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MATHEMATICSCODE - 6

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w w w .p a g e a c a d e m y.c o m w w w.b its a tp a g e .c o m2 of 23

[CODE-6]

PART -III(MATHEMATICS)Section 1 (Maximum Marks : 15)

• This section contains FIVE questions.

• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options iscorrect.

• For each question, darken the bubble corresponding to the correct option in the ORS.

• For each question, marks will be awarded in one of the following categories:Full Markls : +3 If only the bubble corresponding to the correct option is darkened.

Zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : -1 In all other cases.

37. A computer producting factory has only two plants 1 2.T and T Plant 1T produces 20% and plant

2T produces 80% of the total computers produced. 7% of computers produced in the factory

turn out to be defective. It is known that

P (computer turns out to be defective given that it is produced in plant 1T )

= 10 P (computer turns out to be defective given that it is produced in plant 2T ),

Where ( )P E denotes the probability of an event E. A computer produced in the factory israndomly selected and it does not turn out to be defective. Then the probability that it is produced

in plant 2T is.

A)36

73B)

47

79C)

78

93D)

75

83

Sol : C

Let D be event of getting defective computers

( ) ( ) ( )1 220 80 7

, ,100 100 100

P T P T P D= = =

Given ( ) ( ) ( )1 27

..... 1100

P D T P D T∩ + ∩ =

1 2

10.D D

P PT T

=

( ) ( )( ) ( )1

1 22

10. .P T

P D T P D TP t

∩ = ∩

www.aieeepage.comPAGEwww.aieeepage.comwww.pageacademy.comwww.bitsatpage.com





[CODE-6]

( )220

10. .80

P D T= ∩

( ) ( ) ( )1 25

.... 22

P D T D T∩ = ∩

From (1) & (2) ( ) ( )1 25 2

,100 100

P D T P D T∩ = ∩ =

Required probabiltiy( )

( )22

P T DTP

D P D

∩ = =

( ) ( )( )

2 2

1

P T P T D

P D

− ∩=

−

80 2100 100

71

100

−=

−

78

93=

38. A debate club consists of 6 girls and 4 boys. A team of 4 member is to be selected from this clubincluding the selection of a captain (from among these 4 members) for the team. If the team has toinclude at most one boy, then the number of ways of selecting the team is

A) 380 B) 320 C) 260 D) 95

Sol :

Case (i) 3 Girls, 1 Boy.

No.of ways of selecting 3 Girls and 1 Boy 6 43 1 80C C= × =

No.of ways of selecting a captain 3 1 1 4C= + =

No.of ways of selecting 3G, 1B (including captian) 80 4 320= × =Case (ii) 4 Girls, No boy

No.of ways of selecting 4 Girls 6 4 15C= =

No.of ways of selecting a captain 4 1 4C= =






[CODE-6]

No.of ways of selecting 4 Girls (includins captain) 15 4 60= × =

∴ Required combination 320 60 380= + =

39. The least value of R∈ for which2 14 1,x

x + ≥ for all 0,x > is

A)1

64B)

1

32C)

1

27D)

1

25

Sol : C

2 14 1xx

+ ≥

2 14 1xx

≥ −

2 3

1 14

x x ≥ −

( )3

14

x

x

−≥

( ) ( )31x

f xx

−=

( ) ( )3 2

6

1 3'

x x xf x

x

− −=

[ ]26

3 30

x x x

x

− += =

2 3

3/ 2

x

x

==

( )

31

24

27 / 8

− ≥

41

2 ≥ 8×

27






[CODE-6]

1

27 ≥

Least value of1

27 =

40. Let .6 12

− < < − Suppose 1 1and are the roots of the equation 2 2 sec 1 0x x − + = and

2 2and are the roots of the equation 2 2 tan 1 0.x x + − = If 1 1 2 2 ,and > > then

1 2 + equals

A) ( )2 sec tan − B) 2sec C) 2 tan− D) 0Sol : B

Let6 12

− < < −

1 1, are the roots of 2 2 sec 1 0x x − + = 2 2, are the roots of 2 2 tan 1 0x x + − =

22sec 4sec 4

2x

± −=22 tan 4 tan 4

2x

− ± +=

secx Tan = ± tan sec = − ±

1 1 ∴ > 2 2 ∴ >

1 sec Tan ∴ = − ( )2 sec Tan ∴ = −

1 sec Tan = + 2 sec Tan =− −

1 2 2sec + = .

41. Let ( ), : 0, .2

S x x = ∈ − ≠ ±

The sum of all distinct solutions of the equation

( )3 sec cos 2 tan cot 0x ecx x x+ + − = in the set S is equal to

A)7

9

− B)2

9

− C) 0 D)5

9






[CODE-6]

Sol : C

( ), : 0,2

S x x = ∈ − ≠ ±

( )3 sec cos 2 tan cot 0x x x x+ + − =

( )cos 23 12 0

cos sin sin cos

x

x x x x

−+ + =

2 3 sin cos 2cos 2 0x x x+ − =

3 1sin cos cos 2

2 2x x x+ =

cos cos 23

x x − + =

2 2 ,3

x n x n z − + = ± ∈

2 23

n x x − = ± +

2 3 , 23 3

n x n x − = − = −

2

9 3

nx = −

09

n x= ⇒ = 0,

3n x

= = −

51

9n x

−= ⇒ =

71

9If n then x

= − =

Sum of all distinct solutions5 7

09 3 9 9

= − − + =






[CODE-6]

SECTION 2 (Maximum Marks : 32)

• This section contains EIGHT questions.

• Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE ofthese four option(s) is (are) correct.

• For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.

• For each question, marks will be awarded in one of the following categories :Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are)

darkened.

Partial Marks : + 1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened.

zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : -2 In all other cases.

For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three willresult in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) willresult in -2 marks, as a wrong option is also darkened.

42. Let ( ): 0,f R∞ → be a differentiable function such that ( ) ( )' 2 f xf xx

= − for all ( )0,x∈ ∞

and ( )1 1.f ≠ Then

A)0

1lim ' 1x

fx→ +

= B)

0

1lim 2x

xfx→ +

=

C) ( )20

lim ' 0x

x f x→ +

= D) ( ) ( )2 0, 2f x for all x≤ ∈Sol : A

( ) ( )' 2 f xf xx

= −

( ) ( )' 2x f x x f x= −

( ) ( )' 2xf x f x x+ = on integration

( ) 2xf x x C= +

( ) cf x xx

= +






[CODE-6]

( )1 1Since f ≠

1 1c⇒ + ≠

0c⇒ ≠

( ) 2' 1c

f xx

= −

21' 1f cxx

= −

( )0

1' 1 0 1

xLt f A

x+→ = − =

1 1f Cx

x x = +

21 1xf cxx

= +

0

11 0 1

xLt x f

x+→ = + =

( )2 2'x f x x c= −

( )20

' 0xLt x f x c→

= − ≠

43. The circle 2 21 : 3,C x y+ = with centre at O, intersects the parabola2 2x y= at the point P in the

first quadrant. Let the tangent to the circle 1C at P touches other two circles 2 3C and C at

2 3R and R , respectively. Suppose 2 3C and C have equal radii 2 3 and centres 2 3,Q and Q

respectively. If 2 3Q and Q lie on the y-axis, then

A) 2 3 12Q Q = B) 2 3 4 6R R =

C) area of the triangle 2 3 6 2OR R is D) area of the triangle 2 3 4 2PQ Q is

Sol : A, B, C

2 2 23, 2x y x y+ = =






[CODE-6]

2 2 3 0y y+ − =

( )( )3 1 0y y+ − =1, 3y = −

2x = ±

P.O.I in first quadrant is ( )2,1PEquation of tangent at ( ) 2 22,1 3 2 3 0P to x y is x y+ = + − =Let ( )0, y be point on y-axis.

32 3

3

y −∴ =

3 6y − =

3 6 9, 3y = ± = −

( ) ( )2 30, 9 , 0, 3Q Q∴ = = −

2 3 12Q Q =

2R is foot of the perpendicular from 3Q on 2 3 0x y+ − =

Then0 9 6

1 32

h k− − −= =

( ) ( )2 , 2 2, 7R h k = −3R is foot of the perpendicular from 3Q on 2 3 0x y+ − =

Then0 3 6

1 32

h k− += =

( ) ( )3 , 2 2, 1R h k = −2 3 32 64 96 4 6R R = + = =






[CODE-6]

Area of 2 31

2 2 14 2 6 22

OR R∆ = − =

Area of ( )2 31

2 9 3 6 22

PQ Q∆ = + =

44. A solution curve of the differential equation ( )2 24 2 4 0, 0,dyx xy x y y xdx

+ + + + − = > passes

through the point (1, 3). Then the solution curve

A) intersects 2y x= + exactly at one point

B) intersects 2y x= + exactly at two point

C) intersects ( )22y x= +

D) does NOT intersect ( )23y x= +Sol : A,D

( )2 24 2 4 dyx xy x y ydx

+ + + + =

( ) ( )2

2 2

dy y

dx x x y=

+ + +

Let 2t x= +

dt dx

dy dy=

( ) 22 2

t y tdt yt t

dy y y

+ += =

Let t vy=

dv dtv y

dy dy+ =

2dvv y v vdt

+ = +

2

1 1dv dy

v y=∫ ∫






[CODE-6]

1logc y

v− =

log log2

y yy c y c

t x+ = ⇒ + =

+

Given the curve passes through (1, 3)

1 log 3 c+ =

( )log 1 log 3 log 32

yy e

x∴ + = + =

+

2, 1 log 1 log 3If y x y= + + = +

3y⇒ =

2y x∴ = + intersects the curve at (1, 3)

( ) ( )22 , 2 log log 3If y x x y e= + + + =

( )23

2 log2

ex

x+ =

+

( )22. 2 4 0xe x e for x+ + > >∴ The curve does not intersect the curve when x . 0.

When ( ) ( )2

2 3 33 , log

2

x ey x

x y

+ = + = +

( )( )23

2 23 . 3x

xx e e+++ =

( )( )23

2 9 / 2 423 . 9. 3. . 0x

xx e e e e when x+++ > = >

∴ The curve does not intersect the curve when 0x > .






[CODE-6]

45. In a triangle XYZ, let x, y, z be the lengths of sides opposite to the angles X,Y, Z, respectively,

and 2 .s x y z= + + If4 3 2

s x s y s z− − −= = and area of incircle of the triangle XYZ is8

,3

then

A) area of the triangle XYZ is 6 6

B) the radius of circumcircle of the triangle XYZ is35

66

C)4

sin sin sin2 2 2 35

X Y Z =

D)2 3sin

2 5

X Y+ =

Sol : A, C, D

Given 2s x y z= + +

( )4 3 2

s x s y s zk say

− − −= = =

4 , 3 , 2s x k s y k s z k− = − = − =

4 3 2s x s y s z k k k− + − + − = + +

( )3 9s x y z k− + + =

3 2 9s s k− =

9s k=

4 9 4 5

3 9 3 6

2 9 2 7

x s k k k k

y s k k k k

z s k k k k

= − = − == − = − == − = − =

: : 5:6:7x y z =

Given area of incircle is8

3

2 2 883 3

r r = ⇒ =






[CODE-6]

8

3r =

( )( )( ) 9.4.3.2 6 6s s x s y s z∆ = − − − = =

40

xyzR =

567 35

4.6 6 4. 6R = =

4 sin sin .sin2 2 2

x y zr R=

84

3= 35

4× sin sin sin

2 2 26

x y z

8 6 4sin .sin .sin

2 2 2 3 35 35

x y z⇒ = × =

2 2 2 25 36 49 12 1cos

2 2 5 6 2 30 5

x y zz

xy

+ − + −= = = =× × ×

( )21

11 cos 1 cos 6 35sin2 2 2 2 10 5

x yx y z +− ++ += = = = =

46. Let RS be the diameter of the circle 2 2 1,x y+ = where S is the point (1, 0). Let P be a variablepoint (other than R and S) on the circle and tangents to the circle at S and O meet at the point Q.The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then thelocus of E passes through the point(s)

A)1 1

,3 3

B)1 1

,4 2

C)1 1

,3 3

− D)

1 1,

4 2 −

Sol : C

Tangent at S is 1x =

Tangent at P is cos sin 1 0x y + − =






[CODE-6]

1, tan2

Q ∴ =

Normal at ‘P’ is ( )sin cos 0 ..... 1x y − =The line passing through ' ' and parallel to x-axis is

( )tan .... 22

y=

P.I of (1) & (2) is ( )1 1 cot . tan , tan2 2E x y =

1 1cot . tan , tan2 2x y

= =

21 tan2

2

−=

211

2

y−=

21 11 2y x= −

Locus is 2 1 2y x= −

(A)1 1

,3 3

(C)1 1

,3 3

−

lies on this curve

47. Let

3 1 2

2 0 ,

3 5 0

P − −

= −

where .R∈ Suppose ijQ q = is a matrix such that ,PQ kI=

where , 0k R k∈ ≠ and I is the identity matrix of order 3. If ( )2

23 det ,8 2

k kq and Q= − = then

A) 0, 8k = = B) 4 8 0k − + =

C) ( )( ) 9det 2P adj Q = D) ( )( ) 13det 2Q adj P =






[CODE-6]

Sol : B, C

3 1 2

2 0 ,

3 5 0

P − −

= −

12 20P ⇒ = +

3PQ KI k= =

3P Q k=

( )2

312 20 .2

Kk + =

6 10k = +

11 12 13

21 22

31 32 33

8

q q q

kQ q q

q q q

= −

PQ KI=

11 12 13

21 22

31 32 33

3 1 2 0 0

2 0 0 08

3 5 0 0 0

q q qk

kq q k

kq q q

− − − = −

13 33 13 33 13

53 2 0 2 0 3

8 8

k kq q q q q k+ − = + = + =

13

53

8

kq k

−⇒ = +

13

13

5

8

8

kq k

kq

−⇒ = +

⇒ =

33

32

8 8

k kq+ =






[CODE-6]

1333

33

22 8, 14

4

kqk

qkq

− ×−⇒ = = = = −

1 10 6 4k = − ⇒ = − =

168 & 12 20 8

2Q P = = = + =

( ) 2.P adjQ P Q= 2 3 6 98 8 2 2 2= × = × =

( ) 2 2 9. 8 8 2Q adj P Q P= = × =

4 8 4 4 8 0k − + = − − + =

48. Let : , : :f g and h→ → → be differentiable functions such that

( ) ( )( ) ( )( )( )3 3 2,f x x x g f x x and h g g x x for all x= + + = = ∈ . Then

A) ( ) 1' 215

g = B) ( )' 1 666h = C) ( )0 16h = D) ( )( )3 36h g =Sol : B, C

( ) 3 3 2f x x x= + +

( )g f x x=

( )( )h g g x x = 3 3 2g x x x + + =

( )( ) ( )h g g f x f x = ( ) ( )h g x f x=

( ) ( )h g f x f f x =

( ) ( )h x f f x=






[CODE-6]

(A) ( )g f x x= ( )0 2f =

( ) ( )' ' 1g f x f x = ( ) 2' 3 3f x x= +

[ ] ( )' 2 ' 0 1g f = ( )' 0 3f =

( ) 1' 23

g =

(B) ( ) ( ) ( )' ' 'h x f f x f x=

( ) ( ) ( )' 1 ' 1 ' 1h f f f=

( ) ( )' 6 ' 1f f=

( )( )111 6 666= =

(C) ( ) ( ) ( )0 2 8 6 2 16h f f x f= = = + + =

(D) ( ) ( )3 3 27 9 2 38h g f= = + + =

49. Consider a pyramid OPQRS located in the first octant ( )0, 0, 0x y z≥ ≥ ≥ with O as origin, andOP and OR along the x-axis and the y-axis, respectively. The base OPQR of the pyramid is asquare with 3.OP = The point S is directly above the mid-point T of diagonal OQ such that

3.TS = Then

A) the acute angle between OQ and OS is3

B) the equation of the plane containing the triangle OQS is 0x y− =

C) the length of the perpendicular from P to the plane containing the triangle OQS is3

2

D) the perpendicular distance from O to the straight line containing RS is15

2

Sol : B,C,D

(A) D.R’s of OQ are 3, 3, 0

D.R’s of OS are3 3

, ,32 2






[CODE-6]

( ) 19 92 2, cos

9 99 9 9

4 4

OQ OS −

+

= + + +

1 9cos−=9 2

33

2×

1 1cos3 3

−

=

(B)

3 3 0 9 1 1 0

3 3 1 13 3

2 2 2 2

i j k i j k

OQ OS× = =

9 i j = −

The equation of the plane with normal ( )i j− is 0x y− =(C)The perpendicular distance from P (3, 0, 0) to the triangle OQS 0x y− = is

3 0 3

2 2

− =

(D) D.R’s of RS are ( )3 3, ,3 1, 1, 22 2

or− −

Any point on RS ( )3 ,3 , 21 1 2

x y zt is A t t t

−= = = −−

D.R’s of OA are t, 3 - t, 2t

RS lr OA⊥

3 4 0t t t− + + =

6 3t =






[CODE-6]

1

2t =

1 5, ,1

2 2A is

∴

1 25 30 151

4 4 4 2OA = + + = =

SECTION 3 (Maximum Marks : 15)

• This section contains FIVE questions.

• The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.

• For each question, darken the bubble corresponding to the correct integer in the ORS.

• For each question, marks will be awarded in one of the following categories:Full Marks : +3 If only the bubble corresponding to the correct answer is darkened.

Zero marks : 0 In all other cases.

50. Let1 3

,2

iz

− += where 1,i = − and { }, 1,2,3 .r s∈ Let ( )2

2

r s

s r

z zP

z z

−=

and I be the

identity matrix of order 2. Then the total number of ordered pairs ( ),r s for which 2P I= − isSol : 1

1 3iz w

z

− += =

( ) 22

r s

s r

z zP

z z

−=

( ) ( )2 222 2

1 0

0 1

r rs s

s r s r

z z z zP I P

z z z z

− − −= − ⇒ = = −

( )2 4 1r sz z∴ − + = − ( ) 2 2. . 0r s s rz z z z− + =

2 4 1r sz z+ = − ( )2. 1 1 0rr sz z ⇒ + − =

4 2 1sz z+ = − 4 6 1sz z+ = −4 2 1sw w+ = − 1, 3r =

4 21sw w= − − 4 1 1sw + = −






[CODE-6]

4sw w= + 4 2sw = −

( )4 sw w= Not possiblesw w=

1s⇒ =

( ) ( ), 1, 1r s =

51. Let m be the smallest positive integer such that the coefficient of 2x in the expansion of

( ) ( ) ( ) ( ) ( )2 3 49 50 51 31 1 ... 1 1 3 1x x x mx is n C+ + + + + + + + + for some positive integer n. Thenthe value of n is

Sol : 5

( ) ( ) ( ) ( )2 3 49 501 1 ... 1 1x x x mx+ + + + + + + +

( )( )

( )48

2 501 1

1 11 1

xx mx

x

+ − = + + ++ −

( ) ( ) ( )50 2

501 11

x xmx

x

+ − += + +

coefficient of ( )2 51 33 1x x C= +50 50 2 51 51

3 2 3 3. 3 .C C m n C C+ = +

( )50 2 51 50 512 3 3 3. 3 .C m C C n C= − +

50 2 50 502 2 2

51. 3 . .

3C m C n C= +

2 1m n= +

5n = satisfies

52. The total number of distinct [ ]0, 1x∈ for which2

40

2 11

x tdt x

t= −

+∫ is

Sol : 1






[CODE-6]

2

40

( )1

x tf x dt

t=

+∫

1 2

40

(0) 0, (1)1

tf f dt

t= =

+∫

2

4

10

1 2

t

t< ≤

+

1 2

40

1 10 0 (1)

1 2 2

tf

t⇒ < ≤ ⇒ < ≤

+∫

Let ( ) 2 1g x x= −

0

1y=2 -1x

y= )f(x

1

-1

1

2

1

2

∴ No. of solutions = 1

53. The total number of distinct x R∈ for which

2 3

2 3

2 3

1

2 4 1 8 10

3 9 1 27

x x x

x x x

x x x

++ =

+ is

Sol : 2

2 3

2 3

2 3

1

2 4 1 8 10

3 9 1 27

x x x

x x x

x x x

++ =

+






[CODE-6]

3

3 3

3

1 1 1 1 1

2 4 8 2 4 1 10

3 9 27 3 9 1

x

x x

x

+ =

6 3

1 1 1 1 1 1

2 4 8 2 4 1 10

3 9 27 3 9 1

x x+ =

6 3

1 0 0 1 0 0

2 2 9 2 2 1 10

3 6 18 3 6 2

x x+ − =−

( ) ( )6 336 24 4 6 10x x− + − + =6 32 2 10 0x x+ − =6 36 5 0x x+ − =6 3 36 6 5 5 0x x x+ − − =

( ) ( )3 3 36 1 5 1 0x x x− − + =

3 51,6

x = −

51,3

6x = −

54. Let , R ∈ be such that( )2

0

sinlim 1.

sinxx x

x x

→

=−

Then ( )6 + equals

Sol : 7

( )20

sin1

sinxx x

Ltx x

→

=−

[ ]230

1

6

x

x xLt

xx x

→

=− +






[CODE-6]

1, 6 1 = =

1

6 =

( )6 6 6 6 1 7 + = + = + =

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JEE ADVANCED - 2016 Paper - I MATHEMATICS...JEE ADVANCED - 2016 Paper - I MATHEMATICS CODE - 6 PAGE - An exclusive institution for JEE & BITSAT training PAGE PAGE 2 of 23 …