19 JEE MAIN TEST SERIES FFIIIITTJJEEEE JJEEEE … · 21 JEE MAIN TEST SERIES –2_22/01/2018...
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Transcript of 19 JEE MAIN TEST SERIES FFIIIITTJJEEEE JJEEEE … · 21 JEE MAIN TEST SERIES –2_22/01/2018...
19 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
FFIIIITTJJEEEE JJEEEE MMAAIINN TTEESSTT SSEERRIIEESS (Full Test –2)
ANSWER KEY
PHYSICS CHEMISTRY MATHEMATICS
1 A 1 B 1 D
2 A 2 D 2 B
3 B 3 C 3 C
4 C 4 C 4 B
5 B 5 C 5 C
6 A 6 D 6 C
7 C 7 D 7 B
8 D 8 D 8 D
9 B 9 B 9 B
10 A 10 C 10 C
11 A 11 C 11 A
12 B 12 B 12 D
13 D 13 B 13 D
14 A 14 B 14 B
15 B 15 B 15 A
16 B 16 C 16 D
17 B 17 A 17 A
18 D 18 A 18 B
19 D 19 D 19 A
20 D 20 B 20 B
21 D 21 C 21 A
22 D 22 B 22 D
23 A 23 C 23 C
24 B 24 B 24 B
25 B 25 B 25 D
26 D 26 A 26 C
27 D 27 A 27 C
28 A 28 D 28 C
29 D 29 B 29 D
30 C 30 C 30 B
20 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
Hint & Solution Physics
1. (A) Area of sphere = A
eA(T4 – T0
4) = 300 W
4 4 4 40 0
A Ae T T T T 600
2 2
e 13e 1 12 2 2 e
e 2 2 3
2. (A) PV = RT(for one mole)
W = RT
PdV dVV
Given V = CT2/3
1/32dV CT dT
3
or dV 2 dT
V 3 T
2
1
T
2 1T
2 dT 2W RT R T T 166.2J
3 T 3
3. (B)
V1 sin = V2 sin . 4. (C) Temp. at C is 3T0.
So network done = R(1.5 T0) + 3
2 RT0 – RT0 – 0
0
T3 5R RT
2 2 4
5. (B)
Fractions of volume submerged at t1 and t2 are 1
1
d 1 t
and
2 1 12
2 2
d 1 t 1 t.
1 t
6. (A)
23 V
g5 5
V2 = 30
H =
2
2y
430
V 30 16 485m 96cm
2g 20 25 20 50
16 × 6 = 96 N = 6 7. (C)
When length × is hanging the force pulling chain down = m
gxl
Work done in pulling it by further length dx is dW = mgx
dxl
Total work done in pulling chain through length l = KE gained by chain 0
21 mgmV dW dx
2
l
l l
Work done =
0
0
2 2202 2 2 2 2
0 0
2m mg x mg 1 mgg dx ; mV or V
2 2 2 2
ll
ll
l ll l l l
l l l l l
8. D
F = Fu = dV
m vdt
or Vdv = P P ds
dtM M V
2 PV dv ds
M
Integrating it, we get
1/3v v sS2 3 3 3
u 0 0
3PP P 3Pv dV vdt ds ; V u s V u
M M M M
21 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
9. B Since no force is placed along the surface of plate, so, momentum conservation principle for ball is
applicable along the surface of the plate.
mv sin 1 = mv1 sin2
or v sin1 = mv1 sin2 ........ (1)
1 2 1 2
1
V cos V cose
Vcos Vcos
(Let 1 = )
V1 cos 2 = eV cos ......... (2)
1 2
1 2
V sin V sin tan
V cos eV cos e
or 2
tantan
e
or 1
2
tantan
e
10. A According to law of conservation of momentum I = constant When viscous fluid of mass, m is dropped and start spreading out then its moment of inertia increases
and angular velocity decreases. 11. A
G R and ev gR
1
2
g2
g 1
3
v3
v
1
3
g3
g
22 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
23 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
24 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
25 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
CHEMISTRY
26 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
27 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
28 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
29 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
MATHEMATICS 1 D
The lines are x + 7y = 7 2 and x + 7y = 3 2
Distance between them = 10 2
25 2
2. B
x y
1p q
X xcos y sin x Xcos sin
Y x sin ycos y Xsin Ycos
cos sin cos sin
x Y 1p q q p
cos sin cos sin
p q q p
( equal intercepts)
3. C
Slope of AD = b 0 2b
0 a/ 2 a
Slope of BE = b / 2 b
a / 2 a
Since AD and BE are perpendicular
2b b
. 1a a
a2 = 2b
2
Slope of AC is –b/a
Equation of the line through (a, b) perpendicular to AC is
a
y b x ab
ax – by = a2 – b
2
± 22bx by b from (i)
± 2x y b y = 2 x b
So an equation of required line is y 2 x b
4. B
y
y = 1
x 13
3,0 3,0
x
1y 0 x 3
3
5. C
Centre is (3, 4) radius = 3 8 4
3 55
6. C (x – 2) (x – 1) is a factor of x
4 – ax
2 + b
1 – a + b = 0 and 16 – 4a + b = 0
a = 5 and b = 4
30 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
7. B Eliminating x from x + y + z = 4 and x
2 + y
2 + z
2 = 6, we get
y2 + z
2 + (4 – y – z)
2 = 6
2y2 – 2(4 – z)y + (4 – z)
2 + z
2 – 6 = 0
As y is real, we get
(4 – z)2 – 2[(4 – z)
2 + z
2 – 6] 0
3z2 – 8z + 4 0
(3z – 2) (z – 2) 0
2/3 z 2 Thus, maximum value of z is 2 8. D |B| = |A
T.A| = |A
T|.|A| = |A|
2
Let p1(x) = ax2 + bx + c
p2(x) = cx2 + dx + e
p3(x) = fx2 + gx + h
|A| =
2
2
2
independent of x
ax bx c 2ax b 2a c b 2a
cx dx e 2cx d 2c e d 2c
h g 2ffx gx h 2fx g 2f
9. B
Using C1 C1 – C3 in , we get
=
0 sin cos
0 cos sin
2cos sin cos
= 2cos(sin2 - cos
2)
= -2cos cos2
= 0 for = /4 (0, /2)
This is the only value of lying in (0, /2) for which = 0 The system of linear equations will have a non-trivial solution if and only if
1 =
1 sin cos
1 cos sin 0
1 sin cos
Using R2 R2 + R, we get
1 =
2 0 0
1 cos sin 0
1 sin cos
2(-cos2 + sin
2) = 0
-2cos2 = 0
This is true for only one value of (0, /2)
viz, = /4 Thus, statement-1 is also true However statement-2 is not a correct reason for statement-1 10. C We have A
2 + B
2 = (BA)
2 + (AB)
2
= (BA) (BA) + (AB) (AB) = B(AB)A + A(BA)B = B(BA) + A(AB) = BA + AB = A + B 11. A We know that det(A) = a11A11 + a12A21 + a13A31
= (1) (-2) + (3) (-1) + (2) (3)
= 6 - 5
And (det(A))2 = det(Adj A) = -10 + 17 - 6
2
Thus, (6 - 5)2 = -10 + 17 - 6
2
422 – 77 + 35 = 0
= 1, 5/6
31 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
Therefore, a possible value of det(A) is 1. 12. D
1 1 1 1 1 1 1 1 1
1. .... 2. .... 3. ...1 2 3 20 2 3 20 3 20
1 1 1 1 1
20. 1 1 2 1 2 3 .... 1 2 3 .... 2020 1 2 3 20
= 1 1 2 1 3 1 20 1 20 21
... 10 1152 2 2 2 4
13. D
S = 2017 + 2 2016
1 1 1.2016 .2015 .... .1
4 4 4
2 2016 2017
2 2016 2017
1 1 1 1 1.S .2017 .2016 .... .2 .1
4 4 4 4 43 1 1 1 1
.S 2017 ....4 4 4 4 4
(subtracting)
2017
1 11
3 4 4S 2017
141
4
S = 2017
4 4 12017 1
3 9 4
14. B
h 0 h 0
f a f a h f a h f alim lim f ' a
h h
h 0 h 0
f a 2h f a f a h f af a 2h f a hlim lim
2h 2h
=
h 0 h 0
f a 2h f a f a h f a1 1 1lim lim f ' a f ' a f ' a
2h 2 h 2 2
h 0 h 0
f a 2h f a f a 2h f alim 2 lim 2f ' a
h 2h
15. A
2
2 2
2cos 1 1 2sin .cos sin cos
sin sinsin sin
= |1 + cot | = -(1 + cot) 3
4
16. B Squaring and adding the given relations we get 16 + 9 + 24sin(P + Q) = 37
sin(P + Q) = 1/2
sinR = 1/2
R = /6 or 5/6
If R = 5/6, then P < /6
3sinP < 3/2
3sinP + 4cosQ < 3/2 + 4 < 6
So R 5/617. A
AB = 2asin2
In ABC,
AC
tanAB
AC = 2a sin2
tan
32 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
18. B
du dv
f ''' f ' sin f ''' f ' cosd d
2 2
2du dvf ''' f '
d d
I = f ''' f ' d f '' f c
19. A
fn(x) = t, 1 2 n 1
dt 1
dx xf x f x .....f x
1
1 2 n n 1
dtxf x f x ......f x dx logt c f x c
dx
20. B
f(x) = 2ex [ f(x) = f(x), f(0) = 2]
(x) = x + 2ex
1 11 1 0
x 2x 6 x x 2x 2 2
00 0
f x x dx 2 xe 2e dx 2 xe e e 2 e e 1 e 1 2e
21. A 22. D
1 1
1
1 100 0
p x dx 1.p x dx xp x I p 1 I
Where I1 = 1 1 1 1
10 0 0 0
xp' x dx 1 x p' 1 x dx 1 x p' x dx p' x dx I
2I1 = 1
0p x p 1 p 0
Thus 1
0
1 1 1p x dx p 1 p 1 p 0 p 1 p 0 41 1 21
2 2 2
23. C 24. B
x x
x x
10 10y
10 10
2x 1 y
101 y
10
1 1 yx log
2 1 y
f-1
(y) = 10
1 1 ylog
2 1 y
25. D
Period is equal to
226
2 / 6
26. C
1 1 1 1 2
2
xtan x sin ,sin x cos 1 x
1 x
.
So, required value is 2
2
2
x 1011 x
1171 x
27. C
1 1 2 1 2
2
1cos sin x x 1 cos x x 1
21 x x
2
2
1x x 1
1 x x
x2 + x + 1 = 1 x = -1, 0
33 JEE MAIN TEST SERIES–2_22/01/2018
FIITJEE Ltd. Property No.80, Behind Sales Tax Office, Mangal Pandey Nagar, University Road, Meerut. Tel: 0121–4054378, 7351004567 Web: www.fiitjee.com
28. C
1 1x 1 xx tan tan
x 2 x 2
1
x 1 x
x 2 x 2x tanx x 1
1x 2
1
2
x 2x tan
2x 5x 4
2 2
x x 2 x 20 as x
2x 5x 4 2x 5x 4
1
2
29. B Since ‘f’ is continuous and can take on rational values. ‘f’ must be a constant function
f(1.5) = f(2) = 10 30. B u(x) = h(f(g(x))) = h((x
2)) = h(sin x
2) = log sin x
2
Hence u(x) = 2x cot x2 and u(x) = 2 cot x
2 – 4x
2 cosec
2x
2