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

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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



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



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



CHEMISTRY



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



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



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



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



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

19 jee main test series ffiiiittjjeeee jjeeee … · 21 jee main test series –2_22/01/2018...

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