Download Solution Home Assignment # 05

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
–1
(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
å tan 3 tan 3 = –3.
Ans. (B)
R = log2 sin æç p ö÷ + log2 sin æç 2p ö÷ + log2 sin æç 3p ö÷
è 5ø
è 5ø
è 5ø
+ log2 sin æç 4p ö÷ – log25
è 5ø
R = log2 æç sin p .sin 2p .sin 3p .sin 4p ö÷ – log25
è
5
5
5ø
5
= log2 æç sin2 p .sin2 2p ö÷ – log25
è
5
5ø
= log2
1
4
= log2
1
{(1 – cos 72°) (1 – cos144°)} – log25
4
= log2
1
{(1 – sin18°)(1 + cos 36°)}–log25
4
= log2
1
4
æ
2 p
2 2p ö
çè 2 sin 5 .2 sin 5 ÷ø
– log25
5 – 1ö æ
5 + 1ö ïü
ïìæ
1+
íç 1 –
ý
÷
ç
4 øè
4 ÷ø ïþ
îïè
– log25
æ
ö æ
ö
= log2 1 ç 5 - 5 ÷ ç 5 + 5 ÷ – log25
4
4
4
è
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 – 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.
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
ø è
ø
1 æ 20 ö
– log25 = log2
4 çè 4 ´ 4 ÷ø
= log25 – log216 – log25 = – 4.
1
æ 5ö
çè 16 ÷ø
– log25
JEE-Mathematics
Ans. (D)
6cos5q – 6cos4q – 5cos3q + 5cos2q + cosq – 1 = 0
Þ 6cos4q(cosq – 1) – 5cos2q(cosq – 1)
+ (cosq – 1) = 0
Þ (cosq – 1){6cos4q – 5cos2q +1} = 0
Þ (cosq – 1)(3cos2q – 1)(2cos2q – 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 – 3sin 2x + sin3x = cosx – 3cos2x + cos3x
Þ 2sin2xcosx – 3sin2x = 2cos2x cosx – 3cos2x
Þ sin2x(2cosx – 3) = cos2x(2cosx – 3)
Þ (2cosx – 3)(sin2x – cos2x) = 0
\ cosx =
sinx = at x = a, b; a Î (0, p/2)
3
& b Î (p/2,p)
\ Number of solutions in [–p, 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
2sinæç A - B ö÷cos æç A + B ö÷ = 0
è 2 ø è 2 ø
2sin çæ A + B ÷ö sin çæ B - A ÷ö = 0
è 2 ø è 2 ø
Þ
sin æç A - B ö÷ = 0
è 2 ø
2
\
2
x=
Ans. (B)
p
p
with 2x ¹ (2k + 1)
4
2
np p
+
2 8
1
1
y + y ³ 2 Þ y+ y ³ 2
sinx + cosx = 2 is only possible case. When y = 1
Þ cosx
1
2
1
+sinx
2
=1
pö
14.
cannot be zero
simultaneously.
2
p
= 2np
4
Þ
x-
\
at n = 0, x = , y = 1
Þ x = 2np +
p
4
p
4
Ans. (A)
AH2 + BC2 = 4R2 cos2 A + 4R cos2 A = 4R2
Þ
as sin A + B & cos A + B
or tan2x = 1 with cos2x ¹ 0
Þ 2x = np +
æ
&
3
2
Þ cos çè x - ÷ø = cos0
4
Ans. (A)
sin A = sin B and cos A = cos B
Þ
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 = – p, 0, p
9.
A
q1 + q2 + q3 = 2p
1
Þ 8 solutions
Ans. (D)
3sin2x – sinx + ln (sgn(cot–1x)) = 0
Q cot–1 x Î (0, p)
\ sgn(cot–1x) = 1 Þ ln (sgn(cot–1x) = 0
Now, 3sin2x – 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
Þ tan B =
+ tanC =
=1 20.
3
tanC
4
3
4t
+
1
tanC =0
4
21.
4t2 + (tanC)t + 3 = 0
t is real
Þ
D³0
tan2C – 4 (3)(4) ³ 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 – 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 – (Dearrangement)
4! – (Dearrangement of 4 objects)
24 – 9 = 15
Ans. (B)
=
2
3
3
3
3
ø
N(N - 3)
2
Total time they sing =
23.
No. of digits
è
= (360)4 = 1440
Ans. (C)
4
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æ 6! ö
D, H, I, K= ç 2! ÷ .4
³ 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
Þ 2 cos 2x - 1 + 1 - cos2x = 1 or cos2x = 1
\ 2x = 2np, n Î 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 – 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
Þ on solving N = 7 or – 4
Þ N=7
(as N > 0)
Ans. (D)
x = np – tan–13 Þ tanx = –3
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 = –
3
1
10
satisfies
JEE-Mathematics
the equation, so equation holds true for tanx
= –3 and cosx = –
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 ³ 1
x+y£3
x, y ³ 0
x+y+z=3
x, y, z ³ 0
number of non-negative integral solutions of
the above equation is 3+3–1C
5
3–1 = 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
Þ -1 £ cos ( p sin x ) £ 1
30.
2
1
=
10 5
Þ
æp
ö
-1 £ sin ç ( cos p sin x ) ÷ £ 1
2
è
ø
Þ
æp
ö
-p £ p sin ç cos ( p sin x ) ÷ £ p
è2
ø
Þ –1 £ y £ 1
Ans. (C)
1
sinx ³
Ans. (A)
ƒ (q) = 3 + 2sin2q – 3sin2q
= 3 + 1 – cos2q – 3sin2q
= 4 – (cos2q + 3sin2q)
5 +1
Þ sin x ³
5 -1
p
Þ sin x ³ 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
é p 9p ù
xÎê , ú
ë10 10 û
) ƒ(q) £ 1
Ans. (A)
31.
tan 500° + 2 tan 470°
tan140° + 2 tan110°
=
1 + tan 500° tan 490° 1 + tan140° tan130°
1 æ 2 tan 20°
2 ö
æ
ö
+ ÷
+
= - ç
= -2ç
÷
2
2
tø
2 è 1 - tan 20° tan 20° ø
è1 - t
2
2
2
2
t+
Ans. (C)
A>B
3sinx – 4sin3x = k Þ sin3x = k
A & B are roots of sin3x = k
Þ 3B = sin–1 k, 3A = p – sin–1 k
Now C = p – (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– sin2x + 2sin2x
= 1 – (sin2x – 2sin2x)
= 3 – (sin2x + 2cos2x)
- tan 40° - 2cot20°
2t
p
9p ù
é
ê2kp + 10 , 2kp + 10 ú
ë
û
Ans. (A)
3 sin 2 x +2 cos
= 1 + (- tan 40°)( - cot 40°)
1
general solution
1
1
27
= 28
t
t2 – 28t + 27 = 0
t = 1, 27
\ sin2x + 2cos2x = 0, 3
Þ 2sinx cosx + 2cos2x = 0, 3
Case-I : 2cosx(sinx + cosx) = 0
Þ cosx = 0, tanx = –1
Case-II : 2cosx(sinx + cosx) = 3
Þ cos2x + sin2x = 2 which is not possible
2p
æ
ö
= p - ç - sin -1 k + sin -1 k ÷ =
3
3
3
3
è
ø
4
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log 4 +
HS
JEE-Mathematics
from case-I general solution is
p
2
ƒ (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}
p
4
x = (2K + 1) , Kp - , K Î 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
æ
= – çç t è
36.
2A A
3
3
Þ g(t) £ g(0)
Þ g(t) £ 1
Þ ƒ (x) £ 1 & 1 + 2x4 ³ 1
Þ Inequality holds only for x = 0.
Ans. (A)
We have tan(A + B) tan(A – B) = 1
Þ
B
D
2A
3 = sin B
AD
BD
.......... (i)
37.
A
= 3
CD
sin
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HS
34.
35.
Þ
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
Þ
1 + tan B 1 - tan B
+
=4
1 - tan B 1 + tan B
=
sin
p
Þ
.......... (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 =
æ
ö
æ
ö
Also tan ç 4 + B ÷ + tan ç 4 - B ÷ = 4
è
ø
è
ø
C
Using sine rule in DABD
sin
4 - 7 öæ
4+ 7ö
t–
֍
÷
֍
3 øè
3 ÷ø
A
3
5
8R
³ 8.2 = 4
r
(Q R ³ 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 ³ 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
———
JEE-Mathematics
40.
72
Ans. (C)
cos 2 ( px) - sin2 ( py) =
1
2
Þ cosp(x + y) cosp(x – y) =
p
3
Þ cosp(x + y) cos =
Þ cosp(x + y) = 1
\
1
,
6
for n = –1, (x, y)
42.
(Q x – y =
1
)
3
Þ x + y = 2n, n Î I
x + y = 2n & x – y =
Þ x=n+
41.
1
2
1
2
44.
1
3
1
, nÎI
6
7ö
æ 5
=ç- , - ÷
6ø
è 6
y=n–
Ans. (D)
First of all select 1 + 2 + 3 + 4 = 10 pens out of
25 identical pens and distribute them as desired.
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 ³ 0).
As now any one can get any number of pens.
\ non-negative integral solution of
45.
x1 + x2 + x3 + x4 = 15 will be the number
of ways so 15 + 4 – 1C4 – 1 = 18C3.
Ans. (B)
The prime digits are 2, 3, 5, 7.
If we fix 2 at first place, then other eleven places
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.
2
2
2
2
2
6
So, favourable cases = 2
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
Hence number of vectors ar are
40 + 10 = 50
Ans. (C)
Without any loss of generality assume the wts
to be 1, 2, 3, 4, 5, 6
It is obvious that 1 should be at the top of pyramid.
If 2, 3 make second row then
1
2
4
5
Þ 12 ways
6-------3!
If 2, 4 make second row
1
2
4-------2
3
5
Þ 4 ways
6-------2
Total number of ways = 16
Ans. (B)
x, y, z, w = k
Q
2
2sin
x
p
2
3cos
2
y
Î [0, 10]
4sin
2
z
5cos
2
w
³ 120
Þ 2sin x 3cos y 4sin z 5cos w ³ 2. 3. 4. 5
Taking logarithm both sides we have
Þ sin2 x log 2 + cos2 y log 3 + sin2 z log 4
+ cos2 w log5 ³ log2 + log3 + log4 + log5
Þ cos2 x log 2 + sin2 y log 3 + cos2 z log 4
2
2
2
2
+ sin2 w log 5 £ 0
which is possible only when
p
2
cos2 x = 0 Þ x = mp + ,m Î I
26
1
\ Required probability = 22 = 16
2
2
43.
3-------2!
sin2 y = 0 Þ y = np, n Î I
Ans. (C)
p
2
r r
a.b = 0
cos2 z = 0 Þ z = rp + , r Î I
Þ x1 – 2x2 + x3 = 0 Þ 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
5C + 5C = 20 ways.
2
2
6
sin2 w = 0 Þ w = pp , p Î I
Q
x, y, z, w Î [0, 10]
Þ x=
p 3p 5p
,
,
2 2
2
Þ y = 0, p, 2p, 3p
(three solutions)
(four solutions)
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Total
HS
JEE-Mathematics
(three solutions)
=2.
Þ w = 0, p, 2p, 3p (four solutions)
Hence the number of ordered 4-tuple (x, y, z,
w) is 3 . 4 . 3 . 4 = 144
B(4,4)
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HS
ò
xdx
O
-p/2
4´4
2 3/ 2
16
éë x ùû
1
3
=
= 3 =
16
16 3
1
2
49.
=
Ans. (B)
50.
or ´ 0 ´ = 9 + 9 = 18
O
||
|
=
p2
4
Þ
P (B1) =
1
;
2
P (B2) =
1
;
6
1
3
Let A = accident occurs
P (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
6
1
+
+
=
=
=
30
30
30
30
5
Þ (A) is correct
=
D
F
p
E
3p/2 2p
–1
ƒ (x) = max(cosx,
A2 1
=
A1 3
1
2
1
P (A B1 ) = ; P (A B2 ) = ; P (A B3 ) =
5
5
10
P (A) = P(A Ç B1) + P(A Ç B2) + P(A Ç B3)
28
111
B
A
p/2
p
2
Ans. (A,B,C)
Ec = F; E Ç 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
C
p/2
1
2
= . AC . BD = . p .
(b) without zero 3C1 . 9 . 8 = 216
Total 216 + 18 + 9 + 9 = 252
252
999
||
A2 = Area of triangle ABCDA
Ans. (B)
1324 2143 3142 4132
51.
1432 2413 3214 4113
.
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.
p=
||
g(x) = min(sinx, sin–1(sinx))
x
0
´ ´ 0
D
L
A(4,0)
2p
|
K
||
C
(0,4)
C
B
||
=
(a)
A
||
||
Area OALKO
Area OABC
4
48.
p
O
||
144
74
Ans. (D)
Probability =
47.
p/2
||
46.
3p2
4
. (AB + CF).AF =
||
And probability =
1
2
||
p 3p 5p
,
,
2 2
2
||
Þ z=
1 1
·
1 5
1
· =
P (B1 A ) = 2 5 =
10 1
2
1
5
Þ (B) is correct
cos –1(cosx))
A1 = Area of ABCDEFA
= 2 . area of trapezium ABCFA
7
JEE-Mathematics
54.
1 2
·
2 5 1
5
6
· =
P (B2 A ) = 1 =
30 1 3
5
Þ (C) is incorrect
2sinx – 2cosx – sin2x – cosx + cos3x + 2cosx = 0
2sinx(1 – cosx) + 4cos3x – 3cosx
– 2(2cos2x – 1) + cosx = 0
1 1
·
3 10 = 1 · 5 = 1
P (B3 A ) =
1
30 1
6
5
Þ (D) is incorrect
Ans. (A,B,C,D)
P(A) + P(B) – P(A Ç B) = 0.8
\ P(A Ç B) = 0.5 + 0.4 – 0.8 = 0.1
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
(1–cosx)(sinx + 1)(2sinx – 1) = 0
x = 0, 2p ,
P(A Ç B ) = P(B) – P(A Ç B)
55.
= 0.5 – 0.1 = 0.4 Þ (A) is correct
53.
0.4
P (B Ç A )
0. 4
2
=
=
=
P(B A ) =
1 - P(A )
P( A )
0. 6
3
Þ (B) is correct
P(A) · P(B) = 0.2
Þ P(A Ç B) < P(A) · P(B)
Þ (C) is correct
P(A or B but not both) = 0.9 – 2 × 0.1 = 0.7
Ans. (A,B,C)
cos2x – sin2x = (cosx – sinx) (1+sinx cosx)
cos x – sinx =0 or cos x + sinx= 1+ sinx cosx
cosx = sinx
or (cosx+ sinx) = 1+
tanx = 1
x = np +
(2n) !
Also
56.
1
57.
2
cos æçè x - p ö÷ø =cos
4
x–
1
2
=
(A)
C 4 -1
=
10 + 4 -1
C 4 -1
(D)
P(A Ç B)
=
P(B)
9
C3
42
=
C 3 143
13
9
C3
C3
13
Ans. (A,B,C,D)
9
9
9
81
; P(B) = P(A Ç B) =
´
=
10
10 10 100
9
1
´
æ A ö P(A Ç B) 10 10
=
Pç ÷ =
81
P(B)
èBø
1100
1
2
p
4
P(A È B) = P(A) + P(B) - P(A Ç B) =
58.
p
p
= 2np ±
4
4
x=2np +
2n [1 . 3 . ...... (2 n - 1) ]
n!
Hence (A), (B), (C), (D).
Ans. (A,D)
P(A) =
+ sinx.
2n n ! [1 . 3 . ...... (2 n - 1) ]
(n !)2
6+ 4
t2 –2 t+1 = 0
t =1
sinx + cosx =1
cosx .
cn =
Similarly (B), (D) follows .
2
t2 - 1
2
2n
=
( cos x + sin x )2 - 1
t =1 +
Ans. (A,B,C,D)
Total number of permutations = (n !)2 = 2ncn
Let cosx + sinx = t
p
,n Î I
4
3p p 5p
, ,
2
6
6
p
2
x= 2np
8
9
10
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
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52.
Ans. (C,D)
2sinx – 2cos2x – sin2x – 2sinx sin2x + 2cosx = 0
HS
JEE-Mathematics
59.
Þ 10C1 × 5C1 + 10C1 × 5C1 = 100 ways
(D) 102 = 100 ways.
Ans. (A,B,C,D)
A
r(p – C) = 6 ..... (i)
C1
r(p – B) = 8 ..... (ii)
B1
C
(p–
r(p – A) = 10 ..... (iii)
B
(i) + (ii) + (iii)
4
)
r=
12
p
Angles are in A.P.
R=
60.
A
B
C
sin sin
2
2
2
2C
arc length AB = R(2C)
B
arc length AB = R(2A)
arc length CA = R(2B)
Ans. (B,D)
Total numbers which are not divisible
by 5 are = 6 ! ´ 6 = 4320
NODE6\E_NODE6 (E)\DATA\2013\IIT-JEE\TARGET\MATHS\HOME ASSIGNMENT (Q.BANK)\SOLUTION\HOME ASSIGNMENT # 05
1 or 2 or 3
HS
A
6 2
p sin(15°)
C
1
61.
2
2
22 sin
t+
x - 3sin x +1
+ 23 -(2 sin
2
x - 3 sin x +1)
=9
=t
8
=9Þ
t
sin x = -
62.
× × × ×
t2 – 9t + 8 = 0
1
2
,
sin x =
1
2
, sin x = 1
Ans. (A,B,C,D)
Consider tan20° – 2tan10°
=
sin 20° cos10° - 2 sin10° cos20°
cos10° cos20°
=
(sin 30° + sin10°) - 2(sin 30° - sin10°)
2cos10° cos 20°
=
3sin10° - sin 30°
4 sin3 10°
=
>0
2cos10° cos20° 2cos10° cos20°
Þ tan20° > 2tan10°
\ y>x & z>w
...... (i)
Now 2w – 2y
= 2tan10° + tan70° – tan20° – tan50°
= 2tan10° + tan20° tan50° tan70°
= 2tan10° + tan50°
= 2x which is positive
\ w=x+y &w>y
..... (ii)
number of numbers = 4 ! ´ 4 = 96
\ 1897th number in the list is 4213567.
with 42 ..........
2
x - 3 sin x +1)
Þ (t – 8)(t – 1) = 0
Þ 2sin2x – 3sinx + 1 = 3
or 2sin2x – 3sinx + 1 = 0
4 ways
4
2
number of numbers = 2 ! ´ 2 = 4
[Total = 1992 + 4 = 1996 ]
Now, when first 4 places are , 4 3 1 5 ´ ´ ´
then the remaining 3 places in each case be
filled in 3 ! = 6 ways
which makes total numbers = 2002 and the
(2002)th number is 4315762
Hence (2001)st number is just before it
= 4315726
Ans. (A,D)
2(2 sin
5 can't
come here
Now when 1 or 2 or 3 occupies the 7th place,
then the number of numbers
= 3 ´ 5 ! ´ 5 = 1800
¯
(last can be filled only in 5 ways)
\ 1800th number in the list is 3765421.
when 1st two places are 41 .......... then
4
1
2 ways
p
p
p
; B= ; A=
2
3
6
r = 4R sin
3
C
A1
Þ r(3p – (A + B + C)) = 24 Þ
C=
number of numbers = 4 ! ´ 4 = 96
Total so far = 1800 + 192 = 1992 .
\ 1994th number in the list is 4312576.
1st three places are filled as 4 3 1 2 ..........
× × × ×
4 ways
9
JEE-Mathematics
z>w>y>x
1
2
Also cot
66.
1
2
= tan20° + tan50° - tan20° - tan70°
63.
=
1æ
sin70°
sin20°
ö
ç
2 è cos20° cos 50° cos 50° cos70° ÷ø
&
=
sin140° - sin 40°
=0
4 cos20° cos 50° cos70°
Q
64.
=
c2 + a 2 - b2
2abc
c
= 2 2
cos c
a + c - a2
b2 + c2 – a2 ¹ c2 + a2 – b2
¹ a2 + b2 – c2 for every DABC.
(A) is not correct
\
(D)
29 = 512
39 > 1000
24 61 = 976
34.11 = 891 < 976
54. 2 > 1000
Þ maximum value of M = 24.61 which is
divisible by 2 & 61
Ans. (B,C)
(B)
2 sin A cos A
cos A
= 2R sin A a =
aR
a
\ clearly (D) is not correct.
sin 2A
2
a cos A + b cos B + cos C
a +b+c
=
R (sin 2A + sin 2B + sin 2C)
2s
=
R ´ 4 sin A sin B sin C
2s
(t – [|sinx|])! = 3! 5! 7!
If x =
p
np +
2
4R ´
, (n Î I)
=
then (t – 1)! = 3! 5! 7!
Þ (t – 1)! = 10!
(C)
Þ t – 1 = 10 Þ t = 11
If
65.
p
2
cos A
cos B
cos C
+
+
a
b
c
a 2 + b2 + c 2
=
2abc
A
67.
Ans. (A,B,D)
N = Normal coin
q
q
D
L
E
C
N
E
D = Double headed coin
E = Head on the 2nd toss
9
tan B
AL 2
tan q =
=
= 3 tan B = 3 tan C
9
DL
6
D
E1 = Head on the 1st toss
E2 = Tail on the 1st toss
(A) E = (E Ç N) + (E Ç D)
= P (N) . P (E/N) + P (D) . P (E/D)
Also tan A = tan(180 - 2B) = - tan2B
=
a
b
c
´
´
2R 2R 2R = abc = r
R
4R 2 / s
2s
(b2 + c 2 - a 2 ) + (c 2 + a 2 - b2 ) + (a 2 + b2 - c2 )
=
2abc
x ¹ np + , (n Î I)
then (t – 0)! = 10! Þ t = 10
Ans. (A,B,C,D)
DADE is isosceles
as BD = DE = EC
Þ DABC is isosceles
Þ ÐB = ÐC
B
A
= tan2 q
2
2abc
1
(tan20° + 2 tan 50° - tan70°)
2
p19
9 cot2
2abc
a
b
= 2 2
2 ,
cos A b + c - a
cos B
=
Ans. (A,B)
M must have exactly 10 divisors
So M = p19 or p14 p2
Þ
Ans. (B,C)
(A)
1
1
= tan20° + tan 50° - tan 70°
2
2
A
= tan B
2
2 tan B
6 tan q
=
2
tan B - 1 tan2 q - 9
10
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Combining (i) & (ii)
Also 2y – z
HS
JEE-Mathematics
=
1 é1 1
1 ù
.
+
.1
2 êë 2 2
2 úû
T H
=
69.
1é
1ù
1. ú
2 êë
2û
+
H
H
Þ
3
2 5
+ =
8
8 8
(B) P (E1/E) =
(C) P (E2/E) =
(
P ( E1 Ç E )
P (E)
P (E2 Ç E )
P (E)
=
(
1 1 1
4
5
=
)
´
1
= 2 25 2 =
8
5
NODE6\E_NODE6 (E)\DATA\2013\IIT-JEE\TARGET\MATHS\HOME ASSIGNMENT (Q.BANK)\SOLUTION\HOME ASSIGNMENT # 05
HS
Þ
2
æ
3 ö æb
ö
a
b
+
c
çç
÷
÷ =0
2 ÷ø çè 2
ø
è
\
a
3
=
b
2
)
cos B =
Ans. (A,B,D)
(A) Last two digits should be 08, 16, 36, 60, 68, 80
\ Number of number ending with 08, 60
and 80 = (3 × 2 × 1) × 3 = 18
Number ending with 16, 36, 68
= (2 × 2 × 1) × 3 = 12
\ Total number divisible by 4 =18 +12 = 30
(B) Numbers divisible by 11 is possible only 70.
when
Case-I : 0, 1, 8, occurs alternately and
3, 6, occurs alternately
i.e. odd places are occupied by 0, 1, 8 and
even places are occupied by 3, 6.
But the number has to be greater than
30,000
\ such numbers are 2! × 2! = 4
Case-II : odd places are occupied by 0, 3, 6
and even places are occupied by 1, 8.
\ such numbers are 2 × 2! × 2! = 8
Total required numbers = 8 + 4 = 12
\
3
2
11
&
2
a +c -b
2ac
2
c=
=
b
2
3 2 b2
- b2
b +
4
4
=0
2ac
ÐB = 90°
Similarly ÐA = 60° & ÐC = 30°
\
DABC is right angled but not isosceles
é
1 ù é 4 ù
ú=ê
ú=2
ëétan A + tan C ûù = ê 3 +
3û ë 3û
ë
ì r2 ü ì r tan B / 2 ü ì tan 45° ü
í ý=í
ý=í
ý=
î r1 þ î r tan A / 2 þ î tan 30° þ
{ 3} =
3 -1
Ans. (B,D)
æ
2 aö
çè 1 - tan 2 ÷ø
æ
2 aö
çè1 + tan 2 ÷ø
a
2 + (2a – 1)
(a + 2)
2 a
1 + tan
2
2 tan
= 2a + 1
Let tan
a
=t
2
(a + 2) 2t + (2a – 1) (1 – t2) = (2a + 1) (1 + t2)
Þ 4a . t2 – 2t (a + 2) + 2 = 0
æ1
1ö
1
Þ t2 – t çè 2 + a ÷ø +
=0
2a
Þ
æ 1 ö æ 1ö
çè t - 2 ÷ø çè t - a ÷ø
\
tan
a
1
=
2
2
or
tan
If
tan
a
2
or
If
8
2 × 3 × 2 × 1 × 1 = 12
(D) Sum of digits = 18 (divisible by 3)
\ All even numbers formed using the
digits 0, 1, 3, 6, 8 will be divisible by 6.
\ Required numbers = 3! × 3 = 18
2
c 1
=
b 2
&
Þ a=b.
T T
(C)
3
b2
a2 - ab 3 + b2 +
- ca + c2 = 0
4
4
144
42444
3 1442443
2
1 +1
4 4
5
8
1 ´1 = 1
(D) 1
2 2 2
8
68.
Ans. (A,C)
a2 + b2 + c2 = ca + ab 3
=
1
2
=0
a
1
=
2
a
tan
a
2
=
1
a
JEE-Mathematics
Þ x Î [3, 30]
73.
1
2a
a
= 2
tan a =
.
1
a -1
1- 2
a
2´
Þ
74.
Ans. (A,B,C)
P ( C4 / H ) =
=
(C)
2
8
3
8
..... P(H / C7 ) = 1
75.
1 5
´
1
7 8
P ( C4 / H ) =
=
1 é2 3
8ù 7
+ + ....... ú
7 êë 8 8
8û
12
C1 ´ 12 - 4 -1 C4 -1
4
76.
Paragraph for Question 75 to 77
Ans. (A)
Ans. (D)
é 3250 ù é 3250 ù é 3250 ù é 3250 ù é 3250 ù
ê 5 ú + ê 25 ú + ê 125 ú + ê 625 ú + ê 3125 ú
ë
û ë
û ë
û ë
û ë
û
1 2
´
7 8 = 2
5
35
8
77.
= 650 + 130 + 26 + 5 + 1 = 812
Ans. (B)
200
C100 =
;
é 200 ù
2
3G
RGG Þ
GRR
Þ
é 200 ù
é 200 ù
é 200 ù
é 200 ù é 200 ù é 200 ù é 200 ù
+ê 5 ú+ê 6 ú+ê 7 ú+ê 8 ú
ë 2 û ë 2 û ë 2 û ë 2 û
BBBRRG
BBBRGG
2 . 3 C1 . 3 C2 é 3 C1 . 2 C1 . 1C1 . 2 ù
.ê
ú
6
6
C3
C3
ë
û
= 18 . 12
20 20
(100!)
= ê 2 ú + ê 22 ú + ê 23 ú + ê 2 4 ú
ë
û ë
û ë
û ë
û
In the top two cases the probability of the
required event is zero. Hence,
P(E) =
200!
Exponent of 2 in 200!
3R
withdraw
=105
é 300 ù é 300 ù é 300 ù
ê 7 ú + ê 72 ú + ê 73 ú = 42 + 6 + 0 = 48
ë
û ë
û ë
û
1 35 5
´
=
7 8
8
Ans. (A,C)
RRR
Bag
GGG
é 2009 ù
(D) (2 + 1)(1 + 1)(1 + 1)(2 + 1) – 1 = 35
so P(H / C1 ) = , P(H / C2 ) =
72.
é 2009 ù
(B) 9C4 – 4 = 122
i +1
P ( H / Ci ) =
7
P(C1/H) =
Ans. (B,C)
a2 – b2 = c
Þ (a – b)(a + b) = c
Þ a – b = 1 Þ a = 3, b = 2, c = 5 & d = 7
and sum of all possible numbers
= 3!(1111)(2 + 3 + 5 + 7) = 113322
Ans. (A,C,D)
é 2009 ù
P(HC 4 )
P(HC4 )
=
P(H)
P(HC1 È HC2 È ... È HC 7 )
Also P(H) =
27
100
(A) ê 11 ú + ê 112 ú + ê 113 ú
ë
û ë
û ë
û
= 182 + 16 + 1 = 199
P(C 4 ) P(H / C 4 )
å P(Ci )P(H / Ci )
also
Þ Probability =
= 100 + 50 + 25 + 12 + 6 + 3 + 1 + 0 = 197
Exponent of 2 in 100!
é 100 ù
é100 ù
é100 ù
= ê 2 ú + ê 22 ú ........... + ê 27 ú
ë
û ë
û
ë
û
= 50 + 25 + 12 + 6 + 3 + 1 + 0 = 97
54 27
=
=
100 50
12
2197 a
297.297 l
so
200
\
exponent of 2 in 200C100 is 3
C100 =
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\
71.
Also x2 – 33x + 90 £ 0 Þ (x – 3)(x – 30) £ 0
1
2´
2= 4
tan a =
1
3
14
HS
JEE-Mathematics
78.
Paragraph for Question 78 to 80
Ans. (A)
n(S1) is number of positive integral solutions
of the equation x . y. z = 22 . 3. 5
P(R) =
=
1
1
y = 2p .3q .5r
2
2
79.
82.
2
z = 2p .3q .5r
\ p1 + p2 + p3 = 2
......... (i)
q1 + q2 + q3 = 1
......... (ii)
r1 + r2 + r3 = 1
......... (iii)
Non-negative integral solutions of (i) = 4C2
Non-negative integral solutions of (ii) = 3C1
Non-negative integral solutions of (iii) = 3C1
\ n(S1) = 4C2 . 3C1. 3C1 = 54
Ans. (C)
3
3
3
n( S2 Ç S 3) = Number of positive integral
solutions of the inequality 15 < x + y + z < 20
Consider x + y + z < 20
Þ x + y + z £ 19
{Q x, y, z Î I+}
Þ x + y + z + w = 19
\ Number of positive integral solutions = 19C3
Consider x + y + z £ 15
Þ x + y + z + w = 15
\ Number of positive integral solutions = 15C3
NODE6\E_NODE6 (E)\DATA\2013\IIT-JEE\TARGET\MATHS\HOME ASSIGNMENT (Q.BANK)\SOLUTION\HOME ASSIGNMENT # 05
\
HS
80.
=
83.
=
0
1
2
P(1) =
,
P(2) =
4
14
,
1 é 8 + 9 + 8 ù 25 5
=
=
3 êë 15 úû 45 9
84.
Ans. (D)
EEE, AA, LL, RR, C, N
Cases :
1.
3 alike + 2 alike + 1 different
1
2.
C1 ´ 5 C3 ´
C3 ´
6!
= 360
2! 2! 2!
2 alike + 2 alike + 2 different
4
C2 ´ 4 C2 ´
6!
= 6480
2! 2!
2 alike + 4 different
4
6.
6!
= 1200
3!
2 alike + 2 alike + 2 alike
4
5.
6!
= 720
3! 2!
3 alike + 3 different
1
4.
C1 ´ 3 C1 ´ 4 C1 ´
3
C1 ´ 5 C 4 ´
6!
= 7200
2!
All different
6
C6 ´ 6! = 720
Total words = 16680
85. Ans. (A)
P(3) =
17
Paragraph for Question 84 to 86
20
(2 + 2 )(3 + 3 + 3 + 3 )
=
9
54
1
14
2
Ans. (D)
3.
Ans. (C)
Þ a1 = 2, a2 = 3, a3 = 4
If P(i) µ i2
9
3.6
Paragraph for Questions 81 to 83
81.
3
10
10
10
P(all are red) = 2 + 3 10+ 4
1 1
1 a1 + a2 + ....... + a 8
+
+ ....... +
=
a1 a 2
a8
54
1
4
1 é 2 C1 . 4 C1 +3 C1 . 3 C1 +2 C1 . 4 C1 ù
ú
6
C2
ë
û
n( S2 Ç S3) = 19C3 – 15C3 = 514
0
4
P(G) = 14 . 6 + 14 . 6 + 14 . 6 = 42
Ans. (B)
P(different color)
=3ê
Ans. (A)
54 = 2 . 33
\ Total divisors = 8
Let a1, a2, ........ a8 be the total divisors
\
2 + 12 + 36 50 25
=
=
84
84 42
1
Let x = 2p .3q .5r
1
1 2 4 3 9 4
. +
. +
.
14 6 14 6 14 6
5!
6!
´
= 10 ´ 180 = 1800
3! 2! 2! 2!
9
14
13
JEE-Mathematics
91.
6!
Arrangement of consonants : 2! 2!
Selection of 5 gaps for vowels with
arrangement : 7 C5 ´ 5!
hence probability =
3! 2!
92.
15.7!
6!
7!
5!
´
´
Total words =
=
2
2! 2! 5! 2! 3! 2!
Paragraph for Question 87 to 89
87. Ans. (A)
Let a = 2x + 1, b = 2y + 1, c = 2z + 2, where x,
y, z Î whole number
\ a + b + c = 16
Þ 2x + 1 + 2y + 1 + 2z + 2 = 16
Þ x+y+z=6
Þ Number of integral solutions
= 6 + 3 – 1C3–1 = 8C2 = 28
88. Ans. (C)
Possible
values of a
Possible
values of b
1
2
3
2 to 9 3 to 9 4 to 9
=8
=7
=6
....
....
8
only 9
=
=1
Þ
1
(k + 2) have real roots if
4
1
k 2 - 4 .(k + 2) ³ 0
4
k Î [2, 5] hence probability =
(iii) y + 1 – (x + y) > x Þ x <
1
2
14
GH JK
0,
1
95.
E
D
2
GH 12 , 12 JK
C
0 (0,0)
GH JK
1
2
A
(1,0)
x
,0
Paragraph for Questions 93 to 95
Ans. (A)
P(w) = P(E1 I w) + P(E 2 I w)
= P(E1 ).P(w / E1 ) + P(E2 ).P(w / E 2 )
=
1 9 2 8 25 5
.
+ .
=
=
3 10 3 10 30 6
Ans. (C)
P(s ) = P(E1 Ç s) + P(E 2 Ç s )
= P(E1 ).P( s / E1 ) + P(E2 ).P( s / E 2 )
1 8 2 9 26 13
.
+ .
=
=
3 10 3 10 30 15
Ans. (D)
P(E1 ).P(w / E1 )
æE ö
Pç 1 ÷ =
è w ø P(E1 ).P(w / E1 ) + P(E 2 ).P(w / E 2 )
=
3
5
1
2
y
(0,1) B
Area of DCDE
Area of DOAB
=
Þ (k + 1)(k – 2) ³ 0
Þ
1
2
1 1 1
´ ´
1
=2 2 2 =
1
4
´1´1
2
Paragraph for Question 90 to 92
Ans. (C)
D³0
(ii) x + 1 – (x + y) > y Þ y <
Þ Probability
8´9
´ 10 = 360
2
Equation x2 + kx +
Ans. (B)
The given conditions
Þ x > 0, y > 0, 1 – (x + y) > 0
(\ z = 1 – (x + y))
Þ x > 0, y > 0 & x + y < 1
...... (i)
Now condition for formation of a triangle
Þ sum of any two sides is greater than third side
plot on x-y plane
89. Ans. (B)
The number abc can not be formed, if in the 93.
selection of n cards, we get either two of a,b,c
(i.e. ab or ac or ca) or only one of a, b, c (i.e. a
or b or c only)
Þ total ways = 2n + 2n + 2n – 1 – 1 – 1 = 45
(Q only a or b or c are repeated twice)
94.
Þ 3 . 2n = 48
2n = 16
n=4
90.
Area of smaller circle 1
=
Area of bigger circle 4
Þ (i) x + y > 1 – (x + y) Þ x + y <
C can take any value from 0 to 9
Hence total number = (8 + 7 + 6 +.....+ 2 + 1) ×10
=
Ans. (A)
Radii of smaller and bigger circles are 1 and 2
respectively
.
1 9
3 10
1 9
2 8
3 10
3 10
.
+ .
=
9
25
NODE6\E_NODE6 (E)\DATA\2013\IIT-JEE\TARGET\MATHS\HOME ASSIGNMENT (Q.BANK)\SOLUTION\HOME ASSIGNMENT # 05
86. Ans. (C)
HS
JEE-Mathematics
96.
Ans. (A)®(S), (B)®(P), (C)®(Q), (D)®(R)
(A) Rectangles can be of sizes
1 × 1, 1 × 2, 1 × 3 ....... 1 × 8 = 8 numbers
(D)
98.
2 × 2 × 2 × 3 2 × 8 = 7 numbers
3 × 3 ........ 3 × 8 = 6 numbers
M
P ( A / Ei ) =
= (8 + 7 + 6 + ... 1) = 36
(B) 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 25 – 1 = 31
(C) 157500 = 22 32 54 71
Þ Number of divisors divisible by
2, 3, 5 are (2)(2)(4)(1 + 1) = 32
(D) 7! onwards are divisible by 35 hence
remainder of (1! + 2! + 3! + 4! + 5! + 6!),
when divided by 35 is remainder of
=
1
873
35
NODE6\E_NODE6 (E)\DATA\2013\IIT-JEE\TARGET\MATHS\HOME ASSIGNMENT (Q.BANK)\SOLUTION\HOME ASSIGNMENT # 05
HS
(C)
=
1
11
14
C5
3
=
C 5 13
P(E11 ).P ( A / E11 )
P(A)
1 3
.
6
= 15 13 =
65
1
6
P ( B / E i ) = 0 if i = 0, 1, 2 or 13, 14
P (B / Ei ) =
\
245
(B) Using Bay's theorem :
8
)
(D) Let B denotes the probability of 3 black
and 2 white balls, then
= C2 + 50 C1 ´ C1 = 45 + 400 = 89
25
52
C5 + 6 C5 + ..... + 14 C5
C
P ( A / E11 ) =
40
=
5
(C) By Baye's Theorem,
Now to get product divisible by 5, either
select 2 numbers from these 10 numbers
or select 1 from these 10 and 1 from
remaining 40
Þ Probability
5 3
+
9 5
15
(
(B) Clearly P ( A / E11 ) =
Numbers divisible by 5 are 5, 15, 25, .... 95
Þ Total 10 numbers.
Probability =
1
1
.
15 14 C5
6
= 15 . 14 C = 6
5
1, 3, 5,........... 99 Þ Total 50 numbers
5
9
for i ³ 5
i =0
(A) Odd natural numbers < 100 :
50 ´ 49
2
C5
C5
14
14
Ans. (A)®(Q), (B)®(P), (C)®(S), (D)®(R)
C2
i
(A) P(A) = å P(E i ).P ( A / E i )
i.e. = 33.
10
1
(i = 0, 1,.....,14)
15
P ( A / E i ) = 0 for i = 0, 1, 2, 3, 4.
Þ Total rectangles
10
.
Ans. (A)®(Q), (B)®(P), (C)®(R), (D)®(Q)
Let Ei denotes the event that the bag contains i
black and (14 – i) white balls (i = 0, 1, ....., 14)
and A denotes the event that five balls drawn
are all black, then
P ( Ei ) =
8 × 8 = 1 number
97.
50 49 48 2
24
´
´
´
=
52 51 50 49 663
99.
i
P(B) =
C3 .14 -i C2
14
C5
14
å P(E ).P ( B / E )
i =0
i
1 1 3 11
é C3 . C2 + 4C3 .10 C2 + .. + 12C3 .2 C2 ùû
.
15 14 C5 ë
=
5005
1
=
14
15. C5 6
Ans. (A)®(S,T); (B)®(Q); (C)®(R); (D)®(P)
(A) I N D I A T I M E
I
I
56 + 28 + 8 + 1 93
=
256
256
IIINDATME
15
i
=
I
C5 + 8 C6 + 8 C7 + 8 C8
28
for i = 3, 4, ....,12
JEE-Mathematics
Arrangements of I's in 9 positions
= 3C1.2C1.1C1
Arrangement of remaining distinct letters
= 6!
Total required arrangements
= 3× 2× 1× 6! = 6× 6!
(B) F O R T U N A T E
T
T
Type of selection
1
2 Alike, 2different
2
4 different
1
S
S E
E
1
E
3
{
S R G J
1 1 2 3
7!
7!
´ 2! =
12
(1!) 2!2!3!
2
7!
7!
´ 3! =
3
8
(2!) 3!1!
(C) S R G J
7!
7!
´ 3! =
24
(1!) 3!4!
2
2
1
4
2 1
1 1
3
2
\
E S
= P(50p, 10p, 1Re) =
4 5
3
1
´
´
=
12 11 10 22
(B) Required probability is
=
3!
P(10p, 10p, 1Re)
2!
= 3´
A
5
4 3
3
´
´
=
12 11 10 22
(C) Required probability is
=
+
æ 4! .5! ö
number of ways = 4 = ç 45 ÷ .4
è
ø
7
102. Ans. (A)®(R), (B)®(S), (C)®(P), (D)®(Q)
(A) Required probability is
A
AADDCNITE
Arrangements
of
A's = 3C2 × 3 × 2
Arrangements of
If both the D's are kept
in the rows in which
A appears
2
(D) Each toy has 4 options.
SSSEEEWTN
Arrangements of S's = 3 × 2 × 1
Arrangements of remaining (all distinct)
letters =3!
Total required arrangements
= 6× 2× 3!=3× 4!
(D) C A N D I D A T E
D's =
4!
(B) S R G J
E S
S
4!/2!
(3
+24
3 +3
2)
14
If exactly one D appears in
the row in which A doesn't appear
Arrangement of remaining letters (all
distinct) =5!
Total required arrangement = 3× 6 × 11× 5!
= 33 × 6!
100. Ans. (A)®(Q), (B)®(S), (C)®(P), (D)®(R)
(A) Use nCr + nCr–1 = n+1Cr
(B) Number of ways = 4C2 . 4C2 = 36 ways
(C) Available digits =1, 2, 2, 3, 3, 4
16
P(50p, 50p, 50p)
P(10p, 10p, 10p) + P(50p, 50p, 50p) + P(1Re, 1Re, 1Re)
4 3 2
. .
12 11 10
= 5 4 3 4 3 2 3 2 1
. .
+
. .
+
. .
12 11 10 12 11 10 12 11 10
=
24
4
=
60 + 24 + 6 15
(D) Required probability is
3!
P(10p,10p,1Re)
2!
=
1 - 3!P(10p, 50p, 1Re)
5 4 3
3´ . .
12 11 10 = 3
5 4 3 16
1-6´ . .
12 11 10
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S E
C1 C2
(D) (500!) = 2k .7k .I
k2 < k1 & k2 = 82
\ (500!) = (14)82 . I
\ k = 82
101. Ans. (A)®(S), (B)®(P), (C)®(Q), (D)®(R)
(A)
Arrangements of remaining letters
(all distinct) =7!
Total required arrangements = 18 × 7!
(C) S W E E T N E S S
4!/2!2!
3
Number of ways = 6 + 6 × 12 + 24 = 102
TTFORUNAE
Arrangements of T's = 3C2 × 3 × 2
Number of selection Arrangement
2 Alike, 2 others alike
HS
JEE-Mathematics
(C) nC2 = nC3 Þ n = 5
103. Ans. (A)®(Q); (B)®(S); (C)®(P)
(A) D = a2 – (b – c)2
D = a2 – b2 – c2 + 2bc
b2 + c2 – a2 = 2bc – D
(D) S = 5! (100 + 101 + 102 + 103 + 104 + 105 )(1 + 2 + 3)
2!
= 60 (1 1 1 1 1 1)6 = 39999960
sum of digits = 54
105. Ans. (A)®(Q); (B)®(R); (C)®(S)
Interpret from the venn diagram
1
2
2bc cosA = 2bc – bc sinA
4cosA + sinA = 4
Aö
A
A
æ
4 ç 1 - 2 sin2 ÷ + 2 sin cos = 4
2
2
2
è
ø
8sin2
A
A
A
- 2sin cos = 0
2
2
2
Þ 4tan
Þ
H
A
–1
2
25% 15% 10%
=0
A 1
tan =
2 4
50%
8
15
Þ tanA =
(B) tanA = 3k, tanB = 4k, tanC = 5k
In any triangle ABC
tanA + tanB + tanC = tanA tanB tanC
Þ 12k = 60k3
3
tan A =
\
5
NODE6\E_NODE6 (E)\DATA\2013\IIT-JEE\TARGET\MATHS\HOME ASSIGNMENT (Q.BANK)\SOLUTION\HOME ASSIGNMENT # 05
tan B =
5
4
P(A ) + P( B ) = 2 – 0.8 = 1.2 Þ (R)
tan C = 5
5
2
3
; P (Y) = ; P (Z) = p
3
4
E : exactly two bullets hit
(C) P (X) =
2cos2 q - 2sin q cos q
1
=
2
2
2(cos q - sin q)
1 + tan q
a+b=
HS
1
Þ k=
106. (A)®(S) ; (B)®(R); (C)®(P)
(A) P(R) = P(T ; HTT ; HTHTT ; HHTTT)
Þ 11/16
(B) P(AÈB) = 0.6 ; P(AÇB) = 0.2
P(AÈB) = P(A) + P(B) – P(AÇB)
\ P(A) + P(B) = 0.6 + 0.2 = 0.8
2 5
sinA sinB sinC =
7
(C) ƒ (q) =
E
P(E) = P(X Y Z ) + P(Y Z X ) + P(Z X Y )
11
2 3
3 1
2 1
= · (1 - p) + · p + p ·
24
3 4
4 3
3 4
5p
4
tan(a + b) = 1
Þ tana + tanb = 1 – tana tanb
Þ p=
1
2
107. Ans. 168
12 = 2 . 2. 3
1
The number should be of the form a1 b1 c2 ,
= 1 + tan a + tan b + tan a tan b
where a, b, c are prime numbers.
1
1
For number to be least c = 2, b = 3, a = 5
= 1 + 1 - tan a tan b + tan a tan b =
2
\ Least number = 22 . 31 . 51
104. Ans. (A)®(R); (B)®(Q); (C)®(P); (D)®(S)
Sum of all divisors
(A) There are 5 even powers & 6 odd powers
= (20 + 21+ 22)(30 + 31)(50 + 51) = 168
5
6
\ Number of GP's = C2 + C2
108. Ans. 510
(B) Let the common difference be 'd'.
Required number of ways
\ 365 = 5 + (n – 1)d
3
2
3
æ
ö
x
x
6
Þ 360 = (n – 1)d
= coefficient of x in 6!ç 1 + x + + ÷
2! 3! ø
è
\ Total number of AP's = Total number
= coefficient of x6 in
of divisors of 360 = 4 × 3 × 2 = 24
1
1
ƒ (a) . ƒ (b) = 1 + tan a . 1 + tan b
17
JEE-Mathematics
æ
æ x2 x3 ö
6! ç (1 + x)3 + 3(1 + x)2 ç
+
÷
ç
è 2! 3! ø
è
B2 : pack B was selected Þ P(B2) =
2
æ x2 x3 ö æ x2 x 3 ö
+3(1 + x) ç
+
+
÷ +ç
÷
è 2! 3! ø è 2! 3! ø
ö
÷
÷
ø
1 1 1 1
B ¾¯¾¯¾
¯¾¯® 48 cards
A KQJ
= 510
A : two cards drawn all of same rank.
Now A = (A Ç B1) + (A Ç B2)
\ P(A) = P(A Ç B1) + P(A Ç B2)
= P(B1)P(A/B1) + P(B2)P(A/B2)
109. Ans. 1
A
12
P(A/B1) =
D
E
I2
B
48
9
I1
p/2–C
C/2
P(A/B2) =
C
C1 · 4C2
C2
C1 · 4C 2 + 4C1 · 3C2
48
DI1EI2 is right angle triangle at E
\ E is orthocentre
\ in DBEC,
BE
BC
=
C
Cö
æ
sin
sin ç 90 + ÷
2
2ø
è
Þ BE = a tan
C
=1
2
110. Ans. 49
Let H1 and H2 are
orthocentres of
DAEF and DABC
respectively
AH2 = 2RcosA
AH1 = 2R'cosA
\
P(B1/A) =
12
H1
4
E
=
2
C
C1 · 4 C2 + 9C1 · 4C2 + 4C1· 3C2
(12)(6)
12
=
(12)(6) + (9)(6) + (4)(3)
23
Þ m + n = 35 Ans.
=
r2
=5
......(1)
A
tan
2
Area of quadrilateral DIFD is
7 8
7
.
=
8 15
15
111. Ans. 35
=
Let B1 : pack A was selected Þ P(B1) =
C1 · 4 C2
112. Ans. 3
Area of quadrilateral ADIE is
distance d = 2 cos A(R - R ')
7æ
2
1
ö
= 2. ç
8
æ
ö
æ
ö÷
ç 2. ç 15 ÷ 2 ç 15 ÷ ÷
ç
÷
è 8 øø
è è 8 ø
.
12
3
H2
B
P(A Ç B1 )
P (A )
P(B1 )P(A / B1 )
= P(B )P(A / B ) + P(B )P(A / B )
1
1
2
2
A
F
C2
r2
1
;
2
4 aces
Pack A ¾¾¯¾® 48 cards in 12 different
denominations
18
B
tan
2
(2)
(1)
= 10
B
2 =2
A
tan
2
tan
......(2)
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3
3
1 ö
+ +
÷
è 3!3! 3! 2!2!2! ø
æ
= 6! ç
3
1
; Pack
2
HS
JEE-Mathematics
B
A
cos
2
2 =2
A
B 1
sin cos
2
2
Use C & D
sin
æA Bö
sin ç + ÷
è2 2ø=3
æB Aö 1
sin ç - ÷
è2 2ø
C
2
=3
æ B-A ö
sin ç
÷
è 2 ø
cos
115. Ans. 5
(sinx – 1)3 + (cosx – 1)3 + (sinx)3
= (2sinx + cosx – 2)3
Þ (sinx+cosx–2)(cosx+cosx–1)(2sinx–1) = 0
Þ sinx + cosx = 1 & sin x =
p p 5p
Þ x = 0, , , , 2p
6 2 2
116. Ans. 1
sinx – 1 < 0 for all x
Þ sin6x > 0
Þ 2np < 6x < 2np + p
p p
p
æ pö
\ In ç 0, ÷ , 0 < x < , < x <
....(1)
è 2ø
6 3
2
tan 8°
3tan 24°
+
2
1 - 3tan 8° 1 - 3tan 2 24°
NODE6\E_NODE6 (E)\DATA\2013\IIT-JEE\TARGET\MATHS\HOME ASSIGNMENT (Q.BANK)\SOLUTION\HOME ASSIGNMENT # 05
HS
3p
æ pö p
In ç 0, ÷ , £ x £
è 2ø 8
8
tan 8°
1 - 3 tan 2 8°
1 é 8 tan 8°
ù
+ tan 8° - tan 8°ú
= ê
2
8 ë 1 - 3 tan 8°
û
1
[3 tan 24° - tan 8°]
8
1
3
= - tan 8° + tan 24°
8
8
Now given expression reduces to 1
3
3
9
- tan 8° + tan 24° - tan 24° + tan 72°
8
8
8
8
9
27
- tan 72 + tan 216
8
8
27
27 ´ 3
- tan 216° +
tan ( 216 ´ 3) °
8
8
81
1
= tan108° - tan 8°
8
8
81 - 1
= 10
Þ x+y =
8
114. Ans. 3
from given figure r1 + r3 = 2r2 ...(1)
and 2pr22 = pr12 + pr32
...(2)
from (1) & (2) r1 = r2 = r3
=
Also 1 + tan2x -2 2 tan x £ 0
Þ
2 - 1 £ tan x £ 2 + 1
9 tan 72°
27 tan 216°
+
2
1 - 3 tan 72° 1 - 3tan 2 216°
Consider,
np
np p
<x<
+
3
3 6
or
113. Ans. 10
+
1
2
...(2)
é p p ö æ p 3p ù
From (1) & (2), x Î ê , ÷ È ç , ú
ë8 6 ø è 3 8 û
p p p 3p
=p
Þa+b+c+d= + + +
8 6 3 8
117. Ans. 9
2a = 4sin2x + cosec2x + sin2y + 4cosec2y
= (2sinx – cosecx)2 + 4 + (siny – cosecy)2
+ 3cosec2y + 2
= 6 + (2sinx – cosecx)2 + (siny – cosecy)2 +3cosec2y
Þ minimum value of 2a = 6 + 3 = 9,
when 2sinx = cosecx & siny = cosecy
118. Ans. 3
P = ( 3 + tan1°)( 3 + tan 2° ) .... ( 3 + tan 29° )
now
=
3 + tan1° = tan 60° + tan1°
sin 60°.cos1°.sin1°.cos 60°
sin 61 °
=
cos 60°.cos1°
cos 60°.cos1°
hence
P=
sin 61°
sin 62°
sin 63°
.
.
.......
cos 60°.cos1° cos 60°.cos 2° cos 60°.3°
sin 89°
cos 60°.cos 29°
19
JEE-Mathematics
sin 61°.sin 62°.sin 63°......sin 89°
= 2 29
cos1°.cos 2°.cos 3°........cos 29°
r
(as N = Dr)
119. Ans. 2
We have
ìïæ 1 ö 2
÷ +
– íç
ïîè 3 ø
= 229.
(
+ 4 2- 3
= 4(2)(4 + 3) – 5 –
pö
æ 2 p
+ cot 2 ÷
= ç tan
24
24 ø
è
) (
) (
2
)
2 æ 1 ö
2 +1 + ç ÷ +
è 3ø
ü
( 3) ïý
2
þï
\
5p ö
æ 2 5p
+ cot 2 ÷ + 1
+ ç tan
24
24 ø
è
2
2
Applying (tan q + cot q) = 2 + 4 cot 22q
= 2 + 4 cot2
=
(
) (
)
2
2
85
+ 4é 2 + 3 + 2 - 3 ù
êë
ûú
3
p
11p ö æ 2 2p
10p ö
æ
+ tan2
|||ly b = ç tan2 + tan2
÷ - ç tan
÷
24
24 ø è
24
24 ø
è
286 - 253 33
= 111
=
3
3
Hence log(2b – a)(2a – b) = 2
120. Ans. 26
We have (a cotA + b cotB + c cotC)
6p ö
æ
- ç tan 2
÷
24 ø
è
{(
) (
2
2 -1 +
) + (2 + 3 ) }
2
2
)}
2 +1
Ans.
æ cos A ö
æ cos B ö
÷ + 2R sinB ç
÷
= 2R sinA ç
è sin A ø
è sin B ø
æ cos C ö
÷
+ 2R sinC ç
è sin C ø
8p ö æ
5p
7p ö
æ 2 4p
+ tan 2 ÷ + ç tan 2
+ tan 2
– ç tan
÷
24
24 ø è
24
24 ø
è
+
363
= 121
3
æ 143 ö 253
÷and (2b – a) = 2ç
è 3 ø 3
3p
9p ö
æ
+ ç tan 2
+ tan 2
÷
24
24 ø
è
{(
153 - 10 143
=
3
3
=
85
253
+ 4( 2)( 4 + 3) =
=
3
3
p
pö
æ
= ç tan 2 + cot 2 ÷ – 2 - 3
24
24 ø
è
–1
æ 253 ö 143 506 - 143
÷=
(2a – b) = 2ç
3
3
è 3 ø
=
p
70
2 5p
+ 2 + cot
+
+1
12
3
12
2
10
10
= 51 –
3
3
=
2
)
rö
æ
= 2R (cosA + cosB + cosC) = 2R ç1 + ÷
è Rø
= 2(R + r) = 2(10 + 3) = 26.
121. Ans. 5
1
é 1 + cos 2x + 1 + cos 4x + .... + 1 + cos12x ù
û
2 ë
= [|cos x| + |cos 2x| + |cos 3x| + .... + |cos 6x| ]
y=
2
at x =
20
p
5
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) (
)
(
7 p ö æ 2 6p ö
æ 2 5p
+ tan 2
÷ + ç tan
÷
+ ç tan
24
24 ø è
24 ø
è
2 -1 +
5p
5p ö
+ cot 2 ÷ –1
24
24 ø
5p ö
æ1 ö æ
– ç + 3 ÷ + ç 2 + 4 cot 2 ÷ – 1
12 ø
è3 ø è
10
= 2 + 4 2 + 3 2 –14 + 6 –
+2
3
3p
9p ö æ 2 4p
8p ö
æ
+ ç tan 2
+ tan 2
+ tan 2
÷ + ç tan
÷
24
24 ø è
24
24 ø
è
(
ïþ
2
æ
2 p ö
÷ – [2 × (4 + 3)]+ 2(2 + 1)
= ç 2 + 4 cot
12 ø
è
11p ö æ 2 2p
10p ö
æ 2 p
+ tan 2
+ tan 2
a = ç tan
÷ + ç tan
÷
24
24 ø è
24
24 ø
è
ìï
2
2
+í 2- 3 + 2+ 3 +
îï
2ü
( 3 ) ïý + æçè tan
HS
JEE-Mathematics
é
p
= ê cos 5
ë
+ cos
2p
3p
4p
5p
6p ù
+ cos
+ cos
+ cos
+ cos
5
5
5
5
5 úû
=
é
p
2p
p
p ù
2p
= ê cos 5 + cos 5 + cos 5 + cos 5 + cos p + cos 5 ú
û
ë
æ 5 + 1 ö æ 5 -1 ö
5
÷ + 2ç
÷ +1 =
= 3çç
÷ ç
÷
4
è 4 ø è 4 ø
=
5
4
[
p + q2
]
[
]
5 +1
Þ p = 5, q = 1
=
5
5
[ p + q2] =
×6=5
6
6
122. Ans. 2
Using (m,n) theorem in D ABC (1 + 1) cotq
= 1· cot30° – 1· cot45°
3 -1
2
cot q =
A
B
NODE6\E_NODE6 (E)\DATA\2013\IIT-JEE\TARGET\MATHS\HOME ASSIGNMENT (Q.BANK)\SOLUTION\HOME ASSIGNMENT # 05
HS
=
a/2
AD
=
sin 45° sin( q + 45°)
AD
a/2
=
1
(sin q + cos q) 1 / 2
2
11 - 6 3
3 +1
a
=
2
8-2 3
Þ
a
=
2
(
(4 + 2 3 ) (11 - 6 3 )
8-2 3
8-2 3
=1
\a = 2
17C
4
17 ·16·15·14 15·14
–
= (10)(17)(14) – (15)(7)
1· 2· 3· 4
1· 2
20
Number of quadrilateral fromed =
8-2 3
q
– 15C2
= 2380 – 105 = 2275 Ans.
Alternatively :
1
\
8-2 3
124. Ans. 2275
We have number of quadrilateral
C
Using D ADC,
\
)
4 x 1001 x 1000
2
= 2002000
Þ 2002 Ans.
a
2
D
2
=
30° 45°
a
2
)(
3 + 1 11 - 6 3
123. Ans. 2002
f(–1) = 0 Þ a – 2b + c = 0 Þ 2b = a + c
Þ a, b, c in A.P.
Now 1, 3, 5,...........2001 (1001 numbers)
2, 4, 6,..............2002 (1001 numbers)
number of triplets of a, b, c in A.P. are
(1001C2 + 1001C2) = 2 · 1001C2
However a and c can be interchanged
no. of polynomials = 4 · 1001C2
then
Þ
a
=
2
(
8-2 3
2
C1 ´ 15C3
4
= 25 × 91 = 2275
125. Ans. 12
sin x + 2cos a cos x = 2
solution is possible if 1 + 4 cos2a ³ 4
3 -1
Þ cos2a ³
8-2 3
)
Þ np –
3 + 1 11 - 6 3
21
3
4
Þ sin2a £
p
p
£ a £ np +
6
6
1
2
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JEE-Mathematics
22
HS