JEE Main 2021 Mathematics February 26 Shift 2 Paper & Solutions

# JEE Main 2021 Mathematics February 26 Shift 2 Paper & Solutions | JEE Main & Advanced Mock Test Series PDF Download

``` Page 1

JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
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.
1. If the mirror image of the point (1, 3, 5) with
respect to the plane 4x – 5y + 2z = 8 is
( ?, ?, ?), then 5(?? ?? ?? ?? ?) equals :
(1) 43 (2) 47
(3) 41 (4) 39
Sol.
??
?? ?? ?? ? ? ??? ?
?? ??
???
13 5 4153258
2
45 2 16254
?? ? ? ? ?
?? ?
?
13 52
45 2 5
?
?? ? ? ? ? ? ? ?
810 4
1, 3, 5
55 5
?????? ?? ?? ?? 555547
2. A natural number has prime factorization
given by n = 2
x
3
y
5
z
, where y and z are such
that y + z = 5 and y
–1
+ z
–1
=
5
,y z
6
?
. Then the
number of odd divisors of n, including 1, is :
(1) 12 (2) 6x
(3) 11 (4) 6
Sol. y + z = 5 …(i)
?
?? ? ? ?
11 y z 5
yz 6
yz yz 6
…(ii)
Equation with y and z as roots is
x
2
– 5x + 6 = 0
x = 2, 3, y = 3, z = 2 (y > z)
n = 2
x
· 3
3
· 5
2
For odd divisors x = 1 only
No. of odd divisors = 1 × 4 × 3 = 12
3. Let A = {1, 2, 3, ..., 10} and f : A ? A be
defined as
f(k) =
k1 ifkisodd
kifkiseven
? ?
?
?
PART–C : MATHEMATICS
Then the number of possible functions
g : A ? A such that gof = f is :
(1) 5
5
(2) 10
5
(3) 5! (4)
10
C
5
Sol. Not that f(1) = f(2) = 2
f(3) = f(4) = 4
f(5) = f(6) = 6
f(7) = f(8) = 8
f(9) = f(10) = 10
gof(1) = f(1) ? g(2) = f(1) = 2
gof(2) = f(2) ? g(2) = f(2) = 2
gof(3) = f(3) ? g(4) = f(3) = 4
? In function g(x), 2, 4, 6, 8, 10 should be
mapped to 2, 4, 6, 8, 10 respectively. Each
of remaining elements can be mapped to
any of 10 elements.
Number of possible g(x) is 10
5
4. Let slope of the tangent line to a curve at any
point P(x, y) be given by
2
xyy
.
x
?
If the curve
intersects the line x + 2y = 4 at x = –2, then the
value of y, for which the point (3, y) lies on the
curve, is :
(1)
18
35
(2)
4
–
3
(3)
18
–
11
(4)
18
–
19
Sol.
??
2
dy y
y
dx x
??
2
dy y
y
dx x
?? ?
2
1dy 1 1
1
dx x y y
Let
?
1
z
y
?
?
2
1dy dz
dx dx y
?
??
dz 1
z1
dx x
Page 2

JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
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.
1. If the mirror image of the point (1, 3, 5) with
respect to the plane 4x – 5y + 2z = 8 is
( ?, ?, ?), then 5(?? ?? ?? ?? ?) equals :
(1) 43 (2) 47
(3) 41 (4) 39
Sol.
??
?? ?? ?? ? ? ??? ?
?? ??
???
13 5 4153258
2
45 2 16254
?? ? ? ? ?
?? ?
?
13 52
45 2 5
?
?? ? ? ? ? ? ? ?
810 4
1, 3, 5
55 5
?????? ?? ?? ?? 555547
2. A natural number has prime factorization
given by n = 2
x
3
y
5
z
, where y and z are such
that y + z = 5 and y
–1
+ z
–1
=
5
,y z
6
?
. Then the
number of odd divisors of n, including 1, is :
(1) 12 (2) 6x
(3) 11 (4) 6
Sol. y + z = 5 …(i)
?
?? ? ? ?
11 y z 5
yz 6
yz yz 6
…(ii)
Equation with y and z as roots is
x
2
– 5x + 6 = 0
x = 2, 3, y = 3, z = 2 (y > z)
n = 2
x
· 3
3
· 5
2
For odd divisors x = 1 only
No. of odd divisors = 1 × 4 × 3 = 12
3. Let A = {1, 2, 3, ..., 10} and f : A ? A be
defined as
f(k) =
k1 ifkisodd
kifkiseven
? ?
?
?
PART–C : MATHEMATICS
Then the number of possible functions
g : A ? A such that gof = f is :
(1) 5
5
(2) 10
5
(3) 5! (4)
10
C
5
Sol. Not that f(1) = f(2) = 2
f(3) = f(4) = 4
f(5) = f(6) = 6
f(7) = f(8) = 8
f(9) = f(10) = 10
gof(1) = f(1) ? g(2) = f(1) = 2
gof(2) = f(2) ? g(2) = f(2) = 2
gof(3) = f(3) ? g(4) = f(3) = 4
? In function g(x), 2, 4, 6, 8, 10 should be
mapped to 2, 4, 6, 8, 10 respectively. Each
of remaining elements can be mapped to
any of 10 elements.
Number of possible g(x) is 10
5
4. Let slope of the tangent line to a curve at any
point P(x, y) be given by
2
xyy
.
x
?
If the curve
intersects the line x + 2y = 4 at x = –2, then the
value of y, for which the point (3, y) lies on the
curve, is :
(1)
18
35
(2)
4
–
3
(3)
18
–
11
(4)
18
–
19
Sol.
??
2
dy y
y
dx x
??
2
dy y
y
dx x
?? ?
2
1dy 1 1
1
dx x y y
Let
?
1
z
y
?
?
2
1dy dz
dx dx y
?
??
dz 1
z1
dx x
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
???
dz 1
z1
dx x
IF =
1
dx
nx
x
ee x
?
??

?? ? ?
?
zx 1xdx
?
?? ?
2
x
zx c
2
2
xx
c
y2
?
??
…(i)
Putting x = –2 in x + 2y = 4, we get y = 3
Put (–2, 3) in (i)
?
?
4
c
3
(i)??
?
??
2
xx 4
y 23
…(ii)
Put x = 3 in (ii)
?
??
394
y2 3
?
?
18
y
19
5. The triangle of maximum area that can be
inscribed in a given circle of radius ‘r’ is :
(1) An isosceles triangle with base equal to 2r.
(2) An equilateral triangle of height
2r
.
3
(3) A right angle triangle having two of its
sides of length 2r and r.
(4) An equilateral triangle having each of its
side of length 3r.
Sol. Area of triangle ABC
A
BC
x
r
O
r
r
M -
22
rx
?? ?
1
ABCAM
2
?? ? ? ?
22
1
2r x (r x)
2
?? ?
22
A (rx)rx
?? ?
?? ? ? ? ?
??
22 2
22
22 22
dA x r x rx x
rx (rx)
dx
rx rx
?? ?? ?
??
??
22
22 22
rrx2x (xr)(2x r)
rx rx
? ? ?
dA r
0x
dx 2
Sign change of
dA
dx
at ?
r
x
2
? A has
maximum at ?
r
x
2
?? ?
22
BC 2 r x 3r,
?
3
AM r
2
? ?? AB AC 3r
6. A seven digit number is formed using digits
3, 3, 4, 4, 4, 5, 5. The probability, that number
so formed is divisible by 2, is :
(1)
1
7
(2)
6
7
(3)
4
7
(4)
3
7
Sol. For even number, units place should be filled
with 4 only.
????
6!
6! 3! 3
2!2!2!
P
7!
2! 7! 7
2!3!2!
7. Let F
1
(A, B, C) = (A ? ~B) ? [~C ? (A ? B)] ? ~A
and F
2
(A, B) = (A ? B) ? (B ?  ~A) be two
logical expressions. Then :
(1) F
1
and F
2
both are tautologies
(2) Both F
1
and F
2
are not tautologies
(3) F
1
is a tautology but F
2
is not a tautology
(4) F
1
is not a tautology but F
2
is a tautology
Sol.
A
B
C
A ~B ?
Page 3

JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
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.
1. If the mirror image of the point (1, 3, 5) with
respect to the plane 4x – 5y + 2z = 8 is
( ?, ?, ?), then 5(?? ?? ?? ?? ?) equals :
(1) 43 (2) 47
(3) 41 (4) 39
Sol.
??
?? ?? ?? ? ? ??? ?
?? ??
???
13 5 4153258
2
45 2 16254
?? ? ? ? ?
?? ?
?
13 52
45 2 5
?
?? ? ? ? ? ? ? ?
810 4
1, 3, 5
55 5
?????? ?? ?? ?? 555547
2. A natural number has prime factorization
given by n = 2
x
3
y
5
z
, where y and z are such
that y + z = 5 and y
–1
+ z
–1
=
5
,y z
6
?
. Then the
number of odd divisors of n, including 1, is :
(1) 12 (2) 6x
(3) 11 (4) 6
Sol. y + z = 5 …(i)
?
?? ? ? ?
11 y z 5
yz 6
yz yz 6
…(ii)
Equation with y and z as roots is
x
2
– 5x + 6 = 0
x = 2, 3, y = 3, z = 2 (y > z)
n = 2
x
· 3
3
· 5
2
For odd divisors x = 1 only
No. of odd divisors = 1 × 4 × 3 = 12
3. Let A = {1, 2, 3, ..., 10} and f : A ? A be
defined as
f(k) =
k1 ifkisodd
kifkiseven
? ?
?
?
PART–C : MATHEMATICS
Then the number of possible functions
g : A ? A such that gof = f is :
(1) 5
5
(2) 10
5
(3) 5! (4)
10
C
5
Sol. Not that f(1) = f(2) = 2
f(3) = f(4) = 4
f(5) = f(6) = 6
f(7) = f(8) = 8
f(9) = f(10) = 10
gof(1) = f(1) ? g(2) = f(1) = 2
gof(2) = f(2) ? g(2) = f(2) = 2
gof(3) = f(3) ? g(4) = f(3) = 4
? In function g(x), 2, 4, 6, 8, 10 should be
mapped to 2, 4, 6, 8, 10 respectively. Each
of remaining elements can be mapped to
any of 10 elements.
Number of possible g(x) is 10
5
4. Let slope of the tangent line to a curve at any
point P(x, y) be given by
2
xyy
.
x
?
If the curve
intersects the line x + 2y = 4 at x = –2, then the
value of y, for which the point (3, y) lies on the
curve, is :
(1)
18
35
(2)
4
–
3
(3)
18
–
11
(4)
18
–
19
Sol.
??
2
dy y
y
dx x
??
2
dy y
y
dx x
?? ?
2
1dy 1 1
1
dx x y y
Let
?
1
z
y
?
?
2
1dy dz
dx dx y
?
??
dz 1
z1
dx x
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
???
dz 1
z1
dx x
IF =
1
dx
nx
x
ee x
?
??

?? ? ?
?
zx 1xdx
?
?? ?
2
x
zx c
2
2
xx
c
y2
?
??
…(i)
Putting x = –2 in x + 2y = 4, we get y = 3
Put (–2, 3) in (i)
?
?
4
c
3
(i)??
?
??
2
xx 4
y 23
…(ii)
Put x = 3 in (ii)
?
??
394
y2 3
?
?
18
y
19
5. The triangle of maximum area that can be
inscribed in a given circle of radius ‘r’ is :
(1) An isosceles triangle with base equal to 2r.
(2) An equilateral triangle of height
2r
.
3
(3) A right angle triangle having two of its
sides of length 2r and r.
(4) An equilateral triangle having each of its
side of length 3r.
Sol. Area of triangle ABC
A
BC
x
r
O
r
r
M -
22
rx
?? ?
1
ABCAM
2
?? ? ? ?
22
1
2r x (r x)
2
?? ?
22
A (rx)rx
?? ?
?? ? ? ? ?
??
22 2
22
22 22
dA x r x rx x
rx (rx)
dx
rx rx
?? ?? ?
??
??
22
22 22
rrx2x (xr)(2x r)
rx rx
? ? ?
dA r
0x
dx 2
Sign change of
dA
dx
at ?
r
x
2
? A has
maximum at ?
r
x
2
?? ?
22
BC 2 r x 3r,
?
3
AM r
2
? ?? AB AC 3r
6. A seven digit number is formed using digits
3, 3, 4, 4, 4, 5, 5. The probability, that number
so formed is divisible by 2, is :
(1)
1
7
(2)
6
7
(3)
4
7
(4)
3
7
Sol. For even number, units place should be filled
with 4 only.
????
6!
6! 3! 3
2!2!2!
P
7!
2! 7! 7
2!3!2!
7. Let F
1
(A, B, C) = (A ? ~B) ? [~C ? (A ? B)] ? ~A
and F
2
(A, B) = (A ? B) ? (B ?  ~A) be two
logical expressions. Then :
(1) F
1
and F
2
both are tautologies
(2) Both F
1
and F
2
are not tautologies
(3) F
1
is a tautology but F
2
is not a tautology
(4) F
1
is not a tautology but F
2
is a tautology
Sol.
A
B
C
A ~B ?
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
~C (A B ) ?? ?? ?
~A
?
F1 is not a
tautology
F1
?? ? B~A~B~A
A  B ?
~B  ~A = ~(A B) ??
?
=  F
2
A  B ?
~B  ~A = ~(A B) ??
?
=  F
2
? F
2
is a tautology
8. Let f : R ? R be defined as
2
x
2sin – ,if x – 1
2
f(x)|ax x b|,if –1 x 1
sin( x), if x 1
? ? ??
?
?? ?
??
?
?
??? ??
?
?
??
?
?
?
If f(x) is continuous on R, then a + b equals :
(1) –1 (2) –3
(3) 3 (4) 1
Sol.
?
?? f(1) 2
?
?? ? ? f( 1 ) ab 1
?? ? ab 1 2
...(i)
?
?? ? f(1 ) ab1
?
? f(1 ) 0
?? ? ? ?? ? ab1 0 ab1 0
? a + b = –1 ...(ii)
9. Let L be a line obtained from the intersection
of two planes x + 2y + z = 6 and y + 2z = 4.
If point P( ??? ??? ?) is the foot of perpendicular
from (3, 2, 1) on L, then the value of
21(?? ?? ?? ?? ?) equals :
(1) 68 (2) 102
(3) 142 (4) 136
Sol. Direction of line L = ?? ?
ˆ ˆˆ
ij k
ˆ ˆˆ
12 1 3i 2j k
01 2
d.r’s = < 3, –2, 1>
A point on line (–2, 4, 0)
Line =
??
??
?
x2 y 4 z
321
Foot of perpendicular from (3, 2, 1) be (3 ? – 2,
–2 ? + 4, ?)
(3 ? – 5) . 3 + (–2 ? + 2) (–2) + ( ? – 1)1 = 0
9 ? – 15 + 4 ? – 4 + ? – 1 = 0
14 ? – 20  = 0 ? ??
10
7
??
?? ? ?
??
??
16 8 10
(, , ) , ,
77 7
? 21( ? + ? + ?) = (16 + 8 + 10)3 = 102
10. Let f(x) be a differentiable function at x = a
with f ?(a) = 2 and f(a) = 4. Then
xa
xf(a)–af(x)
lim
x–a ?
equals :
(1) 4 – 2a (2) a + 4
(3) 2a – 4 (4) 2a + 4
Sol.
?
? ??
?
??
? ?? xa
xf(a) af(x) 0
Llim form
xa 0
Using L’ Hospital rule we get
?
??
?
xa
f(a) af (x)
Llim
1
?? ? ? f(a) af (a) 4 2a
Page 4

JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
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.
1. If the mirror image of the point (1, 3, 5) with
respect to the plane 4x – 5y + 2z = 8 is
( ?, ?, ?), then 5(?? ?? ?? ?? ?) equals :
(1) 43 (2) 47
(3) 41 (4) 39
Sol.
??
?? ?? ?? ? ? ??? ?
?? ??
???
13 5 4153258
2
45 2 16254
?? ? ? ? ?
?? ?
?
13 52
45 2 5
?
?? ? ? ? ? ? ? ?
810 4
1, 3, 5
55 5
?????? ?? ?? ?? 555547
2. A natural number has prime factorization
given by n = 2
x
3
y
5
z
, where y and z are such
that y + z = 5 and y
–1
+ z
–1
=
5
,y z
6
?
. Then the
number of odd divisors of n, including 1, is :
(1) 12 (2) 6x
(3) 11 (4) 6
Sol. y + z = 5 …(i)
?
?? ? ? ?
11 y z 5
yz 6
yz yz 6
…(ii)
Equation with y and z as roots is
x
2
– 5x + 6 = 0
x = 2, 3, y = 3, z = 2 (y > z)
n = 2
x
· 3
3
· 5
2
For odd divisors x = 1 only
No. of odd divisors = 1 × 4 × 3 = 12
3. Let A = {1, 2, 3, ..., 10} and f : A ? A be
defined as
f(k) =
k1 ifkisodd
kifkiseven
? ?
?
?
PART–C : MATHEMATICS
Then the number of possible functions
g : A ? A such that gof = f is :
(1) 5
5
(2) 10
5
(3) 5! (4)
10
C
5
Sol. Not that f(1) = f(2) = 2
f(3) = f(4) = 4
f(5) = f(6) = 6
f(7) = f(8) = 8
f(9) = f(10) = 10
gof(1) = f(1) ? g(2) = f(1) = 2
gof(2) = f(2) ? g(2) = f(2) = 2
gof(3) = f(3) ? g(4) = f(3) = 4
? In function g(x), 2, 4, 6, 8, 10 should be
mapped to 2, 4, 6, 8, 10 respectively. Each
of remaining elements can be mapped to
any of 10 elements.
Number of possible g(x) is 10
5
4. Let slope of the tangent line to a curve at any
point P(x, y) be given by
2
xyy
.
x
?
If the curve
intersects the line x + 2y = 4 at x = –2, then the
value of y, for which the point (3, y) lies on the
curve, is :
(1)
18
35
(2)
4
–
3
(3)
18
–
11
(4)
18
–
19
Sol.
??
2
dy y
y
dx x
??
2
dy y
y
dx x
?? ?
2
1dy 1 1
1
dx x y y
Let
?
1
z
y
?
?
2
1dy dz
dx dx y
?
??
dz 1
z1
dx x
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
???
dz 1
z1
dx x
IF =
1
dx
nx
x
ee x
?
??

?? ? ?
?
zx 1xdx
?
?? ?
2
x
zx c
2
2
xx
c
y2
?
??
…(i)
Putting x = –2 in x + 2y = 4, we get y = 3
Put (–2, 3) in (i)
?
?
4
c
3
(i)??
?
??
2
xx 4
y 23
…(ii)
Put x = 3 in (ii)
?
??
394
y2 3
?
?
18
y
19
5. The triangle of maximum area that can be
inscribed in a given circle of radius ‘r’ is :
(1) An isosceles triangle with base equal to 2r.
(2) An equilateral triangle of height
2r
.
3
(3) A right angle triangle having two of its
sides of length 2r and r.
(4) An equilateral triangle having each of its
side of length 3r.
Sol. Area of triangle ABC
A
BC
x
r
O
r
r
M -
22
rx
?? ?
1
ABCAM
2
?? ? ? ?
22
1
2r x (r x)
2
?? ?
22
A (rx)rx
?? ?
?? ? ? ? ?
??
22 2
22
22 22
dA x r x rx x
rx (rx)
dx
rx rx
?? ?? ?
??
??
22
22 22
rrx2x (xr)(2x r)
rx rx
? ? ?
dA r
0x
dx 2
Sign change of
dA
dx
at ?
r
x
2
? A has
maximum at ?
r
x
2
?? ?
22
BC 2 r x 3r,
?
3
AM r
2
? ?? AB AC 3r
6. A seven digit number is formed using digits
3, 3, 4, 4, 4, 5, 5. The probability, that number
so formed is divisible by 2, is :
(1)
1
7
(2)
6
7
(3)
4
7
(4)
3
7
Sol. For even number, units place should be filled
with 4 only.
????
6!
6! 3! 3
2!2!2!
P
7!
2! 7! 7
2!3!2!
7. Let F
1
(A, B, C) = (A ? ~B) ? [~C ? (A ? B)] ? ~A
and F
2
(A, B) = (A ? B) ? (B ?  ~A) be two
logical expressions. Then :
(1) F
1
and F
2
both are tautologies
(2) Both F
1
and F
2
are not tautologies
(3) F
1
is a tautology but F
2
is not a tautology
(4) F
1
is not a tautology but F
2
is a tautology
Sol.
A
B
C
A ~B ?
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
~C (A B ) ?? ?? ?
~A
?
F1 is not a
tautology
F1
?? ? B~A~B~A
A  B ?
~B  ~A = ~(A B) ??
?
=  F
2
A  B ?
~B  ~A = ~(A B) ??
?
=  F
2
? F
2
is a tautology
8. Let f : R ? R be defined as
2
x
2sin – ,if x – 1
2
f(x)|ax x b|,if –1 x 1
sin( x), if x 1
? ? ??
?
?? ?
??
?
?
??? ??
?
?
??
?
?
?
If f(x) is continuous on R, then a + b equals :
(1) –1 (2) –3
(3) 3 (4) 1
Sol.
?
?? f(1) 2
?
?? ? ? f( 1 ) ab 1
?? ? ab 1 2
...(i)
?
?? ? f(1 ) ab1
?
? f(1 ) 0
?? ? ? ?? ? ab1 0 ab1 0
? a + b = –1 ...(ii)
9. Let L be a line obtained from the intersection
of two planes x + 2y + z = 6 and y + 2z = 4.
If point P( ??? ??? ?) is the foot of perpendicular
from (3, 2, 1) on L, then the value of
21(?? ?? ?? ?? ?) equals :
(1) 68 (2) 102
(3) 142 (4) 136
Sol. Direction of line L = ?? ?
ˆ ˆˆ
ij k
ˆ ˆˆ
12 1 3i 2j k
01 2
d.r’s = < 3, –2, 1>
A point on line (–2, 4, 0)
Line =
??
??
?
x2 y 4 z
321
Foot of perpendicular from (3, 2, 1) be (3 ? – 2,
–2 ? + 4, ?)
(3 ? – 5) . 3 + (–2 ? + 2) (–2) + ( ? – 1)1 = 0
9 ? – 15 + 4 ? – 4 + ? – 1 = 0
14 ? – 20  = 0 ? ??
10
7
??
?? ? ?
??
??
16 8 10
(, , ) , ,
77 7
? 21( ? + ? + ?) = (16 + 8 + 10)3 = 102
10. Let f(x) be a differentiable function at x = a
with f ?(a) = 2 and f(a) = 4. Then
xa
xf(a)–af(x)
lim
x–a ?
equals :
(1) 4 – 2a (2) a + 4
(3) 2a – 4 (4) 2a + 4
Sol.
?
? ??
?
??
? ?? xa
xf(a) af(x) 0
Llim form
xa 0
Using L’ Hospital rule we get
?
??
?
xa
f(a) af (x)
Llim
1
?? ? ? f(a) af (a) 4 2a
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
11. Let A(1, 4) and B(1, –5) be two points. Let P be
a point on the circle (x – 1)
2
+ (y – 1)
2
= 1 such
that (PA)
2
+ (PB)
2
have maximum value, then
the points, P, A and B lie on :
(1) an ellipse (2) a parabola
(3) a straight line (4) a hyperbola
Sol. Let P be (1 + cos ?, 1 + sin ?)
(PA)
2
+ (PB)
2
=
?? ? ? ?? ?? ?? ? ?
22 2
cos sin 3 cos
?? ???
2
sin 6
= 1 – 6sin ? + 9 + 1 + 12 sin ? + 36
= 45 + 6 sin ? maximum at
?
??
2
? P (1, 2)
? P, A and B are colinear
12. Let A
1
be the area of the region bounded by
the curves y = sinx, y = cosx and y-axis in the
2
be the area of the
region bounded by the curves y = sinx,
y = cosx, x-axis and x =
2
?
Then,
(1) A
1
: A
2
= 1 :
2
and A
1
+ A
2
= 1
(2) A
1
: A
2
= 1 : 2 and A
1
+ A
2
= 1
(3) 2A
1
= A
2
and A
1
+ A
2
= 1 + 2
(4) A
1
= A
2
and A
1
+ A
2
=
2
Sol.
A
1
A
2
X
Y
?
2
A
1
= ??
?
???
?
2
0
cosx sinx dx 2 1
A
2
= ??
? ?
?
?? ?
??
4 2
0
4
sinx dx cosx dx 2 2 1
? A
1
: A
2
=
1: 2
& A
1
+ A
2
= 1
13. Let f(x) =
x
tx
0
ef(t)dt e ?
? be a differentiable
function for all x
? R. Then f(x) equals :
(1)
x
(e –1)
e
(2)
x
e
2e –1
(3)
x
(e –1)
2e –1
(4)
x
e
e–1
Sol. Apply Lebnitz' Rule we get
t
1
(x) = e
x
+ (y) + e
x
?
?
??
x
dy
edx
y1
? ln (y + 1) = e
x
+ c
? (0, 1 )
c = ln
??
??
??
2
e
y + 1 = e
ex
?
2
e
?
? ??
?? ?
??
??
x
e1
y 2e 1
14. If 0 < a, b < 1, and tan
–1
a + tan
–1
b =
4
?
, then
the value of
(a + b) –
22
ab
2
??
?
??
??
??
+
33
ab
3
??
?
??
??
??
–
44
ab
4
??
?
??
??
??
+ ...
is :
(1) e
2
– 1 (2) log
e

e
2
??
??
??
(3) e (4) log
e
2
Sol.
?? ?
?? ??
?? ?
??
?
??
11 1
ab
tan a tan b tan
aab 4
? a + b + ab = 1
? (1 + a)(1 + b) = 2
Given
????
?? ? ? ??? ????
????
????
23 23
aa b b
a ... b ...
23 23
ln(1 + a) + ln(1 + b)
? ln(1 + a)(1 + b) = ln 2
Page 5

JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
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.
1. If the mirror image of the point (1, 3, 5) with
respect to the plane 4x – 5y + 2z = 8 is
( ?, ?, ?), then 5(?? ?? ?? ?? ?) equals :
(1) 43 (2) 47
(3) 41 (4) 39
Sol.
??
?? ?? ?? ? ? ??? ?
?? ??
???
13 5 4153258
2
45 2 16254
?? ? ? ? ?
?? ?
?
13 52
45 2 5
?
?? ? ? ? ? ? ? ?
810 4
1, 3, 5
55 5
?????? ?? ?? ?? 555547
2. A natural number has prime factorization
given by n = 2
x
3
y
5
z
, where y and z are such
that y + z = 5 and y
–1
+ z
–1
=
5
,y z
6
?
. Then the
number of odd divisors of n, including 1, is :
(1) 12 (2) 6x
(3) 11 (4) 6
Sol. y + z = 5 …(i)
?
?? ? ? ?
11 y z 5
yz 6
yz yz 6
…(ii)
Equation with y and z as roots is
x
2
– 5x + 6 = 0
x = 2, 3, y = 3, z = 2 (y > z)
n = 2
x
· 3
3
· 5
2
For odd divisors x = 1 only
No. of odd divisors = 1 × 4 × 3 = 12
3. Let A = {1, 2, 3, ..., 10} and f : A ? A be
defined as
f(k) =
k1 ifkisodd
kifkiseven
? ?
?
?
PART–C : MATHEMATICS
Then the number of possible functions
g : A ? A such that gof = f is :
(1) 5
5
(2) 10
5
(3) 5! (4)
10
C
5
Sol. Not that f(1) = f(2) = 2
f(3) = f(4) = 4
f(5) = f(6) = 6
f(7) = f(8) = 8
f(9) = f(10) = 10
gof(1) = f(1) ? g(2) = f(1) = 2
gof(2) = f(2) ? g(2) = f(2) = 2
gof(3) = f(3) ? g(4) = f(3) = 4
? In function g(x), 2, 4, 6, 8, 10 should be
mapped to 2, 4, 6, 8, 10 respectively. Each
of remaining elements can be mapped to
any of 10 elements.
Number of possible g(x) is 10
5
4. Let slope of the tangent line to a curve at any
point P(x, y) be given by
2
xyy
.
x
?
If the curve
intersects the line x + 2y = 4 at x = –2, then the
value of y, for which the point (3, y) lies on the
curve, is :
(1)
18
35
(2)
4
–
3
(3)
18
–
11
(4)
18
–
19
Sol.
??
2
dy y
y
dx x
??
2
dy y
y
dx x
?? ?
2
1dy 1 1
1
dx x y y
Let
?
1
z
y
?
?
2
1dy dz
dx dx y
?
??
dz 1
z1
dx x
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
???
dz 1
z1
dx x
IF =
1
dx
nx
x
ee x
?
??

?? ? ?
?
zx 1xdx
?
?? ?
2
x
zx c
2
2
xx
c
y2
?
??
…(i)
Putting x = –2 in x + 2y = 4, we get y = 3
Put (–2, 3) in (i)
?
?
4
c
3
(i)??
?
??
2
xx 4
y 23
…(ii)
Put x = 3 in (ii)
?
??
394
y2 3
?
?
18
y
19
5. The triangle of maximum area that can be
inscribed in a given circle of radius ‘r’ is :
(1) An isosceles triangle with base equal to 2r.
(2) An equilateral triangle of height
2r
.
3
(3) A right angle triangle having two of its
sides of length 2r and r.
(4) An equilateral triangle having each of its
side of length 3r.
Sol. Area of triangle ABC
A
BC
x
r
O
r
r
M -
22
rx
?? ?
1
ABCAM
2
?? ? ? ?
22
1
2r x (r x)
2
?? ?
22
A (rx)rx
?? ?
?? ? ? ? ?
??
22 2
22
22 22
dA x r x rx x
rx (rx)
dx
rx rx
?? ?? ?
??
??
22
22 22
rrx2x (xr)(2x r)
rx rx
? ? ?
dA r
0x
dx 2
Sign change of
dA
dx
at ?
r
x
2
? A has
maximum at ?
r
x
2
?? ?
22
BC 2 r x 3r,
?
3
AM r
2
? ?? AB AC 3r
6. A seven digit number is formed using digits
3, 3, 4, 4, 4, 5, 5. The probability, that number
so formed is divisible by 2, is :
(1)
1
7
(2)
6
7
(3)
4
7
(4)
3
7
Sol. For even number, units place should be filled
with 4 only.
????
6!
6! 3! 3
2!2!2!
P
7!
2! 7! 7
2!3!2!
7. Let F
1
(A, B, C) = (A ? ~B) ? [~C ? (A ? B)] ? ~A
and F
2
(A, B) = (A ? B) ? (B ?  ~A) be two
logical expressions. Then :
(1) F
1
and F
2
both are tautologies
(2) Both F
1
and F
2
are not tautologies
(3) F
1
is a tautology but F
2
is not a tautology
(4) F
1
is not a tautology but F
2
is a tautology
Sol.
A
B
C
A ~B ?
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
~C (A B ) ?? ?? ?
~A
?
F1 is not a
tautology
F1
?? ? B~A~B~A
A  B ?
~B  ~A = ~(A B) ??
?
=  F
2
A  B ?
~B  ~A = ~(A B) ??
?
=  F
2
? F
2
is a tautology
8. Let f : R ? R be defined as
2
x
2sin – ,if x – 1
2
f(x)|ax x b|,if –1 x 1
sin( x), if x 1
? ? ??
?
?? ?
??
?
?
??? ??
?
?
??
?
?
?
If f(x) is continuous on R, then a + b equals :
(1) –1 (2) –3
(3) 3 (4) 1
Sol.
?
?? f(1) 2
?
?? ? ? f( 1 ) ab 1
?? ? ab 1 2
...(i)
?
?? ? f(1 ) ab1
?
? f(1 ) 0
?? ? ? ?? ? ab1 0 ab1 0
? a + b = –1 ...(ii)
9. Let L be a line obtained from the intersection
of two planes x + 2y + z = 6 and y + 2z = 4.
If point P( ??? ??? ?) is the foot of perpendicular
from (3, 2, 1) on L, then the value of
21(?? ?? ?? ?? ?) equals :
(1) 68 (2) 102
(3) 142 (4) 136
Sol. Direction of line L = ?? ?
ˆ ˆˆ
ij k
ˆ ˆˆ
12 1 3i 2j k
01 2
d.r’s = < 3, –2, 1>
A point on line (–2, 4, 0)
Line =
??
??
?
x2 y 4 z
321
Foot of perpendicular from (3, 2, 1) be (3 ? – 2,
–2 ? + 4, ?)
(3 ? – 5) . 3 + (–2 ? + 2) (–2) + ( ? – 1)1 = 0
9 ? – 15 + 4 ? – 4 + ? – 1 = 0
14 ? – 20  = 0 ? ??
10
7
??
?? ? ?
??
??
16 8 10
(, , ) , ,
77 7
? 21( ? + ? + ?) = (16 + 8 + 10)3 = 102
10. Let f(x) be a differentiable function at x = a
with f ?(a) = 2 and f(a) = 4. Then
xa
xf(a)–af(x)
lim
x–a ?
equals :
(1) 4 – 2a (2) a + 4
(3) 2a – 4 (4) 2a + 4
Sol.
?
? ??
?
??
? ?? xa
xf(a) af(x) 0
Llim form
xa 0
Using L’ Hospital rule we get
?
??
?
xa
f(a) af (x)
Llim
1
?? ? ? f(a) af (a) 4 2a
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
11. Let A(1, 4) and B(1, –5) be two points. Let P be
a point on the circle (x – 1)
2
+ (y – 1)
2
= 1 such
that (PA)
2
+ (PB)
2
have maximum value, then
the points, P, A and B lie on :
(1) an ellipse (2) a parabola
(3) a straight line (4) a hyperbola
Sol. Let P be (1 + cos ?, 1 + sin ?)
(PA)
2
+ (PB)
2
=
?? ? ? ?? ?? ?? ? ?
22 2
cos sin 3 cos
?? ???
2
sin 6
= 1 – 6sin ? + 9 + 1 + 12 sin ? + 36
= 45 + 6 sin ? maximum at
?
??
2
? P (1, 2)
? P, A and B are colinear
12. Let A
1
be the area of the region bounded by
the curves y = sinx, y = cosx and y-axis in the
2
be the area of the
region bounded by the curves y = sinx,
y = cosx, x-axis and x =
2
?
Then,
(1) A
1
: A
2
= 1 :
2
and A
1
+ A
2
= 1
(2) A
1
: A
2
= 1 : 2 and A
1
+ A
2
= 1
(3) 2A
1
= A
2
and A
1
+ A
2
= 1 + 2
(4) A
1
= A
2
and A
1
+ A
2
=
2
Sol.
A
1
A
2
X
Y
?
2
A
1
= ??
?
???
?
2
0
cosx sinx dx 2 1
A
2
= ??
? ?
?
?? ?
??
4 2
0
4
sinx dx cosx dx 2 2 1
? A
1
: A
2
=
1: 2
& A
1
+ A
2
= 1
13. Let f(x) =
x
tx
0
ef(t)dt e ?
? be a differentiable
function for all x
? R. Then f(x) equals :
(1)
x
(e –1)
e
(2)
x
e
2e –1
(3)
x
(e –1)
2e –1
(4)
x
e
e–1
Sol. Apply Lebnitz' Rule we get
t
1
(x) = e
x
+ (y) + e
x
?
?
??
x
dy
edx
y1
? ln (y + 1) = e
x
+ c
? (0, 1 )
c = ln
??
??
??
2
e
y + 1 = e
ex
?
2
e
?
? ??
?? ?
??
??
x
e1
y 2e 1
14. If 0 < a, b < 1, and tan
–1
a + tan
–1
b =
4
?
, then
the value of
(a + b) –
22
ab
2
??
?
??
??
??
+
33
ab
3
??
?
??
??
??
–
44
ab
4
??
?
??
??
??
+ ...
is :
(1) e
2
– 1 (2) log
e

e
2
??
??
??
(3) e (4) log
e
2
Sol.
?? ?
?? ??
?? ?
??
?
??
11 1
ab
tan a tan b tan
aab 4
? a + b + ab = 1
? (1 + a)(1 + b) = 2
Given
????
?? ? ? ??? ????
????
????
23 23
aa b b
a ... b ...
23 23
ln(1 + a) + ln(1 + b)
? ln(1 + a)(1 + b) = ln 2
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
15. The sum of the series
2
n1
n6n10
(2n 1)!
?
?
??
?
? is equal
to :
(1)
–1
41 19
ee 10
88
??
(2)
–1
41 19
e– e 10
88
?
(3)
–1
41 19
– ee 10
88
??
(4)
–1
41 19
ee 10
88
??
Sol.
??
?
?
??
?
?
2
n1
n6n10
2n 1 !
=
?
?
??
?
?
2
n1
12n 12n20
2(2n1)!
=
?
?
?? ?
?
?
n1
1n(2n 1) 11n 20
2(2n1)!
=
??
??
??
?
?
??
n1 n1
11 29
11 n
1n 1
22

2 (2n)!2 (2n 1 )!
=
?? ?
?? ?
?
?? ?
??
?? ?
n1 n1 n1
12n 111 2n1 29 1

4 (2n)! 2 2 (2n 1)! 4 (2n 1)!
=
?? ?
?? ?
??
??
?? ?
n1 n1 n1
12n 11 129 1
4 (2n 1)!4 2n! 4 (2n 1)!
=
?? ?
?? ? ? ? ?
?? ?
??? ? ?? ? ? ? ?
?? ? ? ? ?
?? ? ? ? ?
11 1
1e e 11e e 29 e e
11
42 4 2 4 2
=
??
?? ? ?
??
?? ?? ? ?
?? ? ?
?? ? ?
11
15 e e 11 e e
10
22 4 2
=
?
??
1
41 19
ee 10
88
16. For x > 0, if f(x) =
x
e
1
log t 1
dt,thenf(e) f
(1 t) e
??
?
??
?
??
? is
equal to :
(1) 1 (2)
1
2
(3) 0 (4) –1
Sol. ?
?
?
x
1
lnt
f(x) dt
1t
then
??
?
??
?
??
?
1/x
1
1lnt
f dt
x1t
Let
? ? ??
2
11
tdt du
u
u
?
?? ??
??
?? ??
?? ??
?
?
x
2 1
1
ln
11
u
fdx
1
x
u
1
u
?? ??
??
??
??
??
??
??
xx
11
1lnu lnt
fdu dt
xu1u t1t
?
??
??
??
??
?? ???
??
??
??
??
??
?
x
1
111
fxf lnt dt
x1tt1t
??
???
??
??
??
?
x
1
11 1
lnt dt
1t t t 1
?? ??
?
x
2
1
lnt 1
dt lnx
t2
?
?? ??
??
?? ?
??
??
2 11 1
fe f lne
e2 2
17. If vectors 12
ˆˆ ˆˆ ˆ ˆ
axi– j kanda i yj zk
??
?? ??? are
collinear, then a possible unit vector parallel to
the vector
ˆ ˆˆ
xi yj zk ?? is :
(1) ??
1
ˆ ˆˆ
i– j k
3
?
(2) ??
1
ˆ ˆˆ
i j–k
3
?
(3) ??
1
ˆˆ
i– j
2
(4) ??
1
ˆ ˆ
–j k
2
?
Sol. ??

21
aa
??
?? ?? ? ?
ˆˆ ˆˆ ˆ
i yj zk xi j k
?? ? ?? ?? 1 x,y ,z
?? ? ????
?
1
ˆˆ ˆˆ ˆ ˆ
xi yj zk i j k
Unit vector
?? ? ?
?
?
?? ??
?
22
2
1
ˆ ˆ
ij k
1
?? ? ?
?
??
22
4
ˆ ˆˆ
ij k
12
Let ?
2
= 1, possible unit vector
??
?
ˆ ˆˆ
ij k
3
```

## JEE Main & Advanced Mock Test Series

366 docs|219 tests

## JEE Main & Advanced Mock Test Series

366 docs|219 tests

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,

,

,

,

,

,

,

,

,

,

,

,

;