JEE - Advance Mathematics 2015 paper 1.pdf

JEE ADVANCED 2015 MATHEMATICS (Paper-1)SECTION - I

This section contains 8 questions. Each question, when workedout will result in one integer from 0 to 9 (both inclusive).

1. If the normals of the parabola y2 = 4x drawn at the end pointsof its latus rectum are tangents to the circle (x – 3)2 + (y + 2)2

= r2, then the value of r2 is

2. Let f : R ® R be a function defined by {[ ], 2( ) 0, 2£= >

x xf x x

where [x] is the greatest integer less than or equal to x, if

2 2

1

( )2 ( 1)

xf xI dxf x

-

=+ +ò , then the value of (4I – 1) is

3. A cylindrical container is to be made from certain solidmaterial with the following constraints: It has a fixed innervolume of V mm3, has a 2 mm thick solid wall and is open atthe top. The bottom of the container is a solid circular discof thickness 2 mm and is of radius equal to the outer radiusof the container.If the volume of the material used to make the container isminimum when the inner radius of the container is 10 mm,

then the value of 250pV

is

4. Let

26

2( ) 2cos ( )

p+

= òx

xF x t dt for all Îx R and

1: 0, [0, )2

é ù ® ¥ê úë ûf be a continuous function. For

10,2

é ùÎ ê úë ûa , if ( ) 2¢ +F a is the area of the region bounded

by x = 0, y = 0, y = f(x) and x = a, then f(0) is5. The number of distinct solutions of the equation

2 4 4 6 65 cos 2 cos sin cos sin 24

+ + + + =x x x x x

in the interval [0, 2p] is

6. Let the curve C be the mirror image of the parabola y2 = 4xwith respect to the line x + y + 4 = 0. If A and B are the pointsof intersection of C with the line y = –5, then the distancebetween A and B is

7. The minimum number of times a fair coin needs to be tossed,so that the probability of getting at least two heads is atleast 0.96, is

8. Let n be the number of ways in which 5 boys and 5 girls canstand in a queue in such a way that all the girls standconsecutively in the queue. Let m be the number of ways inwhich 5 boys and 5 girls can stand in a queue in such a waythat exactly four girls stand consecutively in the queue. Then

the value of mn is

SECTION - IIThis section contains 10 multiple choice questions. Each questionhas 4 choices (a), (b), (c) and (d) out of which ONE or MORETHAN ONE are correct.

9. Which of the following values of a satisfy the equation

2 2 2

2 2 2

2 2 2

(1 ) (1 2 ) (1 3 )

(2 ) (2 2 ) (2 3 ) –648 ?

(3 ) (3 2 ) (3 3 )

+ a + a + a

+ a + a + a = a

+ a + a + a

(a) –4 (b) 9(c) –9 (d) 4

10. In R3, consider the planes P1 : y = 0 and P2 : x + z = 1. Let P3be the plane, different from P1 and P2, which passes throughthe intersection of P1 and P2. If the distance of the point(0, 1, 0) from P3 is 1 and the distance of a point (a, b, g) fromP3 is 2, then which of the following relations is (are) true?(a) 2a + b +2g + 2 = 0 (b) 2a – b +2g + 4 = 0(c) 2a + b –2g – 10 = 0 (d) 2a – b +2g – 8 = 0

11. In R3, let L be a straight line passing through the origin.Suppose that all the points on L are at a constant distancefrom the two planes P1 : x + 2y – z + 1 = 0 and P2 : 2x – y + z– 1 = 0. Let M be the locus of the feet of the perpendiculars

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drawn from the points on L to the plane P1. Which of thefollowing points lie (s) on M?

(a)5 20, – , –6 3

æ öç ÷è ø

(b)1 1 1– , – ,6 3 6

æ öç ÷è ø

(c)5 1– ,0,6 6

æ öç ÷è ø

(d)1 2– ,0,3 3

æ öç ÷è ø

12. Let P and Q be distinct points on the parabola y2 = 2x suchthat a circle with PQ as diameter passes through the vertexO of the parabola. If P lies in the first quadrant and the areaof the triangle DOPQ is 3 2 , then which of the following is(are) the coordinates of P?

(a) (4,2 2) (b) (9,3 2)

(c)1 1,4 2

æ öç ÷è ø

(d) (1, 2)

13. Let y(x) be a solution of the differential equation

(1 ) 1¢+ + =x xe y ye . If y(0) = 2, then which of the followingstatement is (are) true?(a) y(–4) = 0 (b) y(–2) = 0(c) y(x) has a critical point in the interval (–1, 0)(d) y(x) has no critical point in the interval (–1, 0)

14. Consider the family of all circles whose centers lie on thestraight line y = x. If this family of circle is represented by thedifferential equation 1 0Py Qy¢¢ ¢+ + = , where P, Q are func-

tions of x, y and y¢ 2

2here y = ,æ ö

=¢ ¢¢ç ÷è ø

dy d yydx dx , then which of

the following statements is (are) true?(a) P = y + x (b) P = y – x(c) P + Q = 1 – x + y + y¢ + (y')2 (d) P – Q = x + y – y¢ –(y')2

15. Let g : R ® R be a differentiable function with g(0) = 0, g'(0)

= 0 and g'(1) ¹ 0. Let ( ), 0( ) | |

0, 0

ìï ¹= íï =î

x g x xf x xx

and h(x) = e|x| for all Îx R . Let (foh)(x) denote f(h(x)) and(hof)(x) denote h(f(x)). Then which of the following is (are) true?(a) f is differentiable at x = 0(b) h is differentiable at x = 0(c) foh is differentiable at x = 0(d) hof is differentiable at x = 0

16. Let ( ) sin sin sin6 2

æ p p öæ ö= ç ÷ç ÷è øè ø

f x x for all x Î R and g(x) = 2p

sin x for all x Î R. Let (fog)(x) denote f(g(x)) and (gof)(x)denote g(f(x)). Then which of the following is (are) true?

(a) Range of f is 1 1,2 2

é ù-ê úë û

(b) Range of fog is 1 1,2 2

é ù-ê úë û

(c) 0

( )lim( ) 6®

p=

x

f xg x

(d) There is an Îx R such that ( )( ) 1=gof x

17. Let DPQR be a triangle. Let ,= =uuur uuurrra QR b RP and =

uuurrc PQ . If

12=ra , 4 3=

r

b , . 24=r rb c , then which of the following

is (are) true?

(a)2

122

- =r

rca (b)

230

2+ =

rrca

(c) 48 3´ + ´ =rr r ra b c a (d) . –72=

rra b18. Let X and Y be two arbitrary, 3 × 3, non-zero, skew-symmetric

matrices and Z be an arbitrary 3 × 3, non zero, symmetricmatrix. Then which of the following matrices is (are) skewsymmetric?(a) Y3Z4 – Z4Y3 (b) X44 + Y44

(c) X4Z3 – Z3X4 (d) X23 + Y23

SECTION - IIIThis section contains TWO questions. Each question contains two columns, Column I and Column II. Column I has four entries (A),(B), (C) and (D). Column II has five entries (P), (Q), (R), (S) and (T). Match the entries in Column I with the entries in Column II. Oneor more entries in Column I may match with one or more entries in Column II.19. Match the following

Column I Column II(A) In R2, if the magnitude of the projection vector (P) 1

of the vector ˆa + b$i j on ˆ ˆ3 +i j is 3 and

if 2 3a = + b , then possible value of a is/are(B) Let a and b be real numbers such that the (Q) 2

function 2

23 2, 1( )

, 1ì- - <= í

+ ³îax xf x

bx a xif differentiable for all Îx RThen possible value of a is (are)

3


(C) Let w ¹ 1 be a complex cube root of unity. (R) 3If 2 4 3 2 4 3(3 3 2 ) (2 3 3 )+ +- w + w + + w - wn n

2 4 3( 3 2 3 ) 0,++ - + w + w =n then possiblevalue (s) of n is (are)

(D) Let the harmonic mean of two positive real (S) 4numbers a and b be 4. If q is a positivereal number such that a, 5, q, b is an arithmeticprogression, then the value(s) of | q – a | is (are)

(T) 520. Match the following.

Column I Column II(A) In a triangle DXYZ, let a, b, and c be the (P) 1

lengths of the sides opposite to the anglesX, Y and Z, respectively. If 2(a2 – b2) = c2

and sin( )

sin-

l =X Y

Z , then possible values of

n for which cos( ) 0pl =n is (are)(B) In a triangle DXYZ, let a, b and c be the (Q) 2

lengths of the sides opposite to the anglesX, Y, and Z respectively. If 1 + cos 2X – 2cos 2Y

= 2 sin X sin Y, then possible value (s) of ab is (are)

(C) In R2, let ˆ ˆ ˆ3 , 3i j i j+ + and ˆ ˆ(1 )i jb + -b (R) 3be the position vectors of X , Y, and Z withrespect to the origin O, respectively. If thedistance of Z from the bisector of the

acute angle of uuur

OX with uuur

OY is 32

,

then possible value(s) of b is (are)(D) Suppose that F(a) denotes the area of the (S) 5

region bounded by x = 0, x = 2, y2 = 4x and| 1 | | 2 |= a - + a - +ay x x x , where {0, 1}a Î .

Then the value(s) of 8( ) 23

a +F , when

a = 0 and a = 1, is (are)(T) 6

4


1. (2) End points of latus rectum of y2 = 4x are (1, +2)Equation of normal to y2 = 4x at (1, 2) isy – 2 = –1(x – 1) or x + y –3 = 0As it is tangent to circle (x – 3)2 + (y + 2)2 = r2

\3 (–2) – 3

2+

= r Þ r2 = 2

2. (0) I = 2 2

–1

( )2 ( 1)+ +ò

xf xdx

f x

–1 < x < 2 Þ 0 < x2 < 4Also 0 < x2 < 1 Þ f(x2) = [x2] = 0

1 < x2 < 2 Þ f(x2) = [x2] = 12 < x2 < 3 Þ f(x2) = 0 (using definition of f)3 < x2 < 4 Þ f(x2) = 0 (using definition of f)

Also 1 < x2 < 2 Þ 1 < x < 2

Þ 2 < x + 1 < 2 + 1Þ f(x + 1) = 0

\ I =

22 2

1 1

1 2 1 1–2 0 4 4 4 4

é ù´= = =ê ú

+ ê úë ûò

x xdx

Þ 4I = 1 or 4I – 1 = 03. (4) Let r be the internal radius and R be the external radius.

Let h be the internal height of the cylinder.

\ pr2h = V Þ h = 2p

Vr

Also Vol. of material = M = p[(r + 2)2 – r2]h + p(r + 2)2 × 2

or M = 4p(r + 1). 2p

Vr

+ 2p (r + 2)2

Þ M = 4V 21 1é ù

+ê úë ûr r + 2p(r + 2)2

dMdr

= 4V 2 3–1 2–é ù

ê úë ûr r + 4p(r + 2)

For min. value of M, dMdr

= 0

Þ 3–4Vr

(r + 2) + 4p(r + 2) = 0

Þ 34Vr

= 4p or r3 = pV

= 1000

\ V = 1000p

\250p

V = 4

4. (3) F(x) =2 / 6 22cos+p

òxx

t dt

( )a¢F = 2cos2 26pæ öa +ç ÷è ø .2a – 2cos2a

( )a¢F + 2 = 0

( )f x dxa

òÞ ( )F ¢¢ a = f(a)

\ f(a) = 4a.2cos2 2. – sin .2

6 6p é p ùæ ö æ öa + a + aç ÷ ç ÷ê úè ø è øë û

+ 4 cos2 26pæ öa +ç ÷è ø – 4cosa (–sina)

\ f(0) = 4cos26p

= 4 × 34

= 3

5. (8)54

cos22x + cos4x + sin4x + cos6x + sin6x = 2

Þ54

cos22x + 1 – 12

sin22x + 1 – 34

sin22x = 2

Þ54

(cos22x – sin22x) = 0 Þ cos4x = 0

Þ 4x = (2n + 1)2p

or x = (2n + 1)8p

For xÎ[0, 2p], n can take values 0 to 7\ 8 solutions.

6. (4) Let (t2, 2t) be any point on y2 = 4x. Let (h, k) be theimage of (t2, 2t) in the line x + y + 4 = 0. Then

2 22 –2( 2 4)1 1 2

h t k t t t- - + += =

Þ h = –(2t + 4) and k = –(t2 + 4)For its intersection with, y = –5, we have

–(t2 + 4) = –5 Þ t = +1\ A(–6, –5) and B(–2, –5)\ AB = 4.

SOLUTIONSPaper - 1

5


7. (8) P(x > 2) > 0.96Þ 1 – P(x = 0) – P(x = 1) > 0.96Þ P(x = 0) + P(x = 1) < 0.04

Þ1 12 2

n nnæ ö æ ö+ç ÷ ç ÷è ø è ø < 0.04

Þ1 1

252nn +

£

Þ2

1

n

n + > 25

Þ minimum value of n is 8.8. (5) n = 5! × 6!

For second arrangement,5 boys can be made to stand in a row in 5! ways, creating 6alternate space for girls. A group of 4 girls can be selected in5C4 ways. A group of 4 and single girl can be arranged at 2places out of 6 in 6P2 ways. Also 4 girls can arrangethemselves in 4! ways.\ m = 5! × 6P2 × 5C4 × 4!

5! 6 5 5 4!5! 6!

mn

´ ´ ´ ´=

´ = 5

9. (b, c) R2 – R1, R3 – R2

2 2 2(1 ) (1 2 ) (1 3 )2 3 4 3 6 3 –6482 5 4 5 6 5

+ a + a + aa + a + a + = aa + a + a +

R3 – R2

2 2 2(1 ) (1 2 ) (1 3 )2 2 3 4 3 6 3 –648

1 1 1

+ a + a + aa + a + a + = a

C2 – C1, C3 – C2

2(1 ) (3 2) (5 2)2 3 2 2 –324

1 0 0

+ a a a + a a +a + a a = a

Þ 2a2(–2a) = –324aÞ a3 – 81a = 0Þ a = 0, 9, –9

10. (b, d) P3 : x + ly + z – 1 = 0

Also 2

1

2

l -

+ l = 1 Þ l2 – 2l + 1 = l2 + 2

Þ l = –12

And 2

1

2

a + lb + g -

+ l = 2 Þ

1– –12

32

a b + g = + 2

1– –1 32

a b + g = ±

Þ 2a – b + 2g – 2 = +6Þ 2a – b + 2g – 8 = 0 or 2a – b + 2g + 4 = 0

11. (a, b) L : x y za b c

= = = l

Where

1 2

2 – 0.2 0

As L is parallela b cto both P and Pa b c

+ = ìí- + = î

Þ1 –3 –5a b c

= =

\ Any point on line L is (l, –3l, –5l)Equation of line perpendicular to P1 drawn from any pointon L is

3 51 2 –1

x y z- l + l + l= = = µ

\ M(µ + l, 2µ – 3l, –µ – 5l)But M lies on P1,\ µ + l + 4µ – 6l + µ + 5l + 1 = 0

Þ µ = –16

\ M1 –1 1

– , –3 , –56 3 6

æ öl l l +ç ÷è ø

For locus of M,

x = l – 16

, y = –3l – 13

, z = 5l + 16

Þ1/ 6 1/ 3 1/ 61 –3 –5

x y z+ + -= = = l

On checking the given point, we find –5 –2

0, ,6 3

æ öç ÷è ø and

–1 –1 1, ,

6 3 6æ öç ÷è ø satisfy the above eqn.

12. (a, d) Let point P in first quadrant, lying on parabola y2 = 2x

be 2

,2

a aæ öç ÷è ø

. Let Q be the point 2

,2

b bæ öç ÷è ø

. Clearly a > 0.

O

P

2,

2b b

æ öç ÷ç ÷è ø

Q

2,

2a a


6


Q PQ is the diameter of circle through P, O, Q

\ ÐPOQ = 90° Þ 2 2/ 2 / 2a b

a b´ = –1 Þ ab = –4

Þ b is negative.Also ar. DPOQ = 3 2

2

2

0 0 1

11 3 2

2 2

12

aa

b b

Þ =

Þ14

ab (a – b) = + 3 2

Þ a – b = + 3 2 (using ab = –4)

As a is positive and b is negative, we have a – b = 3 2

a + 4a

= 3 2 (using ab = –4)

Þ a2 – 3 2 a + 4 = 0

Þ a2 – 2 2 a – 2 a + 4 = 0

Þ (a – 2 2 ) (a – 2 ) = 0

Þ 2 2, 2a =

\ Point P can be ( )22 2

, 2 22


or( )2

2, 2

2


i.e. (4, 2 2 ) or (1, 2 )

13. (a, c) 1

1 1

x

x xdy e ydx e e

+ =+ +

I.F. = 1 + ex

\ Soln : y(1 + ex) = x + cy(0) = 2 Þ c = 4

\ y = 41x

xe

+

+\ y(–4) = 0

Also 2( 1) – ( 4)

( 1)

x x

xdy e e xdx e

+ +=

+

For critical point dydx

= 0

Þ ex(x + 3) = 1Þ x + 3 = e–x

Its solution will be intersection point of y = x + 3 and y = e–x

y e = –x

y x = + 3

(0, 1)

–1

Clearly there is a critical point in (–1, 0).14. (b, c) Let the equation of circle be

x2 + y2 + 2gx + 2gy + c = 0Þ 2x + 2yy' + 2g + 2gy' = 0Þ x + yy' + g + gy' = 0 ...(i)Differentiating again,

1 + yy" + (y')2 + gy" = 0

Þ g = –21 ( )y yy

yé ù+ +¢ ¢¢ê ú

¢¢ê úë û

Substituting value of g in eqn. (i)

x + yy' – 2 21 ( ) 1 ( )–y yy y yy

y yæ ö+ + + +¢ ¢¢ ¢ ¢¢ç ÷¢¢ ¢¢è ø

y' = 0

Þ xy" + yy'y" – 1– (y')2 – yy" – y' – (y')3 – yy y¢ ¢¢ = 0Þ (x – y)y" – y'(1 + y' + (y')2) = 1or (y – x)y" + [1 + y' + (y')2] y' + 1 = 0

Py" + Qy' + 1 = 0\ P = y – x, Q = 1 + y' + (y')2

Also P + Q = 1 – x + y + y' + (y')2

15. (a, d) f(x) = ( ), 0

0 , 0

x g x xx

x

ì ¹ïíï =î

=

– ( ) , 00 , 0( ) , 0

g x xx

g x x

<ìï =íï >î

f ' (x) =

– ( ) , 00 , 0( ) , 0

g x xx

g x x

<¢ìï =íï >¢î

\ Lf ' (0) = –g'(0) = 0Rf ' (0) = g'(0) = 0

\ f is differentiable at x = 0

h(x) = e|x| = – , 0

, 0

x

x

e x

e x

ì <ïí

³ïî

7


h'(x) = –– , 0

, 0

x

x

e x

e x

ì <ïí

³ïî Þ Lh'(0) = –1, Rh'(0) = 1

\ h is not differentiable at x = 0f oh(x) = f(h(x)) = g(e|x|) as e|x| > 0

=

–( ) 0(1) 0

( ) 0

x

x

g e if xg if x

g e if x

ì <ï

=íï

>î

f ' [h(x)] =

– –– ( ). , 00 , 0

( ) , 0

x x

x x

g e e xx

g e e x

ì ¢ <ï

=íï

¢ >î

Lf ' (h(0)) = –g ' (1)Rf ' (h(0)) = g ' (1)

Q g ' (1) ¹ 0, \ Lf ' (h(0)) ¹ Rf ' (h(0))\ foh is not differentiable at x = 0.

hof(x) = ( ) , 01 , 0

f xe xx

ìï ¹í

=ïî

Lh'(f(0)) = 0

( (0)) – ( (0 – ))limk

h f h f kk®

= (– )

0

1limg k

k

ek®

- =

(– )

0

(– )1lim(– )

g k

k

g keg k k®

-´

= 1 × 0 = 0 0 0

(– ) ( )(0) 0 lim lim 0k k

g k g kgk k® ®

æ ö¢ = Þ = =ç ÷è øQ

Rh'(f(0)) = 0

( (0 )) – ( (0))limk

h f k h fk®

+

= ( ) ( )

0 0

( )–1 –1lim lim( )

g k g k

k k

g ke ek g k k® ®

= ´ = 0

\ hof is differentiable at x = 0.16. (a, b, c)

f(x) = sin sin sin6 2

xæ p p öæ öç ÷ç ÷

è øè ø

–1< sin x <1 Þ – sin2 2 2

xp p p£ £

Þ –1 < sin sin2

xpæ öç ÷è ø

< 1

Þ–6p

< 6p

sin sin2

xpæ öç ÷è ø

< 6p

Þ–12

< sin1sin sin

6 2 2xép p ùæ ö £ç ÷ê úè øë û

\ Range of f = –1 1

,2 2

é ùê úë û

fog(x) = sin sin sin sin6 2 2

xé ùp æ p p öæ öê úç ÷ç ÷

è øè øë û

Range of fog = –1 1

,2 2

é ùê úë û

0

sin sin sin6 2

limsin

2x

x

x®

æ p p öæ öç ÷ç ÷è øè ø

p

= 0

sin sin sin sin sin6 2 6 2lim

sinsin sin26 2

x

x x

xx®

æ p p öæ ö p pæ öç ÷ç ÷ ç ÷è øè ø è ø´

pp pæ öç ÷è ø

= p/6

gof(x) = 2p

sin sin sin sin6 2

xæ öæ p p öæ öç ÷ç ÷ç ÷

è øè øè ø

1 1– sin ( ( )) sin2 2 2 2

g f xp pæ ö æ ö£ £ç ÷ ç ÷è ø è ø

–0.73 < g(f (x)) < 0.73\ gof(x) ¹ 1 for any x ÎR.

17. (a, c, d)

a b c o+ + =r r r r

P

Q Ra

bc

Þ2 2

–b c a+ =r r r

Þ2 2 2

2 .b c b c a+ + =r r r r r

Þ 48 + 2

cr

+ 48 = 144 Þ 2

cr

= 48 Þ cr

= 4 3

\2

48– –12 122 2

ca = =

r

r

2

24 302

ca+ = ¹

r

r

8


Also b c=r r

Þ ÐQ = ÐR

and cos(180 – P) = . 1

2b cb c

=r r

r r

Þ ÐP = 120° \ ÐQ = ÐR = 30°

Again 0 ( ) 0a b c a a b c a b c a+ + = Þ ´ + + = Þ ´ = ´r r r r r r r r r r r r

\ 2a b c a a b´ + ´ = ´r r r r r r

= 2 × 12 × 4 3 × sin150 = 48 3

And . 12 4 3 cos150 –72a b = ´ ´ =r r

18. (c, d)X ' = –X, Y ' = –Y, Z ' = Z(Y 3Z 4 – Z 4Y 3)' = (Z 4)'(Y 3)' – (Y 3)'(Z 4)'= (Z ')4(Y ')3 – (Y ')3(Z ')4

= –Z 4Y 3 + Y 3Z 4 = Y 3Z 4 – Z 4Y 3

\ Symmetric matrix.Similarly X 44 + Y 44 is symmetric matrix and X 4Z 3 – Z 3X 4and X 23 + Y 23 are skew symmetric matrices.

19. [A ® Q; B ® P, Q; C ® P, Q, S, T; D ® Q, T]

(A)3 3

2a + b

= Þ a = 2 3

3- b

\2 3 2 3

3- b

= + b Þ b = 0 Þ a = 2

(B) Lf '(1) = –6a and Rf '(1) = b–6a = b ...(i)Also f is continuous at x = 1,

\ –3a – 2 = b + a2

Þ a2 – 3a + 2 = 0 (using (i))Þ a = 1, 2(C) (3 – 3w + 2w2)4n + 3 + (2 + 3w – 3w2)4n + 3

+ (–3 + 2w + 3w2)4n + 3 = 0

Þ (3 – 3w + 2w2)4n + 3 +

4n 32

22 3 3

+æ öw + - wç ÷ç ÷wè ø

+

4n 32–3 2 3+

æ öw + w +ç ÷wè ø

= 0

Þ (3 – 3w + 2w2)4n + 3 [1 + w4n + 3 + (w2)4n + 3] = 0Þ 4n + 3 should be an integer other than multiple of 3.\ n = 1, 2, 4, 5

(D)2aba b+

= 4 Þ ab = 2a + 2b ...(i)

Also a + q = 10 or a = 10 – qand b + 5 = 2q or b = 2q – 5Putting values of a and b in eqn(i)

q = 4 or 152

Þ a = 6 or52

\ |q – a| = 2 or 5.20. [A®P, R, S; B ® P; C ® P, Q; D®S, T]

(A) 2(a2 – b2) = c2

Þ 2(sin2x – sin2y) = sin2zÞ 2sin(x + y) sin(x – y) = sin2zÞ 2sin(x – y) = sin z (Q sin(x + y) = sin z)

Þsin( – ) 1

sin 2x y

z= = l

\ cos(npl) = 0 Þ cos 2

np = 0 Þ n = 1, 3, 5

(B) 1 + cos2X – 2cos2Y = 2sinXsinYÞ 2cos2X – 2cos2Y = 2sinXsinYÞ 1 – sin2X – 1 + 2sin2Y = sinXsinYÞ sin2X + sinXsinY – 2sin2Y = 0Þ (sinX – sinY) (sinX + 2sinY) = 0

Þsinsin

XY = 1 or –2

\ab = 1.

(C) X( 3 , 1), Y(1, 3 ), Z(b, 1 – b)

y

O x

Y(1, 3)

X( 3,1)

y = x

60°

30°

By symmetry, acute angle bisector of ÐXOY is y = x.\ Distance of Z from bisector

= 1 32 2

b - + b= Þ 2b – 1 = + 3 or b = 2 or –1

\ b = 1, 2

(D) For a = 0, y = 3For a = 1, y = |x – 1| + |x – 2| + x

Case IF(a) is the area bounded by x = 0, x = 2, y2 = 4x and y = 3

9


\ F(a) = 20

(3 – 2 )x dxò

y

y = 3

x

x = 2

=

2

0

4 8 23 – 6 –3 3

x xx =

\ F(a) + 8

23

= 6

Case IIF(a) is the area bounded by x = 0, x = 2, y2 = 4x andy = |x – 1| + |x – 2| + x

3 – ,0 11,1 2x x

x x£ <ì

= í + £ £î

\ F(a) 1 20 1

(3 2 ) ( 1 – 2 )x x dx x x dx= - - + +ò òy

x

x = 2

1

1 22 2

0 1

4 43 – – –2 3 2 3x x xx x x x x

æ ö æ ö= + +ç ÷ ç ÷ç ÷ ç ÷

è ø è ø

1 4 8 2 1 43 2 2 – – –1

2 3 3 2 3= - - + + +

8 25 –

3=

F(a) + 8 2

3 = 5

JEE - Advance Mathematics 2015 paper 1.pdf

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