JEE Mains 10 January 2019 Question Paper Shift 1 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


E 1
JEE (Main) Examination–2019/Morning Session/10-01-2019
1. Consider a triangular plot ABC with sides
AB=7m, BC=5m and CA=6m. A vertical
lamp-post at the mid point D of AC subtends
an angle 30° at B. The height (in m) of the
lamp-post is:
(1) 
73
(2) 
2
21
3
(3) 
3
21
2
(4) 
2 21
Ans. (2)
Sol.
B
A C
D
30°
3 3
5
h
7
BD = hcot30° = h3
So, 7
2
 + 5
2
 = +
22
2(h 3) 3)
Þ 37 = 3h
2
 + 9.
Þ 3h
2
 = 28
Þ
==
282
h 21
33
2. Let f : R®R be a function such that
f(x) = x
3
+x
2
f'(1) + xf''(2)+f'''(3), xÎR.
Then f(2) equal :
(1) 8 (2) –2 (3) –4 (4) 30
Ans. (2)
Sol. ƒ(x) = x
3
 + x
2
ƒ'(1) + xƒ''(2) + ƒ'''(3)
Þ ƒ'(x) = 3x
2
 + 2xƒ'(1) + ƒ''(x) .....(1)
Þ ƒ''(x) = 6x + 2ƒ'(1) .....(2)
Þ ƒ'''(x) = 6 .....(3)
put x = 1 in equation (1) :
ƒ'(1) = 3 + 2ƒ'(1) + ƒ''(2) .....(4)
put x = 2 in equation (2) :
ƒ''(2) = 12 + 2ƒ'(1) .....(5)
from equation (4) & (5) :
–3 – ƒ'(1) = 12 + 2ƒ'(1)
Þ 3ƒ'(1) = –15
Þ ƒ'(1) = –5  Þ  ƒ''(2) = 2  ....(2)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Thursday 10
th
 JANUARY , 2019)    TIME : 9 : 30 AM To 12 : 30 PM
put x = 3 in equation (3) :
ƒ'''(3) = 6
\ ƒ(x) = x
3
 – 5x
2
 + 2x + 6
ƒ(2) = 8 – 20 + 4 + 6 = –2
3. If a circle C passing through the point (4,0)
touches the circle x
2
 + y
2
 + 4x – 6y = 12
externally at the point (1, –1), then the radius
of C is :
(1) 
57
(2) 4 (3) 
25
(4) 5
Ans. (4)
Sol. x
2
 + y
2
 + 4x – 6y – 12 = 0
Equation of tangent at (1, –1)
x – y + 2(x + 1) – 3(y – 1) – 12 = 0
3x – 4y – 7 = 0
\ Equation of circle is
(x
2
 + y
2
 + 4x – 6y – 12) + l(3x – 4y – 7) = 0
It passes through (4, 0) :
(16 + 16 – 12) + l(12 – 7) = 0
Þ 20 + l(5) = 0
Þl = –4
\ (x
2
 + y
2
 + 4x – 6y – 12) – 4(3x – 4y – 7) = 0
or x
2
 + y
2
 – 8x + 10y + 16 = 0
Radius = 
+ -= 16 25 16 5
4. In a class of 140 students numbered 1 to 140,
all even numbered students opted mathematics
course, those whose number is divisible by 3
opted Physics course and  theose whose
number is divisible by 5 opted Chemistry
course. Then the number of students who did
not opt for any of the three courses is :
(1) 102 (2) 42 (3) 1 (4) 38
Ans. (4)
Sol. Let n(A) = number of students opted
Mathematics = 70,
n(B) = number of students opted Physics = 46,
n(C) = number of students opted Chemistry
= 28,
n(A Ç B) = 23,
Page 2


E 1
JEE (Main) Examination–2019/Morning Session/10-01-2019
1. Consider a triangular plot ABC with sides
AB=7m, BC=5m and CA=6m. A vertical
lamp-post at the mid point D of AC subtends
an angle 30° at B. The height (in m) of the
lamp-post is:
(1) 
73
(2) 
2
21
3
(3) 
3
21
2
(4) 
2 21
Ans. (2)
Sol.
B
A C
D
30°
3 3
5
h
7
BD = hcot30° = h3
So, 7
2
 + 5
2
 = +
22
2(h 3) 3)
Þ 37 = 3h
2
 + 9.
Þ 3h
2
 = 28
Þ
==
282
h 21
33
2. Let f : R®R be a function such that
f(x) = x
3
+x
2
f'(1) + xf''(2)+f'''(3), xÎR.
Then f(2) equal :
(1) 8 (2) –2 (3) –4 (4) 30
Ans. (2)
Sol. ƒ(x) = x
3
 + x
2
ƒ'(1) + xƒ''(2) + ƒ'''(3)
Þ ƒ'(x) = 3x
2
 + 2xƒ'(1) + ƒ''(x) .....(1)
Þ ƒ''(x) = 6x + 2ƒ'(1) .....(2)
Þ ƒ'''(x) = 6 .....(3)
put x = 1 in equation (1) :
ƒ'(1) = 3 + 2ƒ'(1) + ƒ''(2) .....(4)
put x = 2 in equation (2) :
ƒ''(2) = 12 + 2ƒ'(1) .....(5)
from equation (4) & (5) :
–3 – ƒ'(1) = 12 + 2ƒ'(1)
Þ 3ƒ'(1) = –15
Þ ƒ'(1) = –5  Þ  ƒ''(2) = 2  ....(2)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Thursday 10
th
 JANUARY , 2019)    TIME : 9 : 30 AM To 12 : 30 PM
put x = 3 in equation (3) :
ƒ'''(3) = 6
\ ƒ(x) = x
3
 – 5x
2
 + 2x + 6
ƒ(2) = 8 – 20 + 4 + 6 = –2
3. If a circle C passing through the point (4,0)
touches the circle x
2
 + y
2
 + 4x – 6y = 12
externally at the point (1, –1), then the radius
of C is :
(1) 
57
(2) 4 (3) 
25
(4) 5
Ans. (4)
Sol. x
2
 + y
2
 + 4x – 6y – 12 = 0
Equation of tangent at (1, –1)
x – y + 2(x + 1) – 3(y – 1) – 12 = 0
3x – 4y – 7 = 0
\ Equation of circle is
(x
2
 + y
2
 + 4x – 6y – 12) + l(3x – 4y – 7) = 0
It passes through (4, 0) :
(16 + 16 – 12) + l(12 – 7) = 0
Þ 20 + l(5) = 0
Þl = –4
\ (x
2
 + y
2
 + 4x – 6y – 12) – 4(3x – 4y – 7) = 0
or x
2
 + y
2
 – 8x + 10y + 16 = 0
Radius = 
+ -= 16 25 16 5
4. In a class of 140 students numbered 1 to 140,
all even numbered students opted mathematics
course, those whose number is divisible by 3
opted Physics course and  theose whose
number is divisible by 5 opted Chemistry
course. Then the number of students who did
not opt for any of the three courses is :
(1) 102 (2) 42 (3) 1 (4) 38
Ans. (4)
Sol. Let n(A) = number of students opted
Mathematics = 70,
n(B) = number of students opted Physics = 46,
n(C) = number of students opted Chemistry
= 28,
n(A Ç B) = 23,
E 2
JEE (Main) Examination–2019/Morning Session/10-01-2019
n(B Ç C) = 9,
n(A Ç C) = 14,
n(A Ç B Ç C) = 4,
Now n(A È B È C)
= n(A) + n(B) + n(C) – n(A Ç B) – n(B Ç C)
– n(A Ç C) + n(A Ç B Ç C)
= 70 + 46 + 28 – 23 – 9 – 14 + 4 = 102
So number of students not opted for any course
= Total – n(A È B È C)
= 140 – 102 = 38
5. The sum of all two digit positive numbers which
when divided by 7 yield 2 or 5 as remainder
is :
(1) 1365 (2) 1256 (3) 1465 (4) 1356
Ans. (4)
Sol.
=
+
å
13
r2
(7r 2)
= 
+
´+´
2 13
7. 6 2 12
2
= 7 × 90 + 24 = 654
=
+ æö
+= ´+ ´ =
ç÷
èø
å
13
r1
1 13
(7r 5) 7 13 5 13 702
2
Total = 654 + 702 = 1356
6. Let 
1
ˆ ˆˆ
a 2i j 3k = +l+
r
, 
2
ˆ ˆˆ
b 4i (3 )j 6k = + -l+
r
 and
3
ˆˆˆ
c 3i 6 j ( 1)k = + + l-
r
 be three vectors such that
b 2a =
r
r
 and 
a
r
 is perpendicular to 
c
r
. Then a
possible value of (l
1
,l
2
,l
3
) is :-
(1) 
1
,4,2
2
æö
-
ç÷
èø
(2) 
1
,4,0
2
æö
-
ç÷
èø
(3) (1,3,1) (4) (1,5,1)
Ans. (2)
Sol. + -l + = + l+
21
ˆ ˆ ˆˆ ˆˆ
4i (3 )j 6k 4i 2 j 6k
Þ 3 – l
2
 = 2l
1
 
Þ 2l
1
 + l
2
 = 3 ....(1)
Given 
=
rr
a.c0
Þ 6 + 6l
1
 + 3(l
3
 – 1) = 0
Þ 2l
1
 + l
3
 = –1 .....(2)
Now (l
1
, l
2
, l
3
) = (l
1
, 3 – 2l
1
, –1 – 2l
1
)
Now check the options, option (2) is correct
7. The equation of a tangent to the hyperbola
4x
2
–5y
2
 = 20 parallel to the line x–y = 2 is :
(1) x–y+9 = 0
(2) x–y+7 = 0
(3) x–y+1 = 0
(4) x–y–3 = 0
Ans. (3)
Sol. Hyperbola -=
22
xy
1
54
slope of tangent = 1
equation of tangent = ±- y x 54
Þ y = x ± 1
Þ y = x + 1 or y = x – 1
8. If the area enclosed between the curves y=kx
2
and x=ky
2
, (k>0), is 1 square unit. Then k is:
(1)
1
3
(2) 
2
3
(3) 
3
2
(4) 
3
Ans. (1)
Sol. Area bounded by y
2
 = 4ax & x
2
 = 4by, a, b ¹ 0
is 
16ab
3
by using formula : ==>
1
4a 4b,k0
k
Area ==
11
16. .
4k 4k
1
3
Þ
=
2
1
k
3
Þ
=
1
k
3
9. Let 
2
max{| x |,x }, | x | 2
f(x)
8 2|x|, 2 |x| 4
ì £ ï
=
í
- <£ ï
î
Let S be the set of points in the interval (–4,4)
at which f is not differentiable. Then S:
(1) is an empty set
(2) equals {–2, –1, 1, 2}
(3) equals {–2, –1, 0, 1, 2}
(4) equals {–2, 2}
Ans. (3)
Sol.
+ - £ <- ì
ï
- £ £-
ï
ï
= - <<
í
ï
££
ï
ï
- <£
î
2
2
82x, 4 x2
x, 2 x1
ƒ(x) |x |, 1 x 1
x, 1 x2
8 2x, 2 x4
y=8+2x
y=x
2
y= –x
–4
–2
–1
1 2 4
y=8–2x
y=x
2
y=x
ƒ(x) is not differentiable at x = {–2,–1,0,1,2}
Þ S = {–2, –1, 0, 1, 2}
Page 3


E 1
JEE (Main) Examination–2019/Morning Session/10-01-2019
1. Consider a triangular plot ABC with sides
AB=7m, BC=5m and CA=6m. A vertical
lamp-post at the mid point D of AC subtends
an angle 30° at B. The height (in m) of the
lamp-post is:
(1) 
73
(2) 
2
21
3
(3) 
3
21
2
(4) 
2 21
Ans. (2)
Sol.
B
A C
D
30°
3 3
5
h
7
BD = hcot30° = h3
So, 7
2
 + 5
2
 = +
22
2(h 3) 3)
Þ 37 = 3h
2
 + 9.
Þ 3h
2
 = 28
Þ
==
282
h 21
33
2. Let f : R®R be a function such that
f(x) = x
3
+x
2
f'(1) + xf''(2)+f'''(3), xÎR.
Then f(2) equal :
(1) 8 (2) –2 (3) –4 (4) 30
Ans. (2)
Sol. ƒ(x) = x
3
 + x
2
ƒ'(1) + xƒ''(2) + ƒ'''(3)
Þ ƒ'(x) = 3x
2
 + 2xƒ'(1) + ƒ''(x) .....(1)
Þ ƒ''(x) = 6x + 2ƒ'(1) .....(2)
Þ ƒ'''(x) = 6 .....(3)
put x = 1 in equation (1) :
ƒ'(1) = 3 + 2ƒ'(1) + ƒ''(2) .....(4)
put x = 2 in equation (2) :
ƒ''(2) = 12 + 2ƒ'(1) .....(5)
from equation (4) & (5) :
–3 – ƒ'(1) = 12 + 2ƒ'(1)
Þ 3ƒ'(1) = –15
Þ ƒ'(1) = –5  Þ  ƒ''(2) = 2  ....(2)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Thursday 10
th
 JANUARY , 2019)    TIME : 9 : 30 AM To 12 : 30 PM
put x = 3 in equation (3) :
ƒ'''(3) = 6
\ ƒ(x) = x
3
 – 5x
2
 + 2x + 6
ƒ(2) = 8 – 20 + 4 + 6 = –2
3. If a circle C passing through the point (4,0)
touches the circle x
2
 + y
2
 + 4x – 6y = 12
externally at the point (1, –1), then the radius
of C is :
(1) 
57
(2) 4 (3) 
25
(4) 5
Ans. (4)
Sol. x
2
 + y
2
 + 4x – 6y – 12 = 0
Equation of tangent at (1, –1)
x – y + 2(x + 1) – 3(y – 1) – 12 = 0
3x – 4y – 7 = 0
\ Equation of circle is
(x
2
 + y
2
 + 4x – 6y – 12) + l(3x – 4y – 7) = 0
It passes through (4, 0) :
(16 + 16 – 12) + l(12 – 7) = 0
Þ 20 + l(5) = 0
Þl = –4
\ (x
2
 + y
2
 + 4x – 6y – 12) – 4(3x – 4y – 7) = 0
or x
2
 + y
2
 – 8x + 10y + 16 = 0
Radius = 
+ -= 16 25 16 5
4. In a class of 140 students numbered 1 to 140,
all even numbered students opted mathematics
course, those whose number is divisible by 3
opted Physics course and  theose whose
number is divisible by 5 opted Chemistry
course. Then the number of students who did
not opt for any of the three courses is :
(1) 102 (2) 42 (3) 1 (4) 38
Ans. (4)
Sol. Let n(A) = number of students opted
Mathematics = 70,
n(B) = number of students opted Physics = 46,
n(C) = number of students opted Chemistry
= 28,
n(A Ç B) = 23,
E 2
JEE (Main) Examination–2019/Morning Session/10-01-2019
n(B Ç C) = 9,
n(A Ç C) = 14,
n(A Ç B Ç C) = 4,
Now n(A È B È C)
= n(A) + n(B) + n(C) – n(A Ç B) – n(B Ç C)
– n(A Ç C) + n(A Ç B Ç C)
= 70 + 46 + 28 – 23 – 9 – 14 + 4 = 102
So number of students not opted for any course
= Total – n(A È B È C)
= 140 – 102 = 38
5. The sum of all two digit positive numbers which
when divided by 7 yield 2 or 5 as remainder
is :
(1) 1365 (2) 1256 (3) 1465 (4) 1356
Ans. (4)
Sol.
=
+
å
13
r2
(7r 2)
= 
+
´+´
2 13
7. 6 2 12
2
= 7 × 90 + 24 = 654
=
+ æö
+= ´+ ´ =
ç÷
èø
å
13
r1
1 13
(7r 5) 7 13 5 13 702
2
Total = 654 + 702 = 1356
6. Let 
1
ˆ ˆˆ
a 2i j 3k = +l+
r
, 
2
ˆ ˆˆ
b 4i (3 )j 6k = + -l+
r
 and
3
ˆˆˆ
c 3i 6 j ( 1)k = + + l-
r
 be three vectors such that
b 2a =
r
r
 and 
a
r
 is perpendicular to 
c
r
. Then a
possible value of (l
1
,l
2
,l
3
) is :-
(1) 
1
,4,2
2
æö
-
ç÷
èø
(2) 
1
,4,0
2
æö
-
ç÷
èø
(3) (1,3,1) (4) (1,5,1)
Ans. (2)
Sol. + -l + = + l+
21
ˆ ˆ ˆˆ ˆˆ
4i (3 )j 6k 4i 2 j 6k
Þ 3 – l
2
 = 2l
1
 
Þ 2l
1
 + l
2
 = 3 ....(1)
Given 
=
rr
a.c0
Þ 6 + 6l
1
 + 3(l
3
 – 1) = 0
Þ 2l
1
 + l
3
 = –1 .....(2)
Now (l
1
, l
2
, l
3
) = (l
1
, 3 – 2l
1
, –1 – 2l
1
)
Now check the options, option (2) is correct
7. The equation of a tangent to the hyperbola
4x
2
–5y
2
 = 20 parallel to the line x–y = 2 is :
(1) x–y+9 = 0
(2) x–y+7 = 0
(3) x–y+1 = 0
(4) x–y–3 = 0
Ans. (3)
Sol. Hyperbola -=
22
xy
1
54
slope of tangent = 1
equation of tangent = ±- y x 54
Þ y = x ± 1
Þ y = x + 1 or y = x – 1
8. If the area enclosed between the curves y=kx
2
and x=ky
2
, (k>0), is 1 square unit. Then k is:
(1)
1
3
(2) 
2
3
(3) 
3
2
(4) 
3
Ans. (1)
Sol. Area bounded by y
2
 = 4ax & x
2
 = 4by, a, b ¹ 0
is 
16ab
3
by using formula : ==>
1
4a 4b,k0
k
Area ==
11
16. .
4k 4k
1
3
Þ
=
2
1
k
3
Þ
=
1
k
3
9. Let 
2
max{| x |,x }, | x | 2
f(x)
8 2|x|, 2 |x| 4
ì £ ï
=
í
- <£ ï
î
Let S be the set of points in the interval (–4,4)
at which f is not differentiable. Then S:
(1) is an empty set
(2) equals {–2, –1, 1, 2}
(3) equals {–2, –1, 0, 1, 2}
(4) equals {–2, 2}
Ans. (3)
Sol.
+ - £ <- ì
ï
- £ £-
ï
ï
= - <<
í
ï
££
ï
ï
- <£
î
2
2
82x, 4 x2
x, 2 x1
ƒ(x) |x |, 1 x 1
x, 1 x2
8 2x, 2 x4
y=8+2x
y=x
2
y= –x
–4
–2
–1
1 2 4
y=8–2x
y=x
2
y=x
ƒ(x) is not differentiable at x = {–2,–1,0,1,2}
Þ S = {–2, –1, 0, 1, 2}
E 3
JEE (Main) Examination–2019/Morning Session/10-01-2019
10. If the parabolas y
2
=4b(x–c) and y
2
=8ax have
a common normal, then which one of the
following is a valid choice for the ordered triad
(a,b,c)
(1) (1, 1, 0) (2) 
1
,2,3
2
æö
ç÷
èø
(3)
1
,2,0
2
æö
ç÷
èø
(4) (1, 1, 3)
Ans. (1,2,3,4)
Sol. Normal to these two curves are
y = m(x – c) – 2bm – bm
3
,
y = mx – 4am – 2am
3
If they have a common normal
(c + 2b) m + bm
3
 = 4am + 2am
3
Now (4a – c – 2b) m = (b – 2a)m
3
We get all options are correct for m = 0
(common normal x-axis)
Ans. (1), (2), (3), (4)
Remark :
If we consider question as
If the parabolas y
2
 = 4b(x – c) and y
2
 = 8ax
have a common normal other than x-axis, then
which one of the following is a valid choice for
the ordered triad (a, b, c) ?
 When m ¹ 0 : (4a – c – 2b) = (b – 2a)m
2
= ->Þ >
--
2
cc
m 202
2ab 2ab
Now according to options, option 4 is correct
11. The sum of all values of 
0,
2
p æö
qÎ
ç÷
èø
 satisfying
24
3
sin 2 cos2
4
q+ q=
 is :
(1) 
2
p
(2) p (3) 
3
8
p
(4) 
5
4
p
Ans. (1)
Sol. sin
2
2q + cos
4
2q =
3
4
, 
p æö
qÎ
ç÷
èø
0,
2
Þ 1 – cos
2
2q + cos
4
2q =
3
4
Þ 4cos
4
2q – 4cos
2
2q + 1 = 0
Þ (2cos
2
2q – 1)
2
 = 0
Þ
p
q==
22
1
cos 2 cos
24
Þ
p
q= p± 2n
4
, n Î I
Þ
pp
q=±
n
28
Þ
ppp
q=- ,
828
Sum of solutions 
p
2
12. Let z
1
 and z
2
 be any two non-zero complex
numbers such that 3|z
1
| = 4 |z
2
|.
If 
12
21
3z 2z
z
2z 3z
=+ then :
(1)
1 17
|z|
22
= (2) Re(z) = 0
(3) 
5
|z|
2
= (4) Im(z) = 0
Ans. (Bonus)
Sol. 3|z
1
| = 4|z
2
|
Þ
=
1
2
|z| 4
|z|3
Þ
=
1
2
|3z|
2
|2z|
Let 
== q+q
1
2
3z
a 2cos 2isin
2z
=+=+
12
21
3z 2z 1
za
2z3za
= q+q
53
cos isin
22
Now all options are incorrect
Remark :
There is a misprint in the problem actual
problem should be :
"Let z
1
 and z
2
 be any non-zero complex
number such that 3|z
1
| = 2|z
2
|.
If 
=+
12
21
3z 2z
z
2z 3z
, then"
Given
3|z
1
| = 2|z
2
|
Now 
=
1
2
3z
1
2z
Page 4


E 1
JEE (Main) Examination–2019/Morning Session/10-01-2019
1. Consider a triangular plot ABC with sides
AB=7m, BC=5m and CA=6m. A vertical
lamp-post at the mid point D of AC subtends
an angle 30° at B. The height (in m) of the
lamp-post is:
(1) 
73
(2) 
2
21
3
(3) 
3
21
2
(4) 
2 21
Ans. (2)
Sol.
B
A C
D
30°
3 3
5
h
7
BD = hcot30° = h3
So, 7
2
 + 5
2
 = +
22
2(h 3) 3)
Þ 37 = 3h
2
 + 9.
Þ 3h
2
 = 28
Þ
==
282
h 21
33
2. Let f : R®R be a function such that
f(x) = x
3
+x
2
f'(1) + xf''(2)+f'''(3), xÎR.
Then f(2) equal :
(1) 8 (2) –2 (3) –4 (4) 30
Ans. (2)
Sol. ƒ(x) = x
3
 + x
2
ƒ'(1) + xƒ''(2) + ƒ'''(3)
Þ ƒ'(x) = 3x
2
 + 2xƒ'(1) + ƒ''(x) .....(1)
Þ ƒ''(x) = 6x + 2ƒ'(1) .....(2)
Þ ƒ'''(x) = 6 .....(3)
put x = 1 in equation (1) :
ƒ'(1) = 3 + 2ƒ'(1) + ƒ''(2) .....(4)
put x = 2 in equation (2) :
ƒ''(2) = 12 + 2ƒ'(1) .....(5)
from equation (4) & (5) :
–3 – ƒ'(1) = 12 + 2ƒ'(1)
Þ 3ƒ'(1) = –15
Þ ƒ'(1) = –5  Þ  ƒ''(2) = 2  ....(2)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Thursday 10
th
 JANUARY , 2019)    TIME : 9 : 30 AM To 12 : 30 PM
put x = 3 in equation (3) :
ƒ'''(3) = 6
\ ƒ(x) = x
3
 – 5x
2
 + 2x + 6
ƒ(2) = 8 – 20 + 4 + 6 = –2
3. If a circle C passing through the point (4,0)
touches the circle x
2
 + y
2
 + 4x – 6y = 12
externally at the point (1, –1), then the radius
of C is :
(1) 
57
(2) 4 (3) 
25
(4) 5
Ans. (4)
Sol. x
2
 + y
2
 + 4x – 6y – 12 = 0
Equation of tangent at (1, –1)
x – y + 2(x + 1) – 3(y – 1) – 12 = 0
3x – 4y – 7 = 0
\ Equation of circle is
(x
2
 + y
2
 + 4x – 6y – 12) + l(3x – 4y – 7) = 0
It passes through (4, 0) :
(16 + 16 – 12) + l(12 – 7) = 0
Þ 20 + l(5) = 0
Þl = –4
\ (x
2
 + y
2
 + 4x – 6y – 12) – 4(3x – 4y – 7) = 0
or x
2
 + y
2
 – 8x + 10y + 16 = 0
Radius = 
+ -= 16 25 16 5
4. In a class of 140 students numbered 1 to 140,
all even numbered students opted mathematics
course, those whose number is divisible by 3
opted Physics course and  theose whose
number is divisible by 5 opted Chemistry
course. Then the number of students who did
not opt for any of the three courses is :
(1) 102 (2) 42 (3) 1 (4) 38
Ans. (4)
Sol. Let n(A) = number of students opted
Mathematics = 70,
n(B) = number of students opted Physics = 46,
n(C) = number of students opted Chemistry
= 28,
n(A Ç B) = 23,
E 2
JEE (Main) Examination–2019/Morning Session/10-01-2019
n(B Ç C) = 9,
n(A Ç C) = 14,
n(A Ç B Ç C) = 4,
Now n(A È B È C)
= n(A) + n(B) + n(C) – n(A Ç B) – n(B Ç C)
– n(A Ç C) + n(A Ç B Ç C)
= 70 + 46 + 28 – 23 – 9 – 14 + 4 = 102
So number of students not opted for any course
= Total – n(A È B È C)
= 140 – 102 = 38
5. The sum of all two digit positive numbers which
when divided by 7 yield 2 or 5 as remainder
is :
(1) 1365 (2) 1256 (3) 1465 (4) 1356
Ans. (4)
Sol.
=
+
å
13
r2
(7r 2)
= 
+
´+´
2 13
7. 6 2 12
2
= 7 × 90 + 24 = 654
=
+ æö
+= ´+ ´ =
ç÷
èø
å
13
r1
1 13
(7r 5) 7 13 5 13 702
2
Total = 654 + 702 = 1356
6. Let 
1
ˆ ˆˆ
a 2i j 3k = +l+
r
, 
2
ˆ ˆˆ
b 4i (3 )j 6k = + -l+
r
 and
3
ˆˆˆ
c 3i 6 j ( 1)k = + + l-
r
 be three vectors such that
b 2a =
r
r
 and 
a
r
 is perpendicular to 
c
r
. Then a
possible value of (l
1
,l
2
,l
3
) is :-
(1) 
1
,4,2
2
æö
-
ç÷
èø
(2) 
1
,4,0
2
æö
-
ç÷
èø
(3) (1,3,1) (4) (1,5,1)
Ans. (2)
Sol. + -l + = + l+
21
ˆ ˆ ˆˆ ˆˆ
4i (3 )j 6k 4i 2 j 6k
Þ 3 – l
2
 = 2l
1
 
Þ 2l
1
 + l
2
 = 3 ....(1)
Given 
=
rr
a.c0
Þ 6 + 6l
1
 + 3(l
3
 – 1) = 0
Þ 2l
1
 + l
3
 = –1 .....(2)
Now (l
1
, l
2
, l
3
) = (l
1
, 3 – 2l
1
, –1 – 2l
1
)
Now check the options, option (2) is correct
7. The equation of a tangent to the hyperbola
4x
2
–5y
2
 = 20 parallel to the line x–y = 2 is :
(1) x–y+9 = 0
(2) x–y+7 = 0
(3) x–y+1 = 0
(4) x–y–3 = 0
Ans. (3)
Sol. Hyperbola -=
22
xy
1
54
slope of tangent = 1
equation of tangent = ±- y x 54
Þ y = x ± 1
Þ y = x + 1 or y = x – 1
8. If the area enclosed between the curves y=kx
2
and x=ky
2
, (k>0), is 1 square unit. Then k is:
(1)
1
3
(2) 
2
3
(3) 
3
2
(4) 
3
Ans. (1)
Sol. Area bounded by y
2
 = 4ax & x
2
 = 4by, a, b ¹ 0
is 
16ab
3
by using formula : ==>
1
4a 4b,k0
k
Area ==
11
16. .
4k 4k
1
3
Þ
=
2
1
k
3
Þ
=
1
k
3
9. Let 
2
max{| x |,x }, | x | 2
f(x)
8 2|x|, 2 |x| 4
ì £ ï
=
í
- <£ ï
î
Let S be the set of points in the interval (–4,4)
at which f is not differentiable. Then S:
(1) is an empty set
(2) equals {–2, –1, 1, 2}
(3) equals {–2, –1, 0, 1, 2}
(4) equals {–2, 2}
Ans. (3)
Sol.
+ - £ <- ì
ï
- £ £-
ï
ï
= - <<
í
ï
££
ï
ï
- <£
î
2
2
82x, 4 x2
x, 2 x1
ƒ(x) |x |, 1 x 1
x, 1 x2
8 2x, 2 x4
y=8+2x
y=x
2
y= –x
–4
–2
–1
1 2 4
y=8–2x
y=x
2
y=x
ƒ(x) is not differentiable at x = {–2,–1,0,1,2}
Þ S = {–2, –1, 0, 1, 2}
E 3
JEE (Main) Examination–2019/Morning Session/10-01-2019
10. If the parabolas y
2
=4b(x–c) and y
2
=8ax have
a common normal, then which one of the
following is a valid choice for the ordered triad
(a,b,c)
(1) (1, 1, 0) (2) 
1
,2,3
2
æö
ç÷
èø
(3)
1
,2,0
2
æö
ç÷
èø
(4) (1, 1, 3)
Ans. (1,2,3,4)
Sol. Normal to these two curves are
y = m(x – c) – 2bm – bm
3
,
y = mx – 4am – 2am
3
If they have a common normal
(c + 2b) m + bm
3
 = 4am + 2am
3
Now (4a – c – 2b) m = (b – 2a)m
3
We get all options are correct for m = 0
(common normal x-axis)
Ans. (1), (2), (3), (4)
Remark :
If we consider question as
If the parabolas y
2
 = 4b(x – c) and y
2
 = 8ax
have a common normal other than x-axis, then
which one of the following is a valid choice for
the ordered triad (a, b, c) ?
 When m ¹ 0 : (4a – c – 2b) = (b – 2a)m
2
= ->Þ >
--
2
cc
m 202
2ab 2ab
Now according to options, option 4 is correct
11. The sum of all values of 
0,
2
p æö
qÎ
ç÷
èø
 satisfying
24
3
sin 2 cos2
4
q+ q=
 is :
(1) 
2
p
(2) p (3) 
3
8
p
(4) 
5
4
p
Ans. (1)
Sol. sin
2
2q + cos
4
2q =
3
4
, 
p æö
qÎ
ç÷
èø
0,
2
Þ 1 – cos
2
2q + cos
4
2q =
3
4
Þ 4cos
4
2q – 4cos
2
2q + 1 = 0
Þ (2cos
2
2q – 1)
2
 = 0
Þ
p
q==
22
1
cos 2 cos
24
Þ
p
q= p± 2n
4
, n Î I
Þ
pp
q=±
n
28
Þ
ppp
q=- ,
828
Sum of solutions 
p
2
12. Let z
1
 and z
2
 be any two non-zero complex
numbers such that 3|z
1
| = 4 |z
2
|.
If 
12
21
3z 2z
z
2z 3z
=+ then :
(1)
1 17
|z|
22
= (2) Re(z) = 0
(3) 
5
|z|
2
= (4) Im(z) = 0
Ans. (Bonus)
Sol. 3|z
1
| = 4|z
2
|
Þ
=
1
2
|z| 4
|z|3
Þ
=
1
2
|3z|
2
|2z|
Let 
== q+q
1
2
3z
a 2cos 2isin
2z
=+=+
12
21
3z 2z 1
za
2z3za
= q+q
53
cos isin
22
Now all options are incorrect
Remark :
There is a misprint in the problem actual
problem should be :
"Let z
1
 and z
2
 be any non-zero complex
number such that 3|z
1
| = 2|z
2
|.
If 
=+
12
21
3z 2z
z
2z 3z
, then"
Given
3|z
1
| = 2|z
2
|
Now 
=
1
2
3z
1
2z
E 4
JEE (Main) Examination–2019/Morning Session/10-01-2019
Let 
== q+q
1
2
3z
a cos isin
2z
=+
12
21
3z 2z
z
2z 3z
=+=q
1
a 2cos
a
\ Im(z) = 0
Now option (4) is correct.
13. If the system of equations
x+y+z = 5
x+2y+3z = 9
x+3y+az = b
has infinitely many solutions, then b–a equals:
(1) 5 (2) 18 (3) 21 (4) 8
Ans. (4)
Sol. = = =a- - =a-
a a-
11 1 111
D 1 2 3 0 1 2 ( 1 ) 4 ( 5)
13 021
for infinite solutions D = 0 Þ a = 5
=Þ=
b
x
511
D 0 9 2 30
35
Þ
- -=
b--
0 01
1 1 30
15 25
Þ +b- = Þb-= 2 15 0 13 0
on b = 13 we get D
y
 = D
z
 = 0
a = 5, b = 13
14. The shortest distance between the point 
3
,0
2
æö
ç÷
èø
and the curve y x,(x 0) => is :
(1) 
5
2
(2) 
5
4
(3) 
3
2
(4) 
3
2
Ans. (1)
Sol. Let points 
æö
ç÷
èø
3
,0
2
, (t
2
, t), t > 0
Distance =
æö
+-
ç÷
èø
2
22
3
tt
2
=
-+
42
9
t 2t
4
=
-+
22
5
(t 1)
4
So minimum distance is 
=
55
42
15. Consider the quadratic equation
(c–5)x
2
–2cx + (c–4) = 0, c¹5. Let S be the set
of all integral values of c for which one root of
the equation lies in the interval (0,2) and its
other root lies in the interval (2,3). Then the
number of elements in S is :
(1) 11 (2) 18 (3) 10 (4) 12
Ans. (1)
Sol.
0
2 3
Let ƒ(x) = (c – 5)x
2
 – 2cx + c – 4
\ ƒ(0)ƒ(2) < 0 .....(1)
& ƒ(2)ƒ(3) < 0 .....(2)
from (1) & (2)
(c – 4)(c – 24) < 0
& (c – 24)(4c – 49) < 0
Þ
<<
49
c 24
4
\ s = {13, 14, 15, ..... 23}
Number of elements in set S = 11
16.
20 20
i1
20 20
i1 i i1
C k
21 CC
-
= -
æö
=
ç÷
+ èø
å , then k equals :
(1) 200 (2) 50 (3) 100 (4) 400
Ans. (3)
Sol.
-
= -
æö
=
ç÷
+
èø
å
3
20 20
i1
20 20
i1 i i1
C k
21 CC
Þ
-
=
æö
=
ç÷
èø
å
3
20 20
i1
21
i1 i
C k
21 C
Þ
=
æö
=
ç÷
èø
å
3
20
i1
ik
21 21
Þ
éù
=
êú
ëû
2
3
1 20(21) k
2 21 (21)
Þ 100 = k
Page 5


E 1
JEE (Main) Examination–2019/Morning Session/10-01-2019
1. Consider a triangular plot ABC with sides
AB=7m, BC=5m and CA=6m. A vertical
lamp-post at the mid point D of AC subtends
an angle 30° at B. The height (in m) of the
lamp-post is:
(1) 
73
(2) 
2
21
3
(3) 
3
21
2
(4) 
2 21
Ans. (2)
Sol.
B
A C
D
30°
3 3
5
h
7
BD = hcot30° = h3
So, 7
2
 + 5
2
 = +
22
2(h 3) 3)
Þ 37 = 3h
2
 + 9.
Þ 3h
2
 = 28
Þ
==
282
h 21
33
2. Let f : R®R be a function such that
f(x) = x
3
+x
2
f'(1) + xf''(2)+f'''(3), xÎR.
Then f(2) equal :
(1) 8 (2) –2 (3) –4 (4) 30
Ans. (2)
Sol. ƒ(x) = x
3
 + x
2
ƒ'(1) + xƒ''(2) + ƒ'''(3)
Þ ƒ'(x) = 3x
2
 + 2xƒ'(1) + ƒ''(x) .....(1)
Þ ƒ''(x) = 6x + 2ƒ'(1) .....(2)
Þ ƒ'''(x) = 6 .....(3)
put x = 1 in equation (1) :
ƒ'(1) = 3 + 2ƒ'(1) + ƒ''(2) .....(4)
put x = 2 in equation (2) :
ƒ''(2) = 12 + 2ƒ'(1) .....(5)
from equation (4) & (5) :
–3 – ƒ'(1) = 12 + 2ƒ'(1)
Þ 3ƒ'(1) = –15
Þ ƒ'(1) = –5  Þ  ƒ''(2) = 2  ....(2)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Thursday 10
th
 JANUARY , 2019)    TIME : 9 : 30 AM To 12 : 30 PM
put x = 3 in equation (3) :
ƒ'''(3) = 6
\ ƒ(x) = x
3
 – 5x
2
 + 2x + 6
ƒ(2) = 8 – 20 + 4 + 6 = –2
3. If a circle C passing through the point (4,0)
touches the circle x
2
 + y
2
 + 4x – 6y = 12
externally at the point (1, –1), then the radius
of C is :
(1) 
57
(2) 4 (3) 
25
(4) 5
Ans. (4)
Sol. x
2
 + y
2
 + 4x – 6y – 12 = 0
Equation of tangent at (1, –1)
x – y + 2(x + 1) – 3(y – 1) – 12 = 0
3x – 4y – 7 = 0
\ Equation of circle is
(x
2
 + y
2
 + 4x – 6y – 12) + l(3x – 4y – 7) = 0
It passes through (4, 0) :
(16 + 16 – 12) + l(12 – 7) = 0
Þ 20 + l(5) = 0
Þl = –4
\ (x
2
 + y
2
 + 4x – 6y – 12) – 4(3x – 4y – 7) = 0
or x
2
 + y
2
 – 8x + 10y + 16 = 0
Radius = 
+ -= 16 25 16 5
4. In a class of 140 students numbered 1 to 140,
all even numbered students opted mathematics
course, those whose number is divisible by 3
opted Physics course and  theose whose
number is divisible by 5 opted Chemistry
course. Then the number of students who did
not opt for any of the three courses is :
(1) 102 (2) 42 (3) 1 (4) 38
Ans. (4)
Sol. Let n(A) = number of students opted
Mathematics = 70,
n(B) = number of students opted Physics = 46,
n(C) = number of students opted Chemistry
= 28,
n(A Ç B) = 23,
E 2
JEE (Main) Examination–2019/Morning Session/10-01-2019
n(B Ç C) = 9,
n(A Ç C) = 14,
n(A Ç B Ç C) = 4,
Now n(A È B È C)
= n(A) + n(B) + n(C) – n(A Ç B) – n(B Ç C)
– n(A Ç C) + n(A Ç B Ç C)
= 70 + 46 + 28 – 23 – 9 – 14 + 4 = 102
So number of students not opted for any course
= Total – n(A È B È C)
= 140 – 102 = 38
5. The sum of all two digit positive numbers which
when divided by 7 yield 2 or 5 as remainder
is :
(1) 1365 (2) 1256 (3) 1465 (4) 1356
Ans. (4)
Sol.
=
+
å
13
r2
(7r 2)
= 
+
´+´
2 13
7. 6 2 12
2
= 7 × 90 + 24 = 654
=
+ æö
+= ´+ ´ =
ç÷
èø
å
13
r1
1 13
(7r 5) 7 13 5 13 702
2
Total = 654 + 702 = 1356
6. Let 
1
ˆ ˆˆ
a 2i j 3k = +l+
r
, 
2
ˆ ˆˆ
b 4i (3 )j 6k = + -l+
r
 and
3
ˆˆˆ
c 3i 6 j ( 1)k = + + l-
r
 be three vectors such that
b 2a =
r
r
 and 
a
r
 is perpendicular to 
c
r
. Then a
possible value of (l
1
,l
2
,l
3
) is :-
(1) 
1
,4,2
2
æö
-
ç÷
èø
(2) 
1
,4,0
2
æö
-
ç÷
èø
(3) (1,3,1) (4) (1,5,1)
Ans. (2)
Sol. + -l + = + l+
21
ˆ ˆ ˆˆ ˆˆ
4i (3 )j 6k 4i 2 j 6k
Þ 3 – l
2
 = 2l
1
 
Þ 2l
1
 + l
2
 = 3 ....(1)
Given 
=
rr
a.c0
Þ 6 + 6l
1
 + 3(l
3
 – 1) = 0
Þ 2l
1
 + l
3
 = –1 .....(2)
Now (l
1
, l
2
, l
3
) = (l
1
, 3 – 2l
1
, –1 – 2l
1
)
Now check the options, option (2) is correct
7. The equation of a tangent to the hyperbola
4x
2
–5y
2
 = 20 parallel to the line x–y = 2 is :
(1) x–y+9 = 0
(2) x–y+7 = 0
(3) x–y+1 = 0
(4) x–y–3 = 0
Ans. (3)
Sol. Hyperbola -=
22
xy
1
54
slope of tangent = 1
equation of tangent = ±- y x 54
Þ y = x ± 1
Þ y = x + 1 or y = x – 1
8. If the area enclosed between the curves y=kx
2
and x=ky
2
, (k>0), is 1 square unit. Then k is:
(1)
1
3
(2) 
2
3
(3) 
3
2
(4) 
3
Ans. (1)
Sol. Area bounded by y
2
 = 4ax & x
2
 = 4by, a, b ¹ 0
is 
16ab
3
by using formula : ==>
1
4a 4b,k0
k
Area ==
11
16. .
4k 4k
1
3
Þ
=
2
1
k
3
Þ
=
1
k
3
9. Let 
2
max{| x |,x }, | x | 2
f(x)
8 2|x|, 2 |x| 4
ì £ ï
=
í
- <£ ï
î
Let S be the set of points in the interval (–4,4)
at which f is not differentiable. Then S:
(1) is an empty set
(2) equals {–2, –1, 1, 2}
(3) equals {–2, –1, 0, 1, 2}
(4) equals {–2, 2}
Ans. (3)
Sol.
+ - £ <- ì
ï
- £ £-
ï
ï
= - <<
í
ï
££
ï
ï
- <£
î
2
2
82x, 4 x2
x, 2 x1
ƒ(x) |x |, 1 x 1
x, 1 x2
8 2x, 2 x4
y=8+2x
y=x
2
y= –x
–4
–2
–1
1 2 4
y=8–2x
y=x
2
y=x
ƒ(x) is not differentiable at x = {–2,–1,0,1,2}
Þ S = {–2, –1, 0, 1, 2}
E 3
JEE (Main) Examination–2019/Morning Session/10-01-2019
10. If the parabolas y
2
=4b(x–c) and y
2
=8ax have
a common normal, then which one of the
following is a valid choice for the ordered triad
(a,b,c)
(1) (1, 1, 0) (2) 
1
,2,3
2
æö
ç÷
èø
(3)
1
,2,0
2
æö
ç÷
èø
(4) (1, 1, 3)
Ans. (1,2,3,4)
Sol. Normal to these two curves are
y = m(x – c) – 2bm – bm
3
,
y = mx – 4am – 2am
3
If they have a common normal
(c + 2b) m + bm
3
 = 4am + 2am
3
Now (4a – c – 2b) m = (b – 2a)m
3
We get all options are correct for m = 0
(common normal x-axis)
Ans. (1), (2), (3), (4)
Remark :
If we consider question as
If the parabolas y
2
 = 4b(x – c) and y
2
 = 8ax
have a common normal other than x-axis, then
which one of the following is a valid choice for
the ordered triad (a, b, c) ?
 When m ¹ 0 : (4a – c – 2b) = (b – 2a)m
2
= ->Þ >
--
2
cc
m 202
2ab 2ab
Now according to options, option 4 is correct
11. The sum of all values of 
0,
2
p æö
qÎ
ç÷
èø
 satisfying
24
3
sin 2 cos2
4
q+ q=
 is :
(1) 
2
p
(2) p (3) 
3
8
p
(4) 
5
4
p
Ans. (1)
Sol. sin
2
2q + cos
4
2q =
3
4
, 
p æö
qÎ
ç÷
èø
0,
2
Þ 1 – cos
2
2q + cos
4
2q =
3
4
Þ 4cos
4
2q – 4cos
2
2q + 1 = 0
Þ (2cos
2
2q – 1)
2
 = 0
Þ
p
q==
22
1
cos 2 cos
24
Þ
p
q= p± 2n
4
, n Î I
Þ
pp
q=±
n
28
Þ
ppp
q=- ,
828
Sum of solutions 
p
2
12. Let z
1
 and z
2
 be any two non-zero complex
numbers such that 3|z
1
| = 4 |z
2
|.
If 
12
21
3z 2z
z
2z 3z
=+ then :
(1)
1 17
|z|
22
= (2) Re(z) = 0
(3) 
5
|z|
2
= (4) Im(z) = 0
Ans. (Bonus)
Sol. 3|z
1
| = 4|z
2
|
Þ
=
1
2
|z| 4
|z|3
Þ
=
1
2
|3z|
2
|2z|
Let 
== q+q
1
2
3z
a 2cos 2isin
2z
=+=+
12
21
3z 2z 1
za
2z3za
= q+q
53
cos isin
22
Now all options are incorrect
Remark :
There is a misprint in the problem actual
problem should be :
"Let z
1
 and z
2
 be any non-zero complex
number such that 3|z
1
| = 2|z
2
|.
If 
=+
12
21
3z 2z
z
2z 3z
, then"
Given
3|z
1
| = 2|z
2
|
Now 
=
1
2
3z
1
2z
E 4
JEE (Main) Examination–2019/Morning Session/10-01-2019
Let 
== q+q
1
2
3z
a cos isin
2z
=+
12
21
3z 2z
z
2z 3z
=+=q
1
a 2cos
a
\ Im(z) = 0
Now option (4) is correct.
13. If the system of equations
x+y+z = 5
x+2y+3z = 9
x+3y+az = b
has infinitely many solutions, then b–a equals:
(1) 5 (2) 18 (3) 21 (4) 8
Ans. (4)
Sol. = = =a- - =a-
a a-
11 1 111
D 1 2 3 0 1 2 ( 1 ) 4 ( 5)
13 021
for infinite solutions D = 0 Þ a = 5
=Þ=
b
x
511
D 0 9 2 30
35
Þ
- -=
b--
0 01
1 1 30
15 25
Þ +b- = Þb-= 2 15 0 13 0
on b = 13 we get D
y
 = D
z
 = 0
a = 5, b = 13
14. The shortest distance between the point 
3
,0
2
æö
ç÷
èø
and the curve y x,(x 0) => is :
(1) 
5
2
(2) 
5
4
(3) 
3
2
(4) 
3
2
Ans. (1)
Sol. Let points 
æö
ç÷
èø
3
,0
2
, (t
2
, t), t > 0
Distance =
æö
+-
ç÷
èø
2
22
3
tt
2
=
-+
42
9
t 2t
4
=
-+
22
5
(t 1)
4
So minimum distance is 
=
55
42
15. Consider the quadratic equation
(c–5)x
2
–2cx + (c–4) = 0, c¹5. Let S be the set
of all integral values of c for which one root of
the equation lies in the interval (0,2) and its
other root lies in the interval (2,3). Then the
number of elements in S is :
(1) 11 (2) 18 (3) 10 (4) 12
Ans. (1)
Sol.
0
2 3
Let ƒ(x) = (c – 5)x
2
 – 2cx + c – 4
\ ƒ(0)ƒ(2) < 0 .....(1)
& ƒ(2)ƒ(3) < 0 .....(2)
from (1) & (2)
(c – 4)(c – 24) < 0
& (c – 24)(4c – 49) < 0
Þ
<<
49
c 24
4
\ s = {13, 14, 15, ..... 23}
Number of elements in set S = 11
16.
20 20
i1
20 20
i1 i i1
C k
21 CC
-
= -
æö
=
ç÷
+ èø
å , then k equals :
(1) 200 (2) 50 (3) 100 (4) 400
Ans. (3)
Sol.
-
= -
æö
=
ç÷
+
èø
å
3
20 20
i1
20 20
i1 i i1
C k
21 CC
Þ
-
=
æö
=
ç÷
èø
å
3
20 20
i1
21
i1 i
C k
21 C
Þ
=
æö
=
ç÷
èø
å
3
20
i1
ik
21 21
Þ
éù
=
êú
ëû
2
3
1 20(21) k
2 21 (21)
Þ 100 = k
E 5
JEE (Main) Examination–2019/Morning Session/10-01-2019
17. Let dÎR, and
2 4 d (sin )2
A 1 (sin)2d
5 (2sin )d ( sin )2 2d
- + q- éù
êú
= q+
êú
êú q- - q+ +
ëû
,
qÎ[0,2p]. If the minimum value of det(A) is 8,
then a value of d is :
(1) –7 (2) ( )
2 22 +
(3) –5 (4) ( )
2 21 +
Ans. (3)
Sol. detA =
- + q-
q+
q- - q++
2 4 d sin2
1sin2d
5 2sind sin2 2d
(R
1
 ® R
1
 + R
3
 – 2R
2
)
=
q+
q- + -q
100
1sin2d
5 2sin d 2 2d sin
= +q + -q- q- (2 sin )(2 2d sin ) d(2sin d)
=4 +
 
4d
 
–
 
2sinq
 
+
 
2sinq+2dsinq
 
–
 
sin
2
q–2dsinq+d
2
=d
2
 + 4d + 4 – sin
2
q
=(d + 2)
2
 – sin
2
q
For a given d, minimum value of
det(A) =  (d + 2)
2
 – 1 = 8
Þ d = 1 or –5
18. If the third term in the binomial expansion of
( )
2
5
logx
1x + equals 2560, then a possible value
of x is :
(1) 
22
(2) 
1
8
(3) 
42
(4) 
1
4
Ans. (4)
Sol.
+
logx 5
2
(1x)
==
logx 52
2
32
T C .(x ) 2560
Þ
=
2logx
2
10.x 2560
Þ
=
2logx
2
x 256
Þ 2(log
2
x)
2
 = log
2
256
Þ 2(log
2
x)
2
 = 8
Þ (log
2
x)
2
 = 4     Þ log
2
x = 2 or –2
=
1
x 4 or
4
19. If the line 3x + 4y – 24 = 0 intersects the x-axis
at the point A and the y-axis at the point B, then
the incentre of the triangle OAB, where O is the
origin, is
(1) (3, 4) (2) (2, 2) (3) (4, 4) (4) (4, 3)
Ans. (2)
Sol.
(0,6)
(0,0) (8,0)
(r,r)
3x+4y=24
+-
=
3r 4r 24
r
5
7r – 24 = ±5r
2r = 24  or 12r + 24
r = 14, r = 2
then incentre is (2, 2)
20. The mean of five observations is 5 and their
variance is 9.20. If three of the given five
observations are 1, 3 and 8, then a ratio of other
two observations is :
(1) 4 : 9 (2) 6 : 7
(3) 5 : 8 (4) 10 : 3
Ans. (1)
Sol. Let two observations are x
1
 & x
2
mean = =
å i
x
5
5
 Þ 1 + 3 + 8 + x
1
 + x
2
 = 25
Þ x
1
 + x
2
 = 13 ....(1)
variance (s
2
) = -=
å
2
i
x
25 9.20
5
Þ  =
å
2
i
x 171
Þ +=
22
12
x x 97 .....(2)
by (1) & (2)
(x
1
 + x
2
)
2
 – 2x
1
x
2
 = 97
or x
1
x
2
 = 36
\ x
1
 : x
2
 = 4 : 9