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04­ April­2015 JEE MAIN ­ 2015 PART B ­ MATHEMATICS 31. The sum of coefficients of integral powers of x in the binomial expansion of (1 – 2Öx) 50 is (1) 1 50 (3 ) 2 1 50 (3 – 1) 2 (2) (3) 1 50 (2 + 1) 2 (4) 1 50 (3 + 1) 2 31. (4) (1 – 2Öx) 50 T r + 1 = (– 1) r 50 C r (2Öx) r = (– 1) r 50 C r 2 r x r / 2 r = 0, 2, 4, ......., 50 Sum of coefficients = 50 C 0 2 0 + 50 C 2 2 2 + 50 C 4 2 4 + .......+ 50 C 5 2 50 =
( 1 + x ) 50 + ( 1 - x ) 50 , where x = 2 2 =
3 50 + 1 . 2 32. Let f (x) be a polynomial of degree four having extreme values at x = 1 and x = 2. If é f ( x ) ù
Lim ê1 + 2 ú = 3 , then f (2) is equal to x ® 0 ë
x û
(1) – 4 (2) 0 (3) 4 (4) – 8 æ f ( x ) ö
32. (2) Let f (x) = ax 4 + bx 3 + cx 2 Q Lim ç1 + 2 ÷ = 3 Þ
x ®0 è
x ø
3 2 f ' (x) = 4ax + 3bx + 2cx f ' (1) = 4a + 3b + 4 = 0 f ' (2) = 32a + 12b + 8 = 0 Þ a = 1 / 2, b = – 2
\ f (x) = (1 / 2) x 4 – 2x 3 + 2x 2
\ f (2) = 0. æ f ( x ) ö
Lim ç 2 ÷ = 2 Þ
x ® 0 è x ø
c = 2 33. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of resultant data, is (1) 16.0 (2) 15.8 (3) 14.0 (4) 16.8 18 16 33. (3)
å x i = 16 ´ 16 = 256 Sum of new observations =
i =1 å xi = 256 - 16 + ( 3 + 4 + 5 ) = 252 i =1 18 å x i \
JEE MAIN ­ 2015 Mean of the resultant data = i =1 = 252 / 18 = 14 . 0 . 18 (04­Apr­15) Question & Solutions ­ 1 ­ www. prernaclasses.com 34. The sum of first 9 tems of the series (1) 96 (2) 142 1 3 1 3 + 2 3 1 3 + 2 3 + 3 3 +
+
+ ...... is 1 1 + 3 1 + 3 + 5 (3) 192 (4) 71 2 34. (1)
æ r ( r + 1 ) ö
÷
r = 9 ç
9 1 è 2 ø = 1 ( r + 1 ) 2 1 2
= 2 + 3 2 + ...... + 10 2 = ( 1 2 + 2 2 + ...... + 10 2 ) - 1 å r 2 å
4 r =1 4 4 r =1 [
] [
]
= (1 / 4) × 384 = 96. 35. Let O be the vertex and Q be any point on the parabola x 2 = 8y. If the point P divides the line segment OQ internally in the ratio 1 : 3, then the locus of P is (1) y 2 = x (2) y 2 = 2x (3) x 2 = 2y (4) x 2 = y 35. (3)
3
P Q(2Ö2t, t2)
O (0, 0)
If P : (h, k) Þ h = 2Ö2t / 4, k = t 2 / 4 Eliminating t, h 2 = 2k Þ Locus : x 2 = 2y. 36. Let a and b be the roots of equation x 2 – 6x – 2 = 0. If a n = an – b n , for n ³ 1, then the value of a10 - 2 a 8 is equal to 2 a 9 (1) – 6 36. (2) (2) 3 (3) – 3 (4) 6 a 10 - 2 a 8 ( a10 - b10 ) - 2 ( a 8 - b8 ) a 8 ( a 2 - 2 ) - b8 ( b2 - 2 ) a 8 6 a - b8 6 b
=
=
=
= 3 2 a 9 2 ( a 9 - b9 ) 2 ( a 9 - b9 ) 2 ( a 9 - b9 ) 37. 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 (1) 55 (2 / 3) 10 (2) 220 (1 / 3) 12 (3) 22 (1 / 3) 11 (4) (55 / 3) (2 / 3) 11 12 37. (4) P =
C 3 ´ 29 3 12 38. A complex number z is said to be unimodular if | z | = 1. Suppose z 1 and z 2 are complex numbers z1 - 2 z 2 such that 2 - z z is unimodular and z 2 is not unimodular. Then the point z 1 lies on a 1 2 (1) straight line parallel to y ­axis (3) circle of radius Ö2 38. (2) Q Þ
JEE MAIN ­ 2015 (2) circle of radius 2 (4) straight line parallel to x­axis Þ ( z1 - 2 z 2 )( z 1 - 2 z 2 ) = ( 2 - 2 z 1 z 2 )( 2 - z 1 z 2 ) | z1 - 2 z 2 | 2 = | 2 - z 1 z 2 | 2 | z 1 | 2 (1 – | z 2 | 2 ) = 4 (1 – | z 2 | 2 ) Þ | z 1 | = 2.
(04­Apr­15) Question & Solutions ­ 2 ­ www. prernaclasses.com dx ó
39. The integral ô 2 4 equals õ x (x + 1 ) 3 / 4 (1) (x 4 + 1) 1 / 4 + c (2) – (x 4 + 1) 1 / 4 + c 1 / 4 (3) 39. (3)
æ x 4 + 1 ö
- ç 4 ÷
ç x ÷
è
ø
1 / 4 + c æ x 4 + 1 ö
(4) çç 4 ÷÷
è x ø
+ c dx ó
ô 5 - 4 3 / 4 õ x (x + 1 ) Let 1 + (1 / x 4 ) = t (– 4 / x 5 ) dx = dt (1 / x 5 ) dx = (– 1 / 4) dt
1 / 4 æ x 4 + 1 ö
3 / 4 4 1 / 4 I = (– 1 / 4) ò (dt / t ) = – (1 + (1 / x )) + c = çç 4 ÷÷
è x ø
Þ
+ c 40. The number of points, having both coordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0) is (1) 861 (2) 820 (3) 780 (4) 901 40. (3) x + y < 41 ; x > 0 y > 0 (1,1)(1,2)............(1,39)= 39 cases (2,1)(2,2)............(2,38)= 38 cases
|
|
(39,1) ......................... = 1 case
\
41. 39 ´ 40 required no. of cases = 1+2+3 ...............39 = = 780. 2 The distance of the point (1, 0, 2) from the point of intersection of the line x - 2 y + 1 z - 2 =
=
and the plane x – y + z = 16, is 3 4 12 (1) 8 (2) (3) 13 (4) 3 21 2 14 41. (3) Any point on the given line can be taken as (3l + 2, 4l – 1, 12l + 2) For point of intersection, point (3l + 2, 4l – 1, 12l + 2) must lie on x – y + z = 16
\ 3l + 2 – 4l + 1 + 12l + 2 = 16 Þ l = 1
\ Point º (5, 3, 14) Reqd. distance = ( 5 - 1 ) 2 + ( 3 - 0 ) 2 + ( 14 - 2 ) 2 = 13 . 42. 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 (1) x + 3y + 6z = – 7 (2) x + 3y + 6z = 7 (3) 2x + 6y + 12z = – 13 (4) 2x + 6y + 12z = 13
JEE MAIN ­ 2015 (04­Apr­15) Question & Solutions ­ 3 ­ www. prernaclasses.com 42. (2) For any point, the given line, z = 0
Þ 2x – 5y = 3 and x + y = 5 Solving x = 4, y = 1 \ The point on the line (4, 1, 0) Reqd. plane parallel to x + 3y + 6z = 1 \ D.R of normal to reqd. plane (1, 3, 6)
\ Eq. of reqd. plane : 1(x – 4) + 3(y – 1) + 6(z – 0) = 0
Þ x + 3y + 6z = 7. 43. The area (in sq. units) of the region described by {(x, y) : y 2 £ 2x and y ³ 4x – 1} is (1) 5 / 64 (2) 15 / 64 (3) 9 / 32 (4) 7 / 32 43. (3)
A 1
(1 / 4, 0)
-1 / 2
B
(0, - 1)
Point of intersection of line and parabola have y­coordinates – 1 / 2 and 1
1 \
ó æ y + 1 y 2 ö
÷dy = 9 / 32 . Reqd. area = ô çç
÷
ô 4 2 õè
ø
-1 / 2 44. 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 geometric means between l and n, then G 1 4 + 2G 2 4 + G 3 4 equals (1) 4 lm 2 n (2) 4 lmn 2 (3) 4 l 2 m 2 n 2 (4) 4 l 2 mn 44. (1) G 1 4 = l 3 . n, G 2 4 = l 2 n 2 , G 3 4 = ln 3 Also, 2m = l + n Now G 1 4 + 2G 2 4 + G 3 4 = ln (l 2 + 2ln + n 2 ) = ln (l + n) 2 = 4 lm 2 n. 45. Locus of the image of the point (2, 3) in the line (2x – 3y + 4) + k (x – 2y + 3) = 0, k Î R, is a (1) straight line parallel to y ­axis (2) circle of radius Ö2 (3) circle of radius Ö3 (4) straight line parallel to x­axis 45. (2) Point of intersection of family of lines are (1, 2). Now, equation of any line passing through (1, 2) is y – 2 = m(x – 1). Let the image be (a, b). Now, mid point lies on the line Þ b – 1 = – ma ........ (i) æ b-3 ö
÷m = -1 ......... (ii) Also, ç
è a - 2 ø
eliminating m, we get a2 + b 2 – 2a – 4b + 3 = 0 Locus is, x 2 + y 2 – 2x – 4y + 3 = 0, circle having radius Ö2. OR Locus is going to be a circle with centre at point of concurrency i.e. at (1, 2) and circle must pass through (2, 3). Hence radius = Ö2.
JEE MAIN ­ 2015 (04­Apr­15) Question & Solutions ­ 4 ­ www. prernaclasses.com 46. The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse (1) 18 46. (3) x 2 y 2 +
= 1 , is 9 5 (2) 27 / 2 (3) 27 (4) 27 / 4 B (0, 3) (ae, b 2 / a) (0, 0) A(9 / 2, 0) a 2 = 9, b 2 = 5 Now 5 = 9 (1 – e 2 )
Þ e = 2 / 3 ae = 2 and b 2 / a = 5 / 3 Now the equation of tangent at (2, 5 / 3) is (2x / 9) + (y / 3) = 1 Now, reqd. area = 4 × (1 / 2) × (9 / 2) × 3 = 27. 47. The number of integers greater than 6, 000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is (1) 192 (2) 120 (3) 72 (4) 216 47. (1) Total number of integers = Number of 4 digited integers + Number of 5 digited integers = 3 × 4 P 3 + 5 P 5 = 192. 48. Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A × B, each having at least three elements is (1) 256 (2) 275 (3) 510 (4) 219 48. (4) Number of subsets of sets A × B, each having at least three elements = 8 C 3 + 8 C 4 + ....... + 8 C 8 = 2 8 – 8 C 0 – 8 C 1 – 8 C 2 = 256 – 1 – 8 – 28 = 219. æ 2 x ö
49. Let tan -1 y = tan -1 x + tan -1 ç
÷ , where | x | < 1 / Ö3. Then a value of y is è 1 - x 2 ø
(1) 49. (4)
3 x + x 3 1 - 3 x 2 (2) 3 x - x 3 1 + 3 x 2 (3) 3 x + x 3 1 + 3 x 2 (4) 3 x - x 3 1 - 3 x 2 æ 3 x - x 3 ö
÷
tan -1 y = tan -1 x + 2 tan -1 x = 3 tan -1 x = tan -1 ç
ç 1 - 3 x 2 ÷ for | x | < 1 / Ö3
è
ø
Þ
JEE MAIN ­ 2015 æ 3 x - x 3 ö
÷
y = ç
ç 1 - 3 x 2 ÷ .
è
ø
(04­Apr­15) Question & Solutions ­ 5 ­ www. prernaclasses.com 4 ó
log x 2 dx is 50. The integral ô
2 2 õ log x + log( 36 - 12 x + x ) 2 (1) 4 (2) 1 (3) 6 (4) 2 4 50. (2)
4 ó log x 2 + log( 6 - x ) 2 2 I = ô
dx =
ò dx = 2 2 2 2 õ log x + log( 6 - x ) I = 1. Þ
2 51. The negation of ~ s Ú (~ r Ù s) is equivalent to (1) s Ù (r Ù ~ s) (2) s Ú (r Ú ~ s) (3) s Ù r (4) s Ù ~ r 51. (3) Negation of ~ s Ú (~ r Ù s) is ~ [~ s Ú (~ r Ù s)]
@ ~ (~ s) Ù ~ (~ r Ù s) @ s Ù (~ (~ r) Ú ~ s) @ s Ù (r Ú ~ s)
@ (s Ù r) Ú (s Ú ~ s) @ (s Ù r) Ú F [Q s Ù ~ s is fallacy]
@ s Ù r. 52. 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) Ö3 : Ö2 (2) 1 : Ö3 (3) 2 : 3 (4) Ö3 : 1 52. (4) E x + y A 60° 30° 45° z B y C x D From the figure, tan 30° = 1 / Ö3 = ED
x + y =
......... (1) x + y + z x + y + z Since BD = ED Þ
x + y + z = 3 Þ
x + y Again tan 60° = Ö3 = x + y x Þ
3x
=
z 3 + 1 Þ
2 x y 3 + 1 Now + =
from (2)
z z 2 Þ z : y = Ö3 : 1
JEE MAIN ­ 2015 1
3 ( 3 - 1 ) Þ
3 + 1 .......... (2) 2 x + y = Ö3x ......... (3)
Þ
x =
z x + y =
z y
=
z (04­Apr­15) Question & Solutions ......... (4) 3 + 1 1 = 1 / 3 2 3 ( 3 - 1 ) ­ 6 ­ www. prernaclasses.com ( 1 - cos 2 x )( 3 + cos x ) is equal to x ® 0
x tan 4 x (1) 3 (2) 2 53. Lim (3) 1 / 2 (4) 4 æ 2 sin 2 x 1 - cos 2 x x 4 ö÷
53. (2) Limit = Lim ( 3 + cos x ) Lim = 4 Lim ç
´
. = 2 . 2 ç
x ® 0
x ® 0 x tan 4 x x ® 0 tan 4 x 4 ÷ø
è x r r r 54. Let a , b and c be three non ­ zero vectors such that no two of them are collinear and r r r 1 r r r
r r (a ´ b ) ´ c = | b || c | a . If q is the angle between vectors b and c , then a value of sin q is 3 (1) – Ö2 / 3 (2) 2 / 3 (3) – 2Ö3 / 3 (4) 2Ö2 / 3 r
v r r v r r 1 54. (4) Here (c . a ) b - ( c . b ) a = b c a 3 v r r r væ
1 ö r
( c . a ) b = b c ç cos q + ÷ a 3 ø
è
r r But b and a are non­ collinear Þ cos q +
1 1 2 2 = 0 Þ cos q = Þ sin q =
3 3 3 é 1 2 2 ù
ê
ú
55. If A = ê2 1 - 2 ú is a matrix satisfying the equation AA T = 9I, where I is 3 × 3 identity matrix, êëa 2 b úû
then the ordered pair (a, b) is equal to (1) (– 2, 1) (2) (2, 1) (3) (– 2, – 1) (4) (2, – 1) 55. (3) Since AA T = 9I Þ a + 4 + 2b = 0 for I st row, III rd column and 2a + 2 – 2b = 0 for 2 nd row, 3 rd column
\ a = – 2, b = – 1. ìk x + 1 , 0 £ x £ 3 56. If the function g ( x ) = í
is differentiable, then the value of k + m is î mx + 2 , 3 < x £ 5 (1) 16 / 5 (2) 10 / 3 56. (4) For continuity at x = 3, 3m + 2 = 2k For differentiability at x = 3, k = 4m
(3) 4 \
(4) 2 k = 8 / 5, m = 2 / 5
Þ
k + m = 2. 57. The set of all values of l for which the system of linear equations: 2x 1 – 2x 2 + x 3 = lx 1 2x 1 – 3x 2 + 2x 3 = lx 2 – x 1 + 2x 2 = lx 3 has a non­trivial solution (1) is a singleton (2) contains two elements (3) contains more than two elements (4) is an empty set
JEE MAIN ­ 2015 (04­Apr­15) Question & Solutions ­ 7 ­ www. prernaclasses.com 57. (2) Making coefficient determinant zero, we get a cubic l3 + l2 – 5l + 3 = 0
\ Two distinct roots. 58. The normal to the curve, x 2 + 2xy – 3y 2 = 0, at (1, 1) (1) meets the curve again in the second quadrant. (2) meets the curve again in the third quadrant. (3) meets the curve again in the fourth quadrant. (4) does not meet the curve again. 58. (3) Q x 2 + 2xy – 3y 2 = 0 ........ (i) dy
x + y dy dy =
– 6y . = 0 Þ
dx 3 y - x dx dx \ Slope of normal at (1, 1) = – 2 / 2 = – 1 Equation of normal : y – 1 = – 1(x – 1)
Þ x + y = 2 .......... (ii) From (i) and (ii); x 2 + 2x (2 – x) – 3 (2 – x) 2 = 0 Þ x 2 – 4x + 3 = 0
Þ (x – 1) (x – 3) = 0
Þ x = 1, 3 and y = 1, – 1. Diff. w.r.t. x
Þ
2x + 2y + 2x 59. 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) 2 (2) 3 (3) 4 (4) 1 59. (2) Q (x – 2) 2 + (y – 3) 2 = 5 2 ; C 1 º (2, 3), r 1 = 5 and (x + 3) 2 + (y + 9) 2 = 8 2 ; C 2 º (– 3, – 9), r 2 = 8
\
C 1 C 2 = 25 + 144 = 13 , r 1 + r 2 = 13
C 1 C 2 = r 1 + r 2 . Þ
60. Let y(x) be the solution of the differential equation (x log x) is equal to (1) 0 60. (2) Q (2) 2 (3) 2e dy + y = 2x log x, (x ³ 1). Then y(e) dx (4) e dy 1
+
. y = 2 ; x ³ 1
dx x log x ó 1
dx ô
õ x log x = e log(log x ) = log x \ I.F = e Hence, solution is y . log (x) = ò 2 . log x dx Þ
If x = 1 then k = 2
\ y(e) = 2 (e – e) + 2
JEE MAIN ­ 2015 (04­Apr­15) Question & Solutions y . log (x) = 2 (x log x – x) + k Þ y(e) = 2.
­ 8 ­ www. prernaclasses.com