SCORE JEE (A) JEE-Maacs - ALLEN

[Pages:22]SCORE JEE (Advanced) JEE-Mathematics

HOME ASSIGNMENT # 05

SOLUTION

1. 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.

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

5.

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

Hence total arrangements = (66 + 32)5! = 5880

2!

2. 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

all are of same suit = 4.

6.

So total ways of such a straight = 210 ? 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 ? 4) = 10200.

3. Ans. (B)

Case-I :

a, b, c having exactly one 5 with : (i) 2 even digits (different or same) :

4

C2.3!

+4

C1.

3! 2!

=

48

(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 . 3! = 6

2!

Hence number of ways = 102

4. Ans. (A) By the graph it is clear that

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HS

1

MATHEMATICS

cos2 < sin3 < cos1< sin1 \ sin1 is greatest

sin1

cos1 sin3 ?1

cos2

(3, sin3)

p/2

p

1

23

Ans. (A) tan q = k

3 tan q - tan3 q

3

3= k

1 - 3 tan2 q

3

tan3 q - 3k tan2 q - 3 tan q + k = 0

3

3

3

?

tan

a 3

tan

b 3

=

?3.

Ans. (B)

R

=

log2

sin

? ??

p 5

? ??

+

log2

sin

? ??

2p 5

? ??

+

log2

sin

? ??

3p 5

? ??

+

log2

sin

? ??

4p? 5 ??

? log25

R

=

log2

? ??

sin

p 5

.sin

2p 5

.sin

3p 5

.sin

4p ? 5 ??

? log25

= log2

? ??

sin2

p 5

.sin2

2p ? 5 ??

? log25

= log2

1 4

??? 2

sin2

p 5

.2

sin2

2p ? 5 ??

? log25

1

= log2 4 {(1 ? cos 72?) (1 ? cos144?)} ? log25

1

= log2 4 {(1 ? sin18?)(1 + cos 36?)}?log25

=

log2

1 4

????????1 ?

5? 4

1? ??

? ??1

+

5 + 1? ??

4

??

? ??

?

log25

= log2

1 4

?5?? 4

5? ??

?5+

? ?

4

5? ? ?

? log25

= log2

1 4

? ??

20 ? 4 ? 4??

?

log25

=

log2

? 5? ?? 16??

? log25

= log25 ? log216 ? log25 = ? 4.

JEE-Mathematics

7. Ans. (D)

11. Ans. (B)

6cos5q ? 6cos4q ? 5cos3q + 5cos2q + cosq ? 1 = 0

q1 + q2 + q3 = 2p

A

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

+ (cosq ? 1) = 0 ? (cosq ? 1){6cos4q ? 5cos2q +1} = 0 ? (cosq ? 1)(3cos2q ? 1)(2cos2q ? 1) = 0

cos q = 1; cos q = ? 1 ; cos q = ? 1

3

2

?

?

?

q1 + q2 + q3 = p 222

q1 P q2

q3

B

C

tan q1 + tan q2 + tan q3 = tan q1 tan q2 tan q3

2

2

2

222

rx

=

s tan q1 ; 2

ry

=

s tan q2 ; 22

rz

=

s tan q3 32

0

4

4

? 8 solutions

rx s

+ 2ry s

+ 3rz s

=

6rx ry rz s3

8. Ans. (D) 3sin2x ? sinx + ln (sgn(cot?1x)) = 0 Q cot?1 x ? (0, p) \ sgn(cot?1x) = 1 ? ln (sgn(cot?1x) = 0 Now, 3sin2x ? sinx = 0

1

sinx = 0 or 3 sinx = 0 at x = ? p, 0, p

rx 6

+ ry + rz 32

=

rx ry rz s2

12. Ans. (B)

sinx ? 3sin 2x + sin3x = cosx ? 3cos2x + cos3x

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

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

? (2cosx ? 3)(sin2x ? cos2x) = 0

1

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

\

cosx

=

3 2

or tan2x = 1 with cos2x ? 0

3

& b ? (p/2,p)

? 2x = np + p with 2x ? (2k + 1) p

\ Number of solutions in [?p, p] is 5

4

2

9. Ans. (A)

\

x =

np + p 28

Let the boxes be B1, B2, B3, B4. Let us assume 13. Ans. (B)

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

1

y+ y ?2?

y+1 ? 2 y

the remaining 18 balls. Thus the probability that the two specific balls

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

have been put in the particular box

P(Bi ) =

18 C3 20 C5

= 5?4 20 ?19

1 =

19

10. Ans. (A)

1

1

? cosx 2 +sinx 2 =1

?

cos

? ??

x

-

p 4

? ??

= cos0

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sin A = sin B and cos A = cos B

?

2sin???

A

2

B

???cos???

A

+ 2

B

???

=

0

&

2sin???

A

+ 2

B

???sin???

B

2

A

???

=

0

?

sin???

A

2

B

???

=

0

?

x - p = 2np 4

?

x = 2np +

p 4

\

at

n = 0,

x

=

p 4

,

y = 1

14. Ans. (A) AH2 + BC2 = 4R2 cos2 A + 4R cos2 A = 4R2

?

1 64

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

as sin A + B & cos A + B cannot be zero

2

2

= 64R6 = R6 = 64.

64

simultaneously.

HS

2

15. Ans. (D) tan A + tanB + tanC = (tanA tanB)tanC

tanA + tanB + tanC =

3 4

tanC

given that

tanA tanB =

3 4

?

tan

B =

3 4 tan A

tanA +

3 4 tan A

+ tanC =

3 4

tanC

Let tanA = t

t +

3 4t

+

1 4

tanC

=0

4t2 + (tanC)t + 3 = 0

t is real ? D ? 0

tan2C ? 4 (3)(4) ? 0

tan2C ? 48 Possible only for D option because

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

16. Ans. (C)

l = tan x - cot y

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

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

tan4 x + cot4 y + 8 = 4 tan2 x + 4 cot2 y - 4l2

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

possible if 17. Ans. (B)

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

Total ? (Dearrangement)

4! ? (Dearrangement of 4 objects)

24 ? 9 = 15 18. Ans. (B)

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No. of digits 2 3 3 3 3 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

HS

JEE-Mathematics

\ Total numbers

= 2 + 6 + 18 + 54 + 162 + 486 = 728 19. Ans. (D)

P(atleast two digit same) = 1 ? P(All digits different)

=

1

-

9.9.8 900

=

7 25

20. Ans. (D)

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

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

21. Ans. (A)

A A D H I K R words starting from A = 6! = 720

words starting from

D,

H,

I,

K=

? ? ?

6! 2!

??.4 ?

= (360)4 =1440

words starting from RAA = 4! = 24

RADA = 3! = 6

RADHA = 2! = 2

RADHIAK = 1

\ Number of words before RADHIKA = 2193

22. Ans. (B)

Each of the N persons from a pair with (N ? 3) person (i.e. excluding the person himself and the adjacent two)

So total number of pairs that can be formed

=

N(N - 3) 2

Total

time

they

sing =

N(N - 3) ? 2 2

= 28

? on solving N = 7 or ? 4

? N=7 23. Ans. (D)

x = np ? tan?13

(as N > 0) ? tanx = ?3

2 tan x 3

Now, tan2x = 1 - tan2 x = 4 and

1

1

cosx = ?

=? 1 + tan2 x

10

on substituting these values in the given

equation, we find only cosx = ? 1 satisfies

10

3

JEE-Mathematics

the equation, so equation holds true for tanx 28. Ans. (D)

1

= ?3 and cosx = ?

10

9! = 27 ? 34 ? 5 ? 7 odd factors of the form 3m + 2 are neither

which is possible if x lies in II quadrant.

multiple of 2 nor multiple of 3. So the factors

So, n must be odd integer. 24. Ans. (C)

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

x + y ? 5

x, y ? 1

form 3m + 2, their sum is 40.

x + y ? 3 x + y + z = 3

x, y ? 0 x, y, z ? 0

29. Ans. (A)

-p ? psin x ? p

? -1 ? cos (psin x) ? 1

number of non-negative integral solutions of

the above equation is 3+3?1C3?1 = 5C2 = 10 number of integral points which lies inside or

?

-1

?

sin

? ??

p 2

(cos

p

sin

x

)

? ??

?

1

?

-p

?

p

sin

? ??

p 2

cos

(p

sin

x

) ???

?

p

on the square is 8.

? ?1 ? y ? 1

i.e (1,1) (1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) 30. Ans. (C)

\

Required probability

is

=

2 10

=

1 5

1

sinx ? 5 + 1

25. Ans. (A) (q) = 3 + 2sin2q ? 3sin2q = 3 + 1 ? cos2q ? 3sin2q

? sin x ? 5 - 1 ? sin x ? sin p

4

10

= 4 ? (cos2q + 3sin2q)

- 10 ? cos2q + 3 sin 2q ? 10 -(cos2q + 3sin2q) ? 10

p

9p

10 10

( 5 - 1) 4

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

log(4+ 10) (q) ? 1

26. Ans. (A)

tan 1+

500? + 2 tan 500?

tan tan

470? 490?

=

tan140? + 2 tan110? 1 + tan140? tan130?

- tan 40? - 2cot20?

= 1 + (- tan 40?)(- cot 40?)

=

-

1 2

? ??

2 tan 20? 1 - tan2 20?

+

2 tan 20?

? ??

=

-

1 2

? ??

1

2t - t2

+

2 t

? ??

1

= t(t2 -1) 27. Ans. (C)

A > B 3sinx ? 4sin3x = k ? sin3x = k

A & B are roots of sin3x = k ? 3B = sin?1 k, 3A = p ? sin?1 k Now C = p ? (A + B)

=

p

-

? ??

p 3

-

1 3

sin-1

k

+

1 3

sin-1

k

? ??

=

2p 3

x

?

???1p0,

9p? 10 ??

general solution

???2kp

+

p, 10

2kp

+

9p? 10??

31. Ans. (A)

3 + 3 = 28 sin2x+2 cos2 x

1-sin 2x +2 sin2 x

1? sin2x + 2sin2x

= 1 ? (sin2x ? 2sin2x)

= 3 ? (sin2x + 2cos2x)

\

( ) 3 + 3 .3 sin2x+2cos2 x

3 - sin 2 x+2 cos2 x

= 28

Put 3 = t sin2x+2cos2 x

t + 27 = 28 t

t2 ? 28t + 27 = 0

t = 1, 27 \ sin2x + 2cos2x = 0, 3 ? 2sinx cosx + 2cos2x = 0, 3

Case-I : 2cosx(sinx + cosx) = 0 ? cosx = 0, tanx = ?1

Case-II : 2cosx(sinx + cosx) = 3 ? cos2x + sin2x = 2 which is not possible

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32. 33.

34. 35.

HS

from case-I general solution is

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

2

4

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

36.

2A A 33

B

D

C

Using sine rule in DABD

sin 2A 3

=

sin B

BD AD

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

Using sine rule in DADC

sin C AD

=

sin A 3

CD

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

37.

from (i) & (ii)

BD sin B CD sin C

sin 2A = sin A

38.

3

3

?

sin B sin C

=

CD BD

sin 2A 3

sin A

=

2 cos

A 3

3

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 even number from odd number. There must be 39.

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

5

(x) = (1 ? sin2x)3 + (sin2x)2 Let sin2x = t g(t) = (1 ? t)3 + t2, where 0 ? t ? 1 g'(t) = 2t ? 3(1 ? t)2 = ?{3t2 ? 8t + 3}

=

?

? ??? t -

43

7

? ???

? ???

t

?

4+ 3

7? ???

? 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 ? B) = 1

?

2A = p or A = p

2

4

Also

tan

? ??

p 4

+

B

? ??

+

tan

? ??

p 4

-

B

? ??

=

4

?

1 + tan B + 1 - tan B = 4 1 - tan B 1 + tan B

?

1 - tan2 B 1 1 + tan2 B = 2

or

cos2B = 1 2

Ans. (C)

\ B=p

6

abc(a + b + c) = 1 4RD.2s

D

D

= 8Rs = 8R ? 8.2 = 4

D

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'.

(i) 99988 5 ? 5! = 10

2! 3!

(ii) 99997 5

? 5! = 5 4!

-------- 15

Ans. (D)

Case

(i) x1x2x3x4x5 (ii) x1x1x2x3x4 (iii) x1x1x1x2x3 (iv) x1x1x2x2x3 (v) x1x1x1x2x2

Number of ways

6C5

= 6

3C1.5C3 = 30

2C1 .5C2 = 20

3C2 .4C1 = 12

2C1.2C1 = 4 ------

JEE-Mathematics

Total 72 40. Ans. (C)

cos2 (px) - sin2 (py) = 1 2 1

? cosp(x + y) cosp(x ? y) = 2

In each way x1 & x3 can be interchanged

\ Number of required triplets are

2 ? 20 = 40

Case-II : When numbers are equal x1 = x2 = x3 then number of triplets = 10

p1

1

? cosp(x + y) cos 3 = 2 (Q x ? y = 3 )

Hence number of vectors ar are 40 + 10 = 50

? cosp(x + y) = 1 ? x + y = 2n, n ? I 44. Ans. (C)

1

\ x + y = 2n & x ? y = 3

Without any loss of generality assume the wts to be 1, 2, 3, 4, 5, 6

1

1

? x=n+ 6, y=n? 6, n?I

It is obvious that 1 should be at the top of pyramid. If 2, 3 make second row then

for n = ?1,

(x,

y)

=

? ??

-

5, 6

-

7 6

? ??

41. Ans. (D)

First of all select 1 + 2 + 3 + 4 = 10 pens out of

25 identical pens and distribute them as desired.

2 4

1 3-------2!

5 6-------3!

? 12 ways

If 2, 4 make second row

It can happen only in one way. Now let x1, x2, x3 and x4 pens are given to them respectively (here x1, x2, x3, x4 ? 0). As now any one can get any number of pens.

1

2

4-------2

? 4 ways

3

5 6-------2

Total number of ways = 16

\ non-negative integral solution of

45. Ans. (B)

x1 + x2 + x3 + x4 = 15 will be the number of ways so 15 + 4 ? 1C4 ? 1 = 18C3. 42. Ans. (B)

The prime digits are 2, 3, 5, 7.

x, y, z, w = k p ? [0, 10]

2

Q ? 120 2 3 4 5 sin2 x cos2 y sin2 z cos2 w

If we fix 2 at first place, then other eleven places

? ? 2. 3. 4. 5 2 3 4 5 sin2 x cos2 y sin2 z cos2 w

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are filled by all four digits. So, total number of ways = 411. Now sum of 2 consecutive digits is prime when consecutive digits are (2,3) or (2, 5), then 2 will be fixed at all alternate places.

222222

So, favourable cases = 26

26 1

\ Required probability = 222 = 216 43. Ans. (C)

ar .br = 0

Taking logarithm both sides we have ? sin2 x log 2 + cos2 y log 3 + sin2 z log 4

+ cos2 w log5 ? log2 + log3 + log4 + log5 ? cos2 x log 2 + sin2 y log 3 + cos2 z log 4

+ sin2 w log 5 ? 0 which is possible only when

cos2 x = 0 ? x = mp + p ,m ? I

2

sin2 y = 0 ? y = np, n ? I

cos2 z = 0 ? z = rp + p , r ? I

2

? x1 ? 2x2 + x3 = 0 ? x1 + x3 = 2x2 x1, x2, x3 are selected from the set {1, 2, 3,.....,10},

Case-I : 1, 3, 5, 7, 9 = 5 numbers

2, 4, 6, 8, 10 = 5 numbers

3 numbers in A.P. can be selected by 5C2 + 5C2 = 20 ways.

6

sin2 w = 0 ? w = pp , p ? I

Q

x, y, z, w ? [0, 10]

?

x =

p , 3p 22

, 5p 2

? y = 0, p, 2p, 3p

(three solutions) (four solutions)

HS

JEE-Mathematics

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46. 47. 48.

49.

HS

?

z

=

p 2

,

3p 2

,

5p 2

(three solutions)

? w = 0, p, 2p, 3p (four solutions)

Hence the number of ordered 4-tuple (x, y, z,

w) is 3 . 4 . 3 . 4 = 144

And

probability

=

144 74

Ans. (D)

Area OALKO

Probability = Area OABC

C (0,4)

B(4,4) KL

4

? xdx

= 0

4?4

x

O

A(4,0)

=

2 3

??

x3

/

2

??

=

16 3

=

1

16 16 3

50.

Ans. (B)

1324 2143 3142 4132

1432 2413 3214 4113

51.

.

2431 3241 4321

Total 11 numbers

Ans. (C)

(i) For two digit number = 9 ? (11, 22,....99) (ii) For three digit number all same = 9

(111, 222, ......)

(iii) For 3 digit when 2 digit same and 1 different.

(a) ? ? 0 or ? 0 ? = 9 + 9 = 18

(b) without zero 3C . 9 . 8 = 216 1

Total 216 + 18 + 9 + 9 = 252

p =

252 999

=

28 111

Ans. (B)

C

p/2

B

D

A O p/2

?1

F

E

p

3p/2 2p

(x) = max(cosx, cos?1(cosx))

A1 = Area of ABCDEFA = 2 . area of trapezium ABCFA

7

= 2 . 1 . (AB + CF).AF = 3p2

2

4

p/2

O -p/2

A

B

C

p

|||||||||||||

|||||||||||||2p

D

g(x) = min(sinx, sin?1(sinx))

A2 = Area of triangle ABCDA

1

=2

.

AC . BD = 1

2

.

p

.

p 2

= p2

4

?

A2 = 1 A1 3

Ans. (A,B,C)

Ec = F; E ? Ec = Q; P(E) + P(not E) = 1; E

and Ec can always be equally likely

Ans. (A,B)

B1 = Journey by car B2 = Journey by motor cycle B3 = Journey on foot

1 P (B1) = 2 ;

1 P (B2) = 6 ;

1 P (B3) = 3 Let A = accident occurs

P (A

B1

)

=

1 5

;

P (A

B2 )

=

2 ; P (A

5

B3 )

1 =

10

P (A) = P(A ? B1) + P(A ? B2) + P(A ? B3)

= P(B1) ? P (A B1)+ P(B2) ? P (A B2 )

+ P(B3) ? P (A B3 )

= 1?1 + 1?2 + 1? 1 2 5 6 5 3 10

3 2 1 61 = 30 + 30 + 30 = 30 = 5 ? (A) is correct

P (B1

A) =

1 ?1 25 1

=

1 ?5 10 1

=

1 2

5 ? (B) is correct

JEE-Mathematics

1 2

P (B2

A) =

? 65

1

=

2 ?5 30 1

=

1 3

5

? (C) is incorrect

P

(B3

A) =

1? 1 3 10

1

=

1 ?5 30 1

=

1 6

5

? (D) is incorrect

52. Ans. (A,B,C,D)

54. Ans. (C,D)

2sinx ? 2cos2x ? sin2x ? 2sinx sin2x + 2cosx = 0

2sinx ? 2cosx ? sin2x ? cosx + cos3x + 2cosx = 0 2sinx(1 ? cosx) + 4cos3x ? 3cosx

? 2(2cos2x ? 1) + cosx = 0 2sinx(1 ? cosx) + 4cos3x ? 4cos2x ? 2cosx + 2 = 0 2sinx(1? cosx)+ 4cos2x(cosx?1) ?2(cosx ?1) = 0 (1?cosx){2sinx +2 ?4 (1?sin2x)} =0 (1?cosx){4sin2x + 2 sinx?2} = 0

P(A) + P(B) ? P(A ? B) = 0.8

(1?cosx)(sinx + 1)(2sinx ? 1) = 0

\ P(A ? B) = 0.5 + 0.4 ? 0.8 = 0.1

P(A ? B) = P(B) ? P(A ? B)

= 0.5 ? 0.1 = 0.4 ? (A) is correct

( ) P(B ? A)

0.4

0.4 2

P B A = P(A) = 1- P(A) = 0.6 = 3

? (B) is correct P(A) ? P(B) = 0.2

? P(A ? B) < P(A) ? P(B)

? (C) is correct

x = 0,

2p

,

3p 2

,

p 6

,

5p 6

55. Ans. (A,B,C,D)

(2n) !

Total number of permutations =

(n !)2

= 2nc n

Also 2nc = 2n n ! [1 . 3 . ...... (2 n - 1) ]

n

(n !)2

2n [1 . 3 . ...... (2 n - 1) ]

=

n!

P(A or B but not both) = 0.9 ? 2 ? 0.1 = 0.7

Similarly (B), (D) follows .

53. Ans. (A,B,C)

Hence (A), (B), (C), (D).

cos2x ? sin2x = (cosx ? sinx) (1+sinx cosx) 56. Ans. (A,D) cos x ? sinx =0 or cos x + sinx= 1+ sinx cosx

cosx = sinx

(cos x + sin x)2 - 1

or (cosx+ sinx) = 1+

2

tanx = 1

Let cosx + sinx = t

(A) C 6+4 4 -1 C 10 +4-1 4 -1

=

9 C3 13 C3

= 42 143

(D)

P(A ? B) P(B)

=

9 C3 13 C3

x = np + p ,n ? I 4

t =1 + t2 - 1

2

t2 ?2 t+1 = 0 t =1 sinx + cosx =1

1

11

cosx . + sinx. =

2

22

57. Ans. (A,B,C,D)

P(A) = 9 ; P(B) = P(A ? B) = 9 ? 9 = 81

10

10 10 100

P

? ??

A B

? ??

=

P(A ? B) P(B)

=

9?1 10 10 1 - 81

100

NODE6\E_NODE6 (E)\DATA\2013\IIT-JEE\TARGET\MATHS\HOME ASSIGNMENT (Q.BANK)\SOLUTION\HOME ASSIGNMENT # 05

cos

? ??

x

-

p? 4 ??

=cos

p 4

x?

p 4

=

2np

?

p 4

x=2np

+

p 2

x= 2np

P(A ? B) = P(A) + P(B) - P(A ? B) = 9 10

58. Ans. (A,D) (A) Digits at unit places of 4m are 4 or 6 Digits at unit places of 3n are 3, 9, 7, 1 For divisibility by 5 ; select (4, 1) or (6, 9) at unit places

HS

8

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