JEE Mains 11 January 2019 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


E 1
JEE (Main) Examination–2019/Evening Session/11-01-2019
1. If the point (2, a, b) lies on the plane which passes
through the points (3, 4, 2) and (7, 0, 6) and is
perpendicular to the plane 2x – 5y = 15, then 2a
– 3b is equal to :-
(1) 5 (2) 17 (3) 12 (4) 7
Ans. (4)
Sol. Normal vector of plane
( )
i jk
ˆˆ ˆ
2 5 0 4 5i 2j 3k
4 44
= - =- +-
-
equation of plane is 5(x–7)+ 2y–3(z– 6) = 0
5x + 2y – 3z = 17
2. Let a and b be the roots of the quadratic equation
x
2
 sin q – x (sin q cos q + 1) + cos q = 0
(0 < q < 45º), and a < b. Then 
n
n
n
n0
( 1)
¥
=
æö -
a+
ç÷
b èø
å
is equal to :-
(1) 
11
1 cos 1 sin
+
- q +q
(2)  
11
1 cos 1 sin
+
+ q -q
(3) 
11
1 cos 1 sin
-
- q +q
(4) 
11
1 cos 1 sin
-
+ q -q
Ans. (1)
Sol. D = (1+ sinq cosq)
2
–4sinqcosq= (1–sinq cosq)
2
Þ roots are b = cosecq and a = cosq
( ) ( )
n
n
nn
n
n0 n0 n 0
1
cos sin
¥¥
= ==
æö
æö
ç÷ Þ a+- = q + -q
ç÷
ç÷
b
èø
èø
å åå
11
1cos 1sin
=+
- q +q
3. Let K be the set of all real values of x where the
function f(x) = sin |x| – |x| + 2(x – p) cos |x| is not
differentiable. Then the set K is equal to:-
(1) {p} (2)  {0}
(3) f (an empty set) (4) {0, p}
Ans. (3)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Friday 11
th
 JANUARY , 2019)    TIME : 2 : 30 PM To 5 : 30 PM
Sol. ƒ(x) = sin|x|–|x| + 2(x – p) cosx
Q sin|x| – |x| is differentiable function at x=0
\ k = f
4. Let the length of the latus rectum of an ellipse with
its major axis along x-axis and centre at the origin,
be 8. If the distance between the foci of this ellipse
is equal to the length of its minor axis, then which
one of the following points lies on it ?
(1) ( ) 4 3,23 (2) ( ) 4 3,22
(3) ( ) 42,22 (4) ( ) 4 2,23
Ans. (2)
Sol.
2
2b
8 and 2ae 2b
a
==
b
e
a
Þ=
 and 1– e
2
 = e
2
 
1
e
2
Þ=
b 42 Þ=
 and a = 8
so equation of ellipse is 
22
xy
1
64 32
+=
5. If the area of the triangle whose one vertex is at
the vertex of the parabola, y
2
 + 4(x – a
2
)= 0 and
the other two vertices are the points of intersection
of the parabola and y-axis, is 250 sq. units, then
a value of 'a' is :-
(1) 
55
(2) (10)
2/3
(3) 5(2
1/3
) (4) 5
Ans. (4)
Sol. Vertex is (a
2
,0)
y
2
 = –(x – a
2
) and x = 0 Þ (0, ±2a)
Area of triangle is ()
2
1
.4a. a 250
2
==
3
a 125 or a 5 Þ==
Page 2


E 1
JEE (Main) Examination–2019/Evening Session/11-01-2019
1. If the point (2, a, b) lies on the plane which passes
through the points (3, 4, 2) and (7, 0, 6) and is
perpendicular to the plane 2x – 5y = 15, then 2a
– 3b is equal to :-
(1) 5 (2) 17 (3) 12 (4) 7
Ans. (4)
Sol. Normal vector of plane
( )
i jk
ˆˆ ˆ
2 5 0 4 5i 2j 3k
4 44
= - =- +-
-
equation of plane is 5(x–7)+ 2y–3(z– 6) = 0
5x + 2y – 3z = 17
2. Let a and b be the roots of the quadratic equation
x
2
 sin q – x (sin q cos q + 1) + cos q = 0
(0 < q < 45º), and a < b. Then 
n
n
n
n0
( 1)
¥
=
æö -
a+
ç÷
b èø
å
is equal to :-
(1) 
11
1 cos 1 sin
+
- q +q
(2)  
11
1 cos 1 sin
+
+ q -q
(3) 
11
1 cos 1 sin
-
- q +q
(4) 
11
1 cos 1 sin
-
+ q -q
Ans. (1)
Sol. D = (1+ sinq cosq)
2
–4sinqcosq= (1–sinq cosq)
2
Þ roots are b = cosecq and a = cosq
( ) ( )
n
n
nn
n
n0 n0 n 0
1
cos sin
¥¥
= ==
æö
æö
ç÷ Þ a+- = q + -q
ç÷
ç÷
b
èø
èø
å åå
11
1cos 1sin
=+
- q +q
3. Let K be the set of all real values of x where the
function f(x) = sin |x| – |x| + 2(x – p) cos |x| is not
differentiable. Then the set K is equal to:-
(1) {p} (2)  {0}
(3) f (an empty set) (4) {0, p}
Ans. (3)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Friday 11
th
 JANUARY , 2019)    TIME : 2 : 30 PM To 5 : 30 PM
Sol. ƒ(x) = sin|x|–|x| + 2(x – p) cosx
Q sin|x| – |x| is differentiable function at x=0
\ k = f
4. Let the length of the latus rectum of an ellipse with
its major axis along x-axis and centre at the origin,
be 8. If the distance between the foci of this ellipse
is equal to the length of its minor axis, then which
one of the following points lies on it ?
(1) ( ) 4 3,23 (2) ( ) 4 3,22
(3) ( ) 42,22 (4) ( ) 4 2,23
Ans. (2)
Sol.
2
2b
8 and 2ae 2b
a
==
b
e
a
Þ=
 and 1– e
2
 = e
2
 
1
e
2
Þ=
b 42 Þ=
 and a = 8
so equation of ellipse is 
22
xy
1
64 32
+=
5. If the area of the triangle whose one vertex is at
the vertex of the parabola, y
2
 + 4(x – a
2
)= 0 and
the other two vertices are the points of intersection
of the parabola and y-axis, is 250 sq. units, then
a value of 'a' is :-
(1) 
55
(2) (10)
2/3
(3) 5(2
1/3
) (4) 5
Ans. (4)
Sol. Vertex is (a
2
,0)
y
2
 = –(x – a
2
) and x = 0 Þ (0, ±2a)
Area of triangle is ()
2
1
.4a. a 250
2
==
3
a 125 or a 5 Þ==
E 2
JEE (Main) Examination–2019/Evening Session/11-01-2019
6. The integral 
/4
55
/6
dx
sin2x(tan x cot x)
p
p
+
ò
equals :-
(1) 
1
11
tan
104 93
-
æö p æö
-
ç÷ ç÷
èø èø
(2) 
1
11
tan
54 33
-
æö p æö
-
ç÷ ç÷
èø èø
(3) 
10
p
(4) 
1
11
tan
20 93
-
æö
ç÷
èø
Ans. (1)
Sol.
( )
/4
55
/6
dx
I
sin 2x tan x cot x
p
p
=
+
ò
( )
/4 42
10
/6
1 tan xsec xdx
I
2 1 tanx
p
p
=
+
ò Put tan
5
x
 
= t
5
1
1
2
1
3
1 dt11
I tan
10 10 4 1t
93
-
æö
ç÷
èø
æö p
= =-
ç÷
+
èø
ò
7. Let (x + 10)
50
 + (x – 10)
50
 = a
0
 + a
1
x + a
2
x
2
 + .....
+ a
50
 x
50
, for all xÎR, then 
2
0
a
a
 is equal to:-
(1) 12.50 (2) 12.00 (3) 12.75 (4) 12.25
Ans. (4)
Sol. (10 + x)
50
 + (10 – x)
50
Þ a
2
 = 2.
50
C
2
10
48
, a
0
 = 2.10
50
50
22
2
0
aC
12.25
a 10
==
8. Let a function f : (0, ¥) ® (0, ¥) be defined by
f(x) = 
1
1
x
-
. Then f is :-
(1) Injective only
(2) Not injective but it is surjective
(3) Both injective as well as surjective
(4) Neither injective nor surjective
Ans. (Bonus)
Sol. ()
1x
0 x1
x1
1
x
ƒx1
x1 xx
x1
x
- ì
<£
ï
-
ï
=-==
í
-
ï
³
ï
î
y
1 ƒ(x)
x
Þ ƒ(x) is not injective
but range of function is [0,¥)
Remark : If co-domain is [0,¥), then ƒ(x) will be
surjective
9. Let S = {1, 2, ...... , 20}. A subset B of S is said
to be "nice", if the sum of the elements of B is 203.
Then the probability that a randomly chosen subset
of S is "nice" is :-
(1) 
20
6
2
(2) 
20
5
2
(3) 
20
4
2
(4) 
20
7
2
Ans. (2)
Sol. 7,
1,6
20
5
P
2
=
2,5
3,4
1,2,4
10. Two lines 
x3 y1 z6
1 31
- +-
==
-
 and
x5y2z3
7 64
+ --
==
-
 intersect at the point R. The
reflection of R in the xy-plane has coordinates :-
(1) (2, 4, 7) (2) (–2, 4, 7)
(3) (2, –4, –7) (4) (2, –4, 7)
Ans. (3)
Sol. Point on L
1
 (l + 3, 3l – 1, –l + 6)
Point on L
2
 (7m – 5, –6m + 2, 4m + 3)
Þ l + 3 = 7m – 5 ...(i)
3l – 1 = –6m + 2 ...(ii) Þ l = –1, m=1
point R(2,–4,7)
Reflection is (2,–4,–7)
Page 3


E 1
JEE (Main) Examination–2019/Evening Session/11-01-2019
1. If the point (2, a, b) lies on the plane which passes
through the points (3, 4, 2) and (7, 0, 6) and is
perpendicular to the plane 2x – 5y = 15, then 2a
– 3b is equal to :-
(1) 5 (2) 17 (3) 12 (4) 7
Ans. (4)
Sol. Normal vector of plane
( )
i jk
ˆˆ ˆ
2 5 0 4 5i 2j 3k
4 44
= - =- +-
-
equation of plane is 5(x–7)+ 2y–3(z– 6) = 0
5x + 2y – 3z = 17
2. Let a and b be the roots of the quadratic equation
x
2
 sin q – x (sin q cos q + 1) + cos q = 0
(0 < q < 45º), and a < b. Then 
n
n
n
n0
( 1)
¥
=
æö -
a+
ç÷
b èø
å
is equal to :-
(1) 
11
1 cos 1 sin
+
- q +q
(2)  
11
1 cos 1 sin
+
+ q -q
(3) 
11
1 cos 1 sin
-
- q +q
(4) 
11
1 cos 1 sin
-
+ q -q
Ans. (1)
Sol. D = (1+ sinq cosq)
2
–4sinqcosq= (1–sinq cosq)
2
Þ roots are b = cosecq and a = cosq
( ) ( )
n
n
nn
n
n0 n0 n 0
1
cos sin
¥¥
= ==
æö
æö
ç÷ Þ a+- = q + -q
ç÷
ç÷
b
èø
èø
å åå
11
1cos 1sin
=+
- q +q
3. Let K be the set of all real values of x where the
function f(x) = sin |x| – |x| + 2(x – p) cos |x| is not
differentiable. Then the set K is equal to:-
(1) {p} (2)  {0}
(3) f (an empty set) (4) {0, p}
Ans. (3)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Friday 11
th
 JANUARY , 2019)    TIME : 2 : 30 PM To 5 : 30 PM
Sol. ƒ(x) = sin|x|–|x| + 2(x – p) cosx
Q sin|x| – |x| is differentiable function at x=0
\ k = f
4. Let the length of the latus rectum of an ellipse with
its major axis along x-axis and centre at the origin,
be 8. If the distance between the foci of this ellipse
is equal to the length of its minor axis, then which
one of the following points lies on it ?
(1) ( ) 4 3,23 (2) ( ) 4 3,22
(3) ( ) 42,22 (4) ( ) 4 2,23
Ans. (2)
Sol.
2
2b
8 and 2ae 2b
a
==
b
e
a
Þ=
 and 1– e
2
 = e
2
 
1
e
2
Þ=
b 42 Þ=
 and a = 8
so equation of ellipse is 
22
xy
1
64 32
+=
5. If the area of the triangle whose one vertex is at
the vertex of the parabola, y
2
 + 4(x – a
2
)= 0 and
the other two vertices are the points of intersection
of the parabola and y-axis, is 250 sq. units, then
a value of 'a' is :-
(1) 
55
(2) (10)
2/3
(3) 5(2
1/3
) (4) 5
Ans. (4)
Sol. Vertex is (a
2
,0)
y
2
 = –(x – a
2
) and x = 0 Þ (0, ±2a)
Area of triangle is ()
2
1
.4a. a 250
2
==
3
a 125 or a 5 Þ==
E 2
JEE (Main) Examination–2019/Evening Session/11-01-2019
6. The integral 
/4
55
/6
dx
sin2x(tan x cot x)
p
p
+
ò
equals :-
(1) 
1
11
tan
104 93
-
æö p æö
-
ç÷ ç÷
èø èø
(2) 
1
11
tan
54 33
-
æö p æö
-
ç÷ ç÷
èø èø
(3) 
10
p
(4) 
1
11
tan
20 93
-
æö
ç÷
èø
Ans. (1)
Sol.
( )
/4
55
/6
dx
I
sin 2x tan x cot x
p
p
=
+
ò
( )
/4 42
10
/6
1 tan xsec xdx
I
2 1 tanx
p
p
=
+
ò Put tan
5
x
 
= t
5
1
1
2
1
3
1 dt11
I tan
10 10 4 1t
93
-
æö
ç÷
èø
æö p
= =-
ç÷
+
èø
ò
7. Let (x + 10)
50
 + (x – 10)
50
 = a
0
 + a
1
x + a
2
x
2
 + .....
+ a
50
 x
50
, for all xÎR, then 
2
0
a
a
 is equal to:-
(1) 12.50 (2) 12.00 (3) 12.75 (4) 12.25
Ans. (4)
Sol. (10 + x)
50
 + (10 – x)
50
Þ a
2
 = 2.
50
C
2
10
48
, a
0
 = 2.10
50
50
22
2
0
aC
12.25
a 10
==
8. Let a function f : (0, ¥) ® (0, ¥) be defined by
f(x) = 
1
1
x
-
. Then f is :-
(1) Injective only
(2) Not injective but it is surjective
(3) Both injective as well as surjective
(4) Neither injective nor surjective
Ans. (Bonus)
Sol. ()
1x
0 x1
x1
1
x
ƒx1
x1 xx
x1
x
- ì
<£
ï
-
ï
=-==
í
-
ï
³
ï
î
y
1 ƒ(x)
x
Þ ƒ(x) is not injective
but range of function is [0,¥)
Remark : If co-domain is [0,¥), then ƒ(x) will be
surjective
9. Let S = {1, 2, ...... , 20}. A subset B of S is said
to be "nice", if the sum of the elements of B is 203.
Then the probability that a randomly chosen subset
of S is "nice" is :-
(1) 
20
6
2
(2) 
20
5
2
(3) 
20
4
2
(4) 
20
7
2
Ans. (2)
Sol. 7,
1,6
20
5
P
2
=
2,5
3,4
1,2,4
10. Two lines 
x3 y1 z6
1 31
- +-
==
-
 and
x5y2z3
7 64
+ --
==
-
 intersect at the point R. The
reflection of R in the xy-plane has coordinates :-
(1) (2, 4, 7) (2) (–2, 4, 7)
(3) (2, –4, –7) (4) (2, –4, 7)
Ans. (3)
Sol. Point on L
1
 (l + 3, 3l – 1, –l + 6)
Point on L
2
 (7m – 5, –6m + 2, 4m + 3)
Þ l + 3 = 7m – 5 ...(i)
3l – 1 = –6m + 2 ...(ii) Þ l = –1, m=1
point R(2,–4,7)
Reflection is (2,–4,–7)
E 3
JEE (Main) Examination–2019/Evening Session/11-01-2019
11. The number of functions f from {1, 2, 3, ..., 20}
onto {1, 2, 3, ....., 20} such that f(k) is a multiple
of 3, whenever k is a multiple of 4, is :-
(1) (15)! × 6! (2) 5
6
 × 15
(3) 5! × 6! (4) 6
5
 × (15)!
Ans. (1)
Sol. ƒ(k) = 3m (3,6,9,12,15,18)
for k = 4,8,12,16,20 6.5.4.3.2 ways
For rest numbers 15! ways
Total ways = 6!(15!)
12. Contrapositive of the statement
"If two numbers are not equal, then their squares
are not equal." is :-
(1) If the squares of two numbers are equal, then
the numbers are equal.
(2) If the squares of two numbers are equal, then
the numbers are not equal.
(3) If the squares of two numbers are not equal,
then the numbers are equal.
(4) If the squares of two numbers are not equal,
then the numbers are not equal.
Ans. (1)
Sol. Contrapositive of p ® q is ~q ® ~p
13. The solution of the differential equation,
2
dy
(x y)
dx
=-
, when y(1) = 1, is :-
(1) 
e
2y
log 2(y 1)
2x
-
=-
-
(2) e
2x
log xy
2y
-
=-
-
(3) e
1 xy
log x y2
1 xy
+-
- = +-
-+
(4) e
1 xy
log 2(x 1)
1 xy
-+
- =-
+-
Ans. (4)
Sol. x – y = t Þ 
dy dt
1
dx dx
=-
2
2
dt dt
1 t 1dx
dx 1t
Þ-=Þ=
-
òò
1 1t
nx
2 1t
+ æö
Þ = +l
ç÷
-
èø
l
1 1 xy
nx
2 1 xy
æö +-
Þ = +l
ç÷
-+
èø
l
  given  y(1) = 1
()
1
n111
2
Þ = +lÞl=- l
( )
1 xy
n 2x1
1 xy
æö +-
Þ =-
ç÷
-+
èø
l
( )
1 xy
n 2x1
1 xy
æö -+
Þ- =-
ç÷
+-
èø
l
14. Let A and B be two invertible matrices of order
3 × 3. If det(ABA
T
) = 8 and det(AB
–1
) = 8, then
det (BA
–1
 B
T
) is equal to :-
(1) 16 (2) 
1
16
(3) 
1
4
(4) 1
Ans. (2)
Sol. |A|
2
.|B| = 8 and 
A
8
B
= Þ |A| = 4 and 
1
B
2
=
\ det(BA
–1
.B
T
) 
1 11
4 4 16
= ´=
15. If 
x1
dx f(x) 2x 1 C
2x1
+
= -+
-
ò , where C is a
constant of integration, then f(x) is equal to :-
(1) 
1
(x 4)
3
+
(2) 
1
(x 1)
3
+
(3) 
2
(x 2)
3
+
(4) 
2
(x 4)
3
-
Ans. (1)
Sol.
2
2x 1 t 2x 1 t 2dx 2t.dt -=Þ -= Þ=
2
2
t1
1
x 1 t3
2
dx t dt dt
t2 2x1
+
+
++
==
-
ò òò
( )
3
2
1tt
3t t 9c
236
æö
= + = ++
ç÷
èø
2x 1 9 x4
2x1 c 2x1 c
63
-++ æ ö æö
=- +=-+
ç ÷ ç÷
è ø èø
()
x4
ƒx
3
+
Þ=
Page 4


E 1
JEE (Main) Examination–2019/Evening Session/11-01-2019
1. If the point (2, a, b) lies on the plane which passes
through the points (3, 4, 2) and (7, 0, 6) and is
perpendicular to the plane 2x – 5y = 15, then 2a
– 3b is equal to :-
(1) 5 (2) 17 (3) 12 (4) 7
Ans. (4)
Sol. Normal vector of plane
( )
i jk
ˆˆ ˆ
2 5 0 4 5i 2j 3k
4 44
= - =- +-
-
equation of plane is 5(x–7)+ 2y–3(z– 6) = 0
5x + 2y – 3z = 17
2. Let a and b be the roots of the quadratic equation
x
2
 sin q – x (sin q cos q + 1) + cos q = 0
(0 < q < 45º), and a < b. Then 
n
n
n
n0
( 1)
¥
=
æö -
a+
ç÷
b èø
å
is equal to :-
(1) 
11
1 cos 1 sin
+
- q +q
(2)  
11
1 cos 1 sin
+
+ q -q
(3) 
11
1 cos 1 sin
-
- q +q
(4) 
11
1 cos 1 sin
-
+ q -q
Ans. (1)
Sol. D = (1+ sinq cosq)
2
–4sinqcosq= (1–sinq cosq)
2
Þ roots are b = cosecq and a = cosq
( ) ( )
n
n
nn
n
n0 n0 n 0
1
cos sin
¥¥
= ==
æö
æö
ç÷ Þ a+- = q + -q
ç÷
ç÷
b
èø
èø
å åå
11
1cos 1sin
=+
- q +q
3. Let K be the set of all real values of x where the
function f(x) = sin |x| – |x| + 2(x – p) cos |x| is not
differentiable. Then the set K is equal to:-
(1) {p} (2)  {0}
(3) f (an empty set) (4) {0, p}
Ans. (3)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Friday 11
th
 JANUARY , 2019)    TIME : 2 : 30 PM To 5 : 30 PM
Sol. ƒ(x) = sin|x|–|x| + 2(x – p) cosx
Q sin|x| – |x| is differentiable function at x=0
\ k = f
4. Let the length of the latus rectum of an ellipse with
its major axis along x-axis and centre at the origin,
be 8. If the distance between the foci of this ellipse
is equal to the length of its minor axis, then which
one of the following points lies on it ?
(1) ( ) 4 3,23 (2) ( ) 4 3,22
(3) ( ) 42,22 (4) ( ) 4 2,23
Ans. (2)
Sol.
2
2b
8 and 2ae 2b
a
==
b
e
a
Þ=
 and 1– e
2
 = e
2
 
1
e
2
Þ=
b 42 Þ=
 and a = 8
so equation of ellipse is 
22
xy
1
64 32
+=
5. If the area of the triangle whose one vertex is at
the vertex of the parabola, y
2
 + 4(x – a
2
)= 0 and
the other two vertices are the points of intersection
of the parabola and y-axis, is 250 sq. units, then
a value of 'a' is :-
(1) 
55
(2) (10)
2/3
(3) 5(2
1/3
) (4) 5
Ans. (4)
Sol. Vertex is (a
2
,0)
y
2
 = –(x – a
2
) and x = 0 Þ (0, ±2a)
Area of triangle is ()
2
1
.4a. a 250
2
==
3
a 125 or a 5 Þ==
E 2
JEE (Main) Examination–2019/Evening Session/11-01-2019
6. The integral 
/4
55
/6
dx
sin2x(tan x cot x)
p
p
+
ò
equals :-
(1) 
1
11
tan
104 93
-
æö p æö
-
ç÷ ç÷
èø èø
(2) 
1
11
tan
54 33
-
æö p æö
-
ç÷ ç÷
èø èø
(3) 
10
p
(4) 
1
11
tan
20 93
-
æö
ç÷
èø
Ans. (1)
Sol.
( )
/4
55
/6
dx
I
sin 2x tan x cot x
p
p
=
+
ò
( )
/4 42
10
/6
1 tan xsec xdx
I
2 1 tanx
p
p
=
+
ò Put tan
5
x
 
= t
5
1
1
2
1
3
1 dt11
I tan
10 10 4 1t
93
-
æö
ç÷
èø
æö p
= =-
ç÷
+
èø
ò
7. Let (x + 10)
50
 + (x – 10)
50
 = a
0
 + a
1
x + a
2
x
2
 + .....
+ a
50
 x
50
, for all xÎR, then 
2
0
a
a
 is equal to:-
(1) 12.50 (2) 12.00 (3) 12.75 (4) 12.25
Ans. (4)
Sol. (10 + x)
50
 + (10 – x)
50
Þ a
2
 = 2.
50
C
2
10
48
, a
0
 = 2.10
50
50
22
2
0
aC
12.25
a 10
==
8. Let a function f : (0, ¥) ® (0, ¥) be defined by
f(x) = 
1
1
x
-
. Then f is :-
(1) Injective only
(2) Not injective but it is surjective
(3) Both injective as well as surjective
(4) Neither injective nor surjective
Ans. (Bonus)
Sol. ()
1x
0 x1
x1
1
x
ƒx1
x1 xx
x1
x
- ì
<£
ï
-
ï
=-==
í
-
ï
³
ï
î
y
1 ƒ(x)
x
Þ ƒ(x) is not injective
but range of function is [0,¥)
Remark : If co-domain is [0,¥), then ƒ(x) will be
surjective
9. Let S = {1, 2, ...... , 20}. A subset B of S is said
to be "nice", if the sum of the elements of B is 203.
Then the probability that a randomly chosen subset
of S is "nice" is :-
(1) 
20
6
2
(2) 
20
5
2
(3) 
20
4
2
(4) 
20
7
2
Ans. (2)
Sol. 7,
1,6
20
5
P
2
=
2,5
3,4
1,2,4
10. Two lines 
x3 y1 z6
1 31
- +-
==
-
 and
x5y2z3
7 64
+ --
==
-
 intersect at the point R. The
reflection of R in the xy-plane has coordinates :-
(1) (2, 4, 7) (2) (–2, 4, 7)
(3) (2, –4, –7) (4) (2, –4, 7)
Ans. (3)
Sol. Point on L
1
 (l + 3, 3l – 1, –l + 6)
Point on L
2
 (7m – 5, –6m + 2, 4m + 3)
Þ l + 3 = 7m – 5 ...(i)
3l – 1 = –6m + 2 ...(ii) Þ l = –1, m=1
point R(2,–4,7)
Reflection is (2,–4,–7)
E 3
JEE (Main) Examination–2019/Evening Session/11-01-2019
11. The number of functions f from {1, 2, 3, ..., 20}
onto {1, 2, 3, ....., 20} such that f(k) is a multiple
of 3, whenever k is a multiple of 4, is :-
(1) (15)! × 6! (2) 5
6
 × 15
(3) 5! × 6! (4) 6
5
 × (15)!
Ans. (1)
Sol. ƒ(k) = 3m (3,6,9,12,15,18)
for k = 4,8,12,16,20 6.5.4.3.2 ways
For rest numbers 15! ways
Total ways = 6!(15!)
12. Contrapositive of the statement
"If two numbers are not equal, then their squares
are not equal." is :-
(1) If the squares of two numbers are equal, then
the numbers are equal.
(2) If the squares of two numbers are equal, then
the numbers are not equal.
(3) If the squares of two numbers are not equal,
then the numbers are equal.
(4) If the squares of two numbers are not equal,
then the numbers are not equal.
Ans. (1)
Sol. Contrapositive of p ® q is ~q ® ~p
13. The solution of the differential equation,
2
dy
(x y)
dx
=-
, when y(1) = 1, is :-
(1) 
e
2y
log 2(y 1)
2x
-
=-
-
(2) e
2x
log xy
2y
-
=-
-
(3) e
1 xy
log x y2
1 xy
+-
- = +-
-+
(4) e
1 xy
log 2(x 1)
1 xy
-+
- =-
+-
Ans. (4)
Sol. x – y = t Þ 
dy dt
1
dx dx
=-
2
2
dt dt
1 t 1dx
dx 1t
Þ-=Þ=
-
òò
1 1t
nx
2 1t
+ æö
Þ = +l
ç÷
-
èø
l
1 1 xy
nx
2 1 xy
æö +-
Þ = +l
ç÷
-+
èø
l
  given  y(1) = 1
()
1
n111
2
Þ = +lÞl=- l
( )
1 xy
n 2x1
1 xy
æö +-
Þ =-
ç÷
-+
èø
l
( )
1 xy
n 2x1
1 xy
æö -+
Þ- =-
ç÷
+-
èø
l
14. Let A and B be two invertible matrices of order
3 × 3. If det(ABA
T
) = 8 and det(AB
–1
) = 8, then
det (BA
–1
 B
T
) is equal to :-
(1) 16 (2) 
1
16
(3) 
1
4
(4) 1
Ans. (2)
Sol. |A|
2
.|B| = 8 and 
A
8
B
= Þ |A| = 4 and 
1
B
2
=
\ det(BA
–1
.B
T
) 
1 11
4 4 16
= ´=
15. If 
x1
dx f(x) 2x 1 C
2x1
+
= -+
-
ò , where C is a
constant of integration, then f(x) is equal to :-
(1) 
1
(x 4)
3
+
(2) 
1
(x 1)
3
+
(3) 
2
(x 2)
3
+
(4) 
2
(x 4)
3
-
Ans. (1)
Sol.
2
2x 1 t 2x 1 t 2dx 2t.dt -=Þ -= Þ=
2
2
t1
1
x 1 t3
2
dx t dt dt
t2 2x1
+
+
++
==
-
ò òò
( )
3
2
1tt
3t t 9c
236
æö
= + = ++
ç÷
èø
2x 1 9 x4
2x1 c 2x1 c
63
-++ æ ö æö
=- +=-+
ç ÷ ç÷
è ø èø
()
x4
ƒx
3
+
Þ=
E 4
JEE (Main) Examination–2019/Evening Session/11-01-2019
16. A bag contains 30 white balls and 10 red balls. 16
balls are drawn one by one randomly from the bag
with replacement. If X be the number of white balls
drawn, the 
meanofX
standard deviation of X
æö
ç÷
èø
 is equal
to :-
(1) 4 (2) 
43
3
(3) 
43
(4) 
32
Ans. (3)
Sol. p (probability of getting white ball) = 
30
40
1
q
4
=
 and n = 16
mean =np = 
3
16. 12
4
=
 and standard diviation
31
npq 16. . 3
44
= ==
17. If in a parallelogram ABDC, the coordinates of A,
B and C are respectively (1, 2), (3, 4) and (2, 5),
then the equation of the diagonal AD is:-
(1) 5x + 3y – 11 = 0 (2) 3x – 5y + 7 = 0
(3) 3x + 5y – 13 = 0 (4) 5x – 3y + 1 = 0
Ans. (4)
Sol. co-ordinates of point D are (4,7)
Þ line AD is 5x – 3y + 1 = 0
18. If a hyperbola has length of its conjugate axis equal
to 5 and the distance between its foci is 13, then
the eccentricity of the hyperbola is :-
(1) 2 (2) 
13
6
(3) 
13
8
(4) 
13
12
Ans. (4)
Sol. 2b = 5 and 2ae = 13
b
2
 = a
2
(e
2
 – 1) Þ 
2
25 169
a
44
=-
a6 Þ=
 
13
e
12
Þ=
19. The area (in sq. units) in the first quadrant bounded
by the parabola, y = x
2
 + 1, the tangent to it at the
point (2, 5) and the coordinate axes is :-
(1) 
14
3
(2) 
187
24
(3) 
37
24
(4) 
8
3
Ans. (3)
Sol.
y
(2,5)
(2,0) (3/4,0) 0
4x–y=3
Area ( ) ()
2
2
0
1 5 37
x1dx5
2 4 24
æö
= + -=
ç÷
èø
ò
20. Let 
ˆˆˆˆ
3i j, i 3j ++ and 
ˆˆ
i (1 )j b + -b respectively
be the position vectors of the points A, B and C
with respect to the origin O. If the distance of C
from the bisector of the acute angle between OA
and OB is 
3
2
, then the sum of all possible values
of b is :-
(1) 2 (2) 1 (3) 3 (4) 4
Ans. (2)
Sol. Angle bisector is x – y = 0
( ) 1
3
22
b- -b
Þ=
2 13 Þ b-=
2or1 Þb=-
21. If 
a b c 2a 2a
2b b c a 2b
2c 2c c ab
--
--
--
= (a + b + c) (x + a + b + c)
2
, x ¹ 0 and
a + b + c ¹ 0, then x is equal to :-
(1) –(a + b + c) (2) 2(a + b + c)
(3) abc (4) –2(a + b + c)
Ans. (4)
Sol.
--
--
--
a b c 2a 2a
2b b c a 2b
2c 2c c ab
R
1
 ® R
1
 + R
2
 + R
3
++ ++ ++
= --
--
ab ca b c ab c
2b b c a 2b
2c 2c c ab
Page 5


E 1
JEE (Main) Examination–2019/Evening Session/11-01-2019
1. If the point (2, a, b) lies on the plane which passes
through the points (3, 4, 2) and (7, 0, 6) and is
perpendicular to the plane 2x – 5y = 15, then 2a
– 3b is equal to :-
(1) 5 (2) 17 (3) 12 (4) 7
Ans. (4)
Sol. Normal vector of plane
( )
i jk
ˆˆ ˆ
2 5 0 4 5i 2j 3k
4 44
= - =- +-
-
equation of plane is 5(x–7)+ 2y–3(z– 6) = 0
5x + 2y – 3z = 17
2. Let a and b be the roots of the quadratic equation
x
2
 sin q – x (sin q cos q + 1) + cos q = 0
(0 < q < 45º), and a < b. Then 
n
n
n
n0
( 1)
¥
=
æö -
a+
ç÷
b èø
å
is equal to :-
(1) 
11
1 cos 1 sin
+
- q +q
(2)  
11
1 cos 1 sin
+
+ q -q
(3) 
11
1 cos 1 sin
-
- q +q
(4) 
11
1 cos 1 sin
-
+ q -q
Ans. (1)
Sol. D = (1+ sinq cosq)
2
–4sinqcosq= (1–sinq cosq)
2
Þ roots are b = cosecq and a = cosq
( ) ( )
n
n
nn
n
n0 n0 n 0
1
cos sin
¥¥
= ==
æö
æö
ç÷ Þ a+- = q + -q
ç÷
ç÷
b
èø
èø
å åå
11
1cos 1sin
=+
- q +q
3. Let K be the set of all real values of x where the
function f(x) = sin |x| – |x| + 2(x – p) cos |x| is not
differentiable. Then the set K is equal to:-
(1) {p} (2)  {0}
(3) f (an empty set) (4) {0, p}
Ans. (3)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Friday 11
th
 JANUARY , 2019)    TIME : 2 : 30 PM To 5 : 30 PM
Sol. ƒ(x) = sin|x|–|x| + 2(x – p) cosx
Q sin|x| – |x| is differentiable function at x=0
\ k = f
4. Let the length of the latus rectum of an ellipse with
its major axis along x-axis and centre at the origin,
be 8. If the distance between the foci of this ellipse
is equal to the length of its minor axis, then which
one of the following points lies on it ?
(1) ( ) 4 3,23 (2) ( ) 4 3,22
(3) ( ) 42,22 (4) ( ) 4 2,23
Ans. (2)
Sol.
2
2b
8 and 2ae 2b
a
==
b
e
a
Þ=
 and 1– e
2
 = e
2
 
1
e
2
Þ=
b 42 Þ=
 and a = 8
so equation of ellipse is 
22
xy
1
64 32
+=
5. If the area of the triangle whose one vertex is at
the vertex of the parabola, y
2
 + 4(x – a
2
)= 0 and
the other two vertices are the points of intersection
of the parabola and y-axis, is 250 sq. units, then
a value of 'a' is :-
(1) 
55
(2) (10)
2/3
(3) 5(2
1/3
) (4) 5
Ans. (4)
Sol. Vertex is (a
2
,0)
y
2
 = –(x – a
2
) and x = 0 Þ (0, ±2a)
Area of triangle is ()
2
1
.4a. a 250
2
==
3
a 125 or a 5 Þ==
E 2
JEE (Main) Examination–2019/Evening Session/11-01-2019
6. The integral 
/4
55
/6
dx
sin2x(tan x cot x)
p
p
+
ò
equals :-
(1) 
1
11
tan
104 93
-
æö p æö
-
ç÷ ç÷
èø èø
(2) 
1
11
tan
54 33
-
æö p æö
-
ç÷ ç÷
èø èø
(3) 
10
p
(4) 
1
11
tan
20 93
-
æö
ç÷
èø
Ans. (1)
Sol.
( )
/4
55
/6
dx
I
sin 2x tan x cot x
p
p
=
+
ò
( )
/4 42
10
/6
1 tan xsec xdx
I
2 1 tanx
p
p
=
+
ò Put tan
5
x
 
= t
5
1
1
2
1
3
1 dt11
I tan
10 10 4 1t
93
-
æö
ç÷
èø
æö p
= =-
ç÷
+
èø
ò
7. Let (x + 10)
50
 + (x – 10)
50
 = a
0
 + a
1
x + a
2
x
2
 + .....
+ a
50
 x
50
, for all xÎR, then 
2
0
a
a
 is equal to:-
(1) 12.50 (2) 12.00 (3) 12.75 (4) 12.25
Ans. (4)
Sol. (10 + x)
50
 + (10 – x)
50
Þ a
2
 = 2.
50
C
2
10
48
, a
0
 = 2.10
50
50
22
2
0
aC
12.25
a 10
==
8. Let a function f : (0, ¥) ® (0, ¥) be defined by
f(x) = 
1
1
x
-
. Then f is :-
(1) Injective only
(2) Not injective but it is surjective
(3) Both injective as well as surjective
(4) Neither injective nor surjective
Ans. (Bonus)
Sol. ()
1x
0 x1
x1
1
x
ƒx1
x1 xx
x1
x
- ì
<£
ï
-
ï
=-==
í
-
ï
³
ï
î
y
1 ƒ(x)
x
Þ ƒ(x) is not injective
but range of function is [0,¥)
Remark : If co-domain is [0,¥), then ƒ(x) will be
surjective
9. Let S = {1, 2, ...... , 20}. A subset B of S is said
to be "nice", if the sum of the elements of B is 203.
Then the probability that a randomly chosen subset
of S is "nice" is :-
(1) 
20
6
2
(2) 
20
5
2
(3) 
20
4
2
(4) 
20
7
2
Ans. (2)
Sol. 7,
1,6
20
5
P
2
=
2,5
3,4
1,2,4
10. Two lines 
x3 y1 z6
1 31
- +-
==
-
 and
x5y2z3
7 64
+ --
==
-
 intersect at the point R. The
reflection of R in the xy-plane has coordinates :-
(1) (2, 4, 7) (2) (–2, 4, 7)
(3) (2, –4, –7) (4) (2, –4, 7)
Ans. (3)
Sol. Point on L
1
 (l + 3, 3l – 1, –l + 6)
Point on L
2
 (7m – 5, –6m + 2, 4m + 3)
Þ l + 3 = 7m – 5 ...(i)
3l – 1 = –6m + 2 ...(ii) Þ l = –1, m=1
point R(2,–4,7)
Reflection is (2,–4,–7)
E 3
JEE (Main) Examination–2019/Evening Session/11-01-2019
11. The number of functions f from {1, 2, 3, ..., 20}
onto {1, 2, 3, ....., 20} such that f(k) is a multiple
of 3, whenever k is a multiple of 4, is :-
(1) (15)! × 6! (2) 5
6
 × 15
(3) 5! × 6! (4) 6
5
 × (15)!
Ans. (1)
Sol. ƒ(k) = 3m (3,6,9,12,15,18)
for k = 4,8,12,16,20 6.5.4.3.2 ways
For rest numbers 15! ways
Total ways = 6!(15!)
12. Contrapositive of the statement
"If two numbers are not equal, then their squares
are not equal." is :-
(1) If the squares of two numbers are equal, then
the numbers are equal.
(2) If the squares of two numbers are equal, then
the numbers are not equal.
(3) If the squares of two numbers are not equal,
then the numbers are equal.
(4) If the squares of two numbers are not equal,
then the numbers are not equal.
Ans. (1)
Sol. Contrapositive of p ® q is ~q ® ~p
13. The solution of the differential equation,
2
dy
(x y)
dx
=-
, when y(1) = 1, is :-
(1) 
e
2y
log 2(y 1)
2x
-
=-
-
(2) e
2x
log xy
2y
-
=-
-
(3) e
1 xy
log x y2
1 xy
+-
- = +-
-+
(4) e
1 xy
log 2(x 1)
1 xy
-+
- =-
+-
Ans. (4)
Sol. x – y = t Þ 
dy dt
1
dx dx
=-
2
2
dt dt
1 t 1dx
dx 1t
Þ-=Þ=
-
òò
1 1t
nx
2 1t
+ æö
Þ = +l
ç÷
-
èø
l
1 1 xy
nx
2 1 xy
æö +-
Þ = +l
ç÷
-+
èø
l
  given  y(1) = 1
()
1
n111
2
Þ = +lÞl=- l
( )
1 xy
n 2x1
1 xy
æö +-
Þ =-
ç÷
-+
èø
l
( )
1 xy
n 2x1
1 xy
æö -+
Þ- =-
ç÷
+-
èø
l
14. Let A and B be two invertible matrices of order
3 × 3. If det(ABA
T
) = 8 and det(AB
–1
) = 8, then
det (BA
–1
 B
T
) is equal to :-
(1) 16 (2) 
1
16
(3) 
1
4
(4) 1
Ans. (2)
Sol. |A|
2
.|B| = 8 and 
A
8
B
= Þ |A| = 4 and 
1
B
2
=
\ det(BA
–1
.B
T
) 
1 11
4 4 16
= ´=
15. If 
x1
dx f(x) 2x 1 C
2x1
+
= -+
-
ò , where C is a
constant of integration, then f(x) is equal to :-
(1) 
1
(x 4)
3
+
(2) 
1
(x 1)
3
+
(3) 
2
(x 2)
3
+
(4) 
2
(x 4)
3
-
Ans. (1)
Sol.
2
2x 1 t 2x 1 t 2dx 2t.dt -=Þ -= Þ=
2
2
t1
1
x 1 t3
2
dx t dt dt
t2 2x1
+
+
++
==
-
ò òò
( )
3
2
1tt
3t t 9c
236
æö
= + = ++
ç÷
èø
2x 1 9 x4
2x1 c 2x1 c
63
-++ æ ö æö
=- +=-+
ç ÷ ç÷
è ø èø
()
x4
ƒx
3
+
Þ=
E 4
JEE (Main) Examination–2019/Evening Session/11-01-2019
16. A bag contains 30 white balls and 10 red balls. 16
balls are drawn one by one randomly from the bag
with replacement. If X be the number of white balls
drawn, the 
meanofX
standard deviation of X
æö
ç÷
èø
 is equal
to :-
(1) 4 (2) 
43
3
(3) 
43
(4) 
32
Ans. (3)
Sol. p (probability of getting white ball) = 
30
40
1
q
4
=
 and n = 16
mean =np = 
3
16. 12
4
=
 and standard diviation
31
npq 16. . 3
44
= ==
17. If in a parallelogram ABDC, the coordinates of A,
B and C are respectively (1, 2), (3, 4) and (2, 5),
then the equation of the diagonal AD is:-
(1) 5x + 3y – 11 = 0 (2) 3x – 5y + 7 = 0
(3) 3x + 5y – 13 = 0 (4) 5x – 3y + 1 = 0
Ans. (4)
Sol. co-ordinates of point D are (4,7)
Þ line AD is 5x – 3y + 1 = 0
18. If a hyperbola has length of its conjugate axis equal
to 5 and the distance between its foci is 13, then
the eccentricity of the hyperbola is :-
(1) 2 (2) 
13
6
(3) 
13
8
(4) 
13
12
Ans. (4)
Sol. 2b = 5 and 2ae = 13
b
2
 = a
2
(e
2
 – 1) Þ 
2
25 169
a
44
=-
a6 Þ=
 
13
e
12
Þ=
19. The area (in sq. units) in the first quadrant bounded
by the parabola, y = x
2
 + 1, the tangent to it at the
point (2, 5) and the coordinate axes is :-
(1) 
14
3
(2) 
187
24
(3) 
37
24
(4) 
8
3
Ans. (3)
Sol.
y
(2,5)
(2,0) (3/4,0) 0
4x–y=3
Area ( ) ()
2
2
0
1 5 37
x1dx5
2 4 24
æö
= + -=
ç÷
èø
ò
20. Let 
ˆˆˆˆ
3i j, i 3j ++ and 
ˆˆ
i (1 )j b + -b respectively
be the position vectors of the points A, B and C
with respect to the origin O. If the distance of C
from the bisector of the acute angle between OA
and OB is 
3
2
, then the sum of all possible values
of b is :-
(1) 2 (2) 1 (3) 3 (4) 4
Ans. (2)
Sol. Angle bisector is x – y = 0
( ) 1
3
22
b- -b
Þ=
2 13 Þ b-=
2or1 Þb=-
21. If 
a b c 2a 2a
2b b c a 2b
2c 2c c ab
--
--
--
= (a + b + c) (x + a + b + c)
2
, x ¹ 0 and
a + b + c ¹ 0, then x is equal to :-
(1) –(a + b + c) (2) 2(a + b + c)
(3) abc (4) –2(a + b + c)
Ans. (4)
Sol.
--
--
--
a b c 2a 2a
2b b c a 2b
2c 2c c ab
R
1
 ® R
1
 + R
2
 + R
3
++ ++ ++
= --
--
ab ca b c ab c
2b b c a 2b
2c 2c c ab
E 5
JEE (Main) Examination–2019/Evening Session/11-01-2019
= ++ - ++
--
1 00
(a b c) 2b (a b c) 0
2c 2c c ab
= (a + b + c)(a + b + c)
2
Þ x = –2(a + b + c)
22. Let S
n
 = 1 + q + q
2
 + ....... + q
n
 and
2n
n
q1 q1 q 1
T 1 ........
222
+++ æ öæö æö
=+ + ++
ç ÷ç÷ ç÷
è øèø èø
where q is a real number and q ¹ 1.
If 
101
C
1
 + 
101
C
2
.S
1
 + ...... + 
101
C
101
.S
100
 = aT
100
,
then a is equal to :-
(1) 2
100
(2) 200 (3) 2
99
(4) 202
Ans. (1)
Sol.
101
C
1
 + 
101
C
2
S
1
 + ..... + 
101
C
101
S
100
= aT
100
101
C
1
 + 
101
C
2
(1 + q) + 
101
C
3
(1 + q + q
2
) +
..... + 
101
C
101
(1 + q + ..... + q
100
)
æö
+ æö
-ç÷
ç÷
èø èø
=a
-
101
1q
1
2
2
(1 q)
Þ 
101
C
1
(1 – q) + 
101
C
2
(1 – q
2
) +
..... + 
101
C
101
(1 – q
101
)
æö
+ æö
= a-ç÷
ç÷
èø èø
101
1q
21
2
Þ (2
101
 – 1) – ((1 + q)
101
 – 1)
æö
+ æö
=a-ç÷
ç÷
èø èø
101
1q
21
2
Þ 
æöæö
++ æö æö
- = a- ç÷ç÷
ç÷ ç÷
èø èø èøèø
101 101
101
1q 1q
2 1 21
22
Þ a = 2
100
23. A circle cuts a chord of length 4a on the x-axis
and passes through a point on the y-axis, distant
2b from the origin. Then the locus of the centre
of this circle, is :-
(1) A hyperbola (2) A parabola
(3) A straight line (4) An ellipse
Ans. (2)
Sol. Let equation of circle is
x
2
 + y
2
 + 2ƒx + 2ƒy + e = 0, it passes through
(0, 2b)
Þ 0 + 4b
2
 + 2g × 0 + 4ƒ + c = 0
Þ 4b
2
 + 4ƒ + c = 0 ...(i)
2
2 g c 4a -= ...(ii)
g
2
 – c = 4a
2
 Þ ( )
22
c g 4a =-
Putting in equation (1)
Þ 4b
2
 + 4ƒ + g
2
 – 4a
2
 = 0
Þ x
2
 + 4y + 4(b
2
 – a
2
) = 0, it represent a parabola.
24. If 19th term of a non-zero A.P. is zero, then its
(49th term) : (29th term) is :-
(1) 3 : 1 (2) 4 : 1 (3) 2 : 1 (4) 1 : 3
Ans. (1)
Sol. a + 18d = 0 ...(1)
a 48d 18d 48d 3
a 28d 18d 28d 1
+ -+
==
+ -+
25. Let 
22 22
x dx
f(x)
a x b (d x)
-
=-
+ +-
, xÎR,
where a, b and d are non-zero real constants.
Then :-
(1) f is a decreasing function of x
(2) f is neither increasing nor decreasing
function of x
(3) f' is not a continuous function of x
(4) f is an increasing function of x
Ans. (4)
Sol. ()
( )
222
2
x dx
ƒx
ax
b dx
-
=-
+
+-
()
( ) ( )
( )
22
3/2 3/2
222
2
ab
ƒ'x 0 xR
ax
b dx
= + > "Î
+
+-
ƒ(x) is an increasing function.
26. Let z be a complex number such that
|z| + z = 3 + i (where i = 
1 -
). Then |z| is equal
to :-
(1) 
5
4
(2) 
41
4
(3) 
34
3
(4) 
5
3
Ans. (4)
Sol. |z| + z = 3 + i
z  = 3 – |z| + i
Let 3 – |z| = a Þ |z| = (3 – a)
Þ 
2
z a i z a1 =+Þ =+
Þ 9 + a
2 
– 6a = a
2
 + 1 Þ 
84
a
63
==
45
|z|3
33
Þ = -=