SCORE JEE (A) JEE-Maacs - ALLEN

SCORE JEE (Advanced) JEE-Mathematics

HOME ASSIGNMENT # 05

SOLUTION

1.

MATHEMATICS

Ans. (B)

Case-I :

Two rows contain 2 letters each and one row

has 1 letter.

Possible ways = 4 ? 3c2 + 4c2 ? 3c1 + 4c2 ? 3c2 ? 2

= 12 + 18 + 36 = 66

Case-II :

One row has 3 letters and two others have 1

letter each.

5.

Possible ways = 4c3 ? 3c1 ? 2c1 + 4c1 ? 2c1

= 4 ? 3 ? 2 + 4 ? 2 = 24 + 8 = 32

Hence total arrangements =

2.

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3.

HS

(66 + 32)5!

= 5880

2!

C2 .3! + 4 C1 .

3!

= 48

2!

(ii) one digit divisible by 4 and one odd :

2

C1.4 C1. 3! = 48

Case-II :

a, b, c having exactly two 5 with :

(i) one digit divisible by 4 : 2 .

p/2

2

1

�C1

(3, sin3)

p

3

cos2

Ans. (A)

tan q = k

q

q

- tan3

3

3=k

2 q

1 - 3 tan

3

tan 3

q

q

q

- 3k tan2 - 3 tan + k = 0

3

3

3

a

3!

=6

2!

b

? tan 3 tan 3 = �C3.

Ans. (B)

R = log2 sin ?? p ?�� + log2 sin ?? 2p ?�� + log2 sin ?? 3p ?��

�� 5?

�� 5?

�� 5?

+ log2 sin ?? 4p ?�� �C log25

�� 5?

R = log2 ?? sin p .sin 2p .sin 3p .sin 4p ?�� �C log25

��

5

5

5?

5

= log2 ?? sin2 p .sin2 2p ?�� �C log25

��

5

5?

= log2

1

4

= log2

1

{(1 �C cos 72��) (1 �C cos144��)} �C log25

4

= log2

1

{(1 �C sin18��)(1 + cos 36��)}�Clog25

4

= log2

1

4

?

2 p

2 2p ?

?�� 2 sin 5 .2 sin 5 ��?

�C log25

5 �C 1? ?

5 + 1? ?��

?��?

1+

��? 1 �C

?

��

?

4 ?��

4 ��? ??

??��

�C log25

?

? ?

?

= log2 1 ? 5 - 5 �� ? 5 + 5 �� �C log25

4

4

4

��

Hence number of ways = 102

4.

sin1

cos1

sin3

3 tan

Ans. (D)

Suppose we consider {Ace, 2, 3, 4, 5} for each

card we have 4 possible suits. So, total ways of

such a straight = (4)5 = 210.

But in this we have counted those cases when

6.

all are of same suit = 4.

So total ways of such a straight = 210 �C 4

But we have 10 such straights - {Ace, 2, 3, 4, 5},

{2, 3, 4, 5, 6} ........ {10, J, Q, K, Ace}

So, total ways = 10 . (210 �C 4) = 10200.

Ans. (B)

Case-I :

a, b, c having exactly one 5 with :

(i) 2 even digits (different or same) :

4

cos2 < sin3 < cos1< sin1

\ sin1 is greatest

Ans. (A)

By the graph it is clear that

= log2

? ��

?

1 ? 20 ?

�C log25 = log2

4 ?�� 4 ? 4 ��?

= log25 �C log216 �C log25 = �C 4.

1

? 5?

?�� 16 ��?

�C log25

JEE-Mathematics

Ans. (D)

6cos5q �C 6cos4q �C 5cos3q + 5cos2q + cosq �C 1 = 0

? 6cos4q(cosq �C 1) �C 5cos2q(cosq �C 1)

+ (cosq �C 1) = 0

? (cosq �C 1){6cos4q �C 5cos2q +1} = 0

? (cosq �C 1)(3cos2q �C 1)(2cos2q �C 1) = 0

cos q = 1; cos q = ��

?

0

8.

1

?

4

3

; cos q = ��

Ans. (B)

q1 q2 q3

+

+

= p

2 2

2

tan

2

q1

q

q

s

s

; ry = tan 2 ; rz = tan 3

2

2

2

3

2

rx ry rz rx ry rz

+ + = 2

6 3 2

s

12.

1

3

Ans. (B)

sinx �C 3sin 2x + sin3x = cosx �C 3cos2x + cos3x

? 2sin2xcosx �C 3sin2x = 2cos2x cosx �C 3cos2x

? sin2x(2cosx �C 3) = cos2x(2cosx �C 3)

? (2cosx �C 3)(sin2x �C cos2x) = 0

\ cosx =

sinx = at x = a, b; a ? (0, p/2)

3

& b ? (p/2,p)

\ Number of solutions in [�Cp, p] is 5

Ans. (A)

Let the boxes be B1, B2, B3, B4. Let us assume 13.

that two specific balls have been put in box Bi

(i = 1,2,3,4).

It means in box Bi we have to put 3 balls from

the remaining 18 balls.

Thus the probability that the two specific balls

have been put in the particular box

P(B i ) =

10.

20

C3

5?4

1

=

=

C 5 20 ? 19 19

2sin?? A - B ?��cos ?? A + B ?�� = 0

�� 2 ? �� 2 ?

2sin ?? A + B ��? sin ?? B - A ��? = 0

�� 2 ? �� 2 ?

?

sin ?? A - B ?�� = 0

�� 2 ?

2

\

2

x=

Ans. (B)

p

p

with 2x ? (2k + 1)

4

2

np p

+

2 8

1

1

y + y ? 2 ? y+ y ? 2

sinx + cosx = 2 is only possible case. When y = 1

? cosx

1

2

1

+sinx

2

=1

p?

14.

cannot be zero

simultaneously.

2

p

= 2np

4

?

x-

\

at n = 0, x = , y = 1

? x = 2np +

p

4

p

4

Ans. (A)

AH2 + BC2 = 4R2 cos2 A + 4R cos2 A = 4R2

?

as sin A + B & cos A + B

or tan2x = 1 with cos2x ? 0

? 2x = np +

?

&

3

2

? cos ?�� x - ��? = cos0

4

Ans. (A)

sin A = sin B and cos A = cos B

?

C

rx 2ry 3rz 6rx ry rz

+

+

=

s

s

s

s3

1

18

B

q1 P q2

q3

q

q

q1

q

q

q

+ tan 2 + tan 3 = tan 1 tan 2 tan 3

2

2

2

2

2

2

rx = s tan

?

4

sinx = 0 at x = �C p, 0, p

9.

A

q1 + q2 + q3 = 2p

1

? 8 solutions

Ans. (D)

3sin2x �C sinx + ln (sgn(cot�C1x)) = 0

Q cot�C1 x ? (0, p)

\ sgn(cot�C1x) = 1 ? ln (sgn(cot�C1x) = 0

Now, 3sin2x �C sinx = 0

sinx = 0 or

11.

=

1

64

(AH2 + BC2) (BH2 + AC2) (CH2 + AB2)

64R 6

= R6 = 64.

64

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7.

HS

JEE-Mathematics

15.

Ans. (D)

tan A + tanB + tanC = (tanA tanB)tanC

tanA + tanB + tanC =

3

4

\

19.

tanC

given that

3

4

tanA tanB =

3

4 tan A

tanA +

3

4 tan A

? tan B =

+ tanC =

=1 20.

3

tanC

4

3

4t

+

1

tanC =0

4

21.

4t2 + (tanC)t + 3 = 0

t is real

?

D?0

tan2C �C 4 (3)(4) ? 0

16.

tan2 750 = (2+ 3 )2 = 7+2 3

words starting from RAA = 4! = 24

RADA = 3! = 6

RADHA = 2! = 2

RADHIAK = 1

\ Number of words before RADHIKA = 2193

Ans. (B)

Each of the N persons from a pair with (N �C 3)

person (i.e. excluding the person himself and

the adjacent two)

So total number of pairs that can be formed

4

2

22.

2

tan x + cot y + 8 = 4 tan x + 4 cot y - 4l

2

(tan2 x - 2)2 + (cot2 y - 2)2 + 4l2 = 0

possible if

tan x = 2 ; cot y = 2 ; l = 0

Ans. (B)

Total �C (Dearrangement)

4! �C (Dearrangement of 4 objects)

24 �C 9 = 15

Ans. (B)

=

2

3

3

3

3

?

N(N - 3)

2

Total time they sing =

23.

No. of digits

��

= (360)4 = 1440

Ans. (C)

4

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? 6! ?

D, H, I, K= ? 2! �� .4

? 48

Possible only for D option because

2 tan x cot y = tan2 x + cot 2 y - l2

HS

or

2(2 cos 2x - 1) + 2(1 - cos 2x)2 = 2

? 2 cos 2x - 1 + 1 - cos2x = 1 or cos2x = 1

\ 2x = 2np, n ? I orx = np

Ans. (A)

AADHIKR

words starting from A = 6! = 720

words starting from

l2 = tan2 x + cot2 y - 2 tan x cot y

18.

Ans. (D)

tan2C

l = tan x - cot y

17.

9.9.8 7

=

900 25

2(2cos2x - 1) + 3 - 4cos2x + (2cos2 2x - 1) = 2

Let tanA = t

t+

Total numbers

= 2 + 6 + 18 + 54 + 162 + 486 = 728

Ans. (D)

P(atleast two digit same) = 1 �C P(All digits

different)

3

Single digit number = 2

Two digit number = 2 �� 3

Three digit number = 2 �� 3 �� 3

Four digit number = 2 �� 3 �� 3 �� 3

Five digit number = 2 �� 3 �� 3 �� 3 �� 3

Six digit number = 2 �� 3 �� 3 �� 3 �� 3 �� 3

N(N - 3)

? 2 = 28

2

? on solving N = 7 or �C 4

? N=7

(as N > 0)

Ans. (D)

x = np �C tan�C13 ? tanx = �C3

Now, tan2x =

cosx = ��

2 tan x

3

=

1 - tan2 x 4

1

2

1 + tan x

=��

and

1

10

on substituting these values in the given

equation, we find only cosx = �C

3

1

10

satisfies

JEE-Mathematics

the equation, so equation holds true for tanx

= �C3 and cosx = �C

24.

25.

10

which is possible if x lies in II quadrant.

So, n must be odd integer.

Ans. (C)

x+y?5

x, y ? 1

x+y?3

x, y ? 0

x+y+z=3

x, y, z ? 0

number of non-negative integral solutions of

the above equation is 3+3�C1C

5

3�C1 = C2 = 10

number of integral points which lies inside or

on the square is 8.

i.e (1,1) (1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2)

\

28.

1

Required probability is =

29.

Ans. (D)

9! = 27 �� 34 �� 5 �� 7

odd factors of the form 3m + 2 are neither

multiple of 2 nor multiple of 3. So the factors

may be 1, 5, 7, 35 of which 5 and 35 are of the

form 3m + 2, their sum is 40.

Ans. (A)

-p ? p sin x ? p

? -1 ? cos ( p sin x ) ? 1

30.

2

1

=

10 5

?

?p

?

-1 ? sin ? ( cos p sin x ) �� ? 1

2

��

?

?

?p

?

-p ? p sin ? cos ( p sin x ) �� ? p

��2

?

? �C1 ? y ? 1

Ans. (C)

1

sinx ?

Ans. (A)

? (q) = 3 + 2sin2q �C 3sin2q

= 3 + 1 �C cos2q �C 3sin2q

= 4 �C (cos2q + 3sin2q)

5 +1

? sin x ?

5 -1

p

? sin x ? sin

4

10

( 5 - 1)

4

- 10 ? cos2q + 3 sin 2q ? 10

p

10

-(cos2q + 3sin2q) ? 10

9p

10

4 - (cos 2q + 3sin2q) ? 4 + 10

(

26.

10

�� p 9p ��

x?�� , ��

?10 10 ?

) ?(q) ? 1

Ans. (A)

31.

tan 500�� + 2 tan 470��

tan140�� + 2 tan110��

=

1 + tan 500�� tan 490�� 1 + tan140�� tan130��

1 ? 2 tan 20��

2 ?

?

?

+ ��

+

= - ?

= -2?

��

2

2

t?

2 �� 1 - tan 20�� tan 20�� ?

��1 - t

2

2

2

2

t+

Ans. (C)

A>B

3sinx �C 4sin3x = k ? sin3x = k

A & B are roots of sin3x = k

? 3B = sin�C1 k, 3A = p �C sin�C1 k

Now C = p �C (A + B)

p

2

x

- sin 2 x + 2 cos x )

\ 3sin 2 x + 2 cos x + 33.3 (

= 28

+

sin

2

x

2

cos

x

Put 3

= t

1

= t(t2 - 1)

27.

2

+ 31- sin 2x +2 sin x = 28

1�C sin2x + 2sin2x

= 1 �C (sin2x �C 2sin2x)

= 3 �C (sin2x + 2cos2x)

- tan 40�� - 2cot20��

2t

p

9p ��

��

��2kp + 10 , 2kp + 10 ��

?

?

Ans. (A)

3 sin 2 x +2 cos

= 1 + (- tan 40��)( - cot 40��)

1

general solution

1

1

27

= 28

t

t2 �C 28t + 27 = 0

t = 1, 27

\ sin2x + 2cos2x = 0, 3

? 2sinx cosx + 2cos2x = 0, 3

Case-I : 2cosx(sinx + cosx) = 0

? cosx = 0, tanx = �C1

Case-II : 2cosx(sinx + cosx) = 3

? cos2x + sin2x = 2 which is not possible

2p

?

?

= p - ? - sin -1 k + sin -1 k �� =

3

3

3

3

��

?

4

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log 4 +

HS

JEE-Mathematics

from case-I general solution is

p

2

? (x) = (1 �C sin2x)3 + (sin2x)2

Let sin2x = t

g(t) = (1 �C t)3 + t2, where 0 ? t ? 1

g'(t) = 2t �C 3(1 �C t)2 = �C{3t2 �C 8t + 3}

p

4

x = (2K + 1) , Kp - , K ? I

32.

33.

Ans. (C)

2079000 = 23 �� 33 �� 53 �� 7 �� 11

For the divisors to be even and divisible by

15; 2, 3 and 5 must occur atleast once.

Therefore the total number of required divisors

are 3 �� 3 �� 3 �� 2 �� 2 = 108

Ans. (D)

A

?

= �C ?? t ��

36.

2A A

3

3

? g(t) ? g(0)

? g(t) ? 1

? ? (x) ? 1 & 1 + 2x4 ? 1

? Inequality holds only for x = 0.

Ans. (A)

We have tan(A + B) tan(A �C B) = 1

?

B

D

2A

3 = sin B

AD

BD

.......... (i)

37.

A

= 3

CD

sin

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HS

34.

35.

?

1 - tan2 B 1

1

=

or cos2B =

1 + tan2 B 2

2

2A

3

A

sin

3

38.

= 2cos

\ B=p

6

Ans. (C)

abc(a + b + c) 1

= 4RD .2s

D

D

BD sin B CD sin C

=

2A

A

sin

sin

3

3

?

1 + tan B 1 - tan B

+

=4

1 - tan B 1 + tan B

=

sin

p

?

.......... (ii)

from (i) & (ii)

sin B CD

=

sin C BD

p

p

or A =

2

4

p

Using sine rule in DADC

sin C

AD

2A =

?

?

?

?

Also tan ? 4 + B �� + tan ? 4 - B �� = 4

��

?

��

?

C

Using sine rule in DABD

sin

4 - 7 ??

4+ 7?

t�C

��?

��

��?

3 ?��

3 ��?

A

3

5

8R

? 8.2 = 4

r

(Q R ? 2r)

Ans. (D)

Number must be divisible by 3 & 5 as the sum

is 48 so every number will be divisible by 3.

For divisibility of 5, unit digit must be '5'.

99988 5

?

5!

= 10

2! 3!

(ii) 99997 5

?

5!

=5

4!

(i)

Ans. (C)

Given set of numbers is {1, 2, ........ 9} is which

4 are even and 5 are odd, so in the given product

it is not possible to arrange to substract only

39.

even number from odd number. There must be

atleast one factor involving substraction of an

odd number from another odd number. So

atleast one of the factor is even. Hence product

is always even.

\ Required probability = 1.

Ans. (D)

cos6x + sin4x ? 1 + 2x4

8Rs

=

D

��������

15

Ans. (D)

Case

(i) x1x2x3x4x5

(ii) x1x1x2x3x4

(iii) x1x1x1x2x3

(iv) x1x1x2x2x3

(v) x1x1x1x2x2

Number of ways

6C

5

3C .5C

1

3

2C .5C

1

2

3C .4C

2

1

2C .2C

1

1

=6

= 30

= 20

= 12

=4

������

 ................
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