JEE Main 2021 Mathematics February 25 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE (MAIN)-2021 Phase-1 (25-02-2021)-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. Let ? be the angle between the lines whose
direction cosines satisfy the equations
? ? ? l m n 0
and ? ? ?
2 2 2
l m n 0 . Then the value
of 
4 4
sin cos ? ? ? is :
(1)
3
4
(2)
5
8
(3)
1
2
(4)
3
8
Answer (2)
Sol. l + m – n = 0  ? l = n – m ...(i)
l
2
 + m
2
 – n
2
 = 0 ...(ii)
Substitute l from (i) into (ii)
? (n – m)
2
 + m
2
 – n
2
 = 0
2m(m – n) = 0
m = 0 or m = n
Case-I
m = 0 ? l = n
l
2
 + m
2
 + n
2
 = 1 ? ? ? ?
2
1 2
1 1 –1
l l , l ,
2
2 2
l = n ? ?
1 2
1 –1
n , n ,
2 2
DCs 
? ? ? ?
? ? ? ?
? ? ? ?
1 1 –1 –1
, 0, or , 0,
2 2 2 2
 are DCs of
same line ? l
1
Case-II
m = n ? l = 0 ? l
1
, l
2
 = 0
l
2
 + m
2
 + n
2
 = 1 ? ? ? ?
2
1 2
1 1 –1
m m , m ,
2
2 2
m = n ? ?
1 2
1 –1
n , n ,
2 2
? ? ? ?
? ? ? ?
? ? ? ?
1 1 –1 –1
DCs 0, , or 0, ,
2 2 2 2
are DCs of l
2
PART–C : MATHEMATICS
? ? ? ? ? ? ? ? ?
1 2 1 2 1 2
1 1
cos l l m m n n 0 0
2 2
? ? ? ? ? ? ? ? ?
2 2 4 4
1 3 5
cos ,sin sin cos
4 4 8
2. If Rolle’s theorem holds for the function
? ? ? ?
3 2
f(x) x ax bx 4 , x [1 , 2] ? with
? ?
? ?
? ?
? ?
4
f 0
3
,
then ordered pair (a, b) is equal to :
(1) ? (5, 8) (2) (5, 8)
(3) ? ? ( 5, 8) (4) ? ( 5, 8)
Answer (2)
Sol. f(x) = x
3
 – ax
2
 + bx – 4
f(1) = f(2)
? 3a – b = 7 ...(i)
f ?(x) = 3x
2
 – 2ax + b
? ?
? ?
? ?
? ?
4
f 0
3
? 8a – 3b = 16 ...(ii)
(i) and (ii)
? a = 5, b = 8
3. All possible values of [0, 2 ] ? ? ? for which h
? ? ? ? sin2 tan2 0
 lie in :
(1)
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
3
0, ,
2 2
(2)
? ? ? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
3 7
0, , ,
2 2 4 6
(3)
? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ?
? ? ? ? ? ?
3 3 11
0, , ,
4 2 4 2 6
(4)
? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
3 5 3 7
0, , , ,
4 2 4 4 2 4
Answer (4)
Sol. sin2 ? + tan2 ? > 0 
sin2 cos2 sin2
0
cos2
? ? ? ? ?
?
?
tan2 ?(1 + cos2 ?) > 0
Page 2


JEE (MAIN)-2021 Phase-1 (25-02-2021)-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. Let ? be the angle between the lines whose
direction cosines satisfy the equations
? ? ? l m n 0
and ? ? ?
2 2 2
l m n 0 . Then the value
of 
4 4
sin cos ? ? ? is :
(1)
3
4
(2)
5
8
(3)
1
2
(4)
3
8
Answer (2)
Sol. l + m – n = 0  ? l = n – m ...(i)
l
2
 + m
2
 – n
2
 = 0 ...(ii)
Substitute l from (i) into (ii)
? (n – m)
2
 + m
2
 – n
2
 = 0
2m(m – n) = 0
m = 0 or m = n
Case-I
m = 0 ? l = n
l
2
 + m
2
 + n
2
 = 1 ? ? ? ?
2
1 2
1 1 –1
l l , l ,
2
2 2
l = n ? ?
1 2
1 –1
n , n ,
2 2
DCs 
? ? ? ?
? ? ? ?
? ? ? ?
1 1 –1 –1
, 0, or , 0,
2 2 2 2
 are DCs of
same line ? l
1
Case-II
m = n ? l = 0 ? l
1
, l
2
 = 0
l
2
 + m
2
 + n
2
 = 1 ? ? ? ?
2
1 2
1 1 –1
m m , m ,
2
2 2
m = n ? ?
1 2
1 –1
n , n ,
2 2
? ? ? ?
? ? ? ?
? ? ? ?
1 1 –1 –1
DCs 0, , or 0, ,
2 2 2 2
are DCs of l
2
PART–C : MATHEMATICS
? ? ? ? ? ? ? ? ?
1 2 1 2 1 2
1 1
cos l l m m n n 0 0
2 2
? ? ? ? ? ? ? ? ?
2 2 4 4
1 3 5
cos ,sin sin cos
4 4 8
2. If Rolle’s theorem holds for the function
? ? ? ?
3 2
f(x) x ax bx 4 , x [1 , 2] ? with
? ?
? ?
? ?
? ?
4
f 0
3
,
then ordered pair (a, b) is equal to :
(1) ? (5, 8) (2) (5, 8)
(3) ? ? ( 5, 8) (4) ? ( 5, 8)
Answer (2)
Sol. f(x) = x
3
 – ax
2
 + bx – 4
f(1) = f(2)
? 3a – b = 7 ...(i)
f ?(x) = 3x
2
 – 2ax + b
? ?
? ?
? ?
? ?
4
f 0
3
? 8a – 3b = 16 ...(ii)
(i) and (ii)
? a = 5, b = 8
3. All possible values of [0, 2 ] ? ? ? for which h
? ? ? ? sin2 tan2 0
 lie in :
(1)
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
3
0, ,
2 2
(2)
? ? ? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
3 7
0, , ,
2 2 4 6
(3)
? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ?
? ? ? ? ? ?
3 3 11
0, , ,
4 2 4 2 6
(4)
? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
3 5 3 7
0, , , ,
4 2 4 4 2 4
Answer (4)
Sol. sin2 ? + tan2 ? > 0 
sin2 cos2 sin2
0
cos2
? ? ? ? ?
?
?
tan2 ?(1 + cos2 ?) > 0
JEE (MAIN)-2021 Phase-1 (25-02-2021)-M
? tan2 ? > 0 and cos2 ? ? –1
? ? ? ?
? ?
? ? ? ?
2 n , n
2 2n 1
2
n
 2n 1  ... ii
, 2n 1  ... i
2
2 4
? ? ?
? ? ? ? ? ?
? ? ? ?
? ?
? ?
?
? ? ? ?
? ? ?
? ? ? ?
? ?
? ?
? ? ? ? ? ? 0, 2
?
? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
3 5 3 7
0, , , , 
4 2 4 4 2 4
4. A man is observing, from the top of a tower, a
boat speeding towards the tower from a certain
point A, with uniform speed. At the point, angle
of depression of the boat with the man’s eye is
30° (Ignore man’s height). After sailing for 20
seconds, towards the base of the tower (which
is at the level of water), the boat has reached
a point B, where the angle of depression is 45°.
Then the time taken (in seconds) by the boat
from B to reach the base of the tower is :
(1)
10 3
(2) 10
(3) ? ?
10 3 1 ?
(4) ? ?
10 3 1 ?
Answer (3)
Sol.
P
45° 30°
h
Q
A
B
h
3h
Let PQ = h
PB = h cot45° = h
PA = h cot30 = 3h
AB = PA – PB
? ?
3 1 h ? ?
Distance
Speed
Time
?
AB PB
20 t
?
? ?
? ?
3 1 h
h 20
t 10 3 1
20 t
3 1
?
? ? ? ? ?
?
5. The value of 
?
?
3
1
2 [x ]
1
x e dx
, where [t] denotes the
greatest integer t, ? is :
(1)
? e 1
3
(2)
1
3e
(3)
? e 1
3e
(4)
? e 1
3e
Answer (3)
Sol.
? ?
? ? ?
3 3 3
1 0 1
2 [x ] 2 [x ] 2 [x ]
–1 –1 0
x e dx x e dx x e dx
? ? ? ?
? ?
0 1
2 –1 2 0
–1 0
x e dx x e dx
? ?
? ?
0 1
2 2
–1 0
1
x dx x dx
e
0 1
3 3
1 0
1 x x
e 3 3
?
? ?
?
? ? ?
1 1 e 1
3e 3 3e
6. The statement ? ? A (B A) is equivalent to :
(1) ? ? A (A B) (2) ? ? A (A B)
(3) ? ? A (A B) (4) ? ? A (A B)
Answer (4)
Sol. B ? A = ~B ? A
Also A ? (B ? A) = A ? (~B ? A) = ~ A ? (~B ? A)
= ~ A ? ~ B ? A = ~ A ? A ? ~ B = t ? ~ B = t
A ? (A ? B)
= ~ A ? (A ? B)
= (~ A ? A) ? B
= t ? B = t
7. Let f, g : 
? N N
 such that ? ? ? ? ? f(n 1 ) f(n) f(1 ) n N
and g be any arbitrary function. Which of the
following statements is NOT true ?
(1) If g is onto, then fog is one-one
(2) If f is onto, then ? ? ? f(n) n n N
(3) f is one-one
(4) If fog is one-one, then g is one-one
Answer (1)
Page 3


JEE (MAIN)-2021 Phase-1 (25-02-2021)-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. Let ? be the angle between the lines whose
direction cosines satisfy the equations
? ? ? l m n 0
and ? ? ?
2 2 2
l m n 0 . Then the value
of 
4 4
sin cos ? ? ? is :
(1)
3
4
(2)
5
8
(3)
1
2
(4)
3
8
Answer (2)
Sol. l + m – n = 0  ? l = n – m ...(i)
l
2
 + m
2
 – n
2
 = 0 ...(ii)
Substitute l from (i) into (ii)
? (n – m)
2
 + m
2
 – n
2
 = 0
2m(m – n) = 0
m = 0 or m = n
Case-I
m = 0 ? l = n
l
2
 + m
2
 + n
2
 = 1 ? ? ? ?
2
1 2
1 1 –1
l l , l ,
2
2 2
l = n ? ?
1 2
1 –1
n , n ,
2 2
DCs 
? ? ? ?
? ? ? ?
? ? ? ?
1 1 –1 –1
, 0, or , 0,
2 2 2 2
 are DCs of
same line ? l
1
Case-II
m = n ? l = 0 ? l
1
, l
2
 = 0
l
2
 + m
2
 + n
2
 = 1 ? ? ? ?
2
1 2
1 1 –1
m m , m ,
2
2 2
m = n ? ?
1 2
1 –1
n , n ,
2 2
? ? ? ?
? ? ? ?
? ? ? ?
1 1 –1 –1
DCs 0, , or 0, ,
2 2 2 2
are DCs of l
2
PART–C : MATHEMATICS
? ? ? ? ? ? ? ? ?
1 2 1 2 1 2
1 1
cos l l m m n n 0 0
2 2
? ? ? ? ? ? ? ? ?
2 2 4 4
1 3 5
cos ,sin sin cos
4 4 8
2. If Rolle’s theorem holds for the function
? ? ? ?
3 2
f(x) x ax bx 4 , x [1 , 2] ? with
? ?
? ?
? ?
? ?
4
f 0
3
,
then ordered pair (a, b) is equal to :
(1) ? (5, 8) (2) (5, 8)
(3) ? ? ( 5, 8) (4) ? ( 5, 8)
Answer (2)
Sol. f(x) = x
3
 – ax
2
 + bx – 4
f(1) = f(2)
? 3a – b = 7 ...(i)
f ?(x) = 3x
2
 – 2ax + b
? ?
? ?
? ?
? ?
4
f 0
3
? 8a – 3b = 16 ...(ii)
(i) and (ii)
? a = 5, b = 8
3. All possible values of [0, 2 ] ? ? ? for which h
? ? ? ? sin2 tan2 0
 lie in :
(1)
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
3
0, ,
2 2
(2)
? ? ? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
3 7
0, , ,
2 2 4 6
(3)
? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ?
? ? ? ? ? ?
3 3 11
0, , ,
4 2 4 2 6
(4)
? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
3 5 3 7
0, , , ,
4 2 4 4 2 4
Answer (4)
Sol. sin2 ? + tan2 ? > 0 
sin2 cos2 sin2
0
cos2
? ? ? ? ?
?
?
tan2 ?(1 + cos2 ?) > 0
JEE (MAIN)-2021 Phase-1 (25-02-2021)-M
? tan2 ? > 0 and cos2 ? ? –1
? ? ? ?
? ?
? ? ? ?
2 n , n
2 2n 1
2
n
 2n 1  ... ii
, 2n 1  ... i
2
2 4
? ? ?
? ? ? ? ? ?
? ? ? ?
? ?
? ?
?
? ? ? ?
? ? ?
? ? ? ?
? ?
? ?
? ? ? ? ? ? 0, 2
?
? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
3 5 3 7
0, , , , 
4 2 4 4 2 4
4. A man is observing, from the top of a tower, a
boat speeding towards the tower from a certain
point A, with uniform speed. At the point, angle
of depression of the boat with the man’s eye is
30° (Ignore man’s height). After sailing for 20
seconds, towards the base of the tower (which
is at the level of water), the boat has reached
a point B, where the angle of depression is 45°.
Then the time taken (in seconds) by the boat
from B to reach the base of the tower is :
(1)
10 3
(2) 10
(3) ? ?
10 3 1 ?
(4) ? ?
10 3 1 ?
Answer (3)
Sol.
P
45° 30°
h
Q
A
B
h
3h
Let PQ = h
PB = h cot45° = h
PA = h cot30 = 3h
AB = PA – PB
? ?
3 1 h ? ?
Distance
Speed
Time
?
AB PB
20 t
?
? ?
? ?
3 1 h
h 20
t 10 3 1
20 t
3 1
?
? ? ? ? ?
?
5. The value of 
?
?
3
1
2 [x ]
1
x e dx
, where [t] denotes the
greatest integer t, ? is :
(1)
? e 1
3
(2)
1
3e
(3)
? e 1
3e
(4)
? e 1
3e
Answer (3)
Sol.
? ?
? ? ?
3 3 3
1 0 1
2 [x ] 2 [x ] 2 [x ]
–1 –1 0
x e dx x e dx x e dx
? ? ? ?
? ?
0 1
2 –1 2 0
–1 0
x e dx x e dx
? ?
? ?
0 1
2 2
–1 0
1
x dx x dx
e
0 1
3 3
1 0
1 x x
e 3 3
?
? ?
?
? ? ?
1 1 e 1
3e 3 3e
6. The statement ? ? A (B A) is equivalent to :
(1) ? ? A (A B) (2) ? ? A (A B)
(3) ? ? A (A B) (4) ? ? A (A B)
Answer (4)
Sol. B ? A = ~B ? A
Also A ? (B ? A) = A ? (~B ? A) = ~ A ? (~B ? A)
= ~ A ? ~ B ? A = ~ A ? A ? ~ B = t ? ~ B = t
A ? (A ? B)
= ~ A ? (A ? B)
= (~ A ? A) ? B
= t ? B = t
7. Let f, g : 
? N N
 such that ? ? ? ? ? f(n 1 ) f(n) f(1 ) n N
and g be any arbitrary function. Which of the
following statements is NOT true ?
(1) If g is onto, then fog is one-one
(2) If f is onto, then ? ? ? f(n) n n N
(3) f is one-one
(4) If fog is one-one, then g is one-one
Answer (1)
JEE (MAIN)-2021 Phase-1 (25-02-2021)-M
Sol. Given f, g : N ? N
& f(n + 1) = f(n) + 1
f(2) 2f(1 )
f(3) 3f(1 )
f(4) 4f(4) f is one one.
.................
f(n) nf(1 )
? ? ?
?
? ?
?
?
? ? ?
?
?
?
? ?
?
Now if f is onto ? f(1) = 1
? f(n) n ?
Also it is clear if fog is one-one ? g will be
one-one.
So only option (1) is not correct.
8. The total number of positive integral solutions
(x, y, z) such that xyz = 24 is :
(1) 36 (2) 30
(3) 45 (4) 24
Answer (2)
Sol. Given xyz = 24 = 2
3
 × 3
So total number of positive integral solutions
(x, y, z)
= 
3 + 3 – 1
C
3 – 1
 × 
1 + 3 – 1
C
3 – 1
= 
5
C
2
 × 
3
C
2
= 10 × 3
= 30
9. When a missile is fired a ship, the probability
that it is intercepted is 
1
3
 and the probability
that the missile hits the target, given that it is
not intercepted, is 
3
4
. If three missiles are fired
independently from the ship, then the
probability that all three hit the target, is  :
(1)
1
27
(2)
3
8
(3)
3
4
(4)
1
8
Answer (4)
Sol. Given P (when it is intercepted) 
1
3
?
? P (being not intercepted)
1 2
1
3 3
? ? ? & also
when it is not intercepted, probability it hits the
target 
3
4
?
So when such 3 missiles launched
then P (all 3 hitting the target)
2 3 2 3 2 3
3 4 3 4 3 4
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
1 1 1
2 2 2
? ? ?
1
8
?
10. Let the lines ? ? ? (2 i)z (2 i)z and  (2 + i)z + (i – 2)
z 4i 0, ? ? 
2
(here i 1 ) ? ? be normal to a circle C. .
If the line ? ? ? ? iz z 1 i 0 is tangent to this circle
C, then its radius is :
(1)
3 2
(2)
3
2
(3)
3
2 2
(4)
1
2 2
Answer (3)
Sol. Given lines are
(2 – i) z = (2 + i) 
z
...(1)
and (2 + i)z + (i – 2)
z
 – 4i = 0
or –i(2 + i)z – i (i – 2)
z
 – 4 = 0
? (1 – 2i)z + (1 + 2i)
z
 – 4 = 0 ...(2)
Let z = x + iy
So from (1) we get the line ?
x
y
2
...(3)
and from (2) (1 – 2i)(x + iy) + (1 + 2i)(x –iy) – 4 = 0
? x + 2y – 2 = 0 ...(4)
On solving (3) and (4) we get x = 1, 
1
y
2
?
? These lines were normal to the circle.
So centre = 
? ?
? ?
? ?
1
1 , 
2
Now the line ? iz z  + 1 + i = 0
or ? ? ? ? ? ? i(1 i)z (1 i)z (1 1 ) 0
? ? ? ? ? ? (1 i)z (1 i)z 2 0
? ? ? ? ? ? (z z) i (z z) 2 0 ? 2x – 2y + 2 = 0
x – y + 1 = 0
? This line is tangent to circle
So,
? ?
? ?
?
1 3
1 1
2 2
r
1 1 2
?
3
r
2 2
Page 4


JEE (MAIN)-2021 Phase-1 (25-02-2021)-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. Let ? be the angle between the lines whose
direction cosines satisfy the equations
? ? ? l m n 0
and ? ? ?
2 2 2
l m n 0 . Then the value
of 
4 4
sin cos ? ? ? is :
(1)
3
4
(2)
5
8
(3)
1
2
(4)
3
8
Answer (2)
Sol. l + m – n = 0  ? l = n – m ...(i)
l
2
 + m
2
 – n
2
 = 0 ...(ii)
Substitute l from (i) into (ii)
? (n – m)
2
 + m
2
 – n
2
 = 0
2m(m – n) = 0
m = 0 or m = n
Case-I
m = 0 ? l = n
l
2
 + m
2
 + n
2
 = 1 ? ? ? ?
2
1 2
1 1 –1
l l , l ,
2
2 2
l = n ? ?
1 2
1 –1
n , n ,
2 2
DCs 
? ? ? ?
? ? ? ?
? ? ? ?
1 1 –1 –1
, 0, or , 0,
2 2 2 2
 are DCs of
same line ? l
1
Case-II
m = n ? l = 0 ? l
1
, l
2
 = 0
l
2
 + m
2
 + n
2
 = 1 ? ? ? ?
2
1 2
1 1 –1
m m , m ,
2
2 2
m = n ? ?
1 2
1 –1
n , n ,
2 2
? ? ? ?
? ? ? ?
? ? ? ?
1 1 –1 –1
DCs 0, , or 0, ,
2 2 2 2
are DCs of l
2
PART–C : MATHEMATICS
? ? ? ? ? ? ? ? ?
1 2 1 2 1 2
1 1
cos l l m m n n 0 0
2 2
? ? ? ? ? ? ? ? ?
2 2 4 4
1 3 5
cos ,sin sin cos
4 4 8
2. If Rolle’s theorem holds for the function
? ? ? ?
3 2
f(x) x ax bx 4 , x [1 , 2] ? with
? ?
? ?
? ?
? ?
4
f 0
3
,
then ordered pair (a, b) is equal to :
(1) ? (5, 8) (2) (5, 8)
(3) ? ? ( 5, 8) (4) ? ( 5, 8)
Answer (2)
Sol. f(x) = x
3
 – ax
2
 + bx – 4
f(1) = f(2)
? 3a – b = 7 ...(i)
f ?(x) = 3x
2
 – 2ax + b
? ?
? ?
? ?
? ?
4
f 0
3
? 8a – 3b = 16 ...(ii)
(i) and (ii)
? a = 5, b = 8
3. All possible values of [0, 2 ] ? ? ? for which h
? ? ? ? sin2 tan2 0
 lie in :
(1)
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
3
0, ,
2 2
(2)
? ? ? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
3 7
0, , ,
2 2 4 6
(3)
? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ?
? ? ? ? ? ?
3 3 11
0, , ,
4 2 4 2 6
(4)
? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
3 5 3 7
0, , , ,
4 2 4 4 2 4
Answer (4)
Sol. sin2 ? + tan2 ? > 0 
sin2 cos2 sin2
0
cos2
? ? ? ? ?
?
?
tan2 ?(1 + cos2 ?) > 0
JEE (MAIN)-2021 Phase-1 (25-02-2021)-M
? tan2 ? > 0 and cos2 ? ? –1
? ? ? ?
? ?
? ? ? ?
2 n , n
2 2n 1
2
n
 2n 1  ... ii
, 2n 1  ... i
2
2 4
? ? ?
? ? ? ? ? ?
? ? ? ?
? ?
? ?
?
? ? ? ?
? ? ?
? ? ? ?
? ?
? ?
? ? ? ? ? ? 0, 2
?
? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
3 5 3 7
0, , , , 
4 2 4 4 2 4
4. A man is observing, from the top of a tower, a
boat speeding towards the tower from a certain
point A, with uniform speed. At the point, angle
of depression of the boat with the man’s eye is
30° (Ignore man’s height). After sailing for 20
seconds, towards the base of the tower (which
is at the level of water), the boat has reached
a point B, where the angle of depression is 45°.
Then the time taken (in seconds) by the boat
from B to reach the base of the tower is :
(1)
10 3
(2) 10
(3) ? ?
10 3 1 ?
(4) ? ?
10 3 1 ?
Answer (3)
Sol.
P
45° 30°
h
Q
A
B
h
3h
Let PQ = h
PB = h cot45° = h
PA = h cot30 = 3h
AB = PA – PB
? ?
3 1 h ? ?
Distance
Speed
Time
?
AB PB
20 t
?
? ?
? ?
3 1 h
h 20
t 10 3 1
20 t
3 1
?
? ? ? ? ?
?
5. The value of 
?
?
3
1
2 [x ]
1
x e dx
, where [t] denotes the
greatest integer t, ? is :
(1)
? e 1
3
(2)
1
3e
(3)
? e 1
3e
(4)
? e 1
3e
Answer (3)
Sol.
? ?
? ? ?
3 3 3
1 0 1
2 [x ] 2 [x ] 2 [x ]
–1 –1 0
x e dx x e dx x e dx
? ? ? ?
? ?
0 1
2 –1 2 0
–1 0
x e dx x e dx
? ?
? ?
0 1
2 2
–1 0
1
x dx x dx
e
0 1
3 3
1 0
1 x x
e 3 3
?
? ?
?
? ? ?
1 1 e 1
3e 3 3e
6. The statement ? ? A (B A) is equivalent to :
(1) ? ? A (A B) (2) ? ? A (A B)
(3) ? ? A (A B) (4) ? ? A (A B)
Answer (4)
Sol. B ? A = ~B ? A
Also A ? (B ? A) = A ? (~B ? A) = ~ A ? (~B ? A)
= ~ A ? ~ B ? A = ~ A ? A ? ~ B = t ? ~ B = t
A ? (A ? B)
= ~ A ? (A ? B)
= (~ A ? A) ? B
= t ? B = t
7. Let f, g : 
? N N
 such that ? ? ? ? ? f(n 1 ) f(n) f(1 ) n N
and g be any arbitrary function. Which of the
following statements is NOT true ?
(1) If g is onto, then fog is one-one
(2) If f is onto, then ? ? ? f(n) n n N
(3) f is one-one
(4) If fog is one-one, then g is one-one
Answer (1)
JEE (MAIN)-2021 Phase-1 (25-02-2021)-M
Sol. Given f, g : N ? N
& f(n + 1) = f(n) + 1
f(2) 2f(1 )
f(3) 3f(1 )
f(4) 4f(4) f is one one.
.................
f(n) nf(1 )
? ? ?
?
? ?
?
?
? ? ?
?
?
?
? ?
?
Now if f is onto ? f(1) = 1
? f(n) n ?
Also it is clear if fog is one-one ? g will be
one-one.
So only option (1) is not correct.
8. The total number of positive integral solutions
(x, y, z) such that xyz = 24 is :
(1) 36 (2) 30
(3) 45 (4) 24
Answer (2)
Sol. Given xyz = 24 = 2
3
 × 3
So total number of positive integral solutions
(x, y, z)
= 
3 + 3 – 1
C
3 – 1
 × 
1 + 3 – 1
C
3 – 1
= 
5
C
2
 × 
3
C
2
= 10 × 3
= 30
9. When a missile is fired a ship, the probability
that it is intercepted is 
1
3
 and the probability
that the missile hits the target, given that it is
not intercepted, is 
3
4
. If three missiles are fired
independently from the ship, then the
probability that all three hit the target, is  :
(1)
1
27
(2)
3
8
(3)
3
4
(4)
1
8
Answer (4)
Sol. Given P (when it is intercepted) 
1
3
?
? P (being not intercepted)
1 2
1
3 3
? ? ? & also
when it is not intercepted, probability it hits the
target 
3
4
?
So when such 3 missiles launched
then P (all 3 hitting the target)
2 3 2 3 2 3
3 4 3 4 3 4
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
1 1 1
2 2 2
? ? ?
1
8
?
10. Let the lines ? ? ? (2 i)z (2 i)z and  (2 + i)z + (i – 2)
z 4i 0, ? ? 
2
(here i 1 ) ? ? be normal to a circle C. .
If the line ? ? ? ? iz z 1 i 0 is tangent to this circle
C, then its radius is :
(1)
3 2
(2)
3
2
(3)
3
2 2
(4)
1
2 2
Answer (3)
Sol. Given lines are
(2 – i) z = (2 + i) 
z
...(1)
and (2 + i)z + (i – 2)
z
 – 4i = 0
or –i(2 + i)z – i (i – 2)
z
 – 4 = 0
? (1 – 2i)z + (1 + 2i)
z
 – 4 = 0 ...(2)
Let z = x + iy
So from (1) we get the line ?
x
y
2
...(3)
and from (2) (1 – 2i)(x + iy) + (1 + 2i)(x –iy) – 4 = 0
? x + 2y – 2 = 0 ...(4)
On solving (3) and (4) we get x = 1, 
1
y
2
?
? These lines were normal to the circle.
So centre = 
? ?
? ?
? ?
1
1 , 
2
Now the line ? iz z  + 1 + i = 0
or ? ? ? ? ? ? i(1 i)z (1 i)z (1 1 ) 0
? ? ? ? ? ? (1 i)z (1 i)z 2 0
? ? ? ? ? ? (z z) i (z z) 2 0 ? 2x – 2y + 2 = 0
x – y + 1 = 0
? This line is tangent to circle
So,
? ?
? ?
?
1 3
1 1
2 2
r
1 1 2
?
3
r
2 2
JEE (MAIN)-2021 Phase-1 (25-02-2021)-M
11. If the curves, 
? ?
2 2
x y
1
a b
 and 
? ?
2 2
x y
1
c d
intersect each other at an angle of 90º, then
which of the following relations is TRUE?
(1) a – c = b + d (2) a + b = c + d
(3) a – b = c – d (4)
?
?
?
c d
ab
a b
Answer (3)
Sol. Given, Curves are ? ?
2 2
x y
1
a b
[Ellipse]
and other curves can be written as
? ?
? ?
?
2 2
x y
1 ,
c d
 Which is a hyperbola
Since these both are orthogonal
So, 
? ? ? a b c d
? a – b = c – d
12. The image of the point (3, 5) in the line x – y +
1 = 0, lies on :
(1) (x – 4)
2
 + (y + 2)
2
 = 16
(2) (x – 4)
2
 + (y – 4)
2
 = 8
(3) (x – 2)
2
 + (y – 2)
2
 = 12
(4) (x – 2)
2
 + (y – 4)
2
 = 4
Answer (4)
Sol. Given the point (3, 5)
and the line x – y + 1 = 0
So, let the image is (x, y)
So, we have
? ? 2 3 5 1
x 3 y 5
1 1 1 1
? ?
? ?
? ? ?
? ?
? x = 4, y = 4
? Point (4, 4)
Which will satisfy the curve
(x – 2)
2
 + (y – 4)
2
 = 4
as (4 – 2)
2
 + (4 – 4)
2
= 4 + 0 = 4
13. The value of the integral
(where c is a constant of integration)
(1)
? ?
? ? ? ? ? ? ?
? ?
3
6 4 2
2
1
9 2cos 3cos 6 cos c
18
(2)
3
2 4 6
2
1
11 18cos 9cos 2cos c
18
? ?
? ? ? ? ? ? ?
? ?
(3)
? ?
? ? ? ? ? ? ?
? ?
3
6 4 2
2
1
9 2sin 3sin 6sin c
18
(4)
? ?
? ? ? ? ? ? ?
? ?
3
2 4 6
2
1
11 18sin 9sin 2 sin c
18
Answer (2)
Sol.
Put sin ? = t
? cos ? d ? = dt
?
? ?
2 6 4 2 4 2
2
t t t t 2t 3t 6
 dt
t
? ? ? ?
?
? ?
? ? ? ?
?
5 3 6 4 2
t t t 2t 3t 6t dt 
Put 2t
6
 + 3t
4
 + 6t
2
 = k
? 12(t
5
 + t
3
 + t) dt = dk
?
1
k dk
12
?
?
3
2
2k
12.3
?
? ?
3
6 4 2
2
1
2sin 3sin 6sin C
18
? ? ? ?
= 
? ?
3
2 4 6
2
1
11 18cos 9cos 2sin C
18
? ? ? ? ? ? ?
14. If a curve passes through the origin and the
slope of the tangent to it at any point (x, y) is
? ? ?
?
2
x 4x y 8
x 2
, then this curve also passes
through the point :
(1) (5, 5) (2) (4, 5)
(3) (4, 4) (4) (5, 4)
Answer (1)
Sol.
? ? ? ?
? ?
2
2
x 2 y 4
dy x 4x y 8
dx x 2 x 2
? ? ?
? ? ?
? ?
? ?
Put x – 2 = t
? dx = dt
Page 5


JEE (MAIN)-2021 Phase-1 (25-02-2021)-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. Let ? be the angle between the lines whose
direction cosines satisfy the equations
? ? ? l m n 0
and ? ? ?
2 2 2
l m n 0 . Then the value
of 
4 4
sin cos ? ? ? is :
(1)
3
4
(2)
5
8
(3)
1
2
(4)
3
8
Answer (2)
Sol. l + m – n = 0  ? l = n – m ...(i)
l
2
 + m
2
 – n
2
 = 0 ...(ii)
Substitute l from (i) into (ii)
? (n – m)
2
 + m
2
 – n
2
 = 0
2m(m – n) = 0
m = 0 or m = n
Case-I
m = 0 ? l = n
l
2
 + m
2
 + n
2
 = 1 ? ? ? ?
2
1 2
1 1 –1
l l , l ,
2
2 2
l = n ? ?
1 2
1 –1
n , n ,
2 2
DCs 
? ? ? ?
? ? ? ?
? ? ? ?
1 1 –1 –1
, 0, or , 0,
2 2 2 2
 are DCs of
same line ? l
1
Case-II
m = n ? l = 0 ? l
1
, l
2
 = 0
l
2
 + m
2
 + n
2
 = 1 ? ? ? ?
2
1 2
1 1 –1
m m , m ,
2
2 2
m = n ? ?
1 2
1 –1
n , n ,
2 2
? ? ? ?
? ? ? ?
? ? ? ?
1 1 –1 –1
DCs 0, , or 0, ,
2 2 2 2
are DCs of l
2
PART–C : MATHEMATICS
? ? ? ? ? ? ? ? ?
1 2 1 2 1 2
1 1
cos l l m m n n 0 0
2 2
? ? ? ? ? ? ? ? ?
2 2 4 4
1 3 5
cos ,sin sin cos
4 4 8
2. If Rolle’s theorem holds for the function
? ? ? ?
3 2
f(x) x ax bx 4 , x [1 , 2] ? with
? ?
? ?
? ?
? ?
4
f 0
3
,
then ordered pair (a, b) is equal to :
(1) ? (5, 8) (2) (5, 8)
(3) ? ? ( 5, 8) (4) ? ( 5, 8)
Answer (2)
Sol. f(x) = x
3
 – ax
2
 + bx – 4
f(1) = f(2)
? 3a – b = 7 ...(i)
f ?(x) = 3x
2
 – 2ax + b
? ?
? ?
? ?
? ?
4
f 0
3
? 8a – 3b = 16 ...(ii)
(i) and (ii)
? a = 5, b = 8
3. All possible values of [0, 2 ] ? ? ? for which h
? ? ? ? sin2 tan2 0
 lie in :
(1)
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
3
0, ,
2 2
(2)
? ? ? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
3 7
0, , ,
2 2 4 6
(3)
? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ?
? ? ? ? ? ?
3 3 11
0, , ,
4 2 4 2 6
(4)
? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
3 5 3 7
0, , , ,
4 2 4 4 2 4
Answer (4)
Sol. sin2 ? + tan2 ? > 0 
sin2 cos2 sin2
0
cos2
? ? ? ? ?
?
?
tan2 ?(1 + cos2 ?) > 0
JEE (MAIN)-2021 Phase-1 (25-02-2021)-M
? tan2 ? > 0 and cos2 ? ? –1
? ? ? ?
? ?
? ? ? ?
2 n , n
2 2n 1
2
n
 2n 1  ... ii
, 2n 1  ... i
2
2 4
? ? ?
? ? ? ? ? ?
? ? ? ?
? ?
? ?
?
? ? ? ?
? ? ?
? ? ? ?
? ?
? ?
? ? ? ? ? ? 0, 2
?
? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
3 5 3 7
0, , , , 
4 2 4 4 2 4
4. A man is observing, from the top of a tower, a
boat speeding towards the tower from a certain
point A, with uniform speed. At the point, angle
of depression of the boat with the man’s eye is
30° (Ignore man’s height). After sailing for 20
seconds, towards the base of the tower (which
is at the level of water), the boat has reached
a point B, where the angle of depression is 45°.
Then the time taken (in seconds) by the boat
from B to reach the base of the tower is :
(1)
10 3
(2) 10
(3) ? ?
10 3 1 ?
(4) ? ?
10 3 1 ?
Answer (3)
Sol.
P
45° 30°
h
Q
A
B
h
3h
Let PQ = h
PB = h cot45° = h
PA = h cot30 = 3h
AB = PA – PB
? ?
3 1 h ? ?
Distance
Speed
Time
?
AB PB
20 t
?
? ?
? ?
3 1 h
h 20
t 10 3 1
20 t
3 1
?
? ? ? ? ?
?
5. The value of 
?
?
3
1
2 [x ]
1
x e dx
, where [t] denotes the
greatest integer t, ? is :
(1)
? e 1
3
(2)
1
3e
(3)
? e 1
3e
(4)
? e 1
3e
Answer (3)
Sol.
? ?
? ? ?
3 3 3
1 0 1
2 [x ] 2 [x ] 2 [x ]
–1 –1 0
x e dx x e dx x e dx
? ? ? ?
? ?
0 1
2 –1 2 0
–1 0
x e dx x e dx
? ?
? ?
0 1
2 2
–1 0
1
x dx x dx
e
0 1
3 3
1 0
1 x x
e 3 3
?
? ?
?
? ? ?
1 1 e 1
3e 3 3e
6. The statement ? ? A (B A) is equivalent to :
(1) ? ? A (A B) (2) ? ? A (A B)
(3) ? ? A (A B) (4) ? ? A (A B)
Answer (4)
Sol. B ? A = ~B ? A
Also A ? (B ? A) = A ? (~B ? A) = ~ A ? (~B ? A)
= ~ A ? ~ B ? A = ~ A ? A ? ~ B = t ? ~ B = t
A ? (A ? B)
= ~ A ? (A ? B)
= (~ A ? A) ? B
= t ? B = t
7. Let f, g : 
? N N
 such that ? ? ? ? ? f(n 1 ) f(n) f(1 ) n N
and g be any arbitrary function. Which of the
following statements is NOT true ?
(1) If g is onto, then fog is one-one
(2) If f is onto, then ? ? ? f(n) n n N
(3) f is one-one
(4) If fog is one-one, then g is one-one
Answer (1)
JEE (MAIN)-2021 Phase-1 (25-02-2021)-M
Sol. Given f, g : N ? N
& f(n + 1) = f(n) + 1
f(2) 2f(1 )
f(3) 3f(1 )
f(4) 4f(4) f is one one.
.................
f(n) nf(1 )
? ? ?
?
? ?
?
?
? ? ?
?
?
?
? ?
?
Now if f is onto ? f(1) = 1
? f(n) n ?
Also it is clear if fog is one-one ? g will be
one-one.
So only option (1) is not correct.
8. The total number of positive integral solutions
(x, y, z) such that xyz = 24 is :
(1) 36 (2) 30
(3) 45 (4) 24
Answer (2)
Sol. Given xyz = 24 = 2
3
 × 3
So total number of positive integral solutions
(x, y, z)
= 
3 + 3 – 1
C
3 – 1
 × 
1 + 3 – 1
C
3 – 1
= 
5
C
2
 × 
3
C
2
= 10 × 3
= 30
9. When a missile is fired a ship, the probability
that it is intercepted is 
1
3
 and the probability
that the missile hits the target, given that it is
not intercepted, is 
3
4
. If three missiles are fired
independently from the ship, then the
probability that all three hit the target, is  :
(1)
1
27
(2)
3
8
(3)
3
4
(4)
1
8
Answer (4)
Sol. Given P (when it is intercepted) 
1
3
?
? P (being not intercepted)
1 2
1
3 3
? ? ? & also
when it is not intercepted, probability it hits the
target 
3
4
?
So when such 3 missiles launched
then P (all 3 hitting the target)
2 3 2 3 2 3
3 4 3 4 3 4
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
1 1 1
2 2 2
? ? ?
1
8
?
10. Let the lines ? ? ? (2 i)z (2 i)z and  (2 + i)z + (i – 2)
z 4i 0, ? ? 
2
(here i 1 ) ? ? be normal to a circle C. .
If the line ? ? ? ? iz z 1 i 0 is tangent to this circle
C, then its radius is :
(1)
3 2
(2)
3
2
(3)
3
2 2
(4)
1
2 2
Answer (3)
Sol. Given lines are
(2 – i) z = (2 + i) 
z
...(1)
and (2 + i)z + (i – 2)
z
 – 4i = 0
or –i(2 + i)z – i (i – 2)
z
 – 4 = 0
? (1 – 2i)z + (1 + 2i)
z
 – 4 = 0 ...(2)
Let z = x + iy
So from (1) we get the line ?
x
y
2
...(3)
and from (2) (1 – 2i)(x + iy) + (1 + 2i)(x –iy) – 4 = 0
? x + 2y – 2 = 0 ...(4)
On solving (3) and (4) we get x = 1, 
1
y
2
?
? These lines were normal to the circle.
So centre = 
? ?
? ?
? ?
1
1 , 
2
Now the line ? iz z  + 1 + i = 0
or ? ? ? ? ? ? i(1 i)z (1 i)z (1 1 ) 0
? ? ? ? ? ? (1 i)z (1 i)z 2 0
? ? ? ? ? ? (z z) i (z z) 2 0 ? 2x – 2y + 2 = 0
x – y + 1 = 0
? This line is tangent to circle
So,
? ?
? ?
?
1 3
1 1
2 2
r
1 1 2
?
3
r
2 2
JEE (MAIN)-2021 Phase-1 (25-02-2021)-M
11. If the curves, 
? ?
2 2
x y
1
a b
 and 
? ?
2 2
x y
1
c d
intersect each other at an angle of 90º, then
which of the following relations is TRUE?
(1) a – c = b + d (2) a + b = c + d
(3) a – b = c – d (4)
?
?
?
c d
ab
a b
Answer (3)
Sol. Given, Curves are ? ?
2 2
x y
1
a b
[Ellipse]
and other curves can be written as
? ?
? ?
?
2 2
x y
1 ,
c d
 Which is a hyperbola
Since these both are orthogonal
So, 
? ? ? a b c d
? a – b = c – d
12. The image of the point (3, 5) in the line x – y +
1 = 0, lies on :
(1) (x – 4)
2
 + (y + 2)
2
 = 16
(2) (x – 4)
2
 + (y – 4)
2
 = 8
(3) (x – 2)
2
 + (y – 2)
2
 = 12
(4) (x – 2)
2
 + (y – 4)
2
 = 4
Answer (4)
Sol. Given the point (3, 5)
and the line x – y + 1 = 0
So, let the image is (x, y)
So, we have
? ? 2 3 5 1
x 3 y 5
1 1 1 1
? ?
? ?
? ? ?
? ?
? x = 4, y = 4
? Point (4, 4)
Which will satisfy the curve
(x – 2)
2
 + (y – 4)
2
 = 4
as (4 – 2)
2
 + (4 – 4)
2
= 4 + 0 = 4
13. The value of the integral
(where c is a constant of integration)
(1)
? ?
? ? ? ? ? ? ?
? ?
3
6 4 2
2
1
9 2cos 3cos 6 cos c
18
(2)
3
2 4 6
2
1
11 18cos 9cos 2cos c
18
? ?
? ? ? ? ? ? ?
? ?
(3)
? ?
? ? ? ? ? ? ?
? ?
3
6 4 2
2
1
9 2sin 3sin 6sin c
18
(4)
? ?
? ? ? ? ? ? ?
? ?
3
2 4 6
2
1
11 18sin 9sin 2 sin c
18
Answer (2)
Sol.
Put sin ? = t
? cos ? d ? = dt
?
? ?
2 6 4 2 4 2
2
t t t t 2t 3t 6
 dt
t
? ? ? ?
?
? ?
? ? ? ?
?
5 3 6 4 2
t t t 2t 3t 6t dt 
Put 2t
6
 + 3t
4
 + 6t
2
 = k
? 12(t
5
 + t
3
 + t) dt = dk
?
1
k dk
12
?
?
3
2
2k
12.3
?
? ?
3
6 4 2
2
1
2sin 3sin 6sin C
18
? ? ? ?
= 
? ?
3
2 4 6
2
1
11 18cos 9cos 2sin C
18
? ? ? ? ? ? ?
14. If a curve passes through the origin and the
slope of the tangent to it at any point (x, y) is
? ? ?
?
2
x 4x y 8
x 2
, then this curve also passes
through the point :
(1) (5, 5) (2) (4, 5)
(3) (4, 4) (4) (5, 4)
Answer (1)
Sol.
? ? ? ?
? ?
2
2
x 2 y 4
dy x 4x y 8
dx x 2 x 2
? ? ?
? ? ?
? ?
? ?
Put x – 2 = t
? dx = dt
JEE (MAIN)-2021 Phase-1 (25-02-2021)-M
?
? ?
?
2
dy t y 4
dt t
?
? ? ?
dy y 4
t
dt t t
?
?
? ?
1
dt
t
1
I.F e
t
?
? ? ?
y 4
t C
t t
y = (x – 2)
2
 – 4 + C(x – 2)
? (0, 0)
C = 0
y = (x – 2)
2
 – 4 also passes through (5, 5)
15. If 
? ?
? ?
?
? ? ? ? ? ? ? ?
? ?
2n 2n
n 0 n 0
0 , ,x cos , y sin
2
 and
?
?
? ? ? ?
?
2n 2n
n 0
z cos sin
 then :
(1) xyz = 4
(2) xy – z = (x + y) z
(3) xy + yz + zx = z
(4) xy + z = (x + y)z
Answer (4)
Sol.
?
?
? ? ? ? ?
?
2n 2 4
n 0
x cos 1 + cos + cos + ...
? ? ?
? ?
2
2
1
cosec
1 cos
2n 2 4
n 0
y sin 1 sin sin ...
?
?
? ? ? ? ? ? ? ?
?
? ? ?
? ?
2
2
1
sec
1 sin
?
?
? ? ?
?
2n 2n
n 0
z cos sin
? ?
2
2 2 2 2
1 cos sin cos sin ... ? ? ? ? ? ? ? ?
2 2
1
1 cos sin
?
? ? ?
?
?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
1
z
1 1
1 1 1
x y
?
? ? ? ? ?
1 1 1 1
1 1
x y xy z
?
? ?
?
x y z xy
xy xyz
? (x + y)z = xy + z
16. The equation of the line through the point
(0, 1, 2) and perpendicular to the line
? ? ?
? ?
?
x 1 y 1 z 1
2 3 2
 is :
(1)
? ?
? ?
?
x y 1 z 2
3 4 3
(2)
? ?
? ?
x y 1 z 2
3 4 3
(3)
? ?
? ?
?
x y 1 z 2
3 4 3
(4)
? ?
? ?
?
x y 1 z 2
3 4 3
Answer (3)
Sol. Let equation of line 
? ?
x y – 1 z – 2
a b c
for being perpendicular to 
x y 1 z – 1
2 3 –2
?
? ?
we get
2a + 3b – 2c = 0
Hence satisfying this equation a : b : c = –3 : 4 : 3
Hence required line is ? ?
x – 1 y – 1 z – 2
–3 4 3
17.
? ?
? ? ?
? ?
?
? ? ? ?
? ?
? ?
? ?
n
2
1 1
1 ......
lim
2 n
1
n
n
 is equal to :
(1)
1
2
(2) 0
(3) 1 (4)
1
e
Answer (3)
Sol.
? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ?
?
? ?
? ?
? ?
n
2
n
1 1 1
1 1 ....
2 3 n
L Lim
n
if n ? ?  
? ? ? ? ?
1 1 1
1 .... n
2 3 n
hence 
? ?
? ? ?
?
2
n
1 1
1 ...
2 n
Lim 0
n
L is of 1
?
 form
n
Lim
2
1 1 1
1 ...
2 3 n
L e n e 1
n
? ?
? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ?
? ?
? ?
?