53664886 AIEEE Practice

1.

AIEEE−CBSE−ENG−03 A function f from the set of natural numbers to integers defined by

 n− 1  2 , whenis odd f (n) =  is n  − , whenn is even  2 (A) one−one but not onto (C) one−one and onto both

(B) onto but not one−one (D) neither one−one nor onto

2.

Let z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z2 form an equilateral triangle, then (A) a2 = b (B) a2 = 2b 2 (C) a = 3b (D) a2 = 4b

3.

If z and ω are two non−zero complex numbers such that |zω| = 1, and Arg (z) − π Arg (ω) = , then zω is equal to 2 (A) 1 (B) − 1 (C) i (D) − i x

4.

5.

1+ i  If   = 1, then  1− i  (A) x = 4n, where n is any positive integer (B) x = 2n, where n is any positive integer (C) x = 4n + 1, where n is any positive integer (D) x = 2n + 1, where n is any positive integer

a If b c

a2 b2 c2

1+a3 1+b3 1+c3

= 0 and vectors (1, a, a2) (1, b, b2) and (1, c, c2) are non−

coplanar, then the product abc equals (A) 2 (B) − 1 (C) 1 (D) 0 6.

7.

If the system of linear equations x + 2ay + az = 0 x + 3by + bz = 0 x + 4cy + cz = 0 has a non−zero solution, then a, b, c (A) are in A. P. (C) are in H.P.

(B) are in G.P. (D) satisfy a + 2b + 3c = 0

If the sum of the roots of the quadratic equation ax 2 + bx + c = 0 is equal to the a b c , sum of the squares of their reciprocals, then and are in c a b (A) arithmetic progression (B) geometric progression (C) harmonic progression (D) arithmetic−geometric−progression

8.

The number of real solutions of the equation x2 − 3 |x| + 2 = 0 is (A) 2 (B) 4 (C) 1 (D) 3

9.

The value of ‘a’ for which one root of the quadratic equation (a2 − 5a + 3) x2 + (3a − 1) x + 2 = 0 is twice as large as the other, is 2 2 (A) (B) − 3 3 1 1 (C) (D) − 3 3

I10.

a If A =  b (A) α = a2 (C) α = a2

b α β and A2 =    , then a β α + b2, β = ab (B) α = a2 + b2, β = 2ab 2 2 2 +b,β=a −b (D) α = 2ab, β = a2 + b2

11.

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is (A) 140 (B) 196 (C) 280 (D) 346

12.

The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by (A) 6! × 5! (B) 30 (C) 5! × 4! (D) 7! × 5!

13.

If 1, ω, ω2 are the cube roots of unity, then n 1 ω ω2n n ω2n 1 is equal to ∆= ω 2n n ω 1 ω (A) 0 (C) ω

(B) 1 (D) ω2

14.

If nCr denotes the number of combinations of n things taken r at a time, then the expression nCr+1 + nCr−1 + 2 × nCr equals (A) n+2Cr (B) n+2Cr+1 n+1 (C) Cr (D) n+1Cr+1

15.

The number of integral terms in the expansion of ( 3 + 8 5 )256 is (A) 32 (B) 33 (C) 34 (D) 35

16.

If x is positive, the first negative term in the expansion of (1 + x)27/5 is (A) 7th term (B) 5th term (C) 8th term (D) 6th term

17.

The sum of the series (A) 2 loge 2

1 1 1 − + − ……… upto ∞ is equal to 1⋅ 2 2 ⋅ 3 3 ⋅ 4 (B) log2 2 − 1

(C) loge 2

 4 (D) loge   e

18.

Let f (x) be a polynomial function of second degree. If f (1) = f (− 1) and a, b, c are in A. P., then f′ (a), f′ (b) and f′ (c) are in (A) A.P. (B) G.P. (C) H. P. (D) arithmetic−geometric progression

19.

If x1, x2, x3 and y1, y2, y3 are both in G.P. with the same common ratio, then the points (x1, y1) (x2, y2) and (x3, y3) (A) lie on a straight line (B) lie on an ellipse (C) lie on a circle (D) are vertices of a triangle

20.

The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a, is  π  π a  (A) a cot   (B) cot  n 2    2n   π  π a   (C) a cot  (D) cot  4  2n   2n 

21.

22.

C  A 3b If in a triangle ABC a cos2   + c cos2   = , then the sides a, b and c 2  2 2 (A) are in A.P. (B) are in G.P. (C) are in H.P. (D) satisfy a + b = c

In a triangle ABC, medians AD and BE are drawn. If AD = 4, ∠ DAB = ABE =

π , then the area of the ∆ ABC is 3

8 3 32 (C) 3 (A)

π and ∠ 6

16 3 64 (D) 3 (B)

23.

The trigonometric equation sin−1 x = 2 sin−1 a, has a solution for 1 1 (A) < |a| < (B) all real values of a 2 2 1 1 (C) |a| < (D) |a| ≥ 2 2

24.

The upper

25.

The real number x when added to its inverse gives the minimum value of the sum at x equal to (A) 2 (B) 1 (C) − 1 (D) − 2

3 3 th portion of a vertical pole subtends an angle tan−1 at point in 4 5 the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is (A) 20 m (B) 40 m (C) 60 m (D) 80 m

26.

If f : R → R satisfies f (x + y) = f (x) + f (y), for all x, y ∈ R and f (1) = 7, then n

∑f(r) is r =1

(A)

7(n +1) 2 7n(n +1) (D) 2

7n 2

(B)

(C) 7n (n + 1)

27.

If f (x) = xn, then the value of f (1) − (A) 2n (C) 0

28.

(B) 2n−1 (D) 1

3 + log10 (x3 − x), is 4 − x2 (B) (− 1, 0) ∪ (1, 2) (D) (− 1, 0) ∪ (1, 2) ∪ (2, ∞)

Domain of definition of the function f (x) = (A) (1, 2) (C) (1, 2) ∪ (2, ∞)

29.

f′(1) f′′(1) f′′′(1) (−1)n f n (1) + − +... + is 1! 2! 3! n!

  x  1− tan  [1− sinx]  2  lim  is x→π / 2   x  3 1+ tan  [ π − 2x]  2   1 8 1 (C) 32 (A)

(B) 0 (D) ∞

log(3 + x) − log(3 − x) = k, the value of k is x→0 x 1 (A) 0 (B) − 3 2 2 (C) (D) − 3 3

30.

If lim

31.

Let f (a) = g (a) = k and their n th derivatives fn (a), gn (a) exist and are not equal f(a)g(x) − f(a) −g(a)f(x) + g(a) for some n. Further if xlim = 4, then the value →a g(x) − f(x) of k is (A) 4 (B) 2 (C) 1 (D) 0

32.

The function f (x) = log (x + (A) an even function (C) a periodic function

x2 +1 ), is

(B) an odd function (D) neither an even nor an odd function

33.

 −  1 + 1    |x| x  , If f (x) =  xe 0 , 

x≠ 0

then f (x) is

x= 0

(A) continuous as well as differentiable for all x (B) continuous for all x but not differentiable at x = 0 (C) neither differentiable nor continuous at x = 0 (D) discontinuous everywhere 34.

If the function f (x) = 2x3 − 9ax2 + 12a2 x + 1, where a > 0, attains its maximum and minimum at p and q respectively such that p2 = q, then a equals (A) 3 (B) 1 1 (C) 2 (D) 2

35.

If f (y) = e , g (y) = y; y > 0 and F (t) =

t

y

∫f (t − y) g (y) dy, then 0

(A) F (t) = 1 − e−t (1 + t) (C) F (t) = t et

(B) F (t) = et − (1 + t) (D) F (t) = t e−t b

36.

If f (a + b − x) = f (x), then

∫x

f (x) dx is equal to

a

b

(A)

b

a +b f(b − x)dx 2 a

∫

(B)

b

a +b f(x)dx 2 a

∫

b

b −a f(x)dx (C) 2 a

a +b f(a + b − x)dx (D) 2 a

∫

∫

x2

∫sec

2

37.

The value of

lim

x→0

t dt

0

is

x sin x

(A) 3 (C) 1

(B) 2 (D) 0 1

38.

The value of the integral I =

∫

x (1 − x)n dx is

0

1 n +1 1 1 (C) − n +1 n +2 (A)

39.

1 n+2 1 1 (D) + n +1 n+2 (B)

1+ 24 + 34 + ......+ n4 1+ 23 + 33 + ......+ n3 is − lim n→∞ n→∞ n5 n5 1 (A) (B) zero 30 lim

(C)

40.

1 4

(D)

 esin x  d  Let F (x) =   x  , x > 0. If dx   possible values of k, is (A) 15 (C) 63

4

3

1 5

∫x e

sin x3

dx = F (k) − F (1), then one of the

1

(B) 16 (D) 64

41.

The area of the region bounded by the curves y = |x − 1| and y = 3 − |x| is (A) 2 sq units (B) 3 sq units (C) 4 sq units (D) 6 sq units

42.

Let f (x) be a function satisfying f′ (x) = f (x) with f (0) = 1 and g (x) be a function 1

∫

that satisfies f (x) + g (x) = x . Then the value of the integral 2

f (x) g (x) dx, is

0

e2 5 − 2 2 e2 3 (C) e − − 2 2

e2 3 − 2 2 2 e 5 (D) e + + 2 2

(A) e −

(B) e +

43.

The degree and order of the differential equation of the family of all parabolas whose axis is x−axis, are respectively (A) 2, 1 (B) 1, 2 (C) 3, 2 (D) 2, 3

44.

The solution of the differential equation (1 + y2) + (x − etan

−1

(A) (x − 2) = k −1

(C) x etan

y

−1

e − tan

y

= tan−1 y + k

−1

(B) 2x e2 tan

−1

(D) x e2 tan

y

y

y

)

dy = 0, is dx

+k −1

= etan

y

+k

45.

If the equation of the locus of a point equidistant from the points (a 1, b1) and (a2, b2) is (a1 − a2) x + (b1 − b2) y + c = 0, then the value of ‘c’ is 1 (A) (a22 + b22 − a12 − b12 ) (B) a12 + a22 + b12 − b22 2 1 (C) (a12 + a22 − b12 − b22 ) (D) a12 + b12 − a22 − b22 2

46.

Locus of centroid of the triangle whose vertices are (a cos t, a sin t), (b sin t, − b cos t) and (1, 0), where t is a parameter, is (A) (3x − 1)2 + (3y)2 = a2 − b2 (B) (3x − 1)2 + (3y)2 = a2 + b2 2 2 2 2 (C) (3x + 1) + (3y) = a + b (D) (3x + 1)2 + (3y)2 = a2 − b2

47.

If the pair of straight lines x2 − 2pxy − y2 = 0 and x2 − 2qxy − y2 = 0 be such that each pair bisects the angle between the other pair, then (A) p = q (B) p = − q (C) pq = 1 (D) pq = − 1

48.

a square of side a lies above the x−axis and has one vertex at the origin. The π side passing through the origin makes an angle α (0 < α < ) with the positive 4 direction of x−axis. The equation of its diagonal not passing through the origin is (A) y (cos α − sin α) − x (sin α − cos α) = a (B) y (cos α + sin α) + x (sin α − cos α) = a (C) y (cos α + sin α) + x (sin α + cos α) = a (D) y (cos α + sin α) + x (cos α − sin α) = a

49.

If the two circles (x − 1)2 + (y − 3)2 = r2 and x2 + y2 − 8x + 2y + 8 = 0 intersect in two distinct points, then (A) 2 < r < 8 (B) r < 2 (C) r = 2 (D) r > 2

50.

The lines 2x − 3y = 5 and 3x − 4y = 7 are diameters of a circle having area as 154 sq units. Then the equation of the circle is (A) x2 + y2 + 2x − 2y = 62 (B) x2 + y2 + 2x − 2y = 47 2 2 (C) x + y − 2x + 2y = 47 (D) x2 + y2 − 2x + 2y = 62

51.

The normal at the point (bt12, 2bt1) on a parabola meets the parabola again in the point (bt22, 2bt2), then 2 2 (A) t2 = − t1 − (B) t2 = − t1 + t1 t1 2 2 (D) t2 = t1 − (D) t2 = t1 + t1 t1

52.

The foci of the ellipse Then the value of b2 is (A) 1 (C) 7

x2 y2 x2 y2 1 coincide. + 2 = 1 and the hyperbola − = 16 b 144 81 25 (B) 5 (D) 9

53.

A tetrahedron has vertices at O (0, 0, 0), A (1, 2, 1), B (2, 1, 3) and C (− 1, 1, 2). Then the angle between the faces OAB and ABC will be  19   17   (A) cos−1  (B) cos−1   35  31 (C) 300 (D) 900

54.

The radius of the circle in which the sphere x2 + y2 + z2 + 2x − 2y − 4z − 19 = 0 is cut by the plane x + 2y + 2z + 7 = 0 is (A) 1 (B) 2 (C) 3 (D) 4

55.

The lines

56.

The two lines x = ay + b, z = cy + d and x = a′y + b′, z = c′y + d′ will be perpendicular, if and only if (A) aa′ + bb′ + cc′ + 1 = 0 (B) aa′ + bb′ + cc′ = 0

x−2 y−3 z − 4 x −1 y − 4 z − 5 = = = = and are coplanar if 1 1 −k k 2 1 (A) k = 0 or − 1 (B) k = 1 or − 1 (C) k = 0 or − 3 (D) k = 3 or − 3

(C) (a + a′) (b + b′) + (c + c′) = 0

(D) aa′ + cc′ + 1 = 0

57.

The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x 2 + y2 + z2 + 4x − 2y − 6z = 155 is 4 (A) 26 (B) 11 13 (C) 13 (D) 39

58.

Two systems of rectangular axes have the same origin. If a plane cuts them at distances a, b, c and a′, b′, c′ from the origin, then

(A)

(C)

59.

111111 +2 2 + 2 + 2 + 2 + 2 a b c a′ b′ c′ 111111 −2 2 − 2 + 2 − 2 − 2 a b c a′ b′ c′

(B)

=0

(D)

=0

=0

          a, b, c are 3 vectors, such that a + b + c = 0 , | a |= 1, | b |= 2, | c | = 3, then

      a ⋅ b + b ⋅ c + c ⋅ a is equal to (A) 0 (C) 7

60.

=0

111111 +2 2 − 2 + 2 + 2 − 2 a b c a′ b′ c′ 111111 +2 2 + 2 − 2 − 2 − 2 a b c a′ b′ c′

(B) − 7 (D) 1

   u, v and are three w        u + v − w) ⋅ (u − v) ×(v − w) equals

If

(A) 0    (C) u ⋅ w × v

non−coplanar

vectors,

then

(

   (B) u ⋅ v × w    (D) 3 u ⋅ v × w

61.

ˆ, ˆ ˆ i −4ˆ j +7k i −6ˆ j +10k Consider points A, B, C and D with position vectors 7ˆ ˆ and 5 ˆ ˆ respectively. Then ABCD is a i −3ˆ j +4k i −ˆ j +5k , −ˆ (A) square (B) rhombus (C) rectangle (D) parallelogram but not a rhombus

62.

ˆ are the sides of a triangle ˆ , and AC =5ˆ The vectors AB = 3ˆ i −2ˆ j +4k i + 4k ABC. The length of the median through A is (A) 18 (B) 72

(C)

(D)

33

288

63.

ˆ and 3ˆ ˆ is displaced i +ˆ j −3k i +ˆ j −k A particle acted on by constant forces 4ˆ ˆ to the point 5ˆ ˆ . The total work done by the i +2ˆ j +3k i +4ˆ j +k from the point ˆ forces is (A) 20 units (B) 30 units (C) 40 units (D) 50 units

64.

ˆ . If n ˆ =0 i +ˆ j, v = ˆ i −ˆ j and w = ˆ i + 2ˆ j +3k Let u = ˆ ˆ is unit vector such that u ⋅ n   ˆ | is equal to ˆ = 0, then | w ⋅ n and v ⋅ n (A) 0 (B) 1 (C) 2 (D) 3

65.

The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set (A) is increased by 2 (B) is decreased by 2 (C) is two times the original median (D) remains the same as that of the original set

66.

In an experiment with 15 observations on x, then following results were available: ∑x2 = 2830, ∑x = 170 One observation that was 20 was found to be wrong and was replaced by the correct value 30. Then the corrected variance is (A) 78.00 (B) 188.66 (C) 177.33 (D) 8.33

67.

Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is 4 3 (A) (B) 5 5 1 2 (C) (D) 5 5

68.

Events A, B, C are mutually exclusive events such that P (A) =





1− x and 4 values of x are in the interval  1 1 (A)  ,  3 2  1 13 (C)  , 3 3  69.





P (C) =

3x +1 , P (B) = 3

1− 2x . The set of possible 2

1 2 (B)  ,  3 3 

(D) [0, 1]

The mean and variance of a random variable having a binomial distribution are 4 and 2 respectively, then P (X = 1) is 1 1 (A) (B) 32 16 1 1 (C) (D) 8 4

70.

71.

72.

     The resultant of forces P and Q is R . If Q is doubled then R is doubled. If   the direction of Q is reversed, then R is again doubled. Then P2 : Q2 : R2 is (A) 3 : 1 : 1 (B) 2 : 3 : 2 (C) 1 : 2 : 3 (D) 2 : 3 : 1 Let R1 and R2 respectively be the maximum ranges up and down an inclined plane and R be the maximum range on the horizontal plane. Then R 1, R, R2 are in (A) arithmetic−geometric progression (B) A.P. (C) G.P. (D) H.P.

   A couple is of moment G and the force forming the couple is P . If P is turned  through a right angle, the moment of the couple thus formed is H . If instead,  the forces P are turned through an angle α, then the moment of couple becomes     (A) G sin α − H cos α (B) H cos α + G sin α     (C) G cos α − H sin α (D) H sin α − G cos α

73.

Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity u and the other from rest with uniform  acceleration f . Let α be the angle between their directions of motion. The relative velocity of the second particle with respect to the first is least after a time u sinα f cosα (A) (B) f u ucosα (C) u sin α (D) f

74.

Two stones are projected from the top of a cliff h meters high, with the same speed u so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected at an angle θ to the horizontal then tan θ equals 2u u (A) (B) 2g gh h (C) 2h

75.

u g

(D) u

2 gh

A body travels a distances s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration f and in the second part with constant retardation r. The value of t is given by 2s  1 1 (A) 2s  +  (B) 1 1 + f r f r (C)

2s(f +r)

(D)

 1 1 2s +  f r

1.

2.

5.

Solutions Clearly both one − one and onto Because if n is odd, values are set of all non−negative integers and if n is an even, values are set of all negative integers. Hence, (C) is the correct answer. z12 + z22 − z1z2 = 0 (z1 + z2)2 − 3z1z2 = 0 a2 = 3b. Hence, (C) is the correct answer. a b c

a2 b2 c2

a2 b2 = 0 c2

1 1 a 1 +1 b 1 1 c

a (1 + abc) b c

a2 b2 c2

1 1 =0 1

⇒ abc = − 1. Hence, (B) is the correct answer 4.

1+ i (1+ i)2 =i = 1− i 2 x

1+ i    = ix  1− i  ⇒ x = 4n. Hence, (A) is the correct answer.

6.

1

2a

a

Coefficient determinant = 1 1

3b 4c

b =0 c

2ac . a +c Hence, (C) is the correct answer ⇒b=

8.

x2 − 3 |x| + 2 = 0 (|x| − 1) (|x| − 2) = 0 ⇒ x = ± 1, ± 2. Hence, (B) is the correct answer

7.

Let α, β be the roots 1 1 α+β= 2 + 2 α β α+β=

α2 + β2 − 2αβ (α + β)

 b  b2 − 2ac −  = c2  a ⇒ 2a2c = b (a2 + bc) a b c , , ⇒ are in H.P. c a b

Hence, (C) is the correct answer 10.

9.

a b A=   b a a b a b A2 =     b a b a a2 + b2 2ab  =   2 a + b2   2ab ⇒ α = a2 + b2, β = 2ab. Hence, (B) is the correct answer.

β = 2α

3a − 1 a − 5a + 3 2 2α2 = 2 a − 5a + 6 3α =

2

(3a − 1)2 1 = 2 2 2 a(a − 5a + 3) a + 5a + 6 2 ⇒a= . 3 Hence, (A) is the correct answer

12.

Clearly 5! × 6! (A) is the correct answer

11.

Number of choices = 5C4 × 8C6 + 5C5 × 8C5 = 140 + 56. Hence, (B) is the correct answer

13.

1+ωn +ω2n n 2n ∆ = 1+ω +ω 1+ωn +ω2n

ωn ω2n 1

ω2n 1 ωn

=0 Since, 1 + ωn + ω2n = 0, if n is not a multiple of 3 Therefore, the roots are identical. Hence, (A) is the correct answer 14.

17.

Cr+1 + nCr−1 + nCr + nCr = n+1Cr+1 + n+1Cr = n+2Cr+1. Hence, (B) is the correct answer n

1 1 1 − + − ……… 1⋅ 2 2 ⋅ 3 3 ⋅ 4 1 1 1 1 1 − + + − =1− − ……… 2 2 3 3 4 1 1 1  =1−2   2 − 3 + 4 − .........   

1 1 1   = 2 1− + − + ......... − 1 2 3 4   = 2 log 2 − log e  4 = log   . e Hence, (D) is the correct answer.

15.

General term = 256Cr ( 3 )256−r [(5)1/8]r From integral terms, or should be 8k ⇒ k = 0 to 32. Hence, (B) is the correct answer.

18.

f (x) = ax2 + bx + c f (1) = a + b + c f (− 1) = a − b + c ⇒ a + b + c = a − b + c also 2b = a + c f′ (x) = 2ax + b = 2ax f′ (a) = 2a2 f′ (b) = 2ab f′ (c) = 2ac ⇒ AP. Hence, (A) is the correct answer.

19.

Result (A) is correct answer.

20.

(B)

21.

26.

 1+ cosC   1+ cosA  3b  + c = a  2 2 2     ⇒ a + c + b = 3b a + c = 2b. Hence, (A) is the correct answer

f (1) = 7 f (1 + 1) = f (1) + f (1) f (2) = 2 × 7 only f (3) = 3 × 7 n

∑f(r) = 7 (1 + 2 + ……… + n) r =1

=7 25.

(B)

23.

−

n(n +1) . 2

π sin2 x π ≤ ≤ 4 2 4 π π − ≤ sin−1 (a) ≤ 4 4 1 1 ≤ |a| ≤ . 2 2 Hence, (D) is the correct answer

n n(n −1) n(n −1)(n − 2) + − + ……… 1! 2! 3! = 1 − nC1 + nC2 − ……… = 0. Hence, (C) is the correct answer

27.

LHS = 1 −

30.

1 1 + 2. lim 3 + x 3 − x = x→ 0 1 3

Hence, (C) is the correct answer. 28.

29.

4 − x2 ≠ 0 ⇒x≠±2 x3 − x > 0 ⇒ x (x + 1) (x − 1) > 0. Hence (D) is the correct answer.

 π x tan − (1− sinx)  4 2 lim x→π / 2  π x 4 − (π − 2x)2  4 2 1 = . 32 Hence, (C) is the correct answer.

32.

f (− x) = − f (x) Hence, (B) is the correct answer.

1.

sin (θ + α) =

x 40

x sin a = 140 ⇒ x = 40. Hence, (B) is the correct answer

3x/4 1

tan− (3/4) x/4 θ 40

34.

f (x) = 0 at x = p, q 6p2 + 18ap + 12a2 = 0 6q2 + 18aq + 12a2 = 0 f″ (x) < 0 at x = p and f″ (x) > 0 at x = q.

30.

Applying L. Hospital’s Rule f(a)g′(a) − g(a)f′(a) lim =4 x→2a g′(a) − f′(a)

k(g′(a) − ff ′(a)) =4 (g′(a) − f ′(a)) k = 4.

Hence, (A) is the correct answer. b

36.

∫x f (x) dx a

b

∫(a + b − x) f (a + b − x) dx.

=

a

Hence, (B) is the correct answer. 33.

37.

f′ (0) f′ (0 − h) = 1 f′ (0 + h) = 0 LHD ≠ RHD. Hence, (B) is the correct answer.

tan(x2 ) x→0 x sinx tan(x2 ) lim = x→0  sinx  x2    x  = 1. Hence (C) is the correct answer. lim

1

38.

1

∫

x (1 − x)n dx =

0

∫x

n

(1 − x)

0

1

1 1 − . n +1 n + 2 0 Hence, (C) is the correct answer. =

∫(x

− xn+1) =

n

t

35.

F (t) =

∫f (t − y) f (y) dy 0

t

=

∫f (y) f (t − y) dy 0 t

=

∫e

y

(t − y) dy

0

= xt − (1 + t). Hence, (B) is the correct answer. 34.

Clearly f″ (x) > 0 for x = 2a ⇒ q = 2a < 0 for x = a ⇒ p = a or p2 = q ⇒ a = 2. Hence, (C) is the correct answer.

40.

F′ (x) = =

3

esin x 3x

∫x e

sin x

dx = F (k) − F (1)

64

=

esin x dx = F (k) − F (1) x 1

∫

64

=

∫F′(x) dx = F (k) − F (1) 1

F (64) − F (1) = F (k) − F (1) ⇒ k = 64. Hence, (D) is the correct answer. 41.

Clearly area = 2 = sq units

2×

2 (−1,2)

(2,1) (1,0)

45.

Let p (x, y) (x − a1)2 + (y − b1)2 = (x − a2)2 + (y − b2)2 1 2 (a1 − a2) x + (b1 − b2) y + (b − b12 + a22 − a12 ) = 0. 2 2 Hence, (A) is the correct answer.

46.

x=

a cost + b sint + 1 a sint − bcost +1 ,y= 3 3 2

1 a2 + b2  .  x −  + y2 = 3 9  Hence, (B) is the correct answer.

43.

Equation y2 = 4a 9x − h) 2yy1 = 4a ⇒ yy1 = 2a yy2 = y12 = 0. Hence (B) is the correct answer. 1

42.

∫f(x) [x

2

− f (x)] dx

0

solving this by putting f′ (x) = f (x). Hence, (B) is the correct answer. 50.

47. 49.

Intersection of diameter is the point (1, − 1) πs2 = 154 ⇒ s2 = 49 (x − 1)2 + (y + 1)2 = 49 Hence, (C) is the correct answer. (D) dx −1 (1 + y2) = (esin y − x) dy −1

x dx esub −y + = dy 1+ yα 1+ y2

x2 52.

2

 12   5

−

y2 2

 9    5 5 ⇒ e1 = 4 ae2 =

1−

=1

b2 ×4=3 16

⇒ b2 = 7. Hence, (C) is the correct answer. 54.

(C)

5

4 3

69.

np = 4 npq = 2 1 1 q= ,p= 2 2 n=8 8

 1 p (x = 1) = 8C1   2 1 = . 32 Hence, (A) is the correct answer.

49.

(x − 1)2 + (y − 3)2 = r2 (x − 4)2 + (y + 2)2 − 16 − 4 + 8 = 0 (x − 4)2 + (y + 2)2 = 12.

67.

Select 2 out of 5 2 = . 5 Hence, (D) is the correct answer.

65.

0≤

3x + 1 1− x 1− 2x + + ≤1 3 4 2 12x + 4 + 3 − 3x + 6 − 12x ≤ 1 0 ≤ 13 − 3x ≤ 12 3x ≤ 13 1 ⇒x≥ 3 13 x≤ . 3 Hence, (C) is the correct answer.

3.

z π Arg   =  ω 2 |zω| = 1 z ω = − i or + i.

FIITJEE

AIEEE − 2004 (MATHEMATICS)

Important Instructions: 1 2

i)

The test is of 1 hours duration.

ii)

The test consists of 75 questions.

iii)

The maximum marks are 225.

iv)

For each correct answer you will get 3 marks and for a wrong answer you will get -1 mark.

1.

Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is (1) a function (2) reflexive (3) not symmetric (4) transitive

2.

The range of the function f(x) = 7- xPx- 3 is (1) {1, 2, 3} (2) {1, 2, 3, 4, 5} (3) {1, 2, 3, 4} (4) {1, 2, 3, 4, 5, 6}

3.

Let z, w be complex numbers such that z +iw = 0 and arg zw = π. Then arg z equals p 5p (1) (2) 4 4 3p p (3) (4) 4 2

4.

If z = x – i y and

1 z3

æx y ö ç + ÷ èp q ø is equal to , then =p +iq p2 +q2

(1) 1 (3) 2 5.

)

(2) -2 (4) -1 2

If z2 - 1 = z +1, then z lies on (1) the real axis (3) a circle

6.

(

(2) an ellipse (4) the imaginary axis.

æ0 0 - 1ö ç ÷ Let A =ç 0 - 1 0 ÷. The only correct statement about the matrix A is ç- 1 0 0 ÷ è ø

AIEEE-PAPERS--2

(1) A is a zero matrix (3) A - 1 does not exist

7.

8.

(2) A 2 =I (4) A =( - 1) I , where I is a unit matrix

æ1 - 1 1 ö æ4 2 2 ö ç ÷ ç ÷ Let A =ç2 1 - 3÷( 10) B =ç- 5 0 a ÷. If B is the inverse of matrix A, then α is ç1 1 1 ÷ ç 1 - 2 3÷ è ø è ø (1) -2 (2) 5 (3) 2 (4) -1

If a1, a2 , a3 , ....,an , .... are in G.P., then the value of the determinant logan logan+1 logan+2 logan+3 logan+4 logan+5 , is logan+6 logan+7 logan+8 (1) 0 (3) 2

(2) -2 (4) 1

9.

Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation (1) x2 +18x +16 =0 (2) x2 - 18x - 16 =0 (3) x2 +18x - 16 =0 (4) x2 - 18x +16 =0

10.

If (1 – p) is a root of quadratic equation x2 +px +( 1- p) =0 , then its roots are (1) 0, 1 (2) -1, 2 (3) 0, -1 (4) -1, 1

11.

Let S(K) =1+3 +5 +... +( 2K - 1) =3 +K 2 . Then which of the following is true? (1) S(1) is correct (2) Principle of mathematical induction can be used to prove the formula (3) S(K) Þ S(K +1) (4) S(K) Þ S(K +1)

12.

How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical order? (1) 120 (2) 480 (3) 360 (4) 240

13.

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is (1) 5 (2) 8 C 3 (3) 38

14.

2

(4) 21

If one root of the equation x2 +px +12 =0 is 4, while the equation x2 +px +q =0 has equal roots, then the value of ‘q’ is 49 (1) (2) 4 4 (3) 3 (4) 12 FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 2651394

AIEEE-PAPERS--3

15.

The coefficient of the middle term in the binomial expansion in powers of x of 4

( 1+ax)

and of ( 1- ax)

6

is the same if α equals

5 3 -3 (3) 10

3 5 10 (4) 3

(1) -

16.

(2)

n

The coefficient of xn in expansion of ( 1+x) ( 1- x) is (1) (n – 1) (3) ( - 1)

17.

n- 1

( n - 1)

n

1

å

If S n =

r=0

n

Cr

2

n

å

and tn =

r =0

r n

Cr

, then

1 (1) n 2

(2) ( - 1)

n

(4) ( - 1)

n- 1

( 1- n) n

tn is equal to Sn 1 n- 1 2 2n - 1 (4) 2

(2)

(3) n – 1 18.

Let Tr be the rth term of an A.P. whose first term is a and common difference is d. 1 1 If for some positive integers m, n, m ≠ n, Tm = and Tn = , then a – d equals n m (1) 0 (2) 1 1 1 1 + (3) (4) mn m n

19.

The sum of the first n terms of the series 12 +2 ×22 +32 +2 ×42 +52 +2 ×62 +... is n( n +1)

2

when n is even. When n is odd the sum is 2 3n( n +1) n2 ( n +1) (1) (2) 2 2 (3)

20.

n( n +1) 4

The sum of series

(e (1)

2

(3)

2

én( n +1) ù (4) ê ú ê 2 û ú ë

2

(

)

-1

2 2 e -1

)

2e

1 1 1 + + +... is 2! 4! 6!

(2)

( e - 1)

(4)

(e

2e

2

2

)

-2 e

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3

AIEEE-PAPERS--4

21.

Let α, β be such that π < α - β < 3π. If sinα + sinβ = then the value of cos

(3) 22.

3

(2)

130

130 6 (4) 65

6 65

If u = a2 cos2 q+b2 sin2 q + a2 sin2 q+b2 cos2 q , then the difference between the maximum and minimum values of u2 is given by

(

2 2 (1) 2 a +b

(3) ( a +b) 23.

a- b is 2

3

(1) -

21 27 and cosα + cosβ = , 65 65

)

(2) 2 a2 +b2

2

(4) ( a - b)

The sides of a triangle are sinα, cosα and

2

1+sin a cos a for some 0 < α <

Then the greatest angle of the triangle is (1) 60o (2) 90o (3) 120o (4) 150o

p . 2

24.

A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is 60o and when he retires 40 meter away from the tree the angle of elevation becomes 30o . The breadth of the river is (1) 20 m (2) 30 m (3) 40 m (4) 60 m

25.

If f : R ® S, defined by f(x) =sinx (1) [0, 3] (3) [0, 1]

26.

The graph of the function y = f(x) is symmetrical about the line x = 2, then (1) f(x + 2)= f(x – 2) (2) f(2 + x) = f(2 – x) (3) f(x) = f(-x) (4) f(x) = - f(-x)

27.

The domain of the function f(x) = (1) [2, 3] (3) [1, 2]

3 cos x +1, is onto, then the interval of S is (2) [-1, 1] (4) [-1, 3]

sin- 1 ( x - 3)

is 9 - x2 (2) [2, 3) (4) [1, 2)

2x

28.

4

æ a bö If lim ç1+ + 2 ÷ =e2 , then the values of a and b, are x®¥ è x x ø (1) a Î R , b Î R (2) a = 1, b Î R (3) a Î R, b =2 (4) a = 1 and b = 2 FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 2651394

AIEEE-PAPERS--5

29.

30.

1- tanx p é pù é pù æp ö , x ¹ , x Î ê0, ú . If f(x) is continuous in ê0, ú, then f ç ÷ is Let f(x) = 4x - p 4 ë 2û ë 2û è4 ø 1 (1) 1 (2) 2 1 (3) (4) -1 2 y +...to ¥

If x =ey+e

, x > 0, then

x 1+x 1- x (3) x

dy is dx 1 x 1+x (4) x

(1)

(2)

31.

A point on the parabola y2 =18x at which the ordinate increases at twice the rate of the abscissa is (1) (2, 4) (2) (2, -4) æ- 9 9 ö æ9 9 ö (3) ç , ÷ (4) ç , ÷ è 8 2ø è8 2 ø

32.

A function y = f(x) has a second order derivative f″(x) = 6(x – 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x – 5, then the function is (1) ( x - 1)

2

(2) ( x - 1)

3

(3) ( x +1)

3

(4) ( x +1)

2

33.

The normal to the curve x = a(1 + cosθ), y = asinθ at ‘θ’ always passes through the fixed point (1) (a, 0) (2) (0, a) (3) (0, 0) (4) (a, a)

34.

If 2a + 3b + 6c =0, then at least one root of the equation ax2 +bx +c =0 lies in the interval (1) (0, 1) (2) (1, 2) (3) (2, 3) (4) (1, 3) n

35.

36.

r

1 n lim e is n®¥ r =1 n (1) e (3) 1 – e

å

If

(2) e – 1 (4) e + 1

sinx

òsin(x - a ) dx =Ax +Blogsin(x - a ) +C , then value of (A, B) is

(1) (sinα, cosα) (3) (- sinα, cosα)

(2) (cosα, sinα) (4) (- cosα, sinα)

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5

AIEEE-PAPERS--6

37.

dx

òcos x - sinx is equal to (1) (3)

1

æx p ö log tan ç - ÷+C 2 è2 8 ø

(2)

1

æx 3p ö log tan ç ÷+C 2 è2 8 ø

(4)

1

æx ö log cotç ÷+C 2 è2 ø

1

æx 3p ö log tan ç + ÷+C 2 è2 8 ø

3

38.

The value of

ò|1- x

2

|dx is

-2

28 (1) 3 7 (3) 3

14 3 1 (4) 3

(2)

p/ 2

39.

The value of I =

(sinx +cos x)2

ò 0

1+sin2x

dx is

(1) 0 (3) 2

(2) 1 (4) 3

p

40.

If

ò

p/ 2

xf(sinx)dx =A

0

òf(sinx) dx, then A is 0

(2) π

(1) 0 p (3) 4 41.

If f(x) =

(4) 2π ex , I1 = 1+ex

f(a)

ò

f(a)

xg{x(1- x)}dx and I2 =

f(- a)

ò g{x(1- x)}dx

then the value of

f(- a)

I2 I1

is (1) 2 (3) –1

(2) –3 (4) 1

42. The area of the region bounded by the curves y = |x – 2|, x = 1, x = 3 and the xaxis is (1) 1 (2) 2 (3) 3 (4) 4 43.

The differential equation for the family of curves x2 +y2 - 2ay =0 , where a is an arbitrary constant is (1) 2(x2 - y2 )y¢=xy (2) 2(x2 +y2 )y¢=xy (3) (x2 - y2 )y¢=2xy

44.

6

(4) (x2 +y2 )y¢=2xy

The solution of the differential equation y dx + (x + x2y) dy = 0 is 1 1 =C +logy =C (1) (2) xy xy FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 2651394

AIEEE-PAPERS--7

(3)

1 +logy =C xy

(4) log y = Cx

45.

Let A (2, –3) and B(–2, 1) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex C is the line (1) 2x + 3y = 9 (2) 2x – 3y = 7 (3) 3x + 2y = 5 (4) 3x – 2y = 3

46.

The equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinate axes whose sum is –1 is x y x y x y x y + =- 1 + =- 1 (1) + =- 1and (2) - =- 1and 2 3 -2 1 2 3 -2 1 x y x y x y x y + =1 (3) + =1and + =1 (4) - =1and 2 3 2 1 2 3 -2 1

47.

If the sum of the slopes of the lines given by x2 - 2cxy - 7y2 =0 is four times their product, then c has the value (1) 1 (2) –1 (3) 2 (4) –2

48.

If one of the lines given by 6x2 - xy +4cy2 =0 is 3x + 4y = 0, then c equals (1) 1 (2) –1 (3) 3 (4) –3

49.

If a circle passes through the point (a, b) and cuts the circle x2 +y2 =4 orthogonally, then the locus of its centre is (1) 2ax +2by +(a2 +b2 +4) =0 (2) 2ax +2by - (a2 +b2 +4) =0 (3) 2ax - 2by +(a2 +b2 +4) =0

50.

A variable circle passes through the fixed point A (p, q) and touches x-axis. The locus of the other end of the diameter through A is (1) (x - p)2 =4qy (2) (x - q)2 =4py (3) (y - p)2 =4qx

51.

(4) x2 +y2 +2x - 2y - 23 =0

The intercept on the line y = x by the circle x2 +y2 - 2x =0 is AB. Equation of the circle on AB as a diameter is (1) x2 +y2 - x - y =0 (2) x2 +y2 - x +y =0 (3) x2 +y2 +x +y =0

53.

(4) (y - q)2 =4px

If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along diameters of a circle of circumference 10π, then the equation of the circle is (1) x2 +y2 - 2x +2y - 23 =0 (2) x2 +y2 - 2x - 2y - 23 =0 (3) x2 +y2 +2x +2y - 23 =0

52.

(4) 2ax - 2by - (a2 +b2 +4) =0

(4) x2 +y2 +x - y =0

If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y2 =4ax and x2 =4ay , then FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 2651394

7

AIEEE-PAPERS--8

54.

(1) d2 +(2b +3c)2 =0

(2) d2 +(3b +2c)2 =0

(3) d2 +(2b - 3c)2 =0

(4) d2 +(3b - 2c)2 =0

The eccentricity of an ellipse, with its centre at the origin, is directrices is x = 4, then the equation of the ellipse is (1) 3x2 +4y2 =1 (2) 3x2 +4y2 =12 (3) 4x2 +3y2 =12

1 . If one of the 2

(4) 4x2 +3y2 =1

55.

A line makes the same angle θ, with each of the x and z axis. If the angle β, which it makes with y-axis, is such that sin2 b =3sin2 q, then cos2 q equals 2 1 (1) (2) 3 5 3 2 (3) (4) 5 5

56.

Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is 3 5 (1) (2) 2 2 7 9 (3) (4) 2 2

57.

A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of the point of intersection are given by (1) (3a, 3a, 3a), (a, a, a) (2) (3a, 2a, 3a), (a, a, a) (3) (3a, 2a, 3a), (a, a, 2a) (4) (2a, 3a, 3a), (2a, a, a)

58.

If the straight lines x = 1 + s, y = –3 – λs, z = 1 + λs and x =

59.

The

t , y = 1 + t, z = 2 – 2 t with parameters s and t respectively, are co-planar then λ equals (1) –2 (2) –1 1 (3) – (4) 0 2

2

intersection 2

of

the

spheres

x2 +y2 +z2 +7x - 2y - z =13

and

2

x +y +z - 3x +3y +4z =8 is the same as the intersection of one of the sphere and the plane (1) x – y – z = 1 (2) x – 2y – z = 1 (3) x – y – 2z = 1 (4) 2x – y – z = 1

60.

8

r r r Let a, b and c be three non-zero vectors such that no two of these are collinear. r r r r r r If the vector a +2b is collinear with c and b +3c is collinear with a (λ being some r r r non-zero scalar) then a +2b +6c equals FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 2651394

AIEEE-PAPERS--9

r (1) l a r (3) l c

r (2) l b (4) 0

61.

A particle is acted upon by constant forces 4iˆ +ˆj - 3kˆ and 3iˆ +ˆj - kˆ which displace it from a point ˆi +2jˆ +3kˆ to the point 5iˆ +4jˆ +kˆ . The work done in standard units by the forces is given by (1) 40 (2) 30 (3) 25 (4) 15

62.

If a, b, c are non-coplanar vectors and λ is a real number, then the vectors a +2b +3c, l b +4c and (2l - 1)c are non-coplanar for (1) all values of λ (2) all except one value of λ (3) all except two values of λ (4) no value of λ

63.

Let u, v, w be such that u =1, v =2, w =3 . If the projection v along u is equal to that of w along u and v, w are perpendicular to each other then u - v +w equals (1) 2 (2) 7 (3) 14 (4) 14

64.

Let a, b and c be non-zero vectors such that (a ´ b) ´ c =

1 b c a . If θ is the acute 3

angle between the vectors b and c , then sin θ equals 1 2 (1) (2) 3 3 2 2 2 (3) (4) 3 3 65.

Consider the following statements: (a) Mode can be computed from histogram (b) Median is not independent of change of scale (c) Variance is independent of change of origin and scale. Which of these is/are correct? (1) only (a) (2) only (b) (3) only (a) and (b) (4) (a), (b) and (c)

66.

In a series of 2n observations, half of them equal a and remaining half equal –a. If the standard deviation of the observations is 2, then |a| equals 1 (1) (2) 2 n 2 (3) 2 (4) n

67.

The probability that A speaks truth is

4 3 , while this probability for B is . The 5 4 probability that they contradict each other when asked to speak on a fact is FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 2651394

9

AIEEE-PAPERS--10

3 20 7 (3) 20

(1)

1 5 4 (4) 5

(2)

68.

A random variable X has the probability distribution: X: 1 2 3 4 5 6 7 8 p(X) 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05 : For the events E = {X is a prime number} and F = {X < 4}, the probability P (E ∪ F) is (1) 0.87 (2) 0.77 (3) 0.35 (4) 0.50

69.

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is 37 219 (1) (2) 256 256 128 28 (3) (4) 256 256

70.

With two forces acting at a point, the maximum effect is obtained when their resultant is 4N. If they act at right angles, then their resultant is 3N. Then the forces are (1) (2 + 2)N and (2 - 2)N (2) (2 + 3)N and (2 - 3)N æ 1 ö æ 1 ö 2 ÷N (3) ç2 + 2 ÷N and ç2 2 2 è ø è ø

71.

72.

73.

10

æ 1 ö æ 1 ö 3 ÷N (4) ç2 + 3 ÷N and ç2 2 2 è ø è ø

In a right angle ∆ABC, ∠A = 90° and sides a, b, c are respectively, 5 cm, 4 cm r and 3 cm. If a force F has moments 0, 9 and 16 in N cm. units respectively r about vertices A, B and C, then magnitude of F is (1) 3 (2) 4 (3) 5 (4) 9 r r r Three forces P, Q and R acting along IA, IB and IC, where I is the incentre of a ∆ r r r ABC, are in equilibrium. Then P : Q : R is A B C A B C (1) cos : cos : cos (2) sin : sin : sin 2 2 2 2 2 2 A B C A B C (3) sec : sec : sec (4) cosec : cosec : cosec 2 2 2 2 2 2

A particle moves towards east from a point A to a point B at the rate of 4 km/h and then towards north from B to C at the rate of 5 km/h. If AB = 12 km and BC = 5 km, then its average speed for its journey from A to C and resultant average velocity direct from A to C are respectively 17 13 13 17 (1) km/h and km/h (2) km/h and km/h 4 4 4 4 FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 2651394

AIEEE-PAPERS--11

17 13 13 17 km/h and km/h (4) km/h and km/h 9 9 9 9 1 A velocity m/s is resolved into two components along OA and OB making 4 angles 30° and 45° respectively with the given velocity. Then the component along OB is 1 1 (1) m/s (2) ( 3 - 1) m/s 8 4 1 1 (3) m/s (4) ( 6 - 2) m/s 4 8

(3) 74.

75.

If t1 and t2 are the times of flight of two particles having the same initial velocity u and range R on the horizontal, then t12 +t22 is equal to u2 (1) g

(3)

u2 2g

4u2 (2) 2 g

(4) 1

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11

AIEEE-PAPERS--12

FIITJEE

AIEEE − 2004 (MATHEMATICS) ANSWERS

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

12

3 1 3 2 4 2 2 1 4 3 4 3 4 1 3

16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

2 1 1 2 2 1 4 3 1 4 2 2 2 3 3

31. 4 32. 2 33. 1 34. 1 35. 2 36. 2 37. 4 38. 1 39. 3 40. 2 41. 1 42. 1 43. 3 44. 2 45. 1

46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60.

4 3 4 2 1 1 1 1 2 3 3 2 1 4 4

61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75.

1 3 3 4 3 3 3 2 4 3 3 1 1 4 2

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AIEEE-PAPERS--13

FIITJEE

AIEEE − 2004 (MATHEMATICS) SOLUTIONS

1.

(2, 3) ∈ R but (3, 2) ∉ R. Hence R is not symmetric.

2.

f(x) =7- xPx- 3 7- x ³ 0 Þ x £7 x - 3 ³ 0 Þ x ³ 3, x£5 and 7 - x ³ x - 3 Þ ⇒ 3 £ x £ 5 ⇒ x = 3, 4, 5 ⇒ Range is {1, 2, 3}.

3.

Here ω =

4.

z =( p +iq) =p p2 - 3q2 - iq q2 - 3p2

(

3

Þ

5.

z 3p æ zö ⇒ arg çz. ÷=p ⇒ 2 arg(z) – arg(i) = π ⇒ arg(z) = . i 4 è iø

)

(

y x =p2 - 3q2 & =q2 - 3p2 Þ p q 2

(

2

)

(

x y + p q p2 +q2

) ( z - 1) ( z - 1) = z 2

z2 - 1 = z +1 Þ

2

2

4

=- 2 .

)

2

+2 z +1

Þ z2 +z 2 +2zz =0 Þ z +z =0 ⇒ R (z) = 0 ⇒ z lies on the imaginary axis.

6.

7.

é1 0 0ù ê ú A.A = ê0 1 0ú =I . ê ë0 0 1ú û

AB = I é1 Þ ê ê2 ê ë1 logan

8.

Þ

A(10 B) = 10 1 ù é4 2 ê 1 - 3ú ú ê- 5 0 1 1ú ûê ë1 - 2

-1

logan+1 logan+2

logan+3 logan+4 logan+5 logan+6

logan+7

C3 → C 3 – logan = logan+3 logan+6 9.

I 2ù é10 0 5 - a ù é1 0 0ù ú ê ú ú a ú =ê 0 10 a - 5ú =10 ê ê0 1 0úif a =5 . ê 3ú û ê ë 0 0 5 +a ú û ë0 0 1ú û

C2, C2 logr logr logr

logan+8

→ C3 – C 1 logr logr = 0 (where r is a common ratio). logr

Let numbers be a, b Þ a +b =18, equation

ab =4

Þ

ab =16 , a and b are roots of the

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13

AIEEE-PAPERS--14

x2 - 18x +16 =0 .

Þ

10.

(3) 2

( 1- p)

+p ( 1- p) +( 1- p) =0 p) = 0) Þ ( 1- p) ( 1- p +p +1) =0

(since (1 – p) is a root of the equation x2 + px + (1 –

2( 1- p) =0 ⇒ (1 – p) = 0 ⇒ p = 1 sum of root is a +b =- p and product ab =1- p =0 Þ a +0 =- 1 Þ a =- 1 Þ Roots are 0, –1 Þ

11.

(where β = 1 – p = 0)

S ( k) =1+3 +5 +........ +( 2k - 1) =3 +k2 S(k + 1)=1 + 3 + 5 +............. + (2k – 1) + (2k + 1)

(

)

2 2 = 3 +k +2k +1 =k +2k +4 [from S(k) = 3 +k2 ] = 3 + (k2 + 2k + 1) = 3 + (k + 1)2 = S (k + 1). Although S (k) in itself is not true but it considered true will always imply towards S (k + 1).

12.

Since in half the arrangement A will be before E and other half E will be before A. 6! Hence total number of ways = = 360. 2

13.

Number of balls = 8 number of boxes = 3 Hence number of ways = 7C2 = 21.

14.

Since 4 is one of the root of x2 + px + 12 = 0 ⇒ 16 + 4p + 12 = 0 ⇒ p = –7 and equation x2 + px + q = 0 has equal roots 49 ⇒ D = 49 – 4q = 0 ⇒ q = . 4

15.

Coefficient of Middle term in ( 1+ax) =t3 =4C 2 ×a 2

4

6

Coefficient of Middle term in ( 1- ax) =t4 =6C3 ( - a )

-3 10 Coefficient of xn in (1 + x)(1 – x)n = (1 + x)(nC0 – nC1x + …….. + (–1)n –1 nCn – 1 xn – 1 + (–1)n nCn xn) 4

16.

C 2a 2 =- 6 C3 .a 3 Þ

- 6 =20a

= (–1)n nCn + (–1)n –1 nCn – 1 n

17.

å

t=

r =0

å

n

14

å

n r +n - r n = Þ n n Cr r=0 r =0 Cr

å

Þ

=( - 1)

n n r n- r n- r = = n n Cr r=0 Cn- r r=0 n Cr

å

2tn =

18.

3

n

( 1- n) .

(Q

tn =

a=

n

Cr =n Cn- r

)

n n 1 n = Sn Þ n 2 r=0 Cr 2

å

1 Tm = =a +( m - 1) d n

tn n = Sn 2

.....(1)

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AIEEE-PAPERS--15

and Tn =

1 =a +( n - 1) d m

.....(2)

from (1) and (2) we get a = Hence a – d = 0

1 1 , d= mn mn

If n is odd then (n – 1) is even ⇒ sum of odd terms =

19.

( n - 1) n2 2

2

+n =

n2 ( n +1) 2

.

ea +e- a a2 a 4 a6 =1+ + + +…….. 2 2! 4! 6! ea +e- a a2 a 4 a6 - 1= + + +....... 2 2! 4! 6! put α = 1, we get

20.

( e - 1) 2e

2

=

1 1 1 ……….. + + + 2! 4! 6!

21 27 and cos α + cos β = . 65 65 Squaring and adding, we get 1170 2 + 2 cos (α – β) = (65)2

sin α + sin β = -

21.

-3 a - bö 9 æa - b ö 2æ = = ⇒ cos ç ⇒ cos ç ÷ ÷ è 2 ø 130 è 2 ø 130

22.

æ p a - b 3p ö çQ 2 < 2 < 2 ÷. è ø

u = a2 cos2 q+b2 sin2 q + a2 sin2 q+b2 cos2 q

=

a2 +b2 a2 - b2 a2 +b2 b2 - a2 + cos 2q + + cos 2q 2 2 2 2 2

2

æa2 +b2 ö æa2 - b2 ö 2 ⇒ u =a +b +2 ç ÷- ç ÷ cos 2q 2 2 è ø è ø 2 2 2 min value of u =a +b +2ab 2

2

2

(

2 2 2 max value of u =2 a +b

)

2

2 2 ⇒ umax - umin =( a - b) .

23.

Greatest side is 1+sin a cos a , by applying cos rule we get greatest angle = 120ο.

24.

tan30° =

h 40 +b ⇒ 3 h =40 +b

h …

..(1)

30°

40

60°

b

tan60° = h/b ⇒ h = 3 b ….(2) ⇒ b = 20 m FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 2651394

15

AIEEE-PAPERS--16

25.

- 2 £ sinx - 3 cos x £ 2 ⇒ - 1 £ sinx ⇒ range of f(x) is [–1, 3]. Hence S is [–1, 3].

26.

If y = f (x) is symmetric about the line x = 2 then f(2 + x) = f(2 – x).

27.

9 - x2 >0 and - 1 £ x - 3 £ 1 ⇒ x Î [2, 3)

28.

æ ç 1 ç b öçç a + b è x x2 2 ÷

2x

æ a bö æ a lim ç1+ + 2 ÷ = lim ç1+ + x ®¥ è x x ø è x x ø

x®¥

29.

3 cos x +1 £ 3

ö ÷ a bö ÷´ 2x´ æ ç + 2÷ ÷ èx x ø ÷ =e2a ø

Þ a =1, b Î R

1- tanx 1- tanx 1 f(x) = Þ lim =p 4x - p 4x - p 2 x® 4

30.

31.

y+ey+.....¥

⇒ x = ey+x dy 1 1- x = - 1= ⇒ lnx – x = y ⇒ . dx x x x =ey+e

æ9 2 ö Any point be ç t , 9t÷; differentiating y2 = 18x è2 ø dy 9 1 1 = = =2 (given) Þ t = . ⇒ dx y t 2 æ9 9 ö ⇒ Point is ç , ÷ è8 2 ø

32.

f″ (x) = 6(x – 1) ⇒ f′ (x) = 3(x – 1)2 + c and f′ (2) = 3 ⇒ c = 0 ⇒ f (x) = (x – 1)3 + k and f (2) = 1 ⇒ k = 0 ⇒ f (x) = (x – 1)3.

33.

Eliminating θ, we get (x – a)2 + y2 = a2. Hence normal always pass through (a, 0).

34.

Let f′(x) = ax2 +bx +c ⇒ f(x) = ⇒ f(x) =

ax3 bx2 + +cx +d 3 2

1 2ax3 +3bx2 +6cx +6d , Now f(1) = f(0) = d, then according to Rolle’s 6

(

)

theorem ⇒ f′(x) = ax2 +bx +c =0 has at least one root in (0, 1) n

35. 36. 16

lim

n®¥

r

1 n e = r =1 n

å

1

òe dx =(e - 1) x

0

Put x – α = t FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 2651394

AIEEE-PAPERS--17

sin(a +t) dt =sin a cottdt +cos a dt sint = cos a ( x - a ) +sin a ln sint +c A = cos a, B = sin a

⇒

37.

ò

ò

1 1 = dx 1 1 dx æ pö æx 3p ö = sec çx + ÷dx = log tan ç + ÷+C 2 cos æx + p ö 4ø cos x - sinx 2 è 2 è2 8 ø ç 4÷ è ø

ò

ò -1

38.

ò

ò( x

2

1

)

- 1 dx +

-2

3

-1

x3 x3 1- x dx + x - 1 dx = - x +x 3 3 1 -2

ò(

-1

p 2

ò

2

)

ò(

2

)

1

3

x3 28 + - x = . 3 3 -1 1

p

2

39.

2 ( sinx +cos x) dx = ( sinx +cos x) dx ò 2 ò 0 ( sinx +cos x) 0

40.

Let I =

p

p

= - cos x +sinx 2 = 2. 0

p

ò

p

ò

ò

xf(sinx)dx = (p - x)f(sinx)dx =p f(sinx)dx - I (since f (2a – x) = f (x))

0 p/ 2

⇒I=π

0

0

òf(sinx)dx ⇒ A = π. 0

41.

f(-a) + f(a) = 1 f(a)

I1 =

ò

f(a)

xg{x(1- x)}dx =

f( - a)

ò ( 1- x) g{x(1-

x)}dx

f( - a)

b æ b ö çQ f ( x) dx = f ( a +b - x) dx ÷ ç ÷ a è a ø

ò

ò

f(a)

2I1 =

ò g{x(1- x)}dx = I

2

⇒ I2 / I1 = 2.

f( - a)

2

42.

Area =

ò

3

y =x – 2

ò

(2 - x)dx + (x - 2)dx = 1.

1

2

y=2 – x

1

43.

2

3

2x + 2yy′ - 2ay′ = 0 x +yy¢ a= (eliminating a) y¢ ⇒ (x2 – y2)y′ = 2xy.

45.

y dx + x dy + x2y dy = 0. FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 2651394

17

AIEEE-PAPERS--18

d(xy)

1 1 + dy =0 ⇒ +logy =C . y xy x y 2 2

45.

46.

If C be (h, k) then centroid is (h/3, (k – 2)/3) it lies on 2x + 3y = 1. ⇒ locus is 2x + 3y = 9. x y 4 3 + =1where a + b = -1 and + =1 a b a b ⇒ a = 2, b = -3 or a = -2, b = 1. x y x y + =1. Hence - =1 and 2 3 -2 1 2c 1 and m1 m2 = 7 7 m1 + m2 = 4m1m2 (given) ⇒ c = 2.

47.

m1 + m2 = -

48.

m1 + m2 =

49.

Let the circle be x2 + y2 + 2gx + 2fy + c = 0 ⇒ c = 4 and it passes through (a, b) ⇒ a2 + b2 + 2ga + 2fb + 4 = 0. Hence locus of the centre is 2ax + 2by – (a2 + b2 + 4) = 0.

50.

Let the other end of diameter is (h, k) then equation of circle is (x – h)(x – p) + (y – k)(y – q) = 0 Put y = 0, since x-axis touches the circle ⇒ x2 – (h + p)x + (hp + kq) = 0 ⇒ (h + p)2 = 4(hp + kq) ⇒ (x – p)2 = 4qy.

1 6 3 , m1m2 = and m1 = - . 4c 4c 4 Hence c = -3.

(D = 0)

51.

Intersection of given lines is the centre of the circle i.e. (1, − 1) Circumference = 10π ⇒ radius r = 5 ⇒ equation of circle is x2 + y2 − 2x + 2y − 23 = 0.

52.

Points of intersection of line y = x with x2 + y2 − 2x = 0 are (0, 0) and (1, 1) hence equation of circle having end points of diameter (0, 0) and (1, 1) is x2 + y2 − x − y = 0.

53.

Points of intersection of given parabolas are (0, 0) and (4a, 4a) ⇒ equation of line passing through these points is y = x On comparing this line with the given line 2bx + 3cy + 4d = 0, we get d = 0 and 2b + 3c = 0 ⇒ (2b + 3c)2 + d2 = 0.

54.

Equation of directrix is x = a/e = 4 ⇒ a = 2 b2 = a2 (1 − e2) ⇒ b2 = 3 Hence equation of ellipse is 3x2 + 4y2 = 12.

18

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AIEEE-PAPERS--19

55.

l = cos θ, m = cos θ, n = cos β cos2 θ + cos2 θ + cos2 β = 1 ⇒ 2 cos2 θ = sin2 β = 3 sin2 θ cos2 θ = 3/5.

(given)

56.

Given planes are 2x + y + 2z − 8 = 0, 4x + 2y + 4z + 5 = 0 ⇒ 2x + y + 2z + 5/2 = 0 | d1 - d2 | | - 8 - 5/ 2 | 7 = . Distance between planes = = 22 +12 +22 2 a2 +b2 +c2

57.

Any point on the line

x y +a z = = =t1 (say) is (t1, t1 – a, t1) and any point on 1 1 1

x +a y z = = =t2 ( say) is (2t2 – a, t2, t2). 2 1 1 Now direction cosine of the lines intersecting the above lines is proportional to (2t2 – a – t1, t2 – t1 + a, t2 – t1). Hence 2t2 – a – t1 = 2k , t2 – t1 + a = k and t2 – t1 = 2k On solving these, we get t1 = 3a , t2 = a. Hence points are (3a, 2a, 3a) and (a, a, a).

the line

y- 1 z - 2 x - 1 y +3 z - 1 x = = =s and = = =t are coplanar then plan 1 -l l 1/ 2 1 -1 passing through these lines has normal perpendicular to these lines a +b - c =0 (where a, b, c are direction ratios of the ⇒ a - bλ + cλ = 0 and 2 normal to the plan) On solving, we get λ = -2.

58.

Given lines

59.

Required plane is S1 – S2 = 0 where S1 = x2 + y2 + z2 + 7x – 2y – z – 13 = 0 and S2 = x2 + y2 + z2 – 3x + 3y + 4z – 8 = 0 ⇒ 2x – y – z = 1.

60.

61.

r

r

r

( a +2br ) =t c

….(1) r r and b +3c =t2a ….(2) r r (1) – 2×(2) ⇒ a ( 1+2t2 ) +c ( - t1 - 6) =0 ⇒ 1+ 2t2 = 0 ⇒ t2 = -1/2 & t1 = -6. r r Since a and c are non-collinear. r r r r Putting the value of t1 and t2 in (1) and (2), we get a +2b +6c =0 . 1

r r r r r r Work done by the forces F1 and F2 is (F1 +F2 ) ×d , where d is displacement r r ˆ +(3iˆ +ˆj - k) ˆ =7iˆ +2jˆ - 4kˆ According to question F1 +F2 = (4iˆ +ˆj - 3k) r r r r ˆ - (iˆ +2jˆ +3k) ˆ =4iˆ +2jˆ - 2kˆ . Hence (F +F ) ×d is 40. and d =(5iˆ +4jˆ +k) 1

2

1 2

63.

3

4 = 0 ⇒ λ = 0, 1/2. Condition for given three vectors to be coplanar is 0 l 0 0 2l - 1 FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 2651394

19

AIEEE-PAPERS--20

Hence given vectors will be non coplanar for all real values of λ except 0, 1/2. Projection of v along u and w along u is

63.

According to question

v ×u w ×u and respectively |u| |u|

v ×u w ×u = ⇒ v ×u =w ×u . and v ×w =0 |u| |u|

| u - v +w |2 =| u |2 +| v |2 +| w |2 - 2u ×v +2u ×w - 2v ×w = 14 ⇒ | u - v +w |= 14 .

r r

r

r r r

r r r

( a ´ b) ´ c = 31 b c a ⇒ ( a ×c) b - ( b ×c) a =31 b c a

64.

r r r æ1 r r 1 ör ⇒ a ×c b =ç b c + b ×c ÷a ⇒ a ×c =0 and b c + b ×c =0 3 3 è ø 1 æ ö 2 2 ⇒ b c ç +cos q÷=0 ⇒ cosθ = –1/3 ⇒ sinθ = . 3 è ø 3

(

)

(

)

(

)

65.

Mode can be computed from histogram and median is dependent on the scale. Hence statement (a) and (b) are correct.

66.

xi =a for i =1, 2, .... ,n and xi =- a for i =n, ...., 2n

1 2n

S.D. =

2n

å ( xi - x) i=1

2

⇒2=

1 2n

2n

å

xi2

i=1

æ çSince è

2n

åx

i

i=1

ö 1 =0÷ ⇒ 2 = ×2na2 ⇒ 2n ø

a =2

67.

E1 : event denoting that A speaks truth E 2 : event denoting that B speaks truth

(

)

(

Probability that both contradicts each other = P E1 Ç E 2 +P E1 Ç E 2

)

=

4 1 1 3 7 × + × = 5 4 5 4 20

68.

P(E È F) =P(E) +P(F) - P ( E Ç F ) = 0.62 + 0.50 – 0.35 = 0.77

69.

Given that n p = 4, n p q = 2 ⇒ q = 1/2 ⇒ p = 1/2 , n = 8 ⇒ p(x = 2) = 2

8

70.

20

6

28 æ1 ö æ1 ö C2 ç ÷ ç ÷ = è2 ø è2 ø 256

æ 1 ö æ 1 ö 2 ÷N . P + Q = 4, P2 + Q2 = 9 ⇒ P = ç2 + 2 ÷N and Q =ç2 2 è ø è 2 ø

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AIEEE-PAPERS--21

71.

F . 3 sin θ = 9 F . 4 cos θ = 16 ⇒ F = 5.

C 4cosθ

θ

A

B

θ 3sinθ

F

72.

By Lami’s theorem r r r P :Q :R = Aö Bö Cö æ æ æ sinç90° + ÷: sin ç90° + ÷: sinç90° + ÷ 2ø 2ø 2ø è è è A B C ⇒ cos : cos : cos . 2 2 2

73.

Time T1 from A to B =

A

90+C/2

90+A/2

B

12 = 3 hrs. 4

C

C 13

5 = 1 hrs. 5 Total time = 4 hrs. 17 Average speed = km/ hr. 4

T2 from B to C =

Resultant average velocity =

74.

Component along OB =

75.

t1 =

90+B/2

5

A

12

B

13 km/hr. 4

1 sin30° 1 4 = sin(45° +30°) 8

(

6-

)

2 m/s.

2usin a 2usinb , t2 = where α + β = 900 g g

2 2 ∴ t1 +t2 =

4u2 . g2

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21

–1–

FIITJEE SOLUTION TO AIEEE-2005 MATHEMATICS 1. 1.

2.

2.

3.

3.

4.

If A2 – A + I = 0, then the inverse of A is (1) A + I (2) A (3) A – I (4) I – A (4) Given A2 – A + I = 0 A–1A2 – A–1A + A–1 – I = A–1⋅0 (Multiplying A–1 on both sides) ⇒ A - I + A-1 = 0 or A–1 = I – A. If the cube roots of unity are 1, ω, ω2 then the roots of the equation (x – 1)3 + 8 = 0, are (1) -1 , - 1 + 2ω, - 1 - 2ω2 (2) -1 , -1, - 1 (3) -1 , 1 - 2ω, 1 - 2ω2 (4) -1 , 1 + 2ω, 1 + 2ω2 (3) (x – 1)3 + 8 = 0 ⇒ (x – 1) = (-2) (1)1/3 ⇒ x – 1 = -2 or -2ω or -2ω2 or n = -1 or 1 – 2ω or 1 – 2ω2. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12} be a relation on the set A = {3, 6, 9, 12}. The relation is (1) reflexive and transitive only (2) reflexive only (3) an equivalence relation (4) reflexive and symmetric only (1) Reflexive and transitive only. e.g. (3, 3), (6, 6), (9, 9), (12, 12) [Reflexive] (3, 6), (6, 12), (3, 12) [Transitive].

(1) 2ab (3) ab 4.

x2 y2 + = 1 is a 2 b2

Area of the greatest rectangle that can be inscribed in the ellipse

(1) Area of rectangle ABCD = (2acosθ) (2bsinθ) = 2absin2θ ⇒ Area of greatest rectangle is equal to 2ab when sin2θ = 1.

(2) ab a (4) b Y (-acosθ, bsinθ) B

A(acosθ, bsinθ) X

(-acosθ, -bsinθ)C

5.

D(acosθ, -bsinθ)

(

)

The differential equation representing the family of curves y2 = 2c x + c , where c > 0, is a parameter, is of order and degree as follows: (1) order 1, degree 2 (2) order 1, degree 1

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–2–

5.

(3) order 1, degree 3 (3) y2 = 2c(x + √c) …(i) 2yy′ = 2c⋅1 or yy′ = c …(ii) ⇒ y2 = 2yy′ (x + yy ′ )

(4) order 2, degree 2

[on putting value of c from (ii) in (i)]

On simplifying, we get …(iii) (y – 2xy′)2 = 4yy′3 Hence equation (iii) is of order 1 and degree 3. 6.

6.

1 2 4 1 1  lim  2 sec 2 2 + 2 sec 2 2 + .... + 2 sec 2 1 equals n n n n n  1 1 (1) (2) cos ec1 sec1 2 2 1 (3) tan1 (4) tan1 2 (4) 1 2 4 3 9 1 1  lim sec 2 2 + 2 sec 2 2 + 2 sec 2 2 + .... + sec 2 1 is equal to n →∞  n2 n n n n n n   n →∞

2 2 r 1 r 2 r 2 r = ⋅ sec lim sec n→∞ n2 n2 n→∞ n n n2

lim

1

⇒ Given limit is equal to value of integral

∫ x sec

2

x 2 dx

0

1

or = 7.

7.

1

1 1 2x sec x 2 dx = ∫ sec 2 tdt ∫ 20 20

[put x2 = t]

1 1 1 ( tan t )0 = tan1 . 2 2

ABC is a triangle. Forces P, Q, R acting along IA, IB and IC respectively are in equilibrium, where I is the incentre of ∆ABC. Then P : Q : R is A B C (1) sinA : sin B : sinC (2) sin : sin : sin 2 2 2 A B C (3) cos : cos : cos (4) cosA : cosB : cosC 2 2 2 (3) A Using Lami’s Theorem A B C ∴ P : Q : R = cos : cos : cos . P 2 2 2 Q B

8.

8.

I

R

C

If in a frequently distribution, the mean and median are 21 and 22 respectively, then its mode is approximately (1) 22.0 (2) 20.5 (3) 25.5 (4) 24.0 (4) Mode + 2Mean = 3 Median ⇒ Mode = 3 × 22 – 2 × 21= 66 – 42= 24.

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–3– 9.

9.

10.

10.

Let P be the point (1, 0) and Q a point on the locus y2 = 8x. The locus of mid point of PQ is (1) y2 – 4x + 2 = 0 (2) y2 + 4x + 2 = 0 2 (4) x2 – 4y + 2 = 0 (3) x + 4y + 2 = 0 (1) P = (1, 0) Q = (h, k) such that k2 = 8h Let (α, β) be the midpoint of PQ h +1 k+0 α= β= , 2 2 2α - 1 = h 2β = k. (2β)2 = 8 (2α - 1) ⇒ β2 = 4α - 2 ⇒ y2 – 4x + 2 = 0. If C is the mid point of AB and P is any point outside AB, then (1) PA + PB = 2PC (2) PA + PB = PC (3) PA + PB + 2PC = 0 (4) PA + PB + PC = 0 (1) PA + AC + CP = 0 PB + BC + CP = 0 Adding, we get PA + PB + AC + BC + 2CP = 0 Since AC = − BC

& CP = −PC ⇒ PA + PB − 2PC = 0 . 11.

11.

A

C

P

B

If the coefficients of rth, (r+ 1)th and (r + 2)th terms in the binomial expansion of (1 + y)m are in A.P., then m and r satisfy the equation (2) m2 – m(4r+1) + 4r2 + 2 = 0 (1) m2 – m(4r – 1) + 4r2 – 2 = 0 2 2 (4) m2 – m(4r – 1) + 4r2 + 2 = 0 (3) m – m(4r + 1) + 4r – 2 = 0 (3) Given m Cr −1, mCr , mCr +1 are in A.P.

2 mCr = mCr −1 + mCr +1 ⇒ 2=

m

Cr −1 m Cr +1 + m Cr Cr

m

r m−r + m −r +1 r +1 ⇒ m2 – m (4r + 1) + 4r2 – 2 = 0. π P Q In a triangle PQR, ∠R = . If tan   and tan   are the roots of 2 2 2 ax2 + bx + c = 0, a ≠ 0 then (1) a = b + c (2) c = a + b (3) b = c (4) b = a + c (2) P Q tan   , tan   are the roots of ax2 + bx + c = 0 2   2 b P Q tan   + tan   = − a 2 2 =

12.

12.

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–4– P Q c tan   tan   = 2   2 a P Q tan   + tan   2    2  = tan  P + Q  = 1 2 2 P Q   1 − tan   tan   2 2 b − a = 1 ⇒ − b = a − c ⇒ −b = a − c ⇒ c a a a 1− a c = a + b. 13.

13.

The system of equations αx + y + z = α - 1, x + αy + z = α - 1, x + y + αz = α - 1 has no solution, if α is (1) -2 (3) not -2 (1) αx + y + z = α - 1 x + αy + z = α - 1 x + y + zα = α - 1 α 1 1 ∆= 1 α 1 1 1 α

(2) either – 2 or 1 (4) 1

= α(α2 – 1) – 1(α - 1) + 1(1 - α) = α (α - 1) (α + 1) – 1(α - 1) – 1(α - 1) ⇒ (α - 1)[α2 + α - 1 – 1] = 0 ⇒ (α - 1)[α2 + α - 2] = 0 [α2 + 2α - α - 2] = 0 (α - 1) [α(α + 2) – 1(α + 2)] = 0 (α - 1) = 0, α + 2 = 0 ⇒ α = –2, 1; but α ≠ 1. 14.

14.

The value of α for which the sum of the squares of the roots of the equation x2 – (a – 2)x – a – 1 = 0 assume the least value is (1) 1 (2) 0 (3) 3 (4) 2 (1) x2 – (a – 2)x – a – 1 = 0 ⇒α+β=a–2 α β = –(a + 1) α2 + β2 = (α + β)2 - 2αβ = a2 – 2a + 6 = (a – 1)2 + 5 ⇒ a = 1.

15.

If roots of the equation x2 – bx + c = 0 be two consectutive integers, then b2 – 4c equals (1) – 2 (2) 3

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–5–

15.

16.

16.

17.

(3) 2 (4) Let α, α + 1 be roots α+α+1=b α(α + 1) = c ∴ b2 – 4c = (2α + 1)2 - 4α(α + 1) = 1.

If the letters of word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number (1) 601 (2) 600 (3) 603 (4) 602 (1) Alphabetical order is A, C, H, I, N, S No. of words starting with A – 5! No. of words starting with C – 5! No. of words starting with H – 5! No. of words starting with I – 5! No. of words starting with N – 5! SACHIN – 1 601. The value of

50

C4 +

6

∑ r =1

17.

(4) 1

56 −r

C3 is

(1) 55C4 (3) 56C3 (4) 50

C4 +

(2) 55C3 (4) 56C4 6

∑

56 −r

r =1

⇒ =

(

⇒

50 50

(

C3

C4 +  55 C3 + 54 C3 + 53 C3 + 52C3 + 51C3 + 50 C3 

) C )+

C4 + 50 C3 + 51C3 + 52C3 + 53 C3 + 54 C3 + 55 C3

51

C4 + 51

3

52

C3 + 53 C3 + 54 C3 + 55 C3

⇒ 55C4 + 55C3 = 56C4. 18.

18.

19.

19.

1 0   1 0 and I =  If A =    , then which one of the following holds for all n ≥ 1, by 1 1  0 1 the principle of mathematical indunction (2) An = 2n-1A – (n – 1)I (1) An = nA – (n – 1)I n (4) An = 2n-1A + (n – 1)I (3) A = nA + (n – 1)I (1) By the principle of mathematical induction (1) is true.   1  If the coefficient of x in ax 2 +     bx    then a and b satisfy the relation (1) a – b = 1 a (3) = 1 b (4) 7

FIITJEE

11

11

  1  equals the coefficient of x in ax 2 −    ,  bx    -7

(2) a + b = 1 (4) ab = 1

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–6– 11

r

11−r  1  1  Tr + 1 in the expansion ax 2 +  = 11Cr ax 2   bx    bx  11 11 – r -r 22 – 2r – r (b) (x) = Cr (a) ⇒ 22 – 3r = 7 ⇒ r = 5 ∴ coefficient of x7 = 11C5(a)6 (b)-5 ……(1) 11 r 1  1  11−r   Again Tr + 1 in the expansion ax − 2  = 11Cr ( ax )  − 2  bx    bx  11 11 – r r -r -2r 11 - r (-1) × (b) (x) (x) = Cr a Now 11 – 3r = -7 ⇒ 3r = 18 ⇒ r = 6 ∴ coefficient of x-7 = 11C6 a5 × 1 × (b)-6

⇒

11

C5 ( a ) ( b ) 6

−5

= 11C6a5 × ( b )

(

)

−6

⇒ ab = 1. 20.

20.

Let f : (-1, 1) → B, be a function defined by f(x) = tan−1 and onto when B is the interval  π (1)  0,   2  π π (3)  − ,   2 2 (4)  2x  Given f(x) = tan−1  for x∈(-1, 1) 2   1− x   π π clearly range of f(x) =  − ,   2 2

2x , then f is both one-one 1 − x2

 π (2) 0,   2  π π (4)  − ,   2 2

 π π ∴ co-domain of function = B =  − ,  .  2 2 21.

21.

22.

22.

If z1 and z2 are two non-zero complex numbers such that |z1 + z2| = |z1| + |z2| then argz1 – argz2 is equal to π (2) - π (1) 2 π (3) 0 (4) 2 (3) |z1 + z2| = |z1| + |z2| ⇒ z1 and z2 are collinear and are to the same side of origin; hence arg z1 – arg z2 = 0. If ω =

z

and |ω| = 1, then z lies on 1 z− i 3 (1) an ellipse (3) a straight line (3)

FIITJEE

(2) a circle (4) a parabola.

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–7– z

As given w =

z−

1 i 3

⇒ |w| =

|z| = 1 ⇒ distance of z from origin and point 1 | z − i| 3

 1  0, 3  is same hence z lies on bisector of the line joining points (0, 0) and (0, 1/3).   Hence z lies on a straight line.

(1 + b ) x (1 + c ) x ) x 1 + b x (1 + c ) x ) x (1 + b ) x 1 + c x

1 + a2 x 23.

23.

( (1 + a

If a2 + b2 + c2 = -2 and f(x) = 1 + a2 2

polynomial of degree (1) 1 (3) 3 (4)

2

2

2

2

2

2

then f(x) is a

(2) 0 (4) 2

( ) (1 + b ) x (1 + c ) x f ( x ) = 1 + ( a + b + c + 2 ) x 1 + b x (1 + c ) x , Applying C 1 + ( a + b + c + 2 ) x (1 + b ) x 1 + c x 1 (1 + b ) x (1 + c ) x = 1 1 + b x (1 + c ) x ∵ a + b + c + 2 = 0 1 (1 + b ) x 1 + c x 1 + a2 + b 2 + c 2 + 2 x

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

0

x −1

f(x) = 0

1− x

1

2

(1 + b ) x 2

2

2

1

→ C1 + C2 + C3

2

0 x − 1 ; Applying R1 → R1 – R2 , R2 → R2 – R3 1 + c2 x

f(x) = (x – 1)2 Hence degree = 2. 24.

24.

The normal to the curve x = a(cosθ + θ sinθ), y = a( sinθ - θ cosθ) at any point ‘θ’ is such that (1) it passes through the origin π (2) it makes angle + θ with the x-axis 2  π  (3) it passes through  a , − a   2  (4) it is at a constant distance from the origin (4) dy Clearly = tan θ ⇒ slope of normal = - cot θ dx Equation of normal at ‘θ’ is y – a(sin θ - θ cos θ) = - cot θ(x – a(cos θ + θ sin θ) ⇒ y sin θ - a sin2 θ + a θ cos θ sin θ = -x cos θ + a cos2 θ + a θ sin θ cos θ ⇒ x cos θ + y sin θ = a Clearly this is an equation of straight line which is at a constant distance ‘a’ from origin.

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–8– 25.

25.

A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched? Interval Function (1) (-∞, ∞) x3 – 3x2 + 3x + 3 (2) [2, ∞) 2x3 – 3x2 – 12x + 6 1  3x2 – 2x + 1 (3)  −∞,  3  (4) (- ∞, -4] x3 + 6x2 + 6 (3) Clearly function f(x) = 3x2 – 2x + 1 is increasing when f′(x) = 6x – 2 ≥ 0 ⇒ x∈[1/3, ∞) Hence (3) is incorrect. 2

26.

Let α and β be the distinct roots of ax + bx + c = 0, then lim

26.

equal to a2 2 (1) (α − β) 2 a2 2 (3) − ( α − β ) 2 (1)

x →α

x →α

=

27.

28.

( x − α)

2

)

is

(2) 0 (4)

1 2 (α − β) 2

 ( x − α )( x − β )  2 sin2  a  2 1 − cos a ( x − α )( x − β )   Given limit = lim = lim 2 2 x →α x →α − α − α x x ( ) ( )

= lim

27.

(

1 − cos ax 2 + bx + c

 ( x − α )( x − β )  sin2  a  2 2 2 a2 ( x − α ) ( x − β )   × × 2 2 4 a2 ( x − α ) ( x − β ) 4

2

(x − α)

a2 ( α − β ) 2

2

2

.

1 Suppose f(x) is differentiable x = 1 and lim f (1 + h ) = 5 , then f′(1) equals h →0 h (1) 3 (2) 4 (3) 5 (4) 6 (3) f (1 + h ) − f (1) ; As function is differentiable so it is continuous as it is given f ′ (1) = lim h→0 h f (1 + h ) that lim = 5 and hence f(1) = 0 h →0 h f (1 + h ) Hence f′(1) = lim =5 h →0 h Hence (3) is the correct answer. Let f be differentiable for all x. If f(1) = - 2 and f′(x) ≥ 2 for x ∈ [1, 6] , then (1) f(6) ≥ 8 (2) f(6) < 8 (3) f(6) < 5 (4) f(6) = 5

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–9– 28.

(1) As f(1) = - 2 & f′(x) ≥ 2 ∀ x ∈ [1, 6] Applying Lagrange’s mean value theorem f ( 6 ) − f (1) = f ′ (c ) ≥ 2 5 ⇒ f(6) ≥ 10 + f(1) ⇒ f(6) ≥ 10 – 2 ⇒ f(6) ≥ 8.

29.

If f is a real-valued differentiable function satisfying |f(x) – f(y)| ≤ (x – y)2, x, y ∈ R and f(0) = 0, then f(1) equals (1) -1 (2) 0 (3) 2 (4) 1 (2) f ( x + h) − f ( x ) f′(x) = lim h→0 h

29.

| f ′ ( x ) |= lim

f ( x + h) − f ( x ) h

h→0

(h) ≤ lim h→0

2

h

⇒ |f′(x)| ≤ 0 ⇒ f′(x) = 0 ⇒ f(x) = constant As f(0) = 0 ⇒ f(1) = 0. 30.

If x is so small that x3 and higher powers of x may be neglected, then 1   − 1 + x  2   1/ 2 (1 − x )

(1 + x ) (1) 1 − (3) − 30.

3/2

3

may be approximated as

3 2 x 8

3 2 x 8 x 3 (4) − x 2 2 8

(2) 3x +

3 2 x 8

(3) 2  3 33  2 1  1   (1 – x) 1 + x +  − 1 x − 1 − 3  x  − 3 ( 2 )  x   22  2   2    2 3  3  = (1 – x)1/2  − x 2  = - x 2 . 8  8 

1/2

31.

If x =

∞

∑a , n =0

31.

n

∞

∞

n=0

n =0

y = ∑ bn , z = ∑ c n where a, b, c are in A.P. and |a| < 1, |b| I1 (3) I3 = I4 (2)

2

0

1

3

0

2

2

x x ∫ 2 dx and I4 = ∫ 2 dx then 2

1

3

1

(2) I1 > I2 (4) I3 > I4 1

1

I1 = ∫ 2 dx , I2 = ∫ 2 dx , I3 = ∫ 2 dx , I4 = x2

0

x3

0

x2

0

1

∫2

x3

dx

0

∀ 0 < x < 1, x2 > x3 1

⇒

x ∫ 2 dx > 2

0

1

∫2

x3

dx

0

⇒ I1 > I2. 36. 36.

The area enclosed between the curve y = loge (x + e) and the coordinate axes is (1) 1 (2) 2 (3) 3 (4) 4 (1) 0

Required area (OAB) =

∫ ln ( x + e )dx

1− e

1

1   x dx  = 1. =  x ln ( x + e ) − ∫ x+e  0 37.

37.

The parabolas y2 = 4x and x2 = 4y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes. If S1, S2, S3 are respectively the areas of these parts numbered from top to bottom; then S1 : S2 : S3 is (1) 1 : 2 : 1 (2) 1 : 2 : 3 (3) 2 : 1 : 2 (4) 1 : 1 : 1 (4) y2 = 4x and x2 = 4y are symmetric about line y = x 4 8 2 ⇒ area bounded between y = 4x and y = x is ∫ 2 x − x dx = 3 0 16 16 and A s1 = A s3 = ⇒ A s2 = 3 3 ⇒ A s1 : A s2 : A s3 :: 1 : 1 : 1.

(

38.

38.

)

dy = y (log y − log x + 1), then the solution of the equation is dx x y (1) y log   = cx (2) x log   = cy y x   If x

y (3) log   = cx x (3) x dy = y (log y – log x + 1) dx dy y  y  =  log   + 1 dx x  x  Put y = vx

FIITJEE

x (4) log   = cy y

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–12– dy x dv =v+ dx dx xdv ⇒ v+ = v ( log v + 1) dx xdv = v log v dx dv dx ⇒ = v log v x put log v = z 1 dv = dz v dz dx ⇒ = z x ln z = ln x + ln c z = cx log v = cx y log   = cx . x 39.

39.

The line parallel to the x−axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx − 2ay − 3a = 0, where (a, b) ≠ (0, 0) is 3 from it (1) below the x−axis at a distance of 2 2 (2) below the x−axis at a distance of from it 3 3 (3) above the x−axis at a distance of from it 2 2 from it (4) above the x−axis at a distance of 3 (1) ax + 2by + 3b + λ(bx – 2ay – 3a) = 0 ⇒ (a + bλ)x + (2b – 2aλ)y + 3b - 3λa = 0 a + bλ = 0 ⇒ λ = -a/b a (bx – 2ay – 3a) = 0 ⇒ ax + 2by + 3b b 2a2 3a2 ⇒ ax + 2by + 3b – ax + =0 y+ b b  2a2  3a2 y  2b + =0  + 3b + b  b 

 2b2 + 2a2   3b2 + 3a2  y  = −  b b     y=

( 2 (b

) = −3 ) 2

−3 a 2 + b 2

y=−

FIITJEE

2

+a

2

3 so it is 3/2 units below x-axis. 2

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–13– 40.

40.

A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness than melts at a rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is 1 1 cm/min (2) cm/min (1) 36π 18π 1 5 cm/min (3) cm/min (4) 54π 6π (2) dv = 50 dt dr = 50 4πr2 dt dr 50 = ⇒ where r = 15 dt 4π (15 )2 =

1 . 16π 2

41.

 (log x − 1)  ∫  (1 + (log x)2  dx is equal to log x (1) +C (log x)2 + 1 (3)

41.

xe x +C 1 + x2

(2)

x +C x +1

(4)

x +C (log x)2 + 1

2

(4)

(log x − 1) dx ∫ 2 2 1 + log x ( ) ( ) 2

  1 2log x   = ∫ − dx 2 2 2 1 + ( log x )   1 + ( log x )    t  e 2t et  − dt put logx = t ⇒ dx = et dt = ∫ 2 2 1 + t 2  1+ t     1 2t  t  − e ∫ 1 + t 2 1 + t 2 2 dt  

(

=

42.

) (

(

)

(

)

)

x et +c +c = 2 2 1+ t 1 + ( log x )

 1  Let f : R → R be a differentiable function having f (2) = 6, f′ (2) =   . Then  48  f ( x) 4t 3 lim ∫ dt equals x →2 x−2 6 (1) 24 (2) 36 (3) 12 (4) 18

FIITJEE

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–14– 42.

(4)

f ( x)

4t 3 dt x→2 ∫ x − 2 0 Applying L Hospital rule 2 lim  4f ( x ) f ′ ( x )  = 4f(2)3 f′(2)  x →2  1 = 18. = 4 × 63 × 48 lim

43.

43.

44.

44.

Let f (x) be a non−negative continuous function such that the area bounded by the π π and x = β > curve y = f (x), x−axis and the ordinates x = 4 4 π  π   is  β sin β + cos β + 2β  . Then f   is 4 2   π  π  (1)  + 2 − 1 (2)  − 2 + 1 4  4  π π     (3)  1 − − 2  (4)  1 − + 2  4 4     (4) β π Given that ∫ f ( x ) dx = β sin β + cos β + 2 β 4 π/4 Differentiating w. r. t β π f(β) = β cosβ + sinβ sinβ + 2 4 π π π  π  f   =  1 −  sin + 2 = 1 − + 2 . 4 2 4 2  The locus of a point P (α, β) moving under the condition that the line y = αx + β is a x2 y2 tangent to the hyperbola 2 − 2 = 1 is a b (1) an ellipse (2) a circle (3) a parabola (4) a hyperbola (4) x2 y2 Tangent to the hyperbola 2 − 2 = 1 is a b y = mx ± a2m2 − b 2 Given that y = αx + β is the tangent of hyperbola ⇒ m = α and a2m2 – b2 = β2 ∴ a2α2 – b2 = β2 Locus is a2x2 – y2= b2 which is hyperbola.

45.

If the angle θ between the line 0 is such that sin θ = (1)

5 3

FIITJEE

x +1 y −1 z − 2 = = and the plane 2x − y + 1 2 2

λz+4=

1 the value of λ is 3 (2)

−3 5

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–15–

(3) 45.

3 4

−4 3

(4)

(1) Angle between line and normal to plane is π  2−2+2 λ where θ is angle between line & plane cos  − θ  = 2  3× 5 + λ

⇒ sinθ =

2 λ 3 5+λ

=

1 3

5 ⇒λ= . 3 46.

The angle between the lines 2x = 3y = − z and 6x = − y = − 4z is (2) 900 (1) 00 0 (4) 300 (3) 45

46.

(2) Angle between the lines 2x = 3y = - z & 6x = -y = -4z is 90° Since a1a2 + b1b2 + c1c2 = 0.

47.

If the plane 2ax − 3ay + 4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres x2 + y2 + z2 + 6x − 8y − 2z = 13 and x2 + y2 + z2 − 10x + 4y − 2z = 8, then a equals (1) − 1 (2) 1 (4) 2 (3) − 2 (3) Plane 2ax – 3ay + 4az + 6 = 0 passes through the mid point of the centre of spheres x2 + y2 + z2 + 6x – 8y – 2z = 13 and x2 + y2 + z2 – 10x + 4y – 2z = 8 respectively centre of spheres are (-3, 4, 1) & (5, - 2, 1) Mid point of centre is (1, 1, 1) Satisfying this in the equation of plane, we get 2a – 3a + 4a + 6 = 0 ⇒ a = -2.

47.

48.

The distance between the line ˆ = 5 is r ⋅ (iˆ + 5ˆj + k)

ˆ r = 2iˆ − 2ˆj + 3kˆ + λ(iˆ − ˆj + 4k)

and the plane

10 10 (2) 9 3 3 3 10 (4) (3) 10 3 (2) Distance between the line r = 2iˆ − 2jˆ + 3kˆ + λ ˆi − ˆj + 4kˆ and the plane r ⋅ ˆi + 5ˆj + kˆ = 5 is (1)

48.

(

)

(

)

equation of plane is x + 5y + z = 5 ∴ Distance of line from this plane = perpendicular distance of point (2, -2, 3) from the plane i.e.

FIITJEE

2 − 10 + 3 − 5 1+ 5 + 1 2

=

10 3 3

.

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–16– 49.

ˆ 2 is equal to For any vector a , the value of (a × ˆi)2 + (a × ˆj)2 + (a × k)

49.

(1) 3a2 (3) 2a2 (3) Let a = xiˆ + yjˆ + zkˆ a × ˆi = zjˆ − ykˆ

(

⇒ a × ˆi

)

2

(2) a2 (4) 4a2

= y 2 + z2

(

similarly a × ˆj

)

2

= x 2 + z2

(a × kˆ ) = x + y ⇒ (a × ˆi ) = y + z similarly ( a × ˆj ) = x + z and ( a × kˆ ) = x + y ⇒ ( a × ˆi ) + ( a × ˆj ) + ( a × kˆ ) = 2 ( x + y + z ) = 2 a 2

and

2

2

2

2

50.

2

2

2

2

2

2

2

2

2

2

2

2

2

2

.

If non-zero numbers a, b, c are in H.P., then the straight line passes through a fixed point. That point is (1) (-1, 2) (3) (1, -2)

x y 1 + + = 0 always a b c

(2) (-1, -2) 1  (4)  1, −  2 

50.

(3) a, b, c are in H.P. 2 1 1 − − =0 ⇒ b a c x y 1 + + =0 a b c x y 1 ⇒ = = ∴ x = 1, y = -2 −1 2 −1

51.

If a vertex of a triangle is (1, 1) and the mid-points of two sides through this vertex are (-1, 2) and (3, 2), then the centroid of the triangle is 7   −1 7  (2)  ,  (1)  −1,  3   3 3  7  1 7 (4)  ,  (3)  1,   3 3 3 (3) A(1, 1) Vertex of triangle is (1, 1) and midpoint of sides through this vertex is (-1, 2) and (3, 2) ⇒ vertex B and C come out to be (-3, 3) and (5, 3) (-1, 2) 1− 3 + 5 1+ 3 + 3 (3, 2) , ∴ centroid is 3 3 ⇒ (1, 7/3)

51.

B

FIITJEE

C

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–17–

52.

52.

53.

53.

If the circles x2 + y2 + 2ax + cy + a = 0 and x2 + y2 – 3ax + dy – 1 = 0 intersect in two distinct points P and Q then the line 5x + by – a = 0 passes through P and Q for (1) exactly one value of a (2) no value of a (3) infinitely many values of a (4) exactly two values of a (2) S1 = x2 + y2 + 2ax + cy + a = 0 S2 = x2 + y2 – 3ax + dy – 1 = 0 Equation of radical axis of S1 and S2 S1 – S2 = 0 ⇒ 5ax + (c – d)y + a + 1 = 0 Given that 5x + by – a = 0 passes through P and Q a c − d a +1 ⇒ = = 1 b −a 2 ⇒ a + 1 = -a a2 + a + 1 = 0 No real value of a. A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centre of the circle is (1) an ellipse (2) a circle (3) a hyperbola (4) a parabola (4) Equation of circle with centre (0, 3) and radius 2 is x2 + (y – 3)2 = 4. Let locus of the variable circle is (α, β) ∵ It touches x-axis. ∴ It equation (x - α)2 + (y - β)2 = β2 Circles touch externally ∴

α2 + (β − 3 ) = 2 + β 2

α2 + (β - 3)2 = β2 + 4 + 4β α2 = 10(β - 1/2) ∴ Locus is x2 = 10(y – 1/2) which is parabola.

54.

54.

55.

(α, β)

If a circle passes through the point (a, b) and cuts the circle x2 + y2 = p2 orthogonally, then the equation of the locus of its centre is (1) x2 + y2 – 3ax – 4by + (a2 + b2 – p2) = 0 (2) 2ax + 2by – (a2 – b2 + p2) = 0 (3) x2 + y2 – 2ax – 3by + (a2 – b2 – p2) = 0 (4) 2ax + 2by – (a2 + b2 + p2) = 0 (4) Let the centre be (α, β) ∵ It cut the circle x2 + y2 = p2 orthogonally 2(-α) × 0 + 2(-β) × 0 = c1 – p2 c1 = p2 Let equation of circle is x2 + y2 - 2αx - 2βy + p2 = 0 It pass through (a, b) ⇒ a2 + b2 - 2αa - 2βb + p2 = 0 Locus ∴ 2ax + 2by – (a2 + b2 + p2) = 0. An ellipse has OB as semi minor axis, F and F′ its focii and the angle FBF′ is a right angle. Then the eccentricity of the ellipse is 1 1 (2) (1) 2 2

FIITJEE

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–18–

(3) 55.

1 4

(4)

(1) ∵ ∠FBF′ = 90o ∴

(

a2 e2 + b2

2

a2 e2 + b2

)

2

= (2ae)2

⇒ 2(a2 e2 + b2) = 4a2e2 ⇒ e2 = b2/a2 Also e2 = 1- b2/a2 = 1 – e2

56.

56.

3 B(0, b)

) +(

⇒ 2e2 = 1, e =

1

1 2

F′(-ae, 0)

O

F(ae, 0)

.

ˆ ˆi + kˆ and Let a, b and c be distinct non-negative numbers. If the vectors aiˆ + ajˆ + ck, ciˆ + cjˆ + bkˆ lie in a plane, then c is (1) the Geometric Mean of a and b (2) the Arithmetic Mean of a and b (3) equal to zero (4) the Harmonic Mean of a and b (1) Vector a ˆi + ajˆ + ckˆ , ˆi + kˆ and ciˆ + cjˆ + bkˆ are coplanar a a c 1 0 1 = 0 ⇒ c2 = ab c c b ∴ a, b, c are in G.P.

57.

If a, b, c are non-coplanar vectors and λ is a real number then

(

57.

)

 λ a + b λ 2b λc  = a b + c b  for     (1) exactly one value of λ (3) exactly three values of λ (2)  λ a + b λ 2b λ c  =  a b + c b      λ λ 0 1 0 0

(

(2) no value of λ (4) exactly two values of λ

)

0 λ2 0 0

0 =0 1 1 λ 0 1 0

⇒ λ4 = -1 Hence no real value of λ.

58.

58.

ˆ b = xiˆ + ˆj + (1 − x ) kˆ and c = yiˆ + xjˆ + (1 + x − y ) kˆ . Then a, b, c  Let a = ˆi − k,   depends on (1) only y (2) only x (3) both x and y (4) neither x nor y (4) a = ˆi − kˆ , b = xiˆ + ˆj + (1 − x ) kˆ and c = yiˆ + xjˆ + (1 + x − y ) kˆ

(

a b c  = a ⋅ b × c  

FIITJEE

)

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–19– ˆi

ˆj

kˆ

b×c = x 1 1 − x = ˆi (1 + x – x –x2) - ˆj (x + x2- xy – y + xy) + kˆ (x2 – y) y x 1+ x − y

(

)

a. b × c = 1

which does not depend on x and y. 59.

59.

60.

60.

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is 2 1 (1) (2) 9 9 8 7 (3) (4) 9 9 (2) For a particular house being selected 1 Probability = 3 1  1 1 1 Prob(all the persons apply for the same house) =  × ×  3 = . 9 3 3 3 A random variable X has Poisson distribution with mean 2. Then P(X > 1.5) equals 2 (1) 2 (2) 0 e 3 3 (3) 1 − 2 (4) 2 e e (3) λk P(x = k) = e−λ k! P(x ≥ 2) = 1 – P(x = 0) – P(x = 1) λ = 1 – e-λ – e-λ  1!    =1-

61.

61.

3 . e2

(

)

1 1 1 , P ( A ∩ B ) = and P A = , 6 4 4 where A stands for complement of event A. Then events A and B are (1) equally likely and mutually exclusive (2) equally likely but not independent (3) independent but not equally likely (4) mutually exclusive and independent (3) 1 1 1 P A ∪ B = , P(A ∩ B) = and P A = 6 4 4 ⇒ P(A ∪ B) = 5/6 P(A) = 3/4 Also P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ⇒ P(B) = 5/6 – 3/4 + 1/4 = 1/3 P(A) P(B) = 3/4 – 1/3 = 1/4 = P(A ∩ B) Let A and B be two events such that P A ∪ B =

(

FIITJEE

)

( )

( )

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–20– Hence A and B are independent but not equally likely. 62.

62.

63.

63.

A lizard, at an initial distance of 21 cm behind an insect, moves from rest with an acceleration of 2 cm/s2 and pursues the insect which is crawling uniformly along a straight line at a speed of 20 cm/s. Then the lizard will catch the insect after (1) 20 s (2) 1 s (3) 21 s (4) 24 s (3) 1 2 2t = 21 + 20t 2 ⇒ t = 21. Two points A and B move from rest along a straight line with constant acceleration f and f′ respectively. If A takes m sec. more than B and describes ‘n’ units more than B in acquiring the same speed then (1) (f - f′)m2 = ff′n (2) (f + f′)m2 = ff′n 1 1 (3) ( f + f ′ ) m = ff ′n2 (4) ( f ′ − f ) n = ff ′m2 2 2 (4) v2 = 2f(d + n) = 2f′d v = f′(t) = (m + t)f eliminate d and m we get 1 (f′ - f)n = ff ′m2 . 2

64.

A and B are two like parallel forces. A couple of moment H lies in the plane of A and B and is contained with them. The resultant of A and B after combining is displaced through a distance 2H H (2) (1) A −B A +B H H (3) (4) A −B 2 ( A + B)

64.

(2) (A + B) = d = H  H  d=  .  A +B

65.

The resultant R of two forces acting on a particle is at right angles to one of them and its magnitude is one third of the other force. The ratio of larger force to smaller one is (1) 2 : 1 (2) 3 : 2

65.

(3) 3 : 2 (4) F′ = 3F cos θ F = 3F sin θ ⇒ F′ = 2 2 F

(4) 3 : 2 2 F 3F

F : F′ : : 3 : 2 2 . F′

FIITJEE

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–21–

66.

The sum of the series 1 + (1) (3)

e −1 e e −1 2 e

1 1 1 + + + ......... ad inf. is 4.2! 16.4! 64.6! e +1 (2) e e +1 (4) 2 e

66.

(4) e x + e− x x2 x4 x6 = 1+ + + + ....... 2 2! 4! 6! putting x = 1/2 we get e +1 . 2 e

67.

The value of

π

cos2 x dx, a > 0, is x ∫ −π 1 + a

(1) a π (3)

(2)

π a

π 2

(4) 2 π

67.

(2) π π cos2 x π 2 = dx ∫−π 1 + ax ∫0 cos x dx = 2 .

68.

The plane x + 2y – z = 4 cuts the sphere x2 + y2 + z2 – x + z – 2 = 0 in a circle of radius (1) 3 (2) 1 (3) 2 (4) 2 (2) 1 1 Perpendicular distance of centre  , 0, −  from x + 2y – 2 = 4 2 2

68.

1 1 + −4 3 2 2 = 2 6 5 3 − = 1. radius = 2 2 69.

69.

If the pair of lines ax2 + 2(a + b)xy + by2 = 0 lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then (1) 3a2 – 10ab + 3b2 = 0 (2) 3a2 – 2ab + 3b2 = 0 2 2 (3) 3a + 10ab + 3b = 0 (4) 3a2 + 2ab + 3b2 = 0 (4) 2

(a + b)

2

a+b

− ab

=1

⇒ (a + b)2 = 4(a2 + b2 + ab) ⇒ 3a2 + 3b2 + 2ab = 0.

FIITJEE

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–22– 70.

Let x1, x2, …,xn be n observations such that

70.

possible value of n among the following is (1) 15 (3) 9 (2)

∑x

 ∑ xi  ≥   n  n  ⇒ n ≥ 16.

71.

71.

2

∑x

2 i

(2) 18 (4) 12

2

i

A particle is projected from a point O with velocity u at an angle of 60o with the horizontal. When it is moving in a direction at right angles to its direction at O, its velocity then is given by u u (1) (2) 3 2 2u u (4) (3) 3 3 (4) u cos 60o = v cos 30o 4 . v= 30o 3 o

60

72.

72.

73.

73.

74.

= 400 and ∑ xi = 80 . Then a

30o

If both the roots of the quadratic equation x2 – 2kx + k2 + k – 5 = 0 are less than 5, then k lies in the interval (1) (5, 6] (2) (6, ∞) (3) (-∞, 4) (4) [4, 5] (3) −b 0 ⇒ k∈(-∞, 4). If a1, a2, a3,…, an,… are in G.P., then the determinant logan logan+1 logan+ 2 ∆ = logan+ 3 logan+ 4 logan+ 5 is equal to logan + 6 logan + 7 logan+ 8 (1) 1 (3) 4 (2) C1 – C2, C2 – C3 two rows becomes identical Answer: 0.

(2) 0 (4) 2

A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to (1) –f(x) (2) f(x) (3) f(a) + f(a – x) (4) f(-x)

FIITJEE

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–23– 74.

(1) f(a – (x – a)) = f(a) f(x – a) – f(0) f(x) = -f(x) ∵ x = 0, y = 0, f ( 0 ) = f 2 ( 0 ) − f 2 ( a ) ⇒ f 2 ( a ) = 0 ⇒ f ( a ) = 0  .

75.

If the equation an xn + an−1x n−1 + ...... + a1x = 0 , a1 ≠ 0, n ≥ 2, has a positive root x = α, then the equation nan x n−1 + (n − 1) an−1x n− 2 + ..... + a1 = 0 has a positive root, which is

75.

(1) greater than α (3) greater than or equal to α (2) f(0) = 0, f(α) = 0 ⇒ f′(k) = 0 for some k∈(0, α).

FIITJEE

(2) smaller than α (4) equal to α

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FIITJEE Solutions to AIEEE−2006

MATHEMATICS PART − A

Ans.

ABC is a triangle, right angled at A. The resultant of the forces acting along AB, AC 1 1 with magnitudes and respectively is the force along AD , where D is the AB AC foot of the perpendicular from A onto BC. The magnitude of the resultant is AB2 + AC2 (AB)(AC) (2) (1) 2 2 AB + AC (AB) (AC) 1 1 1 + (4) (3) AB AC AD (4)

Sol:

Magnitude of resultant

1.

2

C

2

AB2 + AC2  1   1  + =  AB   AC  AB ⋅ AC     BC BC 1 = = = AB ⋅ AC AD ⋅ BC AD

=

D

A

B

2.

Suppose a population A has 100 observations 101, 102, … , 200, and another population B has 100 observations 151, 152, … , 250. If VA and VB represent the V variances of the two populations, respectively, then A is VB (1) 1 (2) 9/4 (3) 4/9 (4) 2/3

Ans.

(1)

Sol:

σ2x =

∑d

2 i

n

. (Here deviations are taken from the mean)

Since A and B both has 100 consecutive integers, therefore both have same standard deviation and hence the variance. ∴

( ∑d

VA = 1 As VB

2 i

)

is same in both the cases .

3.

If the roots of the quadratic equation x2 + px + q = 0 are tan30° and tan15°, respectively then the value of 2 + q − p is (3) 2 (2) 3 (3) 0 (4) 1

Ans. Sol:

(2) x2 + px + q = 0 tan 30° + tan 15° = − p tan 30° ⋅ tan 15° = q

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FIITJEE Solutions to AIEEE−2006 tan 45° =

−p tan 30° + tan15° = =1 1 − tan 30° tan15° 1 − q

⇒−p=1−q ⇒ q − p = 1 ∴ 2 + q − p = 3. 6

4.

The value of the integral,

∫ 3

x 9−x + x

dx is

(1) 1/2 (3) 2 Ans.

(2)

Sol:

I=

(2) 3/2 (4) 1

6

x

∫

9−x + x

∫

9−x + x

3 6

I=

9−x

3

dx

dx

6

2I =

∫ dx = 3 ⇒ I = 3

3 . 2

5.

The number of values of x in the interval [0, 3π] satisfying the equation 2sin2x + 5sinx − 3 = 0 is (1) 4 (2) 6 (3) 1 (4) 2

Ans. Sol:

(1) 2 sin2 x + 5 sin x − 3 = 0 ⇒ (sin x + 3) (2 sin x − 1) = 0 ⇒ sin x =

6.

(

)

1 2

∴ In (0, 3π), x has 4 values

(

)

If a × b × c = a × b × c , where a, b and c are any three vectors such that a ⋅ b ≠ 0 ,

b ⋅ c ≠ 0 , then a and c are (1) inclined at an angle of π/3 between them (2) inclined at an angle of π/6 between them (3) perpendicular (4) parallel Ans. Sol:

(4)

( a × b ) × c = a × ( b × c ) , a ⋅ b ≠ 0, b ⋅ c ≠ 0 ⇒ (a ⋅ c ) b − (b ⋅ c ) a = (a ⋅ c ) b − (a ⋅ b ) c (a ⋅ b) c = (b ⋅ c ) a a c

7.

Let W denote the words in the English dictionary. Define the relation R by :

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FIITJEE Solutions to AIEEE−2006 R = {(x, y) ∈ W × W | the words x and y have at least one letter in common}. Then R is (1) not reflexive, symmetric and transitive (2) reflexive, symmetric and not transitive (3) reflexive, symmetric and transitive (4) reflexive, not symmetric and transitive Ans. Sol:

(2) Clearly (x, x) ∈ R ∀ x ∈ W. So, R is reflexive. Let (x, y) ∈ R, then (y, x) ∈ R as x and y have at least one letter in common. So, R is symmetric. But R is not transitive for example Let x = DELHI, y = DWARKA and z = PARK then (x, y) ∈ R and (y, z) ∈ R but (x, z) ∉ R.

8.

If A and B are square matrices of size n × n such that A2 − B2 = (A − B) (A + B), then which of the following will be always true ? (1) A = B (2) AB = BA (3) either of A or B is a zero matrix (4) either of A or B is an identity matrix

Ans. Sol:

(2) A2 − B2 = (A − B) (A + B) A2 − B2 = A2 + AB − BA − B2 ⇒ AB = BA.

9.

The value of

10



∑  sin k =1

2kπ 2kπ  is + icos 11 11 

(1) i (3) −1 Ans.

(4) 10

Sol:

(2) 1 (4) −i

2kπ 2kπ    sin 11 + i cos 11  =  k =1 

∑

10

∑ k =1

10

sin

2kπ 2kπ + i cos 11 11 k =1

∑

= 0 + i (− 1) = − i. 10.

All the values of m for which both roots of the equations x2 − 2mx + m2 − 1 = 0 are greater than −2 but less than 4, lie in the interval (2) m > 3 (1) −2 < m < 0 (3) −1 < m < 3 (4) 1 < m < 4

Ans. Sol:

(3) Equation x2 − 2mx + m2 − 1 = 0 (x − m)2 − 1 = 0 (x − m + 1) (x − m − 1) = 0 x = m − 1, m + 1 − 2 < m − 1 and m + 1 < 4

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FIITJEE Solutions to AIEEE−2006 m > − 1 and m < 3 − 1 < m < 3. 11.

A particle has two velocities of equal magnitude inclined to each other at an angle θ. If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then θ is (1) 90° (2) 120° (3) 45° (4) 60°

Ans.

(2)

Sol:

u sin θ θ tan = 2 4 u + u cos θ 2 θ 1 θ θ 1 ⇒ sin + sin cos θ = sin θ cos 4 2 4 2 4 θ θ 3θ 3 θ ∴ 2 sin = sin = 3 sin − 4 sin 4 4 4 4 θ 1 θ ∴ sin2 = ⇒ = 30° or θ = 120°. 4 4 4

R1

R2

u θ/4 θ/4 θ/2 u/2

u

12.

At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10-minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is 6 5 (2) (1) e 6 5 6 6 (4) 5 (3) 55 e

Ans.

(4)

Sol:

P (X = r) =

e −m mr r!

P (X ≤ 1) = P (X = 0) + P (X = 1) = e−5 + 5 × e−5 =

6 e5

.

13.

A body falling from rest under gravity passes a certain point P. It was at a distance of 400 m from P, 4s prior to passing through P. If g = 10 m/s2, then the height above the point P from where the body began to fall is (1) 720 m (2) 900 m (3) 320 m (4) 680 m

Ans.

(1)

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FIITJEE Solutions to AIEEE−2006 Sol:

1 2 1 gt and h + 400 = g(t + 4)2 . 2 2 Subtracting we get 400 = 8g + 4gt ⇒ t = 8 sec 1 ∴ h = × 10 × 64 = 320m 2 ∴ Desired height = 320 + 400 = 720 m.

We have h =

h Q(t) 400m P(t+4)

π

14.

∫ xf(sin x)dx is equal to 0

π

π

(1) π∫ f(cos x)dx

(2) π∫ f(sin x)dx

0 π/2

(3) Ans. Sol:

π 2

0 π/2

∫ f(sin x)dx

(4) π ∫ f(cos x)dx

0

0

(4) I=

π

π

0 π

0

∫ xf(sin x) dx = ∫ (π − x) f(sin x) dx ∫

= π f(sin x) dx − I 0

π

∫

2I = π f(sin x) dx 0 π

I=

π f(sin x) dx = π 2

∫

0 π/2

=π

π/2

∫ f(sin x) dx 0

∫ f(cos x)dx . 0

15.

A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A. Its equation is (1) x + y = 7 (2) 3x − 4y + 7 = 0 (3) 4x + 3y = 24 (4) 3x + 4y = 25

Ans. Sol:

(3) The equation of axes is xy = 0 ⇒ the equation of the line is x⋅4 + y⋅3 = 12 ⇒ 4x + 3y = 24. 2

16.

The two lines x = ay + b, z = cy + d; and x = a′y + b′, z = c′y + d′ are perpendicular to each other if (1) aa′ + cc′ = −1 (2) aa′ + cc′ = 1 a c a c (3) + =1 + = −1 (4) a′ c ′ a′ c ′

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FIITJEE Solutions to AIEEE−2006 Ans.

(1)

Sol:

Equation of lines

x −b z−d =y= a c

x − b′ z − d′ =y= a′ c′

Lines are perpendicular ⇒ aa′ + 1 + cc′ = 0. 17.

The locus of the vertices of the family of parabolas y = 105 64 35 (3) xy = 16

3 4 64 (4) xy = 105

(!) xy =

Ans.

(1)

Sol:

Parabola: y =

a3 x 2 a 2 x + − 2a is 3 2

(2) xy =

a3 x 2 a 2 x + − 2a 3 2

Vertex: (α, β)  a4  a3  1 8 − + 4⋅ ⋅ 2a  −  +  a4  4  2 3 −a / 2 3  4 3 =− α= ,β=  =− 3 3 4 3 4a 2a / 3 a a 4 3 3 35 a 35 ×3 = − a = − 12 4 16 3  35  105 − a= αβ = − . 4a  16  64

18.

The values of a, for which the points A, B, C with position vectors ˆ ˆi − 3ˆj − 5kˆ and aiˆ − 3ˆj + kˆ respectively are the vertices of a right-angled 2iˆ − ˆj + k, π triangle with C = are 2 (1) 2 and 1 (2) −2 and −1 (3) −2 and 1 (4) 2 and −1

Ans. Sol:

(1) ⇒ BA = ˆi − 2ˆj + 6kˆ CA = (2 − a)iˆ + 2ˆj CB = (1 − a)iˆ − 6kˆ

CA ⋅ CB = 0 ⇒ (2 − a) (1 − a) = 0 ⇒ a = 2, 1.

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FIITJEE Solutions to AIEEE−2006 −π / 2

∫ ( x + π )

19.

3

−3 π / 2

+ cos2 ( x + 3π )  dx is equal to 

π4 32 π (3) 2

π4 π + 32 2 π (4) − 1 4

(2)

(1)

Ans.

(3)

Sol:

I=

−π / 2

∫

−3 π / 2

(x + π)3 + cos2 (x + 3π) dx  

Put x + π = t π/2

∫

I=

−π / 2 π/2

=

π/2

 t 3 + cos2 t  dt = 2  

∫ cos

2

t dt

0

π

∫ (1 + cos 2t) dt = 2 + 0 . 0

20.

If x is real, the maximum value of (1) 1/4 (3) 1

Ans.

(2)

Sol:

y=

3x 2 + 9x + 17 is 3x 2 + 9x + 7 (2) 41 (4) 17/7

3x 2 + 9x + 17 3x 2 + 9x + 7

3x2(y − 1) + 9x(y − 1) + 7y − 17 = 0 D ≥ 0 ∵ x is real 81(y − 1)2 − 4x3 ( y − 1)( 7y − 17 ) ≥ 0

⇒ (y − 1) (y − 41) ≤ 0 ⇒ 1 ≤ y ≤ 41.

21.

In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is 3 1 (B) (1) 5 2 4 1 (C) (D) 5 5

Ans. Sol:

(1) 2ae = 6 ⇒ ae = 3 2b = 8 ⇒ b = 4 b2 = a2(1 − e2) 16 = a2 − a2e2 a2 = 16 + 9 = 25 a=5 ∴e =

3 3 = a 5

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FIITJEE Solutions to AIEEE−2006 22.

Ans. Sol:

 1 2 a 0 Let A =   and B =   , a , b ∈ N. Then 3 4 0 b (1) there cannot exist any B such that AB = BA (2) there exist more than one but finite number of B’s such that AB = BA (3) there exists exactly one B such that AB = BA (4) there exist infinitely many B’s such that AB = BA

(4) 1 2 A= B=  3 4   a 2b  AB =   3a 4b 

a 0  0 b   

a 0   1 2   a 2a  BA =   =  0 b  3 4  3b 4b 

AB = BA only when a = b 23.

The function f(x) = (1) x = 2 (3) x = 0

x 2 + has a local minimum at 2 x (2) x = −2 (4) x = 1

Ans.

(1)

Sol:

x 2 1 + is of the form x + ≥ 2 & equality holds for x = 1 2 x x

24.

Angle between the tangents to the curve y = x2 − 5x + 6 at the points (2, 0) and (3, 0) is π π (1) (2) 2 2 π π (4) (3) 6 4

Ans.

(2) dy = 2x − 5 dx ∴ m1 = (2x − 5)(2, 0) = −1, m2 = (2x − 5)(3, 0) = 1 ⇒ m1m2 = −1

Sol:

25.

Let a1, a2, a3, … be terms of an A.P. If 41 11 2 (3) 7 (1)

Ans.

a1 + a2 + ⋅ ⋅ ⋅ap a1 + a2 + ⋅ ⋅ ⋅ + aq

=

a p2 , p ≠ q , then 6 equals 2 a21 q

7 2 11 (4) 41

(2)

(4)

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FIITJEE Solutions to AIEEE−2006 Sol:

26.

p  2a + ( p − 1) d 2a + ( p − 1) d p p2 2 1 = 2 ⇒ 1 = q 2a1 + ( q − 1) d q  2a1 + ( q − 1) d q 2  p − 1 a1 +  d  2  =p q  q − 1 a1 +  d 2   a6 a 11 For , p = 11, q = 41 → 6 = a21 a21 41

The set of points where f(x) = (1) (−∞, 0) ∪ (0, ∞) (3) (−∞, ∞)

Ans.

(3)

Sol:

 x 1 − x , f (x) =   x , 1 + x

x 0 and y = 3x, x > 0, then a 2

(2) (3, ∞) 1  (4)  −3, −  2 

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FIITJEE Solutions to AIEEE−2006 Ans.

(3)

Sol:

a2 − 3a < 0 and a2 −

36.

The image of the point (−1, 3, 4) in the plane x − 2y = 0 is  17 19  (1)  − , − , 4  (2) (15, 11, 4) 3  3   17 19  (3)  − , − , 1 (4) (8, 4, 4) 3   3

Sol:

If (α, β, γ) be the image then

a 1 >0 ⇒ 0, ∀x ∈ R ∴ f(x) is strictly increasing function ∴ It is one-one Clearly, f(x) is a continuous function and also increasing on R, f ( x ) = −∞ and Lt f ( x ) = ∞ Lt x → −∞ x→∞ ∴ f(x) takes every value between −∞ and ∞ . Thus, f(x) is onto function.

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21

FIITJEE Solutions to AIEEE - 2009 72.

1  In a binomial distribution B  n, p =  , if the probability of at least one success is greater than or 4   9 equal to , then n is greater than 10 1 1 (1) (2) 4 3 4 log10 − log10 log10 + log103

(3)

Sol:

log10

4

9 − log103

(4)

log10

4

4 − log103

(1) n

1 − qn ≥

*73.

9 1 1 3 ⇒  ≤ ⇒ n ≥ − log 3 10 ⇒ n ≥ 4 10 10 log − log103 4 10 4

If P and Q are the points of intersection of the circles

x 2 + y 2 + 3x + 7y + 2p − 5 = 0

and

x + y + 2x + 2y − p = 0 , then there is a circle passing through P, Q and (1, 1) for (1) all values of p (2) all except one value of p (3) all except two values of p (4) exactly one value of p 2

Sol:

2

2

(1) Given circles S = x 2 + y 2 + 3x + 7y + 2p − 5 = 0 S' = x 2 + y 2 + 2x + 2y − p2 = 0

Equation of required circle is S + λS' = 0 As it passes through (1, 1) the value of λ =

− ( 7 + 2p )

(6 − p ) 2

If 7 + 2p = 0, it becomes the second circle ∴it is true for all values of p 74.

The projections of a vector on the three coordinate axis are 6, - 3, 2 respectively. The direction cosines of the vector are 6 3 2 (1) 6, − 3, 2 (2) , − , 5 5 5 6 3 2 6 3 2 (3) , − , (4) − , − , 7 7 7 7 7 7

Sol:

(3) Projection of a vector on coordinate axis are x 2 − x1, y 2 − y1, z2 − z1 x 2 − x1 = 6, y 2 − y1 = −3, z 2 − z1 = 2

( x 2 − x1 )

2

+ ( y 2 − y1 ) + ( z 2 − z1 ) = 36 + 9 + 4 = 7 2

The D.C’s of the vector are

*75.

If Z − (1)

2

6 3 2 ,− , 7 7 7

4 = 2 , then the maximum value of Z is equal to z

3 +1

(3) 2

Sol:

(2)

5 +1

(4) 2 + 2

(2) 4 4 4 4  Z = Z −  + ⇒ Z = Z− + Z Z Z Z  4 4 4 ⇒ Z ≤ Z− + ⇒ Z ≤ 2+ Z Z Z

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22

FIITJEE Solutions to AIEEE - 2009 2

⇒ Z −2 Z −4 ≤0

( Z −(

) ) ( Z − (1 − 5 )) ≤ 0 ⇒ 1 −

5 +1

5 ≤ Z ≤ 5 +1

*76.

Three distinct points A, B and C are given in the 2 – dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point ( - 1, 0) is equal 1 to . Then the circumcentre of the triangle ABC is at the point 3 5  (2)  , 0  (1) ( 0, 0 ) 4  5   5  (3)  , 0  (4)  , 0  2  3 

Sol:

(2) P = (1, 0 ) ; Q ( −1, 0 )

Let A = ( x, y ) AP BP CP 1 = = = AQ BQ CQ 3

..(1)

⇒ 3AP = AQ ⇒ 9AP2 = AQ2 ⇒ 9 ( x − 1) + 9y 2 = ( x + 1) + y 2 2

2

⇒ 9x 2 − 18x + 9 + 9y 2 = x 2 + 2x + 1 + y 2 ⇒ 8x 2 − 20x + 8y 2 + 8 = 0 5 …(2) x +1= 0 2 ∴ A lies on the circle Similarly B, C are also lies on the same circle ⇒ x2 + y2 −

5  ∴ Circumcentre of ABC = Centre of Circle (1) =  , 0  4 

*77.

The remainder left out when 82n − ( 62 )

2n +1

is divided by 9 is

(1) 0 (3) 7 Sol:

(2) 2 (4) 8

(2) 82n − ( 62 )

2n +1

= (1 + 63 ) − ( 63 − 1)

= (1 + 63 ) + (1 − 63 ) n

(

n

2n +1

2n +1

(

= 1 + nc1 63 + nc 2 ( 63 ) + .... + ( 63 ) 2

= 2 + 63 n c1 + nc 2 ( 63 ) + .... + ( 63 )

n −1

−(

2n +1)

c1 + (

n

2n +1)

) + (1−

( 2n +1)

c 2 ( 63 ) + .... − ( 63 )

c1 63 + (

( 2n )

2n +1)

c 2 ( 63 ) + .... + ( −1)( 63 ) 2

( 2n +1)

)

∴ Reminder is 2 *78.

The ellipse x 2 + 4y 2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is (2) x 2 + 12y 2 = 16 (1) x 2 + 16y 2 = 16 (3) 4x 2 + 48y 2 = 48

Sol:

(4) 4x 2 + 64y 2 = 48

(2)

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)

23

FIITJEE Solutions to AIEEE - 2009 x2 y2 + = 1 ⇒ a = 2, b = 1 ⇒ P = ( 2, 1) 4 1 x2 y2 x2 y2 Required Ellipse is 2 + 2 = 1 ⇒ 2 + 2 = 1 a b 4 b (2, 1) lies on it V’ 4 1 1 1 3 4 ⇒ + = 1 ⇒ 2 = 1 − = ⇒ b2 = 16 b2 4 4 3 b 2 2 2 2 x y x 3y ∴ + = 1⇒ + = 1 ⇒ x 2 + 12y 2 = 16 16  4  16 4 3   2 6 10 14 The sum to the infinity of the series 1 + + 2 + 3 + 4 + ...... is 3 3 3 3 (1) 2 (2) 3 (3) 4 (4) 6 x 2 + 4y 2 = 4 ⇒

*79.

Sol:

P (2, 1) 1 A’

2

A

2

V (4, 0)

(2)

2 6 10 14 + + + + .... 3 32 33 34 1 1 2 6 10 S = + 2 + 3 + 4 + .... 3 3 3 3 3 Dividing (1) & (2) 1 1 4 4 4  S  1 −  = 1 + + 2 + 3 + 4 + .... 3 3 3 3  3

Let S = 1 +

…(1) …(2)

  2 4 4 1 1 2 4 4 1  4 4 3 4 2 6 2 6  S = + 2  1 + + 2 + ......  ⇒ S = + 2  = + 2 = + = ⇒ S= ⇒S=3 1 3 3 3 3 3  3 3 3 3 3 2 3 3 2 3 3   1 −  3 

80.

The differential equation which represents the family of curves y = c1ec 2 x , where c1 and c 2 are arbitrary constants is (2) y " = y ' y (1) y ' = y 2 (4) yy " = ( y ' )

(3) yy " = y ' Sol:

(4) y = c1ec 2 x

…(1)

y ' = c 2 c1e

c2 x

y ' = c2 y y " = c2 y ' From (2) y' c2 = y So, y " =

2

…(2)

( y ') y

2

⇒ yy " = ( y ' )

2

81.

One ticket is selected at random from 50 tickets numbered 00, 01, 02, …., 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals 1 1 (2) (1) 14 7 1 5 (3) (4) 50 14

Sol:

(1)

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FIITJEE Solutions to AIEEE - 2009 S = { 00, 01, 02, …., 49 } Let A be the even that sum of the digits on the selected ticket is 8 then A = { 08, 17, 26, 35, 44 } Let B be the event that the product of the digits is zero B = { 00, 01, 02, 03, …., 09, 10, 20, 30, 40 } A ∩ B = {8} 1 1 = 50 = Required probability = P ( A / B ) = 14 14 P (B ) 50 2x Let y be an implicit function of x defined by x − 2x x cot y − 1 = 0 . Then y ' (1) equals P ( A ∩ B)

82.

(1) – 1 (3) log 2 Sol:

(2) 1 (4) – log 2

(1) x 2x − 2x x cot y − 1 = 0 Now x = 1,

…(1)

1 – 2 coty – 1 = 0 ⇒ coty = 0 ⇒ y =

π 2

Now differentiating eq. (1) w.r.t. ‘x’ dy   2x 2x (1 + log x ) − 2  x x ( − c osec 2 y ) + cot y x x (1 + log x )  = 0 dx    π Now at  1,   2    dy  2 (1 + log1) − 2  1( −1)   + 0 = 0    dx 1, π   2    dy   dy  ⇒ 2 + 2  =0⇒  = −1  dx 1, π   dx 1, π  

83.

2



2

The area of the region bounded by the parabola ( y − 2 ) = x − 1 , the tangent to the parabola at the 2

point (2, 3) and the x-axis is (1) 3 (3) 9 Sol:

(2) 6 (4) 12

(3)

Equation

( y − 2)

2

of

tangent

at

(2,

3)

to 2y = x + 4

= x − 1 is S1 = 0

⇒ x − 2y + 4 = 0 Required Area = Area of ∆OCB + Area of OAPD – Area of ∆PCD 3 1 1 = ( 4 x 2 ) + ∫ y 2 − 4y + 5 dy − (1 x 2 ) 2 2 0

(

)

3

 y3  = 4 +  − 2y 2 + 5y  − 1 = 4 − 9 − 18 + 15 − 1 3  0 = 28 − 19 = 9 sq. units

D (0, 3) P (2, 3) C (0, 2)

B (-4, 0)

A (1, 2) A (1, 2) 0

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25

FIITJEE Solutions to AIEEE - 2009 3

Area =

∫ ( 2y − 4 − y

2

0

84.

)

3

(

3

)

+ 4y − 5 dy = ∫ − y + 6y − 5 dy = − ∫ ( 3 − y ) 2

0

0

3

2

 ( y − 3 )3  27  = dy =  = 9 sq.units 3  3  0

Given P ( x ) = x 4 + ax 3 + bx 2 + cx + d such that x = 0 is the only real root of P' ( x ) = 0 . If P ( −1) < P (1) , then in the interval [ −1, 1] (1) P ( −1) is the minimum and P (1) is the maximum of P (2) P ( −1) is not minimum but P (1) is the maximum of P (3) P ( −1) is the minimum and P (1) is not the maximum of P (4) neither P ( −1) is the minimum nor P (1) is the maximum of P

Sol:

(2) P ( x ) = x 4 + ax 3 + bx 2 + cx + d P ' ( x ) = 4x 3 + 3ax 2 + 2bx + c

Q x = 0 is a solution for P' ( x ) = 0 , ⇒ c = 0

∴ P ( x ) = x 4 + ax 3 + bx 2 + d

…(1)

Also, we have P ( −1) < P (1)

⇒ 1− a + b + d < 1+ a + b + d ⇒ a > 0 Q P ' ( x ) = 0 , only when x = 0 and P(x) is differentiable in ( - 1, 1), we should have the maximum and minimum at the points x = - 1, 0 and 1 only Also, we have P ( −1) < P (1) ∴ Max. of P(x) = Max. { P(0), P(1) } & Min. of P(x) = Min. { P(-1), P(0) } In the interval [ 0 , 1 ], P' ( x ) = 4x 3 + 3ax 2 + 2bx = x 4x 2 + 3ax + 2b

(

)

Q P ' ( x ) has only one root x = 0, 4x + 3ax + 2b = 0 has no real roots. 2

∴ ( 3a ) − 32b < 0 ⇒ 2

3a2 0 Thus, we have a > 0 and b > 0 ∴ P' ( x ) = 4x 3 + 3ax 2 + 2bx > 0, ∀x ∈ ( 0, 1) Hence P(x) is increasing in [ 0, 1 ] ∴ Max. of P(x) = P(1) Similarly, P(x) is decreasing in [-1 , 0] Therefore Min. P(x) does not occur at x = - 1 85.

Sol:

The shortest distance between the line y − x = 1 and the curve x = y 2 is (1)

3 2 8

(2)

2 3 8

(3)

3 2 5

(4)

3 4

(1) x − y +1= 0 x=y

…(1)

2

1 = 2y

dy dy 1 = Slope of given line (1) ⇒ = dx dx 2y 2

1 1 1 1  1  1 1 = 1 ⇒ y = ⇒ y = ⇒ x =   = ⇒ ( x, y ) =  ,  2 2 4 2y 2    4 2

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FIITJEE Solutions to AIEEE - 2009 ∴ The shortest distance is

1 1 − +1 4 2 1+ 1

=

3 4 2

=

26

3 2 8

Directions: Question number 86 to 90 are Assertion – Reason type questions. Each of these questions contains two statements Statement-1 (Assertion) and Statement-2 (Reason). Each of these questions also have four alternative choices, only one of which is the correct answer. You have to select the correct choice

86.

Let f ( x ) = ( x + 1) − 1, x ≥ −1 2

{

}

Statement-1 : The set x : f ( x ) = f −1 ( x ) = {0, − 1} Statement-2 : f is a bijection. (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true Sol:

(3) There is no information about co-domain therefore f(x) is not necessarily onto.

87.

Let f ( x ) = x x and g ( x ) = sin x . Statement-1 : gof is differentiable at x = 0 and its derivative is continuous at that point. Statement-2 : gof is twice differentiable at x = 0. (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true

Sol:

(3) f ( x ) = x x and g ( x ) = sin x − sin x 2 ,x < 0 gof ( x ) = sin ( x x ) =  2 ,x ≥ 0  sin x −2x cos x 2 ,x < 0 ∴ ( gof ) ' ( x ) =  2 ,x ≥ 0  2x cos x Clearly, L ( gof ) ' ( 0 ) = 0 = R ( gof ) ' ( 0 )

∴ gof is differentiable at x = 0 and also its derivative is continuous at x = 0 −2cos x 2 + 4x 2 sin x 2 ,x < 0 Now, ( gof ) " ( x ) =  2 2 2 ,x ≥ 0  2cos x − 4x sin x ∴ L ( gof ) " ( 0 ) = −2 and R ( gof ) " ( 0 ) = 2 ∴ L ( gof ) " ( 0 ) ≠ R ( gof ) " ( 0 )

∴ gof(x) is not twice differentiable at x = 0. *88.

Statement-1 : The variance of first n even natural numbers is Statement-2 : The sum of first n natural numbers is numbers is

n ( n + 1)( 2n + 1)

n ( n + 1) 2

n2 − 1 4

and the sum of squares of first n natural

6 (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false

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27

FIITJEE Solutions to AIEEE - 2009 (4) Statement-1 is false, Statement-2 is true Sol:

(4) Statement-2 is true Statement-1: Sum of n even natural numbers = n (n + 1) n ( n + 1) Mean x = = n +1 n 2 1 2 2 2 1 Variance =  ∑ ( x i )  − x =  22 + 42 + ..... + ( 2n )  − ( n + 1)  n n 

()

()

1 2 2 4 n ( n + 1)( 2n + 1) 2 2 2 1 + 22 + ..... + n2  − ( n + 1) = − ( n + 1) n  n 6 (n + 1) 2 ( 2n + 1) − 3 (n + 1) (n + 1) [ 4n + 2 − 3n − 3] (n + 1)(n − 1) n2 − 1 = = = = 3 3 3 3 ∴ Statement 1 is false. =

89.

Statement-1 : ~ ( p ↔~ q ) is equivalent to p ↔ q . Statement-2 : ~ ( p ↔~ q) is a tautology. (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true

Sol:

(3)

p T T F F

90.

p↔q

q T F T F

T F F T

~q F T F T

p ↔~ q

~ ( p ↔~ q)

F T T F

T F F T

Let A be a 2 x 2 matrix Statement-1 : adj ( adj A ) = A Statement-2 : adj A = A (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true

Sol:

(2)

adj A = A

n −1

= A

adj ( adj A ) = A

n−2

2 −1

= A 0

A= A A=A

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AIEEE− −2010− −20

59.

59. Sol.

A small particle of mass m is projected at an angle θ with the x-axis with an initial velocity v0 in the x-y plane as shown in the v sin θ figure. At a time t < 0 , the angular momentum of the g particle is (1) −mgv 0 t 2 cos θ ˆj (2) mgv 0 t cos θ kˆ

1 1 (3) − mgv 0 t 2 cos θ kˆ (4) mgv 0 t 2 cos θ ˆi 2 2 where ˆi, ˆj and kˆ are unit vectors along x, y and z–axis respectively. 3 L = m(r × v)

1 L = m v 0 cos θt ˆi + (v 0 sin θt − gt 2 )jˆ × v 0 cos θ ˆi + (v 0 sin θ − gt)jˆ 2 1 = mv 0 cos θt − gt kˆ 2 1 = − mgv 0 t 2 cos θkˆ 2 60.

–1

The equation of a wave on a string of linear mass density 0.04 kg m

t x y = 0.02(m)sin 2π − 0.04(s) 0.50(m) 60.

(1) 4.0 N 4

Sol.

T = µv 2 = µ

61.

Let cos(α + β) = (1)

61.

is given by

. The tension in the string is

(2) 12.5 N

(3) 0.5 N

(4) 6.25 N

ω2 (2π / 0.004)2 = = 6.25 N 0.04 k2 (2π / 0.50)2

56 33

π 4 5 and let sin(α – β) = , where 0 ≤ α, β ≤ , then tan 2α = 5 13 4 19 20 25 (2) (3) (4) 12 7 16

1

3 4 5 tan(α – β) = 12 3 5 + 56 tan 2α = tan(α + β + α – β) = 4 12 = 3 5 33 1− 4 12 4 5 5 sin(α – β) = 13 cos (α + β) =

62.

tan(α + β) =

Let S be a non-empty subset of R. Consider the following statement: P: There is a rational number x ∈ S such that x > 0. Which of the following statements is the negation of the statement P ? (1) There is no rational number x ∈ S such that x ≤ 0 (2) Every rational number x ∈ S satisfies x ≤ 0

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AIEEE− −2010− −21

62.

(3) x ∈ S and x ≤ 0 x is not rational (4) There is a rational number x ∈ S such that x ≤ 0 2 P: there is a rational number x ∈ S such that x > 0 ~P: Every rational number x ∈ S satisfies x ≤ 0

63.

Let a = ˆj − kˆ and c = ˆi − ˆj − kˆ . Then vector b satisfying a × b + c = 0 and a ⋅ b = 3 is (1) 2iˆ − ˆj + 2kˆ (2) ˆi − ˆj − 2kˆ (3) ˆi + ˆj − 2kˆ (4) −ˆi + ˆj − 2kˆ

63.

4 c = b×a b⋅c = 0 b1ˆi + b2 ˆj + b3kˆ ⋅ ˆi − ˆj − kˆ = 0

(

)(

)

b1 – b2 – b3 = 0 and a ⋅ b = 3 b2 – b3 = 3 b1 = b2 + b3 = 3 + 2b3 b = ( 3 + 2b3 ) ˆi + ( 3 + b3 ) ˆj + b3kˆ . 64.

The equation of the tangent to the curve y = x +

64.

(1) y = 1 3 Parallel to x-axis

(2) y = 2

4 , that is parallel to the x-axis, is x2 (3) y = 3 (4) y = 0

dy =0 dx

1−

x=2 y=3 Equation of tangent is y – 3 = 0(x – 2) 65. 65.

8 =0 x3

y–3=0

π is 2 (3) tan x = (sec x + c)y (4) sec x = (tan x + c)y

Solution of the differential equation cos x dy = y(sin x – y) dx, 0 < x < (1) y sec x = tan x + c (2) y tan x = sec x + c 4 cos x dy = y(sin x – y) dx dy = y tan x − y 2 sec x dx 1 dy 1 − tan x = − sec x y 2 dx y Let −

1 =t y

1 dy dt = y 2 dx dx

dy dt – t tan x = –sec x + (tan x) t = sec x. dx dx tan x dx I.F. = e = sec x Solution is t(I.F) = (I.F) sec x dx −

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AIEEE− −2010− −22

1 sec x = tan x + c y 66.

The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and x =

66.

(1) 4 2 + 2 4 π 4

(2) 4 2 – 1

(3) 4 2 + 1

5π 4

3π 2

π 4

5π 4

( cos x − sin x ) dx + ( sin x − cos x ) dx + ( cos x − sin x ) = 4

0

cos x

π

(4) 4 2 – 2

2 −2

sin x

π 0

3π is 2

5π

3π

4

2

2π

4

2

67.

If two tangents drawn from a point P to the parabola y = 4x are at right angles, then the locus of P is (1) 2x + 1 = 0 (2) x = –1 (3) 2x – 1 = 0 (4) x = 1 2 The locus of perpendicular tangents is directrix i.e, x = –a; x = –1

68.

ˆ b = 2iˆ + 4ˆj + kˆ and c = λ ˆi + ˆj + µkˆ are mutually orthogonal, then (λ, µ) = If the vectors a = ˆi − ˆj + 2k,

67.

68.

69.

(1) (2, –3) (2) (–2, 3) 4 a ⋅ b = 0, b ⋅ c = 0, c ⋅a = 0 2λ + 4 + µ = 0 λ – 1 + 2µ = 0 Solving we get: λ = –3, µ = 2

(4) (–3, 2)

Consider the following relations: R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; S=

69.

(3) (3, –2)

m p , n q

m, n, p and q are integers such that n, q ≠ 0 and qm = pn . Then

(1) neither R nor S is an equivalence relation (2) S is an equivalence relation but R is not an equivalence relation (3) R and S both are equivalence relations (4) R is an equivalence relation but S is not an equivalence relation 2 xRy need not implies yRx m p S: s ⇔ qm = pn n q m m s reflexive n n

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AIEEE−2010−23

m p s n q m p p r s , s n q q s

p m s symmetric q n qm = pn, ps = rq

ms = rn transitive.

S is an equivalence relation. 70.

Let f: R → R be defined by f(x) =

k − 2x, if x ≤ −1 . If f has a local minimum at x = –1, then a 2x + 3, if x > −1

possible value of k is (2) −

(1) 0 70.

3 f(x) = k – 2x = 2x + 3

1 2

(3) –1

(4) 1

if x ≤ –1 if x > –1 2x + 3

k – 2x 1 –1

lim f(x) ≤ –1

x →−1−

71. 71.

This is true where k = –1

The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is (1) 5 (2) 6 (3) at least 7 (4) less than 4 3 First row with exactly one zero; total number of cases = 6 First row 2 zeros we get more cases Total we get more than 7.

Directions: Questions Number 72 to 76 are Assertion – Reason type questions. Each of these questions contains two statements. Statement-1: (Assertion) and Statement-2: (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.

72.

72.

Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ....., 20}. Statement-1: The probability that the chosen numbers when arranged in some order will form an AP 1 is . 85 Statement-2: If the four chosen numbers from an AP, then the set of all possible values of common difference is {±1, ±2, ±3, ±4, ±5}. (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 2

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AIEEE−2010−24

20

N(S) = C4 Statement-1: common difference is 1; total number of cases = 17 common difference is 2; total number of cases = 14 common difference is 3; total number of cases = 11 common difference is 4; total number of cases = 8 common difference is 5; total number of cases = 5 common difference is 6; total number of cases = 2 17 + 14 + 11 + 8 + 5 + 2 1 Prob. = . = 20 85 C4 73.

73.

74.

Statement-1: The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x – y + z = 5. Statement-2: The plane x – y + z = 5 bisects the line segment joining A(3, 1, 6) and B(1, 3, 4). (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 1 A(3, 1, 6); B = (1, 3, 4) Mid-point of AB = (2, 2, 5) lies on the plane. and d.r’s of AB = (2, –2, 2) d.r’s Of normal to plane = (1, –1, 1). AB is perpendicular bisector ∴ A is image of B Statement-2 is correct but it is not correct explanation. 10

Let S1 =

j =1

j ( j − 1) 10 C j , S2 =

10

j

10

j =1

C j and S3 =

10

j2

j =1

10

Cj .

9

74.

Statement-1: S3 = 55 × 2 8 8 Statement-2: S1 = 90 × 2 and S2 = 10 × 2 . (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 2 10 10 10! 8! S1 = j ( j − 1) = 90 = 90 ⋅ 28 . j j − 1 j − 2 ! 10 − j ! j − 2 ! 8 − j − 2 ! ( )( ) ( ) ) ( ( )) j =1 j= 2 (

S2 =

10 j =1

S3 =

10 j =1

j

10 10! 9! = 10 = 10 ⋅ 29 . j ( j − 1)! ( 9 − ( j − 1) )! j − 1 ! 9 − j − 1 ! ) ( ( )) j =1 (

j ( j − 1) + j

10! = j! (10 − j )!

10 j =1

8

j ( j − 1) 10 C j = 8

10 j =1

8

j 10 C j = 90 . 2 + 10 . 2 8

9

9

= 90 . 2 + 20 . 2 = 110 . 2 = 55 . 2 . 75.

2

Let A be a 2 × 2 matrix with non-zero entries and let A = I, where I is 2 × 2 identity matrix. Define Tr(A) = sum of diagonal elements of A and |A| = determinant of matrix A. Statement-1: Tr(A) = 0 Statement-2: |A| = 1 (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false

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AIEEE−2010−25

75.

(3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 2 a b Let A = , abcd ≠ 0 c d a b a b ⋅ c d c d

2

A = 2

A =

a2 + bc ab + bd ac + cd bc + d2

a2 + bc = 1, bc + d2 = 1 ab + bd = ac + cd = 0 c ≠ 0 and b ≠ 0 a+d=0 Trace A = a + d = 0 |A| = ad – bc = –a2 – bc = –1. 76.

Let f: R → R be a continuous function defined by f(x) =

1 . e + 2e − x x

1 , for some c ∈ R. 3 1 Statement-2: 0 < f(x) ≤ , for all x ∈ R 2 2 (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 4 1 ex f(x) = x = e + 2e− x e2x + 2 Statement-1: f(c) =

76.

f′(x) =

(e

2x

)

+ 2 e x − 2e2x ⋅ e x

(e )

2x + 2 2

2x

f′(x) = 0 2x e =2

e + 2 = 2e x e = 2

maximum f(x) = 0 < f(x) ≤

2 1 = 4 2 2

1

2 2 1 1 Since 0 < < 3 2 2 1 f(c) = 3

77.

2x

∀x∈R for some c ∈ R

For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement among the following is r 1 r 2 (1) There is a regular polygon with = (2) There is a regular polygon with = R R 3 2

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AIEEE−2010−26

(3) There is a regular polygon with 77.

r 3 = R 2

(4) There is a regular polygon with

r 1 = R 2

2

a π cot 2 n ‘a’ is side of polygon. a π R = cosec 2 n π cot r n = cos π = π R n cosec n π 2 for any n ∈ N. cos ≠ n 3 r=

78. 78.

2

2009

If α and β are the roots of the equation x – x + 1 = 0, then α (1) –1 (2) 1 (3) 2 2 2

x –x+1=0

1± 3 i 2 1 3 α = +i , 2 2 π π α = cos + isin , 3 3

x=

+β

2009

= (4) –2

1± 1− 4 2

x=

1 i 3 − 2 2 π π β = cos − isin 3 3 π 2009 2009 α +β = 2cos2009 3 2π 2π = 2cos 668π + π + = 2cos π + 3 3 2π 1 = −2 − =1 = −2cos 3 2 79. 79.

80.

80.

β=

The number of complex numbers z such that |z – 1| = |z + 1| = |z – i| equals (1) 1 (2) 2 (3) ∞ (4) 0 1 Let z = x + iy |z – 1| = |z + 1| Re z = 0 x=0 x=y |z – 1| = |z – i| |z + 1| = |z – i| y = –x Only (0, 0) will satisfy all conditions. Number of complex number z = 1 A line AB in three-dimensional space makes angles 45° and 120° with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle θ with the positive z-axis, then θ equals (1) 45° (2) 60° (3) 75° (4) 30° 2

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AIEEE−2010−27

= cos 45° =

1 2

m = cos 120° = −

1 2

n = cos θ where θ is the angle which line makes with positive z-axis. 2 2 2 Now + m + n = 1

1 1 2 + + cos θ = 1 2 4 1 2 cos θ = 4 1 cos θ = 2 π θ= . 3 81.

The line L given by the equation (1)

81.

17

(θ Being acute)

x y + = 1 passes through the point (13, 32). The line K is parallel to L and has 5 b

x y + = 1. Then the distance between L and K is c 3 17 23 (2) (3) 15 17

(4)

23 15

3

b 5 3 Slope of line K = − c Line L is parallel to line k. b 3 bc = 15 = 5 c (13, 32) is a point on L. 13 32 32 8 + =1 =− 5 b b 5 3 b = –20 c= − 4 Equation of K: y – 4x = 3 52 − 32 + 3 23 Distance between L and K = = 17 17

Slope of line L = −

82.

82.

th

A person is to count 4500 currency notes. Let an denote the number of notes he counts in the n minute. If a1 = a2 = ...... = a10 = 150 and a10, a11, ...... are in A.P. with common difference –2, then the time taken by him to count all notes is (1) 34 minutes (2) 125 minutes (3) 135 minutes (4) 24 minutes 1 th Till 10 minute number of counted notes = 1500 n 3000 = [2 × 148 + (n – 1)(–2)] = n[148 – n + 1] 2 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004

AIEEE−2010−28

2

n – 149n + 3000 = 0 n = 125, 24 n = 125 is not possible. Total time = 24 + 10 = 34 minutes. 83.

Let f: R → R be a positive increasing function with lim

x →∞

(1) 83.

2 3

(2)

3 2

f(3x) f(2x) = 1. Then lim = x →∞ f(x) f(x)

(3) 3

(4) 1

4 f(x) is a positive increasing function 0 < f(x) < f(2x) < f(3x) f(2x) f(3x) < 0