – (2a + 6) = –2
4. The maximum slope of the curve
? ? ? ?
4 3 2
1
y x 5x 18x 19x
2
 occurs at the point :
(1) (2, 2) (2) (0, 0)
(3)
? ?
? ?
? ?
21
3,
2
(4) (2, 9)
Answer (1*)
Sol.
4 3 2
1
y x 5x 18x 19x
2
? ? ? ?
Slope = y ? = 2x
3
 – 15x
2
 + 36x – 19 = g(x) say
g ?(x) = 6x
2
 – 30x + 36 = 6(x – 2)(x – 3)
g ?(x) = 0 ? x = 2, 3
Slope g(x) has local maximum at x = 2
x = 2 ? y = 2
Local maximum at (2, 2)
[Note : Overall maximum (Absolute maximum)
value of slope is far greater than that at (2, 2)].
JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
5. In an increasing geometric series, the sum of
the second and the sixth term is 
25
2
and the
product of the third and fifth term is 25. Then,
the sum of 
th th
4 ,6
and 
th
8 term is equal to :
(1) 26 (2) 35
(3) 30 (4) 32
Answer (2)
Sol.
2 6
25
a a
2
? ?
2
3 5 2 6 4
a a 25 a a a ? ? ? ? ?
2
4 4
a 25 a 5 ? ? ?
a
2
 & a
6
 are roots of 
2
25
x x 25 0
2
? ? ?
5
x , 10
2
?
2 6
5
a , a 10
2
? ? ( ? GP is increasing)
a
4
 = 5
a
4
 =a
2
r
2 2 2
5
5 r r 2
2
? ? ? ?
a
8
 = a
6
r
2
 = 10 × 2 = 20
a
4
 + a
6
 + a
8
 = 5 + 10 + 20 = 35
6. The number of seven digit integers with sum of
the digits equal to 10 and formed by using the
digits 1, 2 and 3 only is :
(1) 77 (2) 42
(3) 82 (4) 35
Answer (1)
Sol. Combination of digits
3, 2, 1, 1, 1, 1, 1 ?
7!
42
5!
?
2, 2, 2, 1, 1, 1, 1 ?
7!
35
4! 3!
?
Total = 42 + 35 = 77
7. The sum of infinite series
? ? ? ? ?
2 3 4 5
2 7 12 17 22
1
3
3 3 3 3
+......is equal to :
(1)
13
4
(2)
9
4
(3)
11
4
(4)
15
4
Answer (1)
Sol.
2 3 4 5
2 3 4 5
2 3 4 5
2 7 12 17 22
S 1 ............
3
3 3 3 3
1 1 2 7 12 17
S ...........
3 3
3 3 3 3
2 1 5 5 5 5
S 1 .......
3 3
3 3 3 3
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ?
? ? ? ? ? ? ?
? ?
? ?
5 5
4 5
9 9
1 2
3 3
1
3 3
? ? ? ?
?
?
2 4 5 13
S
3 3 6 6
? ? ?
13
S
4
?
8. Consider the three planes
? ? ?
1
P : 3x 15y 21z 9 ,
? ? ?
2
P : x 3y z 5, and
? ? ?
3
P : 2x 10y 14z 5
Then, which one of the following is true ?
(1)
2 3
P and P are parallel
(2)
1 3
P and P are parallel
(3)
1 2
P and P are parallel
(4)
1 2
P ,P and 
3
P all are parallel
Answer (2)
Sol. Ratios of DRs of normals of P
1
 & P
3
 are
3 15 21
2 10 14
? ?
? Normals are parallel
? P
1
 || P
3
9. Let A be a symmetric matrix of order 2 with
integer entries. If the sum of the diagonal
elements of A
2
 is 1, then the possible number
of such matrices is :
(1) 6 (2) 1
(3) 4 (4) 12
Answer (3)
Sol. Let 
a c
A
c b
? ?
?
? ?
? ?
JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
2 2
2
2 2
a c a c a c ac bc
A
c b c b
ac bc c b
? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ?
a
2
 + b
2
 + 2c
2
 = 1 as a, b, c ?z
c = 0 and a, b = ±1
Total 4 matrices are possible
10. The maximum value of the term independent of
‘t’ in the expansion of 
? ?
? ?
?
? ? ?
? ?
? ?
? ?
10
1
1
10
5
(1 x)
tx
t
where ? x (0,1 ) is :
(1)
2
2.10!
3(5!)
(2)
2
2.10!
3 3(5!)
(3)
2
10!
3(5!)
(4)
2
10!
2(5!)
Answer (2)
Sol.
r
1
1
10
10 10 r
5
r 1 r
(1 x)
T C (tx )
t
?
?
? ?
? ?
?
? ? ?
? ?
? ?
? ?
For term independent of f
10 – r – r = 0  ?  r = 5
1
10
2
6 5
T C x(1 x) f(x) ? ? ?
(Let)
?
1
10
2
5
1
2
x
f (x) C (1 x) 0
2(1 x)
? ?
? ?
? ? ? ? ? ? ?
? ?
? ?
?
? ?
2
2 2x x x
3
? ? ? ?
f ??(x) < 0  at  
2
x
3
?
1
2
10
6(max) 5
2
2 1 2.10!
T C .
3 3
(5!) 3 3
? ?
? ?
? ?
? ?
11. A fair coin is tossed a fixed number of times. If
the probability of getting 7 heads is equal to
probability of getting 9 heads, then the
probability of getting 2 heads is :
(1)
13
15
2
(2)
12
15
2
(3)
8
15
2
(4)
14
15
2
Answer (1)
Sol. Let n number of tosses
Given,
7 n 7 9 n 9
n n
7 9
1 1 1 1
C C
2 2 2 2
? ?
? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? n = 16
? Probability of getting 2 heads = 
16
2
1
16C
2
? ?
? ?
? ?
 
13
15
2
?
12. The value of 
? ? ?
?
?
?
?
n
100
x x
n 1
n 1
e dx,
 where [x] is the
greatest integer ? x, is :
(1) 100(e – 1) (2) 100(1 – e)
(3) 100e (4) 100(1 + e)
Answer (1)
Sol.
? ?
? ?
n 1
x x x
n 1 0
e dx e dx e 1
?
?
? ? ?
? ?
?
? ? ? ?
100
n 1
e 1 100 e 1
?
? ? ?
?
13. In the circle given below, let OA = 1 unit,
OB = 13 unit and 
? PQ OB
. Then, the area of
the triangle PQB (in square units) is :
P
A
Q
B x O
y
(1) 24 2 (2) 24 3
(3) 26 3 (4) 26 2
Answer (2)
Sol. Assume that OB is diameter of the given circle
Using Ptolemy’s Theorem,
OP·QB + OQ·PB = PQ × OB
? 2OP·PB = 13PQ
Also PA
2
 = OP
2
 – 1 = PB
2
 – 12
2
? PB
2
 – OP
2
 = 143
and OP
2
 + PB
2
 = 13
2
Page 4


JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. If 
? ?
a
 and 
? ?
b
 are perpendicular, then
? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ? ? ?
a a a a b
is equal to :
(1) ?
? ? ? ?
a b (2)
?
0
(3)
? ? ? ?
4
1
a b
2
(4)
? ? ? ?
4
a b
Answer (4)
Sol. Let ˆ c be a unit vector in the direction of  ?
?
?
a b .
? ? ? ? ? ? ?
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ a b c, b c a & c a b
? ? ? ?
? ?
? ?
ˆ a b a b c
? ?
2
ˆ
a a b a b b ? ? ? ?
? ?
? ? ?
? ? ? ?
3
ˆ a a a b a b c ? ? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ? ?
? ?
? ? ? ? ?
4
ˆ
a a a a b a b b
?
?
?
4
a b
2. The value of 
?
?
?
?
?
2 2
x
2
cos x
dx
1 3
 is :
(1)
?
4
(2)
? 2
(3)
?
2
(4)
? 4
Answer (1)
Sol. ? ?
a a
a 0
f(x)dx f(x) f(a x) dx
?
? ? ?
? ?
2 2 2 2 2
x x x
0
2
cos x cos x cos ( x)
dx dx
1 3 1 3 1 3
? ?
?
? ?
?
? ?
? ? ?
? ?
PART–C : MATHEMATICS
x 2 2
2 2
x x
0 0
1 3
cos x dx cos x dx
1 3 1 3
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
? ?
? ?
2
0
1
1 cos2x dx
2 4
?
?
? ? ?
?
3. The value of 
? ? ?
? ? ?
? ? ?
(a 1 ) (a 2) a 2 1
(a 2) (a 3) a 3 1
(a 3) (a 4) a 4 1
 is :
(1) 0 (2) ? ? ? (a 2)(a 3)(a 4)
(3) –2 (4) ? ? ? (a 1 )(a 2)(a 3)
Answer (3)
Sol. Given determinant is
2
2
a 3a 2 a 2 1
D a 5a 6 a 3 1
a 7a 12 a 4 1
? ? ?
? ? ? ?
? ? ?
R
3
 ? R
3
 – R
2
; R
2
 ? R
2
 – R
1
2
a 3a 2 a 2 1
2a 4 1 0
2a 6 1 0
? ? ?
? ?
?
Expanding by C
3
D = (2a + 4) – (2a + 6) = –2
4. The maximum slope of the curve
? ? ? ?
4 3 2
1
y x 5x 18x 19x
2
 occurs at the point :
(1) (2, 2) (2) (0, 0)
(3)
? ?
? ?
? ?
21
3,
2
(4) (2, 9)
Answer (1*)
Sol.
4 3 2
1
y x 5x 18x 19x
2
? ? ? ?
Slope = y ? = 2x
3
 – 15x
2
 + 36x – 19 = g(x) say
g ?(x) = 6x
2
 – 30x + 36 = 6(x – 2)(x – 3)
g ?(x) = 0 ? x = 2, 3
Slope g(x) has local maximum at x = 2
x = 2 ? y = 2
Local maximum at (2, 2)
[Note : Overall maximum (Absolute maximum)
value of slope is far greater than that at (2, 2)].
JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
5. In an increasing geometric series, the sum of
the second and the sixth term is 
25
2
and the
product of the third and fifth term is 25. Then,
the sum of 
th th
4 ,6
and 
th
8 term is equal to :
(1) 26 (2) 35
(3) 30 (4) 32
Answer (2)
Sol.
2 6
25
a a
2
? ?
2
3 5 2 6 4
a a 25 a a a ? ? ? ? ?
2
4 4
a 25 a 5 ? ? ?
a
2
 & a
6
 are roots of 
2
25
x x 25 0
2
? ? ?
5
x , 10
2
?
2 6
5
a , a 10
2
? ? ( ? GP is increasing)
a
4
 = 5
a
4
 =a
2
r
2 2 2
5
5 r r 2
2
? ? ? ?
a
8
 = a
6
r
2
 = 10 × 2 = 20
a
4
 + a
6
 + a
8
 = 5 + 10 + 20 = 35
6. The number of seven digit integers with sum of
the digits equal to 10 and formed by using the
digits 1, 2 and 3 only is :
(1) 77 (2) 42
(3) 82 (4) 35
Answer (1)
Sol. Combination of digits
3, 2, 1, 1, 1, 1, 1 ?
7!
42
5!
?
2, 2, 2, 1, 1, 1, 1 ?
7!
35
4! 3!
?
Total = 42 + 35 = 77
7. The sum of infinite series
? ? ? ? ?
2 3 4 5
2 7 12 17 22
1
3
3 3 3 3
+......is equal to :
(1)
13
4
(2)
9
4
(3)
11
4
(4)
15
4
Answer (1)
Sol.
2 3 4 5
2 3 4 5
2 3 4 5
2 7 12 17 22
S 1 ............
3
3 3 3 3
1 1 2 7 12 17
S ...........
3 3
3 3 3 3
2 1 5 5 5 5
S 1 .......
3 3
3 3 3 3
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ?
? ? ? ? ? ? ?
? ?
? ?
5 5
4 5
9 9
1 2
3 3
1
3 3
? ? ? ?
?
?
2 4 5 13
S
3 3 6 6
? ? ?
13
S
4
?
8. Consider the three planes
? ? ?
1
P : 3x 15y 21z 9 ,
? ? ?
2
P : x 3y z 5, and
? ? ?
3
P : 2x 10y 14z 5
Then, which one of the following is true ?
(1)
2 3
P and P are parallel
(2)
1 3
P and P are parallel
(3)
1 2
P and P are parallel
(4)
1 2
P ,P and 
3
P all are parallel
Answer (2)
Sol. Ratios of DRs of normals of P
1
 & P
3
 are
3 15 21
2 10 14
? ?
? Normals are parallel
? P
1
 || P
3
9. Let A be a symmetric matrix of order 2 with
integer entries. If the sum of the diagonal
elements of A
2
 is 1, then the possible number
of such matrices is :
(1) 6 (2) 1
(3) 4 (4) 12
Answer (3)
Sol. Let 
a c
A
c b
? ?
?
? ?
? ?
JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
2 2
2
2 2
a c a c a c ac bc
A
c b c b
ac bc c b
? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ?
a
2
 + b
2
 + 2c
2
 = 1 as a, b, c ?z
c = 0 and a, b = ±1
Total 4 matrices are possible
10. The maximum value of the term independent of
‘t’ in the expansion of 
? ?
? ?
?
? ? ?
? ?
? ?
? ?
10
1
1
10
5
(1 x)
tx
t
where ? x (0,1 ) is :
(1)
2
2.10!
3(5!)
(2)
2
2.10!
3 3(5!)
(3)
2
10!
3(5!)
(4)
2
10!
2(5!)
Answer (2)
Sol.
r
1
1
10
10 10 r
5
r 1 r
(1 x)
T C (tx )
t
?
?
? ?
? ?
?
? ? ?
? ?
? ?
? ?
For term independent of f
10 – r – r = 0  ?  r = 5
1
10
2
6 5
T C x(1 x) f(x) ? ? ?
(Let)
?
1
10
2
5
1
2
x
f (x) C (1 x) 0
2(1 x)
? ?
? ?
? ? ? ? ? ? ?
? ?
? ?
?
? ?
2
2 2x x x
3
? ? ? ?
f ??(x) < 0  at  
2
x
3
?
1
2
10
6(max) 5
2
2 1 2.10!
T C .
3 3
(5!) 3 3
? ?
? ?
? ?
? ?
11. A fair coin is tossed a fixed number of times. If
the probability of getting 7 heads is equal to
probability of getting 9 heads, then the
probability of getting 2 heads is :
(1)
13
15
2
(2)
12
15
2
(3)
8
15
2
(4)
14
15
2
Answer (1)
Sol. Let n number of tosses
Given,
7 n 7 9 n 9
n n
7 9
1 1 1 1
C C
2 2 2 2
? ?
? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? n = 16
? Probability of getting 2 heads = 
16
2
1
16C
2
? ?
? ?
? ?
 
13
15
2
?
12. The value of 
? ? ?
?
?
?
?
n
100
x x
n 1
n 1
e dx,
 where [x] is the
greatest integer ? x, is :
(1) 100(e – 1) (2) 100(1 – e)
(3) 100e (4) 100(1 + e)
Answer (1)
Sol.
? ?
? ?
n 1
x x x
n 1 0
e dx e dx e 1
?
?
? ? ?
? ?
?
? ? ? ?
100
n 1
e 1 100 e 1
?
? ? ?
?
13. In the circle given below, let OA = 1 unit,
OB = 13 unit and 
? PQ OB
. Then, the area of
the triangle PQB (in square units) is :
P
A
Q
B x O
y
(1) 24 2 (2) 24 3
(3) 26 3 (4) 26 2
Answer (2)
Sol. Assume that OB is diameter of the given circle
Using Ptolemy’s Theorem,
OP·QB + OQ·PB = PQ × OB
? 2OP·PB = 13PQ
Also PA
2
 = OP
2
 – 1 = PB
2
 – 12
2
? PB
2
 – OP
2
 = 143
and OP
2
 + PB
2
 = 13
2
JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
then PB
2
 = 156 and OP
2
 = 13
So, 
2 13· 156
PQ 4 3
13
? ?
Area of 
1
PQB ·4 3·12 24 3
2
? ? ?
14. The value of
 
? ?
?
? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ?
?
? ?
? ?
? ?
h 0
3 sin h cos h
6 6
lim 2
3h 3 cos h sin h
 is
(1)
4
3
(2)
3
4
(3)
2
3
(4)
2
3
Answer (1)
Sol. h 0
3 1
sin h cos h
2 6 2 6
lim 2
3 1
3h cosh sinh
2 2
?
? ?
? ? ? ? ? ?
? ?
? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ?
? ?
?
? ?
? ? ? ?
? ? ? ?
? ?
h 0
sin h
2 2 4
lim 2 ·
3
3 3
3h sin h
3
?
? ?
? ?
? ?
? ?
? ?
? ? ?
? ?
?
? ?
? ?
? ? ? ?
15. Let R = {(P, Q) | P and Q are at the same
distance from the origin} be a relation, then the
equivalence class of (1, –1) is the set :
(1) S = {(x, y) | x
2
 + y
2
 = 2}
(2) S = {(x, y) | x
2
 + y
2
 = 1}
(3) S = {(x, y) | x
2
 + y
2
 = 
2
}
(4) S = {(x, y) | x
2
 + y
2
 = 4}
Answer (1)
Sol. ? R = {(P, Q) | P and Q are at the same
distance from the origin}.
Then equivalence class of (1, –1) will
contain all such points which lies on
circumference of the circle of centre at
origin and passing through point (1, –1).
i.e., radius of circle 
2 2
1 1 2 ? ? ?
? Required equivalence class of (S)
= {(x, y) | x
2
 + y
2
 = 2}.
16. The rate of growth of bacteria in a culture is
proportional to the number of bacteria present
and the bacteria count is 1000 at initial time t
= 0. The number of bacteria is increased by
20% in 2 hours. If the population of bacteria is
2000 after 
? ?
? ?
? ?
e
k
6
log
5
 hours, then 
? ?
? ?
? ?
2
e
k
log 2
 is
equal to :
(1) 16 (2) 4
(3) 8 (4) 2
Answer (2)
Sol. At t = 0 ?
?
B 1000
?
dB
B
dt
?
?
?
? ?
?
?
12B 2
B 0
dB
kt
B
[Given]
? ? ?
?
? ?
? ?
?
?
1 2B
ln 2k
B
? ? ?
1
k ln(1 2)
2
To find time when B = 2000
?
? ?
? ?
?
?
2B t
B 0
dB 1
ln(1 2) dt
B 2
? ?
1
ln2 ln(1 2)t
2
? ?
? ?
? ?
? ?
ln4
t hrs.
6
ln
5
? R = ln = 4
Thus 
? ?
? ?
? ?
? ?
2
2
K
2 4
ln
17. Let f be any function defined on R and let it
satisfy the condition :
? ? ? ? ? ? ? ? ? ? ? ? ?
2
f x f y x y , x,y R
If ? ? ? f 0 1 , then :
(1) ? ? f x can take any value in R
(2) ? ? ? ? ? f x 0,  x R
(3) ? ? ? ? ? f x 0,  x R
(4) ? ? ? ? ? f x 0,  x R
Answer (4)
Page 5


JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. If 
? ?
a
 and 
? ?
b
 are perpendicular, then
? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ? ? ?
a a a a b
is equal to :
(1) ?
? ? ? ?
a b (2)
?
0
(3)
? ? ? ?
4
1
a b
2
(4)
? ? ? ?
4
a b
Answer (4)
Sol. Let ˆ c be a unit vector in the direction of  ?
?
?
a b .
? ? ? ? ? ? ?
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ a b c, b c a & c a b
? ? ? ?
? ?
? ?
ˆ a b a b c
? ?
2
ˆ
a a b a b b ? ? ? ?
? ?
? ? ?
? ? ? ?
3
ˆ a a a b a b c ? ? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ? ?
? ?
? ? ? ? ?
4
ˆ
a a a a b a b b
?
?
?
4
a b
2. The value of 
?
?
?
?
?
2 2
x
2
cos x
dx
1 3
 is :
(1)
?
4
(2)
? 2
(3)
?
2
(4)
? 4
Answer (1)
Sol. ? ?
a a
a 0
f(x)dx f(x) f(a x) dx
?
? ? ?
? ?
2 2 2 2 2
x x x
0
2
cos x cos x cos ( x)
dx dx
1 3 1 3 1 3
? ?
?
? ?
?
? ?
? ? ?
? ?
PART–C : MATHEMATICS
x 2 2
2 2
x x
0 0
1 3
cos x dx cos x dx
1 3 1 3
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
? ?
? ?
2
0
1
1 cos2x dx
2 4
?
?
? ? ?
?
3. The value of 
? ? ?
? ? ?
? ? ?
(a 1 ) (a 2) a 2 1
(a 2) (a 3) a 3 1
(a 3) (a 4) a 4 1
 is :
(1) 0 (2) ? ? ? (a 2)(a 3)(a 4)
(3) –2 (4) ? ? ? (a 1 )(a 2)(a 3)
Answer (3)
Sol. Given determinant is
2
2
a 3a 2 a 2 1
D a 5a 6 a 3 1
a 7a 12 a 4 1
? ? ?
? ? ? ?
? ? ?
R
3
 ? R
3
 – R
2
; R
2
 ? R
2
 – R
1
2
a 3a 2 a 2 1
2a 4 1 0
2a 6 1 0
? ? ?
? ?
?
Expanding by C
3
D = (2a + 4) – (2a + 6) = –2
4. The maximum slope of the curve
? ? ? ?
4 3 2
1
y x 5x 18x 19x
2
 occurs at the point :
(1) (2, 2) (2) (0, 0)
(3)
? ?
? ?
? ?
21
3,
2
(4) (2, 9)
Answer (1*)
Sol.
4 3 2
1
y x 5x 18x 19x
2
? ? ? ?
Slope = y ? = 2x
3
 – 15x
2
 + 36x – 19 = g(x) say
g ?(x) = 6x
2
 – 30x + 36 = 6(x – 2)(x – 3)
g ?(x) = 0 ? x = 2, 3
Slope g(x) has local maximum at x = 2
x = 2 ? y = 2
Local maximum at (2, 2)
[Note : Overall maximum (Absolute maximum)
value of slope is far greater than that at (2, 2)].
JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
5. In an increasing geometric series, the sum of
the second and the sixth term is 
25
2
and the
product of the third and fifth term is 25. Then,
the sum of 
th th
4 ,6
and 
th
8 term is equal to :
(1) 26 (2) 35
(3) 30 (4) 32
Answer (2)
Sol.
2 6
25
a a
2
? ?
2
3 5 2 6 4
a a 25 a a a ? ? ? ? ?
2
4 4
a 25 a 5 ? ? ?
a
2
 & a
6
 are roots of 
2
25
x x 25 0
2
? ? ?
5
x , 10
2
?
2 6
5
a , a 10
2
? ? ( ? GP is increasing)
a
4
 = 5
a
4
 =a
2
r
2 2 2
5
5 r r 2
2
? ? ? ?
a
8
 = a
6
r
2
 = 10 × 2 = 20
a
4
 + a
6
 + a
8
 = 5 + 10 + 20 = 35
6. The number of seven digit integers with sum of
the digits equal to 10 and formed by using the
digits 1, 2 and 3 only is :
(1) 77 (2) 42
(3) 82 (4) 35
Answer (1)
Sol. Combination of digits
3, 2, 1, 1, 1, 1, 1 ?
7!
42
5!
?
2, 2, 2, 1, 1, 1, 1 ?
7!
35
4! 3!
?
Total = 42 + 35 = 77
7. The sum of infinite series
? ? ? ? ?
2 3 4 5
2 7 12 17 22
1
3
3 3 3 3
+......is equal to :
(1)
13
4
(2)
9
4
(3)
11
4
(4)
15
4
Answer (1)
Sol.
2 3 4 5
2 3 4 5
2 3 4 5
2 7 12 17 22
S 1 ............
3
3 3 3 3
1 1 2 7 12 17
S ...........
3 3
3 3 3 3
2 1 5 5 5 5
S 1 .......
3 3
3 3 3 3
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ?
? ? ? ? ? ? ?
? ?
? ?
5 5
4 5
9 9
1 2
3 3
1
3 3
? ? ? ?
?
?
2 4 5 13
S
3 3 6 6
? ? ?
13
S
4
?
8. Consider the three planes
? ? ?
1
P : 3x 15y 21z 9 ,
? ? ?
2
P : x 3y z 5, and
? ? ?
3
P : 2x 10y 14z 5
Then, which one of the following is true ?
(1)
2 3
P and P are parallel
(2)
1 3
P and P are parallel
(3)
1 2
P and P are parallel
(4)
1 2
P ,P and 
3
P all are parallel
Answer (2)
Sol. Ratios of DRs of normals of P
1
 & P
3
 are
3 15 21
2 10 14
? ?
? Normals are parallel
? P
1
 || P
3
9. Let A be a symmetric matrix of order 2 with
integer entries. If the sum of the diagonal
elements of A
2
 is 1, then the possible number
of such matrices is :
(1) 6 (2) 1
(3) 4 (4) 12
Answer (3)
Sol. Let 
a c
A
c b
? ?
?
? ?
? ?
JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
2 2
2
2 2
a c a c a c ac bc
A
c b c b
ac bc c b
? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ?
a
2
 + b
2
 + 2c
2
 = 1 as a, b, c ?z
c = 0 and a, b = ±1
Total 4 matrices are possible
10. The maximum value of the term independent of
‘t’ in the expansion of 
? ?
? ?
?
? ? ?
? ?
? ?
? ?
10
1
1
10
5
(1 x)
tx
t
where ? x (0,1 ) is :
(1)
2
2.10!
3(5!)
(2)
2
2.10!
3 3(5!)
(3)
2
10!
3(5!)
(4)
2
10!
2(5!)
Answer (2)
Sol.
r
1
1
10
10 10 r
5
r 1 r
(1 x)
T C (tx )
t
?
?
? ?
? ?
?
? ? ?
? ?
? ?
? ?
For term independent of f
10 – r – r = 0  ?  r = 5
1
10
2
6 5
T C x(1 x) f(x) ? ? ?
(Let)
?
1
10
2
5
1
2
x
f (x) C (1 x) 0
2(1 x)
? ?
? ?
? ? ? ? ? ? ?
? ?
? ?
?
? ?
2
2 2x x x
3
? ? ? ?
f ??(x) < 0  at  
2
x
3
?
1
2
10
6(max) 5
2
2 1 2.10!
T C .
3 3
(5!) 3 3
? ?
? ?
? ?
? ?
11. A fair coin is tossed a fixed number of times. If
the probability of getting 7 heads is equal to
probability of getting 9 heads, then the
probability of getting 2 heads is :
(1)
13
15
2
(2)
12
15
2
(3)
8
15
2
(4)
14
15
2
Answer (1)
Sol. Let n number of tosses
Given,
7 n 7 9 n 9
n n
7 9
1 1 1 1
C C
2 2 2 2
? ?
? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? n = 16
? Probability of getting 2 heads = 
16
2
1
16C
2
? ?
? ?
? ?
 
13
15
2
?
12. The value of 
? ? ?
?
?
?
?
n
100
x x
n 1
n 1
e dx,
 where [x] is the
greatest integer ? x, is :
(1) 100(e – 1) (2) 100(1 – e)
(3) 100e (4) 100(1 + e)
Answer (1)
Sol.
? ?
? ?
n 1
x x x
n 1 0
e dx e dx e 1
?
?
? ? ?
? ?
?
? ? ? ?
100
n 1
e 1 100 e 1
?
? ? ?
?
13. In the circle given below, let OA = 1 unit,
OB = 13 unit and 
? PQ OB
. Then, the area of
the triangle PQB (in square units) is :
P
A
Q
B x O
y
(1) 24 2 (2) 24 3
(3) 26 3 (4) 26 2
Answer (2)
Sol. Assume that OB is diameter of the given circle
Using Ptolemy’s Theorem,
OP·QB + OQ·PB = PQ × OB
? 2OP·PB = 13PQ
Also PA
2
 = OP
2
 – 1 = PB
2
 – 12
2
? PB
2
 – OP
2
 = 143
and OP
2
 + PB
2
 = 13
2
JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
then PB
2
 = 156 and OP
2
 = 13
So, 
2 13· 156
PQ 4 3
13
? ?
Area of 
1
PQB ·4 3·12 24 3
2
? ? ?
14. The value of
 
? ?
?
? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ?
?
? ?
? ?
? ?
h 0
3 sin h cos h
6 6
lim 2
3h 3 cos h sin h
 is
(1)
4
3
(2)
3
4
(3)
2
3
(4)
2
3
Answer (1)
Sol. h 0
3 1
sin h cos h
2 6 2 6
lim 2
3 1
3h cosh sinh
2 2
?
? ?
? ? ? ? ? ?
? ?
? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ?
? ?
?
? ?
? ? ? ?
? ? ? ?
? ?
h 0
sin h
2 2 4
lim 2 ·
3
3 3
3h sin h
3
?
? ?
? ?
? ?
? ?
? ?
? ? ?
? ?
?
? ?
? ?
? ? ? ?
15. Let R = {(P, Q) | P and Q are at the same
distance from the origin} be a relation, then the
equivalence class of (1, –1) is the set :
(1) S = {(x, y) | x
2
 + y
2
 = 2}
(2) S = {(x, y) | x
2
 + y
2
 = 1}
(3) S = {(x, y) | x
2
 + y
2
 = 
2
}
(4) S = {(x, y) | x
2
 + y
2
 = 4}
Answer (1)
Sol. ? R = {(P, Q) | P and Q are at the same
distance from the origin}.
Then equivalence class of (1, –1) will
contain all such points which lies on
circumference of the circle of centre at
origin and passing through point (1, –1).
i.e., radius of circle 
2 2
1 1 2 ? ? ?
? Required equivalence class of (S)
= {(x, y) | x
2
 + y
2
 = 2}.
16. The rate of growth of bacteria in a culture is
proportional to the number of bacteria present
and the bacteria count is 1000 at initial time t
= 0. The number of bacteria is increased by
20% in 2 hours. If the population of bacteria is
2000 after 
? ?
? ?
? ?
e
k
6
log
5
 hours, then 
? ?
? ?
? ?
2
e
k
log 2
 is
equal to :
(1) 16 (2) 4
(3) 8 (4) 2
Answer (2)
Sol. At t = 0 ?
?
B 1000
?
dB
B
dt
?
?
?
? ?
?
?
12B 2
B 0
dB
kt
B
[Given]
? ? ?
?
? ?
? ?
?
?
1 2B
ln 2k
B
? ? ?
1
k ln(1 2)
2
To find time when B = 2000
?
? ?
? ?
?
?
2B t
B 0
dB 1
ln(1 2) dt
B 2
? ?
1
ln2 ln(1 2)t
2
? ?
? ?
? ?
? ?
ln4
t hrs.
6
ln
5
? R = ln = 4
Thus 
? ?
? ?
? ?
? ?
2
2
K
2 4
ln
17. Let f be any function defined on R and let it
satisfy the condition :
? ? ? ? ? ? ? ? ? ? ? ? ?
2
f x f y x y , x,y R
If ? ? ? f 0 1 , then :
(1) ? ? f x can take any value in R
(2) ? ? ? ? ? f x 0,  x R
(3) ? ? ? ? ? f x 0,  x R
(4) ? ? ? ? ? f x 0,  x R
Answer (4)
JEE (MAIN)-2021 Phase-1 (26-02-2021)-M
Sol.
?
2
f(x) – f(y) (x – y)
? ?
f(x) – f(y)
x – y
x – y
?
? ?
?
x y x y
f(x) – f(y)
Lim Lim (x – y)
x – y
? ? ? f (x) 0
? f ?(x) = 0
? f(x) is constant function.
? f(0) = 1 then f(x) = 1
18. If 
? ? ?
? ?
1 1 1
sin x cos x tan y
;
a b c
 0 < x < 1, then the
value of 
? ? ?
? ?
?
? ?
c
cos
a b
 is :
(1) 1 – y
2
(2)
?
2
1 y
y y
(3)
?
?
2
2
1 y
1 y
(4)
?
2
1 y
2y
Answer (3)
Sol. ? ? ? ?
–1 –1 –1
sin x cos x tan y
k (say)
a b c
? sin
–1
 x = ak, cos
–1
x = bk and tan
–1
y = ck
Now,
?
? ?
–1 –1
sin x cos x
2
?
? ? (a b)x
2
?
?
?
?
k
2(a b)
Now 
–1
c
tan y
2(a b)
?
?
?
?
? ? ?
?
? ?
? ?
–1
c
cos cos(2 tan y)
ab
? ? ? ?
? ? ? ? ?
? ? ? ?
?
? ? ? ?
2
–1
2
1– y
cos cos
1 y
[if y > 0]
?
?
2
2
1– y
1 y
19. The intersection of three lines x – y = 0,
x + 2y = 3 and 2x + y = 6 is a :
(1) None of the above
(2) Isosceles triangle
(3) Right angled triangle
(4) Equilateral triangle
Answer (2)
Sol. The given three lines are x – y = 0, x + 2y = 3
and 2x + y = 6 then point of intersection
lines x – y = 0 and x + 2y = 3 is (1, 1)
lines x – y = 0 and 2x + y = 6 is (2, 2)
and lines x + 2y = 3 and 2x + y = 0 is (3, 0)
The triangle ABC has vertices A(1, 1), B(2, 2)
and C(3, 0)
? ? ? ? AB 2 , BC 5 and AC 5
? ?ABC is isosceles
20. If (1, 5, 35), (7, 5, 5), (1, ?, 7) and (2 ?, 1, 2) are
coplanar, then the sum of all possible values of
? ?is
(1)
44
5
(2)
?
44
5
(3)
39
5
(4)
?
39
5
Answer (1)
Sol. Four points (1, 5, 35), (7, 5, 5), (1, ?, 7) and (2 ?,
1, 2) are coplanar then
? ?
?
6 0 –30
0 – 5 –28 0
2 – 1 –4 –33
? ? ? ? ?
? ?
3 3 1
6 0 0
0 – 5 –28 (R (C 5 C ) 0
2 – 1 –4 10 – 38
? ? ? ? ? 6 ( – 5)(10 – 38) – 112 0
? 10 ?
2
 – 88 ? + 78 = 0
? 5 ?
2
 – 44 ? + 39 = 0
? Sum of all possible values of ? ?
44
5