Class : JEE Main 2015 Subject : Mathematics Solution Paper Code : D ur ur
r
31. Let a, b and c be three non­zero vectors such Solution: (3)
that no two of t hem are collin ear and AQ = BQ = CQ 3 ur ur
r 1 ur r ur
3 (a ´ b) ´ c = b c a . If q is the angle between 2 ur
r
\ AQ =3CQ, AC = 2CQ vectors b and c , then a value of sin q is P
2 1. 3 -2 3 2. 3
2 2 3. 3 - 2 4. 3
BC =
(
)
3 - 1 CQ Q 60° C 30° B A Solution: (3)
AB = (3 - 3) CQ = 3( 3 - 1) CQ
r r r
r r r 1 r r r
a.c b - b.c a =
b || c | a 3 ( ) ( )
(
)
1
3
comparing cos q = ­ Þ sin q =
3( 3 - 1) CQ
3
\ AB : BC = ( 3 - 1) CQ = 1 2 2 3
34. The number of points, having both co­ordinates as integers, that lie in the interior of the 32. Let O be the vertex and Q be any point on triangle with vertices (0,0), (0, 41) and (41,0), 2 parabola, x =8y. if the point P divides the line is: segment OQ internally in the ratio 1:3, then the locus of P is : 1. 820 2. 780 1. y 2 = 2x 2. x 2 = 2y 3. x 2 = y 4. y 2 = x 3. 901 4. 861 Solution: (2) Solution: (2) number of points = 1+2+3+ ........ + 39 4 t = t 4
2 t 2
t 2 y =
=
4
2
x =
y x2 =8y Q (4t,2t2 ) 3 =
39 ´ 40 = 780 2 (x,y) P 1 O x (0,41) so, x 2 = 2y is locus of p. 33. If the angles of elevation of the top of a tower from three collinear points, A,B, and C, on a line leading to the foot of the tower, are 30°, 45° and 60° respectively, then the ratio, AB : BC, is : 1 (0,0) 1. 1: 3 2. 2:3 3. 4. 3 :1 3 : 2 (41,0) 35. the equation of the plane containing the line 2x ­ 5y + z = 3; x + y + 4z = 5, and parallel to the plane, x + 3y + 6z = 1 is :
E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com 1. x + 3y + 6z = 7 2. 2x + 6y + 12z = ­13 (1 - cos2x)(3 + cos x ) 38. lim is equal to: x ®0 x tan 4x 3. 2x + 6y + 12z = 13 4. x + 3y + 6z = ­7 Solution: (1) equation of required plane is 1 2 1. 2 2. 3. 4 4. 3 Solution: (1) 2x ­ 5y + z ­ 3 + l (x + y +4z ­ 5) = 0
Þ (2 + l )x+( l ­5)y+(1­4 l )z ­3 ­5 l = 0.......(1) (1 - cos2x) (3 + cos x) ´
tan4 x x 2 x 4 4 = ´ = 2 2 2
lim x ®0 Since, Required plane is parallel to x +3y+6z =1, 2 + l l - 5 1 - 4l 3 + 5 l
=
=
=
So , . 1
3
6
1
Solving any two pair, we get l = 11/2. Putting value of l in equation (1), we get 39. The distance of the point (1,0,2) from the point x - 2 y + 1 z - 2 =
=
3
4
12 and the plane x ­ y + z = 16 is: of intersection of the line 1. 3 21 x + 3y + 6z = 7 2. 13 36. Let A and B be two sets containing four and 3. 2 14 two elements respectively. Then the number of subsets of the set A x B, each having at Solution: (2) least three elements is: 1. 275 2. 510 3. 219 4. 256 Solution: (3) 4. 8 Equation of line is x - 2 y + 1 z - 2 =
=
= l(say) 3
4
12
Þ x = 3 l + 2, y = 4 l ­ 1, z = 12 l + 2 Putting these values in plane x ­ y + z = 16, number of subsets = 2 8 ­{c(8,0)+c(8,1)+c(8,2)} = 256 ­ (1+8+28) we get , l = 1. So,point of intersection (5,3,14) = 219. 37. Locus of the image of the point (2,3) in the line (2x ­ 3y + 4) + k ( x ­ 2y + 3) = 0, distance =
42 + 32 + 12 2 , = 13 40. The sum of coefficients of integral powers of x k e R, is a: (
in the bionomial expansion of 1 - 2 x
1. circle of radius 2 . 2. circle of radius 3 . 1.
1 50 3 - 1 is 2
)
2.
1 50 2 + 1 2
3.
1 50 3 + 1 2
4.
1 50 3 2 (
(
50 )
is )
3. straight line parallel to x­axis. 4. straight line parallel to y­axis. Solution: (1) (
)
( )
Solution: (3) L1 + l L 2 =0 L1 º 2x­3y+y=0 P(2,3) Now, Q(h,k) Ö2 0(1,2) Let x = y. . L2 º x­2y+3=0 (1 - 2 y)50 = a0 + a1y + a2 y2 + ....a50 y50 ......(1) Putting y = 1, in (1) we get Clearly, the locus of Q is a circle. 2 1 = a0 + a1 + a2 + ....a50 .....(2)
E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com putting y = ­1,in (1) we get, 350 = a0 - a1 + a2 - a3 .... + a50 .....(3) Adding (2) and (3), we get 2x 1 ­2x 2 +x 3 =lx 1 2x 1 ­3x 2 +2x 3 =lx 1 350 + 1 = 2( a0 + a2 + .... + a50 )
Þ ( a0 + a2 + .... + a50 ) =
43. The set of all values of l for which the system of linear equations: 350 + 1 2
41. the sum of first 9 terms of the series ­x 1 +2x 3 =lx 3 has a non­trivial solution, 1. contains two elements 2. contains more than two elements. 3. is an empty set 13 13 + 23 13 + 23 + 3 3 +
+
+ ..... is: 1
1+3
1+3+5
1. 14 2. 192 3. 71 4. 96 4. is singleton. Solution : (1) (2 ­ l) x 1 ­ 2x 2 + x 3 =0 2x 1 ­ (3 + l) x 2 + 2x 3 =0 Solution: (4)
­x 1 + 2x 2 ­ lx 3 = 0 (
)
n2 + 2n + 1 (n + 1) 2 T n =
=
4
4 1 é n(n + 1)(2n + 1
(n )(n+ 1) ù
sn = å Tn = ê
+ 2 + n ú
9ë
6
2 û
n =
(2n2 + 9n + 13) 24 9 \ s q =
´ (2 ´ 81 + 81 + 13) = 96 24 42. The area (in sq. units) of the region described by {(x,y) : y 2 < 2x and y > 4x­1} is: 1. 15 64 2. 9
32 3. 7
32 4. 5
64 Solution: (2) 2-l
-2
2
-(3 + l )
1 -1
2
2 = 0 -l
Þ ( l ­1) 2 ( l +3) = 0
Þ l = 1 and l = ­3. So, set contains two elements. 44. a complex number z is said to be unimodular if |z|=1. Suppose z 1 and z 2 are complex number such that z1 - 2z 2 2 - z1 z2 is unimodular and z 2 is not unimodular. Then the point z 1 lies on a : 1. circle of radius 2. 2. circle of radius 2 . 3. straight line parallel to x­axis. 4. straight line parallel to y­axis. Solution : (1) y=4x­1 2 y=1 y =2x O z1 - 2z 2 2 - z1 z2 = 1 Þ | z - 2z |2 = | 2 - z z |2 1
2
1 2 (
) (
Þ ( z1 - 2z2 ) z1 - 2z2 = 2 - z1 z2
) (2 - z z )
1
2 y=­½ 2
2 Þ (| z1 | -4 ) (1- | z2 | ) = 0
Þ | z1 |2 = 4 , Q z 2 is not unimodular
y =1 æ y + 1 y 2 ö
Area = ò ç 4 - 2 ÷ dy = 9/32 ø
y =-1/2 è
3 Þ |z 1 | = 2 which lies on a circle of radius 2.
E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com 45. The number of common tangents to the circles x 2 + y 2 ­ 4x ­ 6y ­ 12 = 0 and x 2 + y 2 + 6x + 18y + 26 = 0 is : 1. 3 2. 4 3. 1 4. 2 Solution : (1) é1 2 2 ù
ê
ú
2 1 -1 ú
48. If A = ê
is a martix satisfying the êë a 2 b úû
equation AA T =91, where I is 3 x 3 identity matrix, then the ordered pair (a,b) is equal to: 1. (2,1) 2. (­2, ­1) 3. (2,­1) 4. (­2,1) Solution:(2) 8 5 C1 (2,3) C2 (­3,­9) é1 2 2 ù é1 2 9 ù
é9 0 0 ù
ê
ú ê
ú
ê
ú
ê2 1 -2ú ê2 1 2 ú = ê0 9 0 ú
êë9 2 6 úû êë2 -2 6 úû
êë0 0 9 úû
Equating the corresponding elements, we get, a +2b = ­4 and a ­ b = ­1. x 2 + y 2 ­ 4x ­ 6y ­ 12 = 0 C 1 (2,3), r 1 = 4 + 9 + 12 = 5 x 2 + y 2 + 6x + 18y + 26=0 C 2 =(­3,­9), r 2 = 9 + 81 - 26 = 8
C1C 2 =
25 + 144 = 13 Solving both equations,we get a = ­2 and b = ­1. 49. If m is the A.M. of two distinct real numbers l and n (l,n>1) and G 1 , G 2 and G 3 are three geomertic means between l and n, then 4 equals. G 14 + 2 G 24 + G 3 1. 4 lmn 2 2. 4 l 2 m 2 n 2 3. 4 l 2 mn 4. 4 lm 2 n 46. The number of integers greater than 6,000 that can be formed, using the digits, 3,5,6, 7 and 8, Solution: (4) without repetition, is: l + n = 2m. 1. 120 2. 72 l,G 1 ,G 2 ,G 3 ,n are in G.P.
3. 216 4. 192 1/4 Solution: (4) Total number of integers = 3(24) + 5! = 72 + 120 = 192. 47. Let y(x) be the solution of the differential equation dy (x logx) +y = 2x logx, (x >1). dx æ n ö
Þ common ratio = r = ç ÷
èlø
Þ G14 = l3n , G24 = l2n2 , G3 4 = ln3 Þ G14 + 2 G24 + G3 4 = l3n + 2 l2n2 + ln3 = ln(l+ n)2 = ln(4 m2 ) = 4 lm2 n . 50. the negation of : s Ù (: r Ú s ) is equivalent to: Then y(e) is equal to: 1. 2 2. 2e 3. e 4. 0 Solution: The question is not theoratically correct. 1. s Ù (r Ú : s) 2. s Ù r
3. s Ù : r
4. s Ù (r Ù : s) Solution:(2)
~ [~ s Ú (~ rÙ s)] = ~ ((~ s Ú ~ r) Ù (~ sÚ s))
= ~ (~ s Ú ~ r) = (sÙ r) .
4 E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com 51. The integral ò
dx equals: x ( x + 1) 3/4 2
1. 4
1 4 4 æ x + 1 ö
2. - ç
÷ + c 4 è x
ø
1 4 1. -(x 4 + 1) + c
3 x - x 3 1 + 3 x 2 2. 3 x - x 3 3. 1 - 3 x 2 3 x + x 3 1 + 3 x 2 3 x + x 3 4. 1 - 3 x 2 Solution : (3) 1 æ x 4 + 1 ö 4 3. ç
÷ + c 4 è x
ø
1 4. ( x 4 + 1) 4 + c
Solution:(2) Þ tan-1 y = tan-1 x + 2 tan-1 x
dx
=
x2 (x 4 + 1) 3/4 ò
=
ò
æ 2x ö
tan-1 y = tan-1 x + tan -1 ç
2 ÷
è1 - x ø
ò
dx 1 x (1 + 4 ) 3/4 x
Þ tan-1 y = 3 tan-1 x
5
(- t3 )dt ì
1
1 4
3 í put 1 + 4 = t Þ 5 dx = -t dt (t)3 x
x
î
1/4 æ x4 + 1 ö
÷
= ­ t + C = - ç
4 è x ø
+ C 52. The normal to the curve, x 2 +2xy­3y 2 =0, at (1,1): 3 -1
-1 æ 3x - x ö
tan
y
=
tan ç
Þ
2 ÷
è 1 - 3x ø
æ 3x - x 3 ö
y =
ç
Þ
2 ÷
è 1 - 3x ø
54. If the function. ìïk x + 1, 0 £
í
ïîmx + 2 , 3 <
x £ 3 x £ 5 1. meets the curve again in the third quadrant. is differentiable, then the value of k+m is: 2. meets the curve again in the fourth quadrant. 1. 10
3 2. 4 3. does not meet the curve again. 4. meets the curve again in the second quadrant. 3. 2 4. 16
5 Solution: (3) Solution:(2) Þ x= y or x + 3y = 0. ïìk x + 1, 0 £ x £ 3 g(x) = í
ïî mx + 2, 3 < x £ 5
(1,1) lies on the line x = y. g(x) is differentiable
Equation of normal to it is y ­ 1 = (­ 1)(x ­ 1)
Þ x + y = 2. æ
k ö
= m ÷
Þ g'(3- ) = g'(3+ ) Þ ç
è 2 x + 1 øx =3 This nomal meets the curve x + 3y = 0 at Þ k = 4m .............(1) (3,­1) which lies in 4th quadrant. Next, g(x) is continuous as it is differentiable. x 2 + 2xy ­ 3y 2 = 0 Þ (x ­ y)(x + 3y) = 0
53. Let So, g(3- ) = g(3+ ) Þ 2k = 3m+2 .........(2) æ 2 x ö
tan-1 y = tan-1 x + tan -1 ç
2 ÷
è 1 - x ø
From (1) and (2), we get m = 2/5 Þ k + m = 2
where |x| < 5 1 . Then a value to y is: 3 E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com 55. The mean of the data set comprising of 16 (a9 - b9 )(a + b) (a + b) observations is 16. if one of the observation = = = 3. 2(a9 - b9 )
2
valued 16 is deleted and three new observations valued 3,4 and 5 are added to the data, then 58. Let f(x) be polynomial of degree four having the mean of the resultant data, is exterme values at x=1 and x=2. 1. 15.8 2. 14.0 f ( x ) ù
é
3. 16.8 4. 16.0 1 + 2 ú =3, then f(2) is equal to : If lim
x ® 0 ê
x û
ë
Solution:(2) 1. 0 2. 4 Sum = S = 16 ´ 16 = 256. 3. ­8 4. ­4 S - 16 + 3 + 4 + 5 256 - 16 + 12 Mean = = = 14. Solution: (1) 18
18
Let f(x) = x 2 (ax 2 +bx+c). 56. The integral 4 log x 2 ò2 log x 2 + log(36 - 12x + x 2 ) dx is euql to: 1. 1 2. 6 3. 2 4. 4 Solution: (1) c + 1 = 3 Þ c = 2. Again,f(x) = x 2 (ax 2 +bx+c) Þ f '(x) = x2 (2 ax+ b) + 2 x(ax2 + bx+ c)
Þ f '(x) = 4a x3 + 3bx2 + 2cx
4 I=
f(x) ù
é
lim ê1 + 2 ú = 3 Þ
x ® 0 x û
ë
log x 2 ò2 log x2 + log(36 - 12 x+ x2 ) dx We have, f '(1) = 0 Þ 4a + 3b + 4 = 0 .......(1) and f '(2) = 0 Þ 32a + 12b + 8 = 0 .........(2) 4 2 log x dx 2 2 log x + log(6 - x)
= ò
2
4 log(6 - x) 2 dx = ò
2
2 2 log(6 - x) + logx
4 Þ 2I = é
ë x ùû Þ I = 1. 2 57. Let a and b be the roots of equation x 2 ­ 6x ­ 2 = 0. If an = a n - b n , for n > 1, then a10 - 2 a 8 the value of is equal to : 2 a9 1. 3 2. ­3 3. 6 4. ­6 From (1) and (2) we have, a = 1/2 and b = ­2
\ f(2) = 0. 59. The area (in sq. units) of the quardrilateral formed by the tangents at the end points of the latera recta to the ellipse x 2 y 2 +
=1 is 9
5 1. 27
2 2. 27 3. 27
4 4. 18 Solution: (2) Solution: (1) B (0,a) 2 (ae, b ) a x 2 ­ 6x ­ 2 = 0 Þ a + b = 6 , ab = -2
a10 - 2a 8 (a10 - b10 ) - 2(a8 - b8 ) =
Now, 2a 9 2(a9 - b9 )
C (­a/e,0) A(a/e,0 o (ae,0)
D(0,­a) (a10 - b10 ) + ab(a8 - b8 ) a 9 (a + b) - b9 (a + b) = Þ
2(a 9 - b9 )
2(a9 - b9 )
6 E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com æ
b 2 ö
ae, ÷
Equation of tangent at ç
, a ø is given by, è
60. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is: 12 ex y aex b2 y + = 1 +
= 1 Þ
2
2 a
a
a
ab
The tangent meets the axes at A(a/e,0) and B(0,a). Similarly we can find out C(­a/e,0) and 11 55 æ 2 ö
3. ç ÷
3 è3ø
11 æ 1 ö
2. 22 ç ÷
è3ø
10 æ 2 ö
4. 55 ç ÷
è3ø
Solution: Theoretically the question is wrong.
D(0,­a). Area of quadrilateral = æ 1 ö
1. 220 ç ÷
è3ø
1
2a 2a 2 ´ 2a ´
= 2
e
e ìQ 9(1 - e2 ) = 5 ï
5 4 ï
2 (2)(9) íÞ e = 1 - = Þ e = 2 / 3 9 9
= = 27 ï
(2 / 3) ïî
7 E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com