Triumph- JEE Advanced Maths - Paper 1

1. MATHEMATICS
[25] IIT JEE ADVANCED – 2013 / Paper-I
PART III MATHEMATICS
Section – 1 (Only One option correct Type)
This section contains 10 multiplechoice questions Each question has four choices (A),(B), (C) and(D), out
of which ONLYONE is correct.
41. Letcomplexnumber  and
1

lieoncircles x x y y r   0
2
0
2 2
b g b g and x x y y r   0
2
0
2 2
4b g b g ,
respectively. If z x iy0 0 0  satisfies the equation 2 20
2 2
z r  ,then  
(A)
1
2
(B)
1
2
(C)
1
7
(D)
1
3
Sol. [C] z z r 0
2 2
   z r0
2 2
   
2
0
2
0
2
2  z z rcos . . . (i)
also z z r 0
2 2
2 
1
40
2
2

 z r 
1
2
1
42 0
2
0
2
 
  z z rcos
 1 2 4
2
0
2
0
2 2
     z z rcos . .. (ii) (i) – (ii)   
2
0
2 2 2 2
1 1 1 4    e j e j e jz r
1 1 1 40
2 2 2 2
   z re je j e j  1
2
2
1 1 4
2
2 2 2

F
HG
I
KJ   
r
r e j e j
       
1
2
1 1 4 1 2 8
2 2 2 2
   e j 7 1
2
   
1
7
42. Four persons independenly solve a certain problem correctlywith probabilities
1
2
3
4
1
4
1
8
, , , . Then the
probabilitythat theproblem is solved correctlybyat least oneof them is
(A)
235
256
(B)
21
256
(C)
3
256
(D)
253
256
Sol. [A] P A B C D P A P B P C P D    b g b g b g b g b g1 ' ' ' '
       1
1
2
1
4
3
4
7
8
1
21
256
235
256

2. MATHEMATICS
[26] IIT JEE ADVANCED – 2013 / Paper-I
43. Let f : ,
1
2
1
L
NM O
QP R (the set of all real numbers)be a positive, non-constant and differentiable function
such that f x f x'b g b g 2 and f
1
2
1
F
HG I
KJ  . Then thevalue of f x dxb g1
2
1
z liesin the interval
(A) 2 1 2e e ,b g (B) e e 1 2 1,b g (C)
e
e


F
HG I
KJ1
2
1, (D) 0
1
2
,
e F
HG I
KJ
Sol. [D] f x f x'b g b g 2  f x f x'b g b g 2 0  e f x e f xx x 
 2 2
2 0'b g b g

d
dx
e f xx
2
0b ge j  e f xx2
b g is strictly decreasing
 e f x e fx

F
HG I
KJ

F
HG I
KJ2
2
1
2 1
2
b g  e f x
e
x
2 1
b g  f x e x
b g 2 1
also f xb g 0  0 2 1
  
f x e x
b g  0
1
2
1
2 1
1
2
1
1
2
1
dx f x dx e dxx
 z zz 
b g
 0
2
2 1
1
2
1
1
2
1
 
L
NMM
O
QPP

zf x dx
e x
b g  0
1
21
2
1
 

zf x dx
e
b g
44. Thenumberofpoints in  ,b g,forwhich x x x x2
0  sin cos , is
(A) 6 (B) 4 (C) 2 (D) 0
Sol. [C] Let f x x x x xb g  2
sin cos
f x x x x x x x x' cos sin sin cosb g b g b g     2 2
+ –
0
f’
f
f 0 1b g 
also f is continuous and lim
x
I x

 b g
 f xb g 0 for exactly two values of x.

3. MATHEMATICS
[27] IIT JEE ADVANCED – 2013 / Paper-I
45. The area enclosed bythe curves y x x sin cos and y x x cos sin over the interval 0
2
,
L
NM O
QP is
(A) 4 2 1d i (B) 2 2 2 1d i (C) 2 2 1d i (D) 2 2 2 1d i
Sol. [B] A x x x x dx   z cos sin cos sinb gc h
0
2

   z cos sin cos sinx x x x dxb gc h
0
4

   z2
0
4
cos sin cos sinx x x x dxb g b gc h

    
F
HG I
KJ  z4 4 4 1
1
2
2 2 2 1
0
4
0
4sin cosx dx x


b g e j
46. A curve passes through the point 1
6
,
F
HG I
KJ. Let the slope of the curve at each point (x, y) be
y
x
y
x
x
F
HG I
KJ sec , 0 . Thenthe equation of thecurve is
(A) sin log
y
x
x
F
HG I
KJ  
1
2
(B) cos logec
y
x
x
F
HG I
KJ   2
(C) sec log
2
2
y
x
x
F
HG I
KJ   (D) cos log
2 1
2
y
x
x
F
HG I
KJ  
Sol. [A]
dy
dx
y
x
y
x
  sec
 x
dt
dx
t sec y txb g
 cost dt
dx
x

 sin lnt x k   sin ln
y
x
y k
F
HG I
KJ  
x y 1
6
,

 k 
1
2
 sin ln
y
x
x
F
HG I
KJ  
1
2

4. MATHEMATICS
[28] IIT JEE ADVANCED – 2013 / Paper-I
47. The value of cot cot


F
HG I
KJF
HG
I
KJ 1
11
23
1 2k
k
n
n
is
(A)
23
25
(B)
25
23
(C)
23
24
(D)
24
23
Sol. [B] cot cot
 
 
F
HG
I
KJ
F
HGG
I
KJJ1
1
23
1
1 2
n k
n
k
  
F
HG
I
KJ   
F
HG
I
KJ

 

 cot cot cot tan tan1
1
23
1 1
1
23
1 1 1
n n
n n n nb gc h b ge j
     
cot tan tan cot tan1 1 1
24 1
23
25
25
23
e j
48. For a b c   0, thedistance between (1, 1)and the point of intersection of thelines ax by c   0
and bx ay c   0 is less than 2 2. Then
(A) a b c   0 (B) a b c   0 (C) a b c   0 (D) a b c   0
Sol. [A] Intersection of given lines




F
HG I
KJc
a b
c
a b
,  2 1 2 2

F
HG I
KJ 
c
a b

c
a b
 1  a b c   a b c   0
49. Perpendiculars aredrawnfrompointsontheline
x y z




2
2
1
1 3
totheplane x y z   3. thefeetof
perpendicularslieontheline
(A)
x y z
5
1
8
2
13





(B)
x y z
2
1
3
2
5





(C)
x y z
4
1
3
2
7





(D)
x y z
2
1
7
2
5





Sol. [D] Intersection of line and plane
x y z



  
  
 

2
2
1
1 3
2 1 3
2 1 3
3
2
i.e. 1
5
2
9
2
, ,
F
HG I
KJ
Foot of perpendicular of point (–2, –1, 0)
x y


  
  
 
2
1
1
1
2
1
2 1 3
1 1 1
i.e. (0, 1, 2)
Equation of projection line is given by
x y z



 



0
1 0
1
5
2
1
2
9
2
2

x y z
2
1
7
2
5






5. MATHEMATICS
[29] IIT JEE ADVANCED – 2013 / Paper-I
50. Let PQ i j k

  3 2   and SQ i j k

    3 4 determine diagonals of a parallelogram PQRS and
PT i j k

    2 3 be another vector. Then the volume of the parallepiped determined by the vectors
PT PQ
 
, and PS

is
(A) 5 (B) 20 (C) 10 (D) 30
Sol. [C] PR PQ PS i j k
  
    3 2   , SQ PQ PS i j k
  
      3 4
 PQ
i j k i j k
i j k


    
  
3 2 3 4
2
2 3
     
  e j e j , PS i j k

    2
volume of parallelopiped

L
NM O
QP   
  
PT PQ PS
1 2 3
2 1 3
1 2 1
10
SECTION – 2 : (One or more options correct Type)
The section contains 5 multiple choice questions. Each questions has four choices (A), (B), (C) and (D) out
of which ONE or MORE are correct
51. Let S kn
k k
k
n
 


 1
1
2
1
4
2
b g
b g
. Then Sn can take value(s)
(A) 1056 (B) 1088 (C) 1120 (D) 1332
Sol. [A, D] S kn
k k
k
n
         


 1 1 2 3 4 5 6
1
2
2
1
4
2 2 2 2 2 2
b g
b g
...
        








 4 1 4 2 4 3 4 4
2 2 2 2
0
1
0
1
0
1
0
1
r r r r
r
n
r
n
r
n
r
n
b g b g b g b g
   

  


 32 20 32
1
2
20 4 4 1
0
1
r
n n
n n n
r
n
b g b g b g b g
S8 1056 S9 1332

6. MATHEMATICS
[30] IIT JEE ADVANCED – 2013 / Paper-I
52. For 3 × 3 matrices M and N, which of thefollowingstatement(s) is(are) NOT correct?
(A) N M NT
is symmetricorskewsymmetric, accordingas M is symmetricorskewsymmetric
(B) M N – N M is skew symmetric for all symmetric matrices M and N
(C) M Nis symmetric for all symmetric matrices M and N
(D) (adj M) (adj N) = adj (M N) for all invertible matrices M and N
Sol. [C, D](A) N M N N M NT T T T
e j  (B) M N N M N M M N N M M N
T T T T T
    b g
(C) M N N M N M
T T T
b g   (D) adjM adjN adj N Mb gb g b g
53. Let f x x x xb g sin , 0.Then forall natural numbers n f x, 'b g vanishes at
(A)auniquepointintheinterval n n, 
F
HG I
KJ1
2
(B) aunique pointintheinterval n n 
F
HG I
KJ1
2
1,
(C)auniquepointintheinterval n n, 1b g (D) twopoints inthe interval n n, 1b g
Sol. [B, C] f A x x x' sin cosb g     0
 tan x x  0
x n n 
F
HG I
KJ,
1
2 , tan  x x  0
x n n  
F
HG I
KJ1
2
1, , tan x x  0 (once)
54. Arectangularsheetoffixedperimeterwithsides havingtheirlengthsintheratio8:15isconvertedintoan
openrectangularboxbyfoldingafterremovingsquaresofequalareafromallfourcorners.Ifthetotalarea
of removed squares is 100, the resultingbox has maximum volume. Thenthe lengths of thesides of the
rectangularsheet are
(A) 24 (B) 32 (C) 45 (D) 60
Sol. [A,C]
55. Alinel passingthroughtheoriginis perpendiculartothelines
1 3 1 2 4 2:   ,          t i t j t k tb g b g b g
2 3 2 3 2 2:   ,         s i s j s k sb g b g b g
Then,thecoordinate(s)ofthepoint(s)on l2 atadistanceof 17 fromthepointofintersectionofland l1
is(are)
(A)
7
3
7
3
5
3
, ,
F
HG I
KJ (B)  1 1 0, ,b g (C) (1, 1, 1) (D)
7
9
7
9
8
9
, ,
F
HG I
KJ

7. MATHEMATICS
[31] IIT JEE ADVANCED – 2013 / Paper-I
Sol. [B, D] d. r of line 
  
  
i j k
i j k1 2 2
2 2 1
2 3 2   
Equation of line 
x y z
2 3 2



The point of intersection of  and 1 is (2, –3, 2).
Any point of 2 is 3 2 3 2 2  s s s, ,b g
3 2 2 3 2 3 2 3 2 17
2 2 2
        s sb g b g b g
 9 28 20 02
s s    9 10 2 0s s  b gb g  s   
10
9
2,
SECTION – 3 : (Integer value correct Tyep)
Integer Answer Type
This section contains 5 questions. The answer to each question is a single digit integer, rangingfrom 0 to 9
(bothinclusive)
56. The coefficients of three consecutive terms of 1
5


x
n
b g arein the ratio 5 : 10 : 14. Thenn =
Sol. [6] n
r
n
r
n
rC C C 


 5 5
1
5
2 5 10 14: : : : 
n
r n r
r n r
n

 

  


5
5
1 4
5
1
2
b g
b g
b gb g
b g
!
! !
! !
!

r
n r

 

1
5
1
2
b g
b g  2 2 5r n r     3 3r n 
n
r
n
r
C
C





5
1
5
2
10
14
n
r n r
r n r
n

  

  


5
1 4
2 3
5
5
7
b g
b gb g
b gb g
b g
!
! !
! !
!

r
n r

 

2
4
5
7
b g
b g  7 14 5 5 20r n r     5 12 6 0n r  
 5 4 3 6 0n n   b g  5 4 12 6 0n n     n  6

8. MATHEMATICS
[32] IIT JEE ADVANCED – 2013 / Paper-I
57. A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the
pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the
removed cards is k, then k  20
Sol. [5]
n n
k k

   
1
2
1224 1
b g b g
By observation n  50

50 51
2
1224 1

   k k  k  25  k  20 5
58. Ofthethreeindependent events E E1 2, and E3 ,theprobabilitythat only E1 occurs is , onlyoccurs in
 and only E3 occurs is . Let the probability p that none of events E E1 2, or E3 occurs satisfy the
equations    2b gp and    3 2b gp . All the given probabilities are assumed to lie in the
interval (0, 1).
Then
Pr
Pr
obability of occurrence of E
obability of occurrence of E
1
3

Sol. [6] Let probability of E E1 2, and E3 is X X1 2, and X3 respectively then
   X X X1 2 31 1b gb g    X X X2 1 31 1b gb g
   X X X3 1 21 1b gb g and P X X X   1 1 11 2 3b gb gb g
Now    2b gP
 X X X X X X X X X1 2 3 2 1 3 1 2 31 1 2 1 1 1 1 1       b gb g b gb gc hb gb gb g
    X X X X X X1 2 3 2 1 31 1 1 1b gb g b gb g
 X X1 22 . . . (i)
l.ly    3 2b gP r
 X X X X X X X X X2 1 3 3 1 2 1 2 31 1 3 1 1 1 1 1       b gb g b gb gb gc hb gb gb g
    2 1 1 1 12 1 3 3 1 2X X X X X Xb gb g b gb g
 X X2 33 . . . (ii)
using(i)and(ii)
X X1 36 
X
X
1
3
6

9. MATHEMATICS
[33] IIT JEE ADVANCED – 2013 / Paper-I
59. Averticallinepassingthroughthepoint(h, 0)intersectstheellipse
x y2 2
4 3
1  atthepointsPandQ.Let
thetangentstotheellipseatPandQmeetatthepointR.If  hb g areaofthetrianglePQR,  1 1
2
1

 
max
h
hb g
and  2 1
2
1

 
min
h
hb g, then
8
5
81 2  
Sol. [9]
O
(4/h, 0)
(h, k)
1½ 2
(h, –k)
 h k
h
h
h h
h h
hb g b g e j  
F
HG I
KJ 
 F
HG
I
KJ  
1
2
2
4 3 4
2
4 3
2
4
2 2
2
3
2
Max     1 2
1
2
1
F
HG I
KJ  , min b g
8
5
8
8
5
3 4
1
4
8
3
2
31 2
3
2 3
2    
F
HG I
KJ   b g
      
8 3
5
15 15
8
8
3
2
3 3 45 36 9
60. Consider the set of eight vectors V ai bj ck a b c       : , , ,1 1l qo t. Threenoncoplanar vectors can
be chosen from V in 2p
ways. Then p is
Sol. [5] Consider acube whose centre is at origin and of side length two unit and faces areparallel to coordi-
nateaxes thenallof thevertices of cuberepresent thegiveneightvectors.Nowthreevectorsofthem will
be coplanerif weselecttwovertices alongbodydiagonal andthirdvertexfrom remaining.
Number of ways to select 3 noncoplanar vectors    8
3
6
1
5
4 32 2C C.