Jee Main Maths Solutions 2015

Head Office : Pushpanjali Place, Boring Road Crossing, Opp. Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri Sai Yamaha, Bailey Road, Patna- 14

Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on -www.iitianstapasya.com-

[ 1 ]IITians TAPASYA...IITians creating IITians

SOLUTION OF JEE MAIN. 2015PART - A [MATHEMATICS]

Q.1. A complex number z is said to be unimoldular if |z| = 1. Suppose z1 and z2 are complex number

such that 1 21 2

z 2z2 z z

is unimodular and z2 is not unimodular. Then the point z1 lies on a :

(1) circle of radius 2 (2) straight line parallel to x-axis

(3) straight line parallel to y-axis (4) circle of radius 2

Sol. (4)2

21 2

21

Z 2Z 12 Z Z

2 2 2 22 21 2 1 1 2 12 4 Z 4ReZ Z 4 Z Z 4ReZ Z

2 2 2 22 2 1 2Z 4 Z 4 Z Z 0

2 21 1Z 4 Z 1 0 |Z1| = 2 & |Z2| = 1 but it is given |Z2| 1

Q. 2. The normal to the curve, x2 + 2xy 3y2 = 0, at (1, 1) :

(1) meets the curve again in the fourth quadrant

(2) does not meet the cruve again

(3) meets the curve again in the second

(4) meets the curve again in the third quadrant

IITians TAPASYA...IITians creating IITians



[ 2 ]IITians TAPASYA...IITians creating IITiansSol. (1) Slope of normal -dx/dy = 1

Eq. of normal (y 1) = 1(x 1) x + y = 2 ...........(i)Now obtain inter section of normal with curve

(x2 y2) + 2 xy 2y2 = 0(x2 y2) + 2y (x y) = 0(x y) (x + y) + 2y (x y) = 0

2(x y) (y + 1) = 0 y = 1, (y = x Not possible)

Now put y = 1 in (i) x = 3.

Q.3. The sum of first 9 terms of the series3 3 3 3 2 31 1 2 1 2 3 .....1 1 3 1 3 5

is :

(1) 192 (2) 71 (3) 96 (4) 142

Sol. (3)3 3 3 3 3 31 1 2 1 2 3 ......1 1 3 1 2 3

Now write general term

Tr =

22

r r 12r

Tr = 21 r 1 2r4 Now using concept of sigma notation

9 9 2r 1 r 1

1Tr r 2r 14

r 1)(2r 111 r(r 1) 94 6 now put r = 9

1 90x19 90 94 6 96

Q.4 Let f (x) be a polynomial of degree four having extreme values at x = 1 and x = 2. If

2x 0

f(x)lim 1 3x

, then f(2) is equal to :(1) 4 (2) 8 (3) 4 (4) 0



[ 3 ]IITians TAPASYA...IITians creating IITiansSol. (4) Given 2

x 0

f(x)1 3xlim

2

2x 0

x f(x) 3xlim

f(o) 0f '(o) 0

But f "(o) 6

It is given that at x = 1 & x = 2 exist extima of function

f(1) = f (2) = 0from above in formation we can write above function in the form

f(x) = a (x) (x 1) (x 2) (x )where a & is arbitrary constant f (2) = 0

Q.5 The negatiion of s ( r s) is equivalent to :(1) s r (2) s r (3) s (r s) (4) s (r s)

Sol. (1)

Make truth table

s r r s r s s r s s r sT T F F F F TT F T F T T FF T F T F T FF F T T F T FNow check the options

s r T F F Fs r F T F F

s r s F F F F s r s T T T T



[ 4 ]IITians TAPASYA...IITians creating IITians

Q.6 If1 2 2

A 2 1 2a 2 b

is a matrix satisfying the equation AAT = 9I, where I 3 x 3 identify matrix, then

the ordered pair (a,b) is equal to :(1) (2, 1) (2) (2, 1) (3) ( 2, 1) (4) (2, 1)

Sol. (1) We know that Det (A) = Det AT

T2

Det AA 729

| A | 729 | A | 27

b 4 2 2b 4 a 6 27 6 a 3 b 1 5

Now check the ordered pair.

Q.7 The integral 2 4 3 / 4dx

x (x 1) equals :

(1)

14 4

4x 1 c

x (2)

14 4

4x 1 c

x (3)

14 4x 1 c (4) 14 4x 1 c

Sol. (1) Given 3 / 45

4

dx1x 1x

Let 1 + 1/x4 = t54x dx dt

1/ 4

3 / 4 4dt 11 c

4t x

Q.8 If m is the A.M. of two distinct real number I and n (I, n > 1) and G1, G2 and G3 are three geometricalmeans between, I and n, then 4 4 41 2 3G 2G G equals.(1) 4 I2m2n2 (2) 4 I2mn (3) 4 Im2n (4) 4 Imn2

Sol. (3)l nm .......(1)

2

given that G1, G2, G3 are three geometrical means between l & n1/ 4 1/ 2 3 / 4

1 2 3n n nG l , G l , Gl l l

Now put the values 4 4 41 2 3G 2G G , Then we get



[ 5 ]IITians TAPASYA...IITians creating IITians 2ln l n now put the value l + n 24lnm

Q.9 Let y(x) be the solution of the differential equation dyx logx y 2x logx, x 1dx . Then y(e) isequal to :(1) 2e (2) e (3) 0 (4) 2

Sol. (4) Dividing both side by (x log x)

dy y 2dx xlogx

I.F = lnx

Solution is y.lnx 2.lnxdx c

y.lnx 2x lnx 2x c

Put x = 1 c 2 ,

Hence equation is 2xy lnx 2x lnx 2x 2 y 2x 2lnx

(But it is not define at x = 1)

If i loosely speaking ( x 1 ) y e 2 (In question x > 1)Q.10 The number of integer greater than 6,000 that can be formed, using the digits, 3,5,6,7 and 8

without repetition is :(1) 72 (2) 216 (3) 192 (4) 120

Sol. (3)

3 x 4 x 3 x 2 = 72 3 x 4 x 3 x 2 x 1 = 120

72 + 120 = 192Q.11 The number of points, having both co-ordinates as integers, that lie in the interior of the triangle

with vertices (0, 0) (0, 41) and (41, 0) is :(1) 780 (2) 901 (3) 861 (4) 820

Sol. (1) (0,41) (41,41)

(41,0)(0,0)

Total vertices in the square = 40 x 40 = 1600

On the line = 40 (1,40)(2,39).....(40,1)



[ 6 ]IITians TAPASYA...IITians creating IITiansIn triangle 1600 40 = 1560Hence in triangle 1560/2 = 780. (Repeat question in IIT JEE 2003) (0, 21) (21, 0)

Q.12 Let and be the roots of equation x2 6x 2 = 0. If n nna , for n 1 , then the value of10 8

9

a 2a2a

is equal to :

(1) 3 (2) 6 (3) 6 (4) 3

Sol. (4) Given equation x2 6x 2 = 0

n nna

10 10 8 810 8

9 99

2a 2a2a 2

8 2 8 2

9 9

2 2

2

2 26 2, 6 2

Now from above expression

9 9

9 9

63

2

Q.13 Let 1 1 1 22xtan y tan x tan

1 x , where

1| x |3

. Then a value of y is :

(1)3

23x x1 3x

(2)

3

23x x1 3x

(3)

3

23x x1 3x

(4)

3

23x x1 3x

Sol. (2) Let 1tan x 1tan y 2

y tan3

3

23 tan tany

1 3 tan

3

23x xy1 3x

Q.14 The distance of the point (1, 0, 2) from the point of intersection of the linex 2 y 1 z 2

3 4 12

and the plane x y + z = 16, is :



[ 7 ]IITians TAPASYA...IITians creating IITians(1) 13 (2) 2 14 (3) 8 (4) 3 21

Sol. (1) Let x 2 y 1 z 2

3 4 12

Now x, y, z lies in the plane x y + z = 16

3 2 4 1 12 2 16 1

Hence point in the plane is (5, 3, 14)

Hence distance is 2 225 1 3 14 2 13 Q.15 The area (in sq. units) of the region described by 2{x,y : y 2x and y 4x 1} is

(1)932 (2)

732 (3)

564 (4)

1564

Sol. (1)

A

B

1 1,8 2

1,12

Area of shaded region1 2

12

y y 1 dy2 4

=932

Q.16 Let O be the vertex and Q be any point on the parabola, x2 = 8y. If the point P divides the linesegment OQ internally in the ratio 1 : 3, then the locus of P is :

(1) x2 = 2y (2) x2 = y (3) y2 = x (4) y2 = 2x

Sol. (1)

O

P (h,k)

Q( )

1

3

4h, 4k



[ 8 ]IITians TAPASYA...IITians creating IITiansNow put and in curve.We get x2 = 2y

Q.17 The mean of the data set comprising of 16 observation is 16. If one of the observation valued 16is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean ofthe resultant data, is :

(1) 14.0 (2) 16.0 (3) 16.0 (4) 15.8

Sol. (1)16

ii 1

x 256

Let x16 = 16

1 2 3 15x x x ......x 240 If new observation are added then new mean is 252/18 = 14

Q.18 The area (in sq. units) of the quadrilateral formed by the tangents at teh end points of the recta to

the ellipse2 2x y 1

9 5 , is :

(1) 27 (2)274 (3) 18 (4)

272

Sol. (1)

A

y

xP

(2,5/3)

B

C

DF1 F2

Now equation of tangent at P isx y 1

9 / 2 3

Area of ABCD = 4 x area of OAB 4 x 1/2 x 9/2 x 3 = 27

Q.19 The equation of the plane containing the line 2x 5y + z = 3 ; x + y + 4z = 5, and parallel to theplane, x + 3y + 6z = 1, is :

(1) 2x + 6y + 12z = 13 (2) 2x + 6y + 12z = 13

(3) x + 3y + 6z = 7 (4) x + 3y + 6z = 7



[ 9 ]IITians TAPASYA...IITians creating IITiansSol. (4) Equation of plane is

2x 5y z 3 x y 4z 5 0

x 2 y 5 z 1 4 3 5

Because plane is parallel to plane x + 3y + 6z = 1.

2 5 1 41 3 6

112

Hence required equation of plane is x + 3y + 6z = 7

Q.20 The number of common tangents to the circles x2 + y2 4x 6y 12 = 0 and x2 + y2+ 6x + 18y + 26= 0, is :

(1) 4 (2) 1 (3) 2 (4) 3

Sol. (4)

2 21 1 1

2 22 2 2

S : x y 4x 6y 12 0 C 2,3 r 5

S : x y 6x 18y 26 0 C 3, 9 r 8

1 2 1 2C C r r Hence circles touch externly. so number of common tangent is 3.Q.21 The set of all values of for which the system of linear equations :

1 2 3 12x 2x x x , 1 2 3 22x 3x 2x x , 1 2 3x 2x x has a non-trivial solution,(1) contains more than two elements (2) is an empty set

(3) is a singleton (4) contains two elements

Sol. (4)

1 2 3

1 2 3

1 2 3

x 2 2x x 0

2x x 3 2x 0x 2x x 0

For non trivial solution2 2 1

2 (3 ) 2 01 2

1, 3

Q.22 If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxescontains exactly 3 balls is :

(1)11122

3 (2)

1155 23 3

(3)1055 2

3 3 (4)

1212203

Sol. (None) No option is correct.



[ 10 ]IITians TAPASYA...IITians creating IITiansQ.23 The sum of coefficients of intgegral power of x in the binomial expansion of 501 2 x is

(1) 501 2 12 (2) 501 3 12 (3) 501 32 (4) 501 3 12 Sol. (2) 50 2 500 1 2 501 2 x C C 2 x C 2 x .....C 2 x .........(1)

50 2 500 1 2 501 2 x C C 2 x C 2 x .....C 2 x .........(2) Now add (1) & (2) and put x = 1.

502 50 50

0 23 1 C 2 C .......2 C

2

Q.24 The integral 4 2

2 22

logx dxlogx log 36 12x x is equal to :

(1) 6 (2) 2 (3) 4 (4) 1

Sol. (4) 4

2

logxIlog x 6 x

......(1)Now using properties

a a

b b4

2

f(x)dx f(a b x)dx

log 6 xI

logx log 6 x

......(2)

add (1) and (2) Hence I = 1.

Q.25 If the function k x 1 , 0 x 3g xmx 2 , 3 x 5

is differentianble, then the value of k + m is :

(1) 4 (2) 2 (3) 16/5 (4) 10/3

Sol. (2) For function to be differentiable it must be continuous

L.H.L R.H.L 2k 3m 2 ....(1) For function to be differentiable L.H.D = R.H.D k 4m m k 2

Q.26 Locus of the image of the point (2,3) in the line (2x 3y + 4) + k (x 2y + 3) = 0,k R , is a(1) circle of radius 3 (2) straight line parallel to x-axis

(3) straight line parallel to y-axis (4) circle of radius 2



[ 11 ]IITians TAPASYA...IITians creating IITiansSol. (4)

O(1, 2)

(2, 3)

(h,k)

Am h+2/2, k+3/2

B

2x-3y+4=0

x-2y+3=0

Now product of slope of AB and OM is 1.

k 3 k 11

h 2 h

2 2h k 2h 4k 3 0 Hence of locus of circle with radius 2 .

Q.27

x 0

1 cos2x 3 cos xlim

x tan4x

is equal to :

(1) 1/2 (2) 4 (3) 3 (4) 2

Sol. (4)

x 0

1 cos2x 3 cos xlim

x tan4x

x 0

2

x 0

1 cos2x 3 cos xlim

x tan4x2sin x 3 cos x

lim 2tan4x4.x.x4x

Q.28 If the angle of elevation of the top of a tower from the three collinear points A, B and C, on a lineleading to the foot of the tower are 30o, 45o and 60o respectively, then the ratio, AB : BC, is :

(1) 2 : 3 (2) 3 :1 (3) 3 : 2 (4) 1: 3

Sol. (2)

ABy

C

D

h

O x 30o45o60

o

Now in triangle ODC

h 3x

In triangle ODB



[ 12 ]IITians TAPASYA...IITians creating IITiansx h In trinagle OAB

x y 3h

AB 3 1 h 3 1 h

BC3

AB 3BC 1

Q.29 Let A and B be two sets containing four and two elements respectvely. Then the number of subsetsof the set A x B, each having at least three elements is :(1) 510 (2) 219 (3) 256 (4) 275

Sol. (2) Total number of subset of (A x B) = 28 = 256

Subset with 0 term = 1 (Null set)

Subset with 1 term = 8

Subset with 2 term = 8C2 = 28

Number of subsets having at least 3 elements = 256 1 8 28 = 219

Q.30 Let a, b and c

be three non-zero vectors such that no two of them are collinear and

1a b c | b || c | a3 . If is the angle between vectors b and c , then a value of sin is :(1) 2 3

3 (2) 2 2

3(3) 2

3 (4) 23

Sol. (2) Given 1a b c | b || c | a3

1c a b | b || c || a |3

| b || c || b || c | cos a c.a b 03

Since a and b are non collinear.Hence their coefficient are zero.

1 2 2cos sin3 3 .

Date post:	26-Sep-2015
Category:	Documents
Upload:	akash-goyal
View:	25 times
Download:	5 times

Jee Main Maths Solutions 2015

Documents