Code : N. Mathematics ( ) ( ) ( ) Q c, a and b are coplanar. x 2 = λ µ... (ii) 1. If (2, 3, 5) is one end of a diameter of the sphere

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1 Mathematics. If (, 3, ) is one end of a diameter of the sphere x + y + z 6x y z + 0 = 0, then the coordinates of the other end of the diameter are () (4, 3, 3) () (4, 9, 3) (3) (4, 3, 3) (4) (4, 3, ) Sol. () Centre of sphere is (3, 6, ) Let other end is (x, y, z ) x + = 3 x = 4 y + 3 = 6 y = 9 + z = z = 3 Other end of diameter is (4, 9, 3) a = ˆi + ˆj+ k,b ˆ = ˆi ˆj+ kˆ and c = xi ˆ+ x ˆj k, ˆ. If the vector c lies in the plane of a and. Let ( ) b, then x equals () () 0 (3) (4) 4 Sol. () r r r Q c, a and b are coplanar r r r c = λ a+µ b ( ) ( ) ( ) xi ˆ + x ˆ j k ˆ = λ ˆ i + ˆ j + k ˆ +µ ˆ i ˆ j + k ˆ x = λ+µ... (i) x = λ µ... (ii) = λ+ µ... (iii) From (i) and (ii) λ = x, µ = From (iii) = x + x = Career Launcher 8 / Solutions AIEEE Test Paper 007 Page

2 Method II = 0 x x ( x+ 4) ( x) + x ( + x) = 0 x = 4 x = 3. Let A(h, k), B(, ) and C(, ) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is, then the set of values which k can take is given by () { 3, } () {, 3} (3) {0, } (4) {, 3} = h k = h ( ) k ( ) + ( ) = ± k = ± k = 3 or 4. Let P = (, 0), Q = (0, 0) and R = ( 3, 3 3 ) be three points. The equation of the bisector of the angle PQR is Sol. () () x+ 3y = 0 () 3x + y = 0 (3) 3 x+ y = 0 (4) 3 x + y = 0 S Y P(3, 3 3) P(, 0) π/3 Q π/3 π/3 X Career Launcher 8 / Solutions AIEEE Test Paper 007 Page

3 Slope of QS, m = tan0 = 3 y = 3x y+ 3x = 0 my + m xy mx = 0 is a bisector of the angle between the lines xy = 0, then m is (). (3) (4) Joint equation of bisector of the lines xy = 0 is y x = 0. If one of the lines of ( ) my + m xy mx = 0 Since ( ) ( y mx)( my+ x) = 0 One of the line is bisector of xy = 0 m = 6. Let F(x) = f(x) + f x, where ( ) x f x = logt dt. Then F(e) equals + t () () (3) 0 (4) Sol. () F(e) = f(e) + f e e /e logt logt = dt + dt + t + t = I + I For I = / e logt dt + t Let t = dt = dz z z When t = z = Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 3

4 t = z = e e e logz z I =. dz + z e logz log t = = dt z z t t dz ( + ) ( + ) e F(e) = I+ I e logt log t = + + t t + t e 0 ( ) dt = logt dt s ds s logt; ds dt, when t. s 0 and t e. s t = = = t = = = = s = = 0 7. Let f:r IR be a function defined by f(x) = Min{x +, x + }. Then which of the following is true? () f(x) is not differentiable at x = 0 () f(x) for all x R (3) f(x) is not differentiable at x =. (4) f(x) is differentiable everywhere f(x) = Min{x +, x + } Y y = x + O X Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 4

5 f(x) = x + x 0 x + x < 0 f(x) is differentiable everywhere. 8. The function f : R {0} R given by Sol. () f( x) = x x e can be made continuous at x = 0 by defining f(0) as () () (3) (4) 0 lim x 0 x x e x e x 0 = lim x 0 x e 0 x ( ) x e 0 = lim x 0 x x e x.e 0 + x 4e = lim x 0 e + e + 4xe 4 = = 4 f(0) = x x x 9. The solution for x of the equation x dt π = is t t () () (3) π (4) 3 Sol. Wrong Question Career Launcher 8 / Solutions AIEEE Test Paper 007 Page

6 0. cos x + 3 sin x equals x π () log tan + C (3) log tan x π C + () log tan x π C + + x π (4) log tan + + C Sol. () I = 3 cos x + sin x = π π sin x cos + cos x sin 6 6 = π sin x + 6 π = cosec x + 6 x π = logtan C + + Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 6

7 . The area enclosed between the curves y = x and y = x is () 3 () 3 (3) (4) 6 y = x Y (,) y = x X' O (0, 0) X Y' y = x Area = ( x ) 0 3/ x x = 3 0 x = = sq. units 3 6. If the difference between the roots of the equation x + ax+ = 0 is less than, then the set of possible values of a is () (, 3) () ( 3, 3) (3) ( 3, ) (4) ( 3, ) Sol. Q α β< Again α+β= a, αβ= ( α β ) < Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 7

8 ( α+β) 4αβ< a 4 < ( ) a 9 a 3,3 <... (i) Also D 0 a 4 0 a (, ) (, )... (ii) From (i) and (ii) a ( 3, ) (,3) 3. In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals () ( ) () ( ) (3) (4) Sol. () Let the GP be a, ar, ar, ar 3... a = ar + ar = r + r r + r = 0 ± + 4 r = r = (Q GP has positive terms) x 4. If sin cos ec π + = 4 then a value of x is () () (3) 3 (4) 4 Sol. (3) sin x 4 π sin + = x 3 π sin cos + = x 3 = x = 3 Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 8

9 . In the binomial expansion of ( a b ) n,n, the sum of th and 6th term is zero, then a b equals () n 4 () n 4 (3) 6 n (4) n 6 Sol. () T + T6 = 0 n n 4 4 n n 4 C a.b C a b = 0 n 4 4 n b C n 4 n 4 a = a b C ( ) ( ) a n! 4! n 4! n 4 = = b! n! n! 6. The set S : = {,, 3..., } is to be partitioned into three sets A, B and C of equal size. Thus, A B C = S, A C = B C = A C = φ. The numbers of ways to partition S is! () ( 3! ) 4 ()! 3! ( 4! ) 3 (3)! 3! ( 3! ) 4! (4) ( 4! ) 3! 3! =! 3 3 Total number of ways = ( 4! ) 3! ( 4! ) π π 7. The largest interval lying in, for which the function x x f ( x) = 4 + cos + log( cos x) is defined, is π () 0, Sol. () x x π π 0, π (3), () [ ] ( ) f(x) = 4 + cos + log cos x π π (4), 4 Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 9

10 For x 4 x R... (i) x For cos x x 0 0 x 4... (ii) For log(cosx) π π cosx > 0 < x <... (iii) From (i), (ii) and (iii) π x 0, 8. A body weighing 3 kg is suspended by two strings m and m long, their other ends being fastened to the extremities of a rod 3 m long. If the rod be so held that the body hands immediately below the middle point. The tension in the strings are () kg and 3 kg () kg and 3 kg (3) kg and kg (4) kg and kg A 90 θ 3 m G θ B T T 90 θ θ m C m 3 kg Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 0

11 3 = + AB = AC + BC ACB = 90 Q G is mid-point of hypotenuse AB. GA = GB = GC GC = 6.m Let GBC = θ, then, GCB = θ By Lami s theorem T T 3 = = sin 80 θ sin 90 +θ sin90 ( ) ( ) T T 3 = = sinθ cosθ T = 3sinθ and T = 3cosθ T = 3 and T = 3 as 3 3 T = kg and T = kg sin θ=, cosθ= A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is Sol. () () 8 43 () 79 (3) 8 9 (4) 8 79 Probability of getting exactly 9 is 9 and probability of not getting 9 is 8 = Required probability = C ! 8 =! = = Career Launcher 8 / Solutions AIEEE Test Paper 007 Page

12 0. Consider a family of circles which are passing through the point (, ) and are tangent to x-axis. If (h, k) are the coordinates of the centre of the circles, then the set of values of k is given by the interval () k () 0 < k < Sol. (3) (3) k (4) k (, ) C(h, k) k ( ) ( ) h+ + k = k h + h+ + k k+ = k ( ) h + h+ k = 0 D 0 4 4( k) k 0 8k 4 k. Let L be the line of intersection of the planes x + 3y + z = and x + 3y + z =. If L makes an angle α with the positive x-axis, then cos α equals. (a) () 3 (3) (4) Sol. () Given planes are x + 3y + z = 0... (i) x+ 3y+ x = 0... (ii) Let l, m, n be the direction cosines of line of intersection of plane (i) and (ii). l + 3m + n = 0... (iii) l+ 3m+ n = 0... (iv) Career Launcher 8 / Solutions AIEEE Test Paper 007 Page

13 Solving (iii) and (iv), we get m = l, n = l l + m + n = 3l = l = ± 3. The differential equation of all circles passing through the origin and having their centres on the x-axis is () dy y = x xy () dy x = y + xy (3) dy x = y + 3xy (4) dy y = x + xy General equation of circle x + y + gx+ fy+ c = 0 As centre is on x-axis, f = 0 As circle is passing through origin, c = 0 Equation of required circle will be x + y + gx = 0... (i) Differentiating w.r.t. x, we get dy x + y + g = 0... (ii) Eliminating g from (i) and (ii) dy x + y + x x y = 0 dy y = x + xy Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 3

14 3. If p and q are positive real numbers such that p + q =, then the maximum value of (p + q) is () () (3) Sol. () Given p + q =... (i) From (i), we can say 0 p and 0 q Put p = sinθ q = cosθ p + q = sinθ + cosθ Maximum value of sinθ + cosθ = Maximum value of p + q = 4. A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB (= a) subtends an angle of 60 at the foot of the tower, and the angle of elevation of the top of the tower from A or B is 30. The height of the tower is (4) () a 3 () a 3 (3) a 3 (4) a 3 P h O A a B OA = OB = radii In OAB, OA = OB = AB = a In POB tan30 = h a Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 4

15 h = 3 a h = a 3. The sum of the series 0 C 0 0 C + 0 C 0 C C 0 is () 0 C 0 () 0 C 0 (3) 0 C 0 (4) 0 Sol. (3) Given series is x = 0 C 0 0 C + 0 C 0 C C 0 x = 0 C 0 0 C + 0 C C 0 ( 0 C0 0 C0) ( 0 C 0 C9) ( 0 C 0 C 8)... ( 0 C0 0 C0) x ( C C C... C... C C C ) C = = As C C + C C = 0 x = C x = C The normal to a curve at P(x, y) meets the x-axis at G. If the distance of G from the origin is twice the abscissa of P, then the curve is a () hyperbola () ellipse (3) parabola (4) circle Sol. (, ) Let y = f(x) be a curve dy = slope of tangent = slope of normal dy Equation of normal X Y y = ( x ) dy G x+ y,0 Career Launcher 8 / Solutions AIEEE Test Paper 007 Page

16 Given dy x+ y = x dy y = x ydy = x or dy y = = 3x y x = + K or ydy = 3x y 3x x y y x = K or = + K or + = K /3 Curve is hyperbola or ellipse. 7. If z + 4 3, then the maximum value of z + is () 0 () 4 (3) 0 (4) 6 Given z+ 4 3 z+ = z+ 4+ ( 3) z z+ z z z+ 6 Maximum value of z+ is The resultant of two forces P N and 3 N is a force of 7 N. If the direction of the 3 N force were reversed, the resultant would be 9 N. The value of P is () 4 N () N (3) 6 N (4) 3 N Sol. () F θ F uur F = PN Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 6

17 uur F = 3N R = 3 + P + 6Pcosθ = 7 40 P cosθ=... (i) 6P F F θ π θ F ( ) R = 9 + P + 6Pcos π θ = 9 P 6Pcosθ = 0 ( ) = [(from i)] P 40 P 0 P = 0 P = N 9. Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0., respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is () 0.7 () 0.06 (3) 0.4 (4) 0. Sol. (3) Let A is the event of the plane I hit the target correctly. B is the event of the plane II hit the target correctly. c P(A) =.3 P( A ) =.7 c P(B) =. PB ( ) =.8 c Probability that the target is hit by the second plane = P( A ).P( B ) =.7. =.4 Assume second plane hit the target only one time. Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 7

18 30. If D = + x + y Sol. (3) for x 0, y 0, then D is () divisible by y but not x () divisible by neither x nor y (3) divisible by both x and y (4) divisible by x but not y D = + x + y C C C,C C C3 0 0 D = x x = xy 0 y + y D is divisible by both x and y. 3. x y For the hyperbola =, which of the following remains constant when α varies? cos α sin α () Abscissae of foci () Eccentricity (3) Directrix (4) Abscissae of vertices Sol. () x cos y = α sin α ( ) sin α= cos α e cos α e = ecosθ = ± Here, a = cosα, b = sinα Abscissae of foci = ±ae = ±ecosα = ± e = cosα a x =± =± cos α e (depends on α) Abscissae of vertices = ±a = ±cosα Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 8

19 3. If a line makes an angle of 4 π with the positive directions of each of x-axis and y-axis, then the angle that the line makes with the positive direction of z-axis is π π π () () (3) 6 3 (4) 4 π Sol. () π l= cos = 4 π m = cos = 4 l + m + n = n + + = n = 0 n = 0 π cos γ = 0 γ = 33. A value of C for which the conclusion of Mean Value Theorem holds for the function f(x) = log e x on the interval [, 3] is () log e 3 () log 3 e (3) log e 3 Sol. () Given function f(x) = loge x Mean Value Theorem for [, 3] f(3) f() f'(c) = 3 loge3 loge = = log e 3 c (4) log3 e c = = log3 e log 3 e Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 9

20 34. The function f(x) tan ( sinx cosx) = + is an increasing function in π π (), Sol. (3) f(x) = tan sinx + cosx ( ) π π (), 4 Let Z = sinx + cosx f(x) = tan (Z) f(x) is increasing only when Z increases π Z = sin x+ 4 π π From options Z increases only when < x < 4 π π (3), 4 π (4) 0, 3. Let If α α A = 0 α α 0 0 A =, then α equals () () (3) (4) α α A = 0 α α 0 0 A = A = A = ± α α A = 0 α α 0 0 α = ± α = ± α = = ( α 0) = α Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 0

21 36. The sum of the series +... up to infinity is! 3! 4! Sol. (3) () e + () e (3) e (4) e +...! 3! 4! 3 x x x x e = !! 3! e = !! 3! 4! = e! 3! 4! 37. If uandv ˆ ˆ are unit vectors and θ is the acute angle between them, then uˆ 3 vˆ is a unit vector for () Exactly one value of θ () Exactly two values of θ (3) More than two values of θ (4) No value of θ Sol. () v θ u uˆ 3vˆ = 6uˆ vsin ˆ θnˆ Where ˆn unit vector perpendicular to û and ˆv = uˆ 3vˆ = 6 sin θnˆ 6sinθ= sinθ= 6 Here is one and only one value of θ between 0 and 90 for which sinθ= 6 Career Launcher 8 / Solutions AIEEE Test Paper 007 Page

22 38. A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point projection. The angle of projection is () bc b bc tan () tan (3) 4 (4) tan a ac ac a S(a,b) ( ) u L θ b a T M(c, 0) c Let u = velocity of projection θ = angle of projection g y = xtanθ x u cos θ As M is on the trajectory. g 0 = ctanθ c u cos θ g tanθ = u cos g = θ u cos θ From (i) and (ii), we get bc tanθ= ac a c... (i) θ b atan a ( )... (ii) bc θ= tan ac a ( ) 39. The average marks of boys in a class is and that of girls is 4. The average marks of boys and girls combined is 0. The percentage of boys in the class is () 60 () 40 (3) 0 (4) 80 Let x = Number of boys in the class y = Number of girls in the class Sum of marks of all boys = x Career Launcher 8 / Solutions AIEEE Test Paper 007 Page

23 Sum of marks of all girls = 4y Average of boys and girls combined marks = 0 x + 4y 0 = x+ y x = 4y Percentage of boys in the class = x 4y 00 = 00 = 80% x+ y y 40. The equation of a tangent to the parabola y = 8x is y = x+. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is () (, 0) () (, ) (3) (0, ) (4) (, 4) Sol. () Given parabola is y = 8x Given tangent y = x +...(i) As second tangent is perpendicular to (i), so that pair is on the directrix as directrix is the director circle. Equation of direcrix x =...(ii) Solving (i) and (ii), we get (, 0) Career Launcher 8 / Solutions AIEEE Test Paper 007 Page 3

Time : 3 hours 02 - Mathematics - July 2006 Marks : 100 Pg - 1 Instructions : S E CT I O N - A

Time : 3 hours 02 - Mathematics - July 2006 Marks : 100 Pg - 1 Instructions : S E CT I O N - A Time : 3 hours 0 Mathematics July 006 Marks : 00 Pg Instructions :. Answer all questions.. Write your answers according to the instructions given below with the questions. 3. Begin each section on a new