Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -1 -AIEEE 2005 MATHEMATICS 1. The differential equation representing the family of curves ( ) 2 2 y c x C = + , where C > 0, is a parameter, is of order and degree as follows : (1) order 1, degree 3 (2) order 2, degree 2 (3) order 1, degree 2 (4) order 1, degree 1 Ans. (1) Sol.: The given equation is differentiated once and c is eliminated which gives y2 + 4x2y´2 – 4xyy´ = 4yy´3. Hence the order is 1 and degree is 3. 2. Area of the greatest rectangle that can be inscribed in the ellipse 2 2 2 2 1 x y a b + = is (1) ab (2) ab (3) 2ab (4) ab Ans. (3) Sol.: From the figure, it is clear that area of the rectangle = 4ab cos q sin q P a b ( cos , sin ) q q = 2ab sin 2q Hence, maximum area = 2ab. 3. 2 2 2 2 2 2 2 1 1 2 4 1 sec sec sec 1 nLimn n n n n ®¥é ù + + + ê ú ë û L equals (1) tan 1 (2) 1 tan1 2 (3) 1 sec1 2 (4) 1 cosec1 2 Ans. (2) Sol.: The general term 2 2 2 sec r r r T n næ ö = ç ÷ è ø . Put r x n = and 1 dx n = which gives the value of limit 1 2 2 0 sec x x dx ò = tan1 2 4. If the cube root of unity are 1, w, w2 then roots of equation (x – 1)3 + 8 = 0, are (1) –1, 1 – 2w, 1 – 2w2 (2) –1, 1 + 2w, 1 + 2w2 (3) –1, –1 + 2w, –1 – 2w2 (4) –1, –1, –1 Ans. (1)Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -2 -Mathematics AIEEE- 2005 Sol.: The given equation is (x – 1)3 + 8 = 0 This implies 3 1 1 2 x - æ ö = - ç ÷ è ø Þ 1 1, 2 x - æ ö=- -w ç ÷ è ø and –w2 Þ x = –1, 1 – 2w and 1 – 2w2 5. If A2 – A + I = 0, then the inverse of A is (1) A – I (2) I – A (3) A + I (4) A Ans. (2) Sol.: Multiplying the given equation by A–1, we get A – I + A–1 = 0 Þ A–1 = I – A. 6. 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}. The relation is (1) an equivalence relation (2) reflexive and symmetric only (3) reflexive and transitive only (4) reflexive only Ans. (3) Sol.: Since (3, 3), (6, 6), (9, 9), (12, 12) are the members of R Þ R is reflexive. Again, (a, b) Î R and (b, c) Î R Þ (a, c) Î R Þ R is transitive. 7. If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately (1) 25.5 (2) 24.0 (3) 22.0 (4) 20.5 Ans. (3) Sol.: mean 2median mode 3 + = = 21 2 22 21.66 22 3 + ´ = ; . 8. 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) x2 + 4y + 2 = 0 (2) x2 – 4y + 2 = 0 (3) y2 – 4x + 2 = 0 (4) y2 + 4x + 2 = 0 Ans. (3) Sol.: Let the point Q be (2t2, 4t) and mid point of PQ be (h, k) then 2 1 22 t h + = and 0 4 2 t k + = Eliminating t, we get y2 – 4x + 2 = 0. 9. If C is the mid point of AB and P is any point outside AB, then (1) 2 0 PA PB PC + + = uuur uuur uuur r(2) 0 PA PB PC + + = uuur uuur uuur r (3) 2 PA PB PC + = uuur uuur uuur (4) PA PB PC + = uuur uuur uuur Ans. (3) Sol.: The position vector of mid point of AB is 2 PA PB PC + = uuur uuur uuurAmity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -3 -AIEEE-2005 Mathematics 10. ABC is a triangle. Forces P, Q, R acting along IA, IB and IC respectively are in equilibrium, where I is the incentre of DABC. Then P : Q : R is (1) cos :cos :cos 2 2 2 A B C (2) cos A : cos B : cos C (3) sin A : sin B : sin C (4) sin :sin :sin 2 2 2 A B C Ans. (1) Sol.: A B C I PQ R 2 B C BIC + æ ö Ð = p -ç ÷ è ø, 2 A B AIB + æ ö Ð = p -ç ÷ è ø, 2 A C AIC + æ ö Ð = p -ç ÷ è ø Applying Lami’s theorem, we get, sin sin sin 2 2 2 P Q R B C A C A B = = + + + Þ cos cos cos 2 2 2 P Q C A B C = = 11. In a triangle PQR, . 2 R p Ð = If tan 2P æ ö ç ÷ è ø and tan 2Q æ ö ç ÷ è ø are the roots of ax2 + bx + c = 0, a ¹ 0 then (1) b = c (2) b = a + c (3) a = b + c (4) c = a + b Ans. (4) Sol.: tan tan 2 2 P Q ba + = - and tan tan 2 2 P Q Ca = , tan 1 2 P Q + æ ö= ç ÷ è ø gives a + b = c. 12. If the coefficient of rth, (r + 1)th and (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 (1) m2 – m(4r + 1) + 4r2 – 2 = 0 (2) m2 – m(4r – 1) + 4r2 + 2 = 0 (3) m2 – m(4r – 1) + 4r2 – 2 = 0 (4) m2 – m(4r + 1) + 4r2 + 2 = 0 Ans. (1) Sol.: Since 1 1 , , m m m r r r C C C - + are in A.P. Þ 1 1 2m m m r r r C C C - + = + which gives 2 2 (4 1) 4 2 0 m m r r - + + - =Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -4 -Mathematics AIEEE-2005 13. Let f : (–1, 1) ® B, be a function defined by 1 2 2 ( ) tan , 1 x f x x - = - then f is both one-one and onto when B is interval (1) , 2 2 p p é ù -ê ú ë û (2) , 2 2 p p æ ö -ç ÷ è ø (3) , 0 2p æ ö ç ÷ è ø (4) 0, 2p é ö÷ êë ø Ans. (2) Sol.: Since 1 2 2 tan 1 xx - - is an increasing function so the range B will be , 2 2 p p æ ö -ç ÷ è ø. 14. If the coefficient of x7 in 11 2 1 ax bx é ù æ ö +ç ÷ ê ú è ø ë û equals the coefficient of x–7 in 11 2 1 ax bx é ù æ ö -ç ÷ ê ú è ø ë û , then a and b satisfy the relation (1) 1 ab = (2) ab = 1 (3) a – b = 1 (4) a + b = 1 Ans. (1) Sol.: The coefficient of x7 in the expansion of 11 2 1 ax bx é ù æ ö +ç ÷ ê ú è ø ë û = 11C5 a6 b–5 and coefficient of x–7 in the expansion of 11 2 1 ax bx é ù æ ö -ç ÷ ê ú è ø ë û = 11C6 a5 b–6. On equating, we get a/b = 1. 15. If 13 z w z i = - and |w| = 1, then z lies on (1) a straight line (2) a parabola (3) an ellipse (4) a circle Ans. (1) Sol.: | | 1 | | 1 3 3 3 z z i z z i z i z w= Þ = Þ = - - - which is bisector of the line joining origin and the point (i/3). 16. If a2 + b2 + c2 = –2 and 2 2 2 2 2 2 2 2 2 1 (1 ) (1 ) ( ) (1 ) 1 (1 ) , (1 ) (1 ) 1 ax b x c x fx a x bx c x a x b x c x + + + = + + + + + + then f(x) is a polynomial of degree (1) 3 (2) 2 (3) 1 (4) 0 Ans. (2)Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -5 -AIEEE-2005 Mathematics Sol.: 2 2 2 2 2 2 2 2 2 1 (1 ) (1 ) ( ) (1 ) 1 (1 ) (1 ) (1 ) 1 ax b x c x fx a x bx c x a x b x c x + + + = + + + + + + , Applying C1 ® C1 + C2 + C3, we get 2 2 2 2 2 2 2 11 1 (1 ) 1 1 x b x x c x b x x cx x x bx c x + + + + = - + + 17. If z1 and z2 are two non-zero complex numbers such that |z1 + z2| = |z1| + |z2|, then arg z1 – arg z2 is equal to (1) 0 (2) 2 -p (3) 2p (4) –p Ans. (1) Sol.: Given that |z1 + z2| = |z1| + |z2| which is possible when Arg z1 = Arg z2 Þ Arg z1 – Arg z2 = 0. 18. The value of a 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) 3 (2) 2 (3) 1 (4) 0 Ans. (3) Sol.: Let a and b be roots of equation, then a2 + b2 = (a + b)2 – 2ab = (a – 1)2 + 5 which gives a = 1. 19. If the roots of the equation x2 – bx + c = 0 be two consecutive integers, then b2 – 4c equals (1) 2 (2) 1 (3) –2 (4) 3 Ans. (2) Sol.: The difference of roots = 1 Þ b2 – 4c = 1. 20. The system of equations ax + y + z = a – 1 x + ay + z = a – 1 x + y + az = a – 1 has no solution, if a is (1) not –2 (2) 1 (3) –2 (4) either –2 or 1 Ans. (3) Sol.: The value of the coefficient determinent D = 1 1 1 1 0 1 1 a a = a which gives (a – 1)2 [a + 2] = 0. So, a = 1 or a = –2 but for a = 1 there are infinite number of solutions of the given system.Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -6 -Mathematics AIEEE-2005 21. The value of 6 50 56 4 3 1 r r C C - = +å is (1) 56C3 (2) 56C4 (3) 55C4 (4) 55C3 Ans. (2) Sol.: 6 50 56 4 3 1 r r C C - = +å = 50C4 + 55C3 + 54C3 + 53C3 + 52C3 + 51C3 + 50C3. = 50C3 + 50C4 + 51C3 + 52C3 + 53C3 + 54C3 + 55C3 = 56C4. 22. If 1 0 1 1 A é ù =ê ú ë û and 1 0 , 0 1 I é ù =ê ú ë û then which one of the following holds for all n ³ 1, by the principle of mathematical induction (1) An = nA + (n – 1)I (2) An = 2n–1A + (n – 1)I (3) An = nA – (n – 1)I (4) An = 2n–1A – (n – 1)I Ans. (3) Sol.: 0 0 1 0 A I = + . The two matrices on the right commute, hence by the Binomial theorem 0 0 1 0 n A I n = + because powers higher than or equal to 2 of the matrix 0 0 1 0 are 0. 23. 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) 603 (2) 602 (3) 601 (4) 600 Ans. (3) Sol.: The alphabatical order of letters is A, C, H, I, N, S. The number of words which begin with any of the five letters A, C, H, I, N are 120 × 5 = 600. The next word will be SACHIN so its rank. is 601. 24. If 0 , n n x a ¥= =å 0 , n n y b ¥= =å 0 n n z c ¥= =å where a, b, c are in A.P. and |a| < 1, |b| < 1, |c| < 1 then x, y, z are in (1) Arithmetic – Geometric Progression (2) HP (3) GP (4) AP Ans. (2) Sol.: 0 , n n x a ¥= =å 0 , n n y b ¥= =å 0 n n z c ¥= =å Þ 1 1 1 , and 1 1 1 x y z a b c = = = - - - Since a, b, c are in A.P. x, y, z are in A.P.Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -7 -AIEEE-2005 Mathematics 25. If x is so small that x3 and higher powers of x may be neglected, then 3 32 12 1 (1 ) 1 2 (1 ) x x xæ ö + - + ç ÷ è ø - may be approximated as (1) 2 38 x - (2) 2 3 2 8 x x - (3) 2 3 1 8 x - (4) 2 3 3 8 x x + Ans. (1) Sol.: Upto terms of order x2 3 32 12 1 (1 ) 1 2 (1 ) x x xæ ö + - + ç ÷ è ø - = 2 2 2 2 3 3 3 3 1 3 3 1 1 1 2 8 2 4 2 8 8 x x x x x x x é ùé ù + + - - - + + = - ê úê ú ë ûë û 26. If 1 1 cos cos , 2y x - - - =a then 4x2 – 4xy cosa + y2 is equal to (1) 4sin2a (2) –4sin2a (3) 2sin2a (4) 4 Ans. (1) Sol.: Given that 1 1 cos cos , 2y x - - - =a Þ cos(cos–1x – cos–1y/2) = cos a Þ 2 2 1 1 /4 cos 2 xy x y + - - = a Þ 4x2 + y2 – 4xy cos a = 4 sin2a 27. If in a DABC, the altitudes from the vertices A, B, C on opposite sides are in H.P., then sin A, sin B, sin C are in (1) Arithmetic – Geometric Progression (2) H.P. (3) G.P. (4) A.P. Ans. (4) Sol.: The altitudes of the triangle are 2 2 2 , , a b c D D D are given in H.P.. Þ sin A, sin B, sin C are in A.P. 28. In a triangle ABC, let . 2 C p Ð = If r is the inradius and R is the circumradius of the triangle ABC, then 2(r + R) equals (1) a + b + c (2) c + a (3) b + c (4) a + b Ans. (4) Sol.: For a right angled triangle R = c/2 and ab r a b c = + + Þ 2( ) ( ) r R a b + = +Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -8 -Mathematics AIEEE-2005 29. A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched? (1) 1 , 3 æ ù -¥ ç ú è û (2) ( ] , 4 -¥ - (3) ( , ) -¥ ¥ (4) [ ) 2, ¥ Ans. (1) Sol.: Given that f(x) = 3x2 – 2x + 1 Þ 1 ( ) 6 2 0 3 f x x x ¢ = - ³ Þ ³ 30. Let a and b be the distinct roots of ax2 + bx + c = 0, then 2 2 1 cos( ) ( ) x ax bx c Lim x ®a- + + -a is equal to (1) 2 2 ( ) 2a - a - b (2) 2 1( ) 2 a-b (3) 2 2 ( ) 2 a a-b (4) 0 Ans. (3) Sol.: 2 2 1 cos( ) ( ) x ax bx c Lim x ®a- + + -a = 2 1 cos ( )( ) lim ( ) x ax x x ®a- -a -b -a = ( ) 2 2 2 a a-b because 2 01 cos 1 lim 2 x x x ® - = . 31. The normal to the curve x = a (cosq + qsinq), y = a (sinq + qcosq) at any point ‘q’ is such that (1) it passes through , 2 a a p æ ö - ç ÷ è ø (2) it is at a constant distance from the origin (3) it passes through the origin (4) it makes angle 2p + q with the x-axis Ans. (2) Sol.: The equation of normal to the curve at any point is y sin q + x cos q = a which is at a constant distance from origin. 32. Let f be differentiable for all x. If f(1) = –2 and f ¢(x) ³ 2 for x Î [1, 6], then (1) f(6) < 5 (2) f(6) = 5 (3) f(6) ³ 8 (4) f(6) < 8 Ans. (3) Sol.: Applying Lagrange’s mean value theorem, (6) (1) () (1,6) 6 1 f f f c c - ¢ = " Î - Þ (6) 2 2 (6) 8 5 f f + ³ Þ ³Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -9 -AIEEE-2005 Mathematics 33. If f is a real-valued differentiable function satisfying |f(x) – f(y)| £ (x – y)2, x, y Î R and f(0) = 0, then f(1) equals (1) 2 (2) 1 (3) –1 (4) 0 Ans. (4) Sol.: |f(x) – f(y)| £ (x – y)2, x, y Î R taking the limit as x ® y, we get | ( )| 0 ( ) 0 ( ) f x f x f x ¢ ¢ £ Þ = Þ is a constant function Þ f(1) = 0 34. Suppose f(x) is differentiable at x = 1 and 0 1 lim (1 ) 5, h f h h ® + = then f ¢(1) equals (1) 5 (2) 6 (3) 3 (4) 4 Ans. (1) Sol.: Since 0 (1 ) (1) lim h f h f h ® + - exists and 0 1 lim (1 ) h f h h ® + exists, it follows that 0 (1) lim h f h ® exists Hence, f(1) = 0 and 0 0 (1 ) (1) (1 ) (1) lim lim 5 h h f h f f h f h h ® ® + - + ¢ = = = . 35. 2 2 (log 1) 1 (log ) x dx x ì ü - í ý + î þ ò is equal to (1) 2 1 x xe C x + + (2) 2 (log ) 1 x C x + + (3) 2 log (log ) 1 x C x + + (4) 2 1 x C x + + Ans. (2) Sol.: Let I = 2 2 (log 1) 1 (log ) x dx x ì ü - í ý + î þ ò . Put log x = t, then I = ( ) ( )2 2 2 1 2 1 1 t t e dt t t æ ö ç ÷ - ç ÷ + + è ø ò = 2 2 1 1 (log ) te x C C t x + = + + + 36. A spherical ball 10 cm in radius is coated with a layer of ice of uniform thickness that 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. 54p (2) 5 cm/min. 6p (3) 5 cm/min. 36p (4) 1 cm/min. 18p Ans. (4) Sol.: Let 3 2 4 4 3 dv dr V r r dt dt = p Þ = p Þ 50 4 (225) dr dt = p Þ 1 cm/m 18 dr dt = pAmity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -10 -Mathematics AIEEE-2005 37. Let f(x) be a non-negative continuous function such that the area bounded by the curve y = f(x), xaxxi and the ordinates 4 x p = and 4 x p = b > is sin cos 2 . 4p æ ö b b+ b+ b ç ÷ è ø Then 2 f p æ ö ç ÷ è ø is (1) 1 2 4p æ ö - - ç ÷ è ø (2) 1 2 4p æ ö - + ç ÷ è ø (3) 2 1 4pæ ö + - ç ÷ è ø (4) 2 1 4pæ ö - + ç ÷ è ø Ans. (2) Sol.: Let /4 () sin cos 2 4 A f x dx b p p = =b b+ b+ b ò . Hence ( ) sin cos sin 2 4 dA f d p = b = b +b b- b + b Þ 1 2 2 4 f p p æ ö= - + ç ÷ è ø 38. Let f : R ® R be a differentiable function having f(2) = 6, 1 (2) . 48 f æ ö ¢ =ç ÷ è ø Then ( ) 3 2 6 4 lim 2 f x x t dt x ® - ò equals (1) 12 (2) 18 (3) 24 (4) 36 Ans. (2) Sol.: By L’Hospital’s rule [ ] ( ) 3 3 2 2 64 4 lim lim ( )4 ( ) 216 18 2 48 f x x x t dt f x f x x ® ® ¢ = = ´ = - ò 39. The area enclosed between the curve y = loge (x + e) and the coordinate axes is (1) 3 (2) 4 (3) 1 (4) 2 Ans. (3) Sol.: Required area = Area of the sector OAB = -e OB A x y [ ] 0 1 1 1 ln( ) ln ln 1 e e e x e dx udu u u u - + = = - = ò ò 40. 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) 2 : 1 : 2 (2) 1 : 1 : 1 (3) 1 : 2 : 1 (4) 1 : 2 : 3 Ans. (2) Sol.: Oy (4,4) x S1 S2 S3 4 2 3 0 64 4 12 xSdx = = ò , 4 2 1 0 64 4 12 ySdy = = ò , 2 1 3 128 64 16 16 12 12 S S S = - - = - =Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -11 -AIEEE-2005 Mathematics 41. If 2 1 1 02 , x I dx = ò 3 1 2 02 , x I dx = ò 2 2 3 12 , x I dx = ò and 3 2 4 1 2x I dx = ò then (1) I3 = I4 (2) I3 > I4 (3) I2 > I1 (4) I1 > I2 Ans. (4) Sol.: 2 3 2 2 x x > if 0 < x < 1. Hence I1 > I2. 42. 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) ¹ (0, 0) is (1) above the x-axis at a distance of 32 from it (2) above the x-axis at a distance of 23 from it (3) below the x-axis at a distance of 32 from it (4) below the x-axis at a distance of 23 from it Ans. (3) Sol.: The equation of the required line can be written as ax + 2by + 3b + l(bx – 2ay – 3a) = 0. The coefficient of x must be zero which implies l equals –a/b. Substituting this value the equation of the line is y = –3/2. 43. If (log log 1), dy x y y x dx= - + then the solution of the equation is (1) log y cx x æ ö= ç ÷ è ø (2) log x cy y æ ö= ç ÷ è ø (3) log x y cy y æ ö= ç ÷ è ø (4) log y x cy x æ ö= ç ÷ è ø Ans. (1) Sol.: The given equation can be written as ln 1 dy y y dx x x æ ö = + ç ÷ è ø . Substituting y u x = gives the differential equation ln du x u u dx = whose solution is ln u = cx. 44. If a vertex of a triangle is (1, 1) and the mid points of two sides through the vertex are (–1, 2) and (3, 2), then the centriod of the triangle is (1) 7 1, 3 æ ö ç ÷ è ø (2) 1 7 , 3 3 æ ö ç ÷ è ø (3) 7 1, 3 æ ö -ç ÷ è ø (4) 1 7 , 3 3 -æ ö ç ÷ è ø Ans. (1) Sol.: The coordinates of the other two vertices are obtained using section formula and are (–3, 3) and (5, 3). Hence the centroids has coordinates (1, 7/3).Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -12 -Mathematics AIEEE-2005 45. 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) infinitely many values of a (2) exactly two values of a (3) exactly one value of a (4) no value of a Ans. (4) Sol.: The equatin of the common chord is 5ax + (c – d)y + a + 1 = 0. This should be the same as 5x + by – a = 0. Hence 5 1 5a c d a b a - + = =- which gives the quadratic a2 + a + 1 = 0 which has no real roots. 46. If non-zero numbers a, b, c are in H.P., then the straight line 1 0 x y a b c + + = always passes through a fixed point. That point is (1) (1, –2) (2) 1 1, 2 æ ö - ç ÷ è ø (3) (–1, 2) (4) (–1, –2) Ans. (1) Sol.: 1 1 2 0 a c b + - = . Comparing this with the equation 1 0 x y a b c + + = the required point is (1, –2) 47. 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 – 2ax – 3by + (a2 – b2 p2) = 0 (2) 2ax + 2by – (a2 + b2 + p2) = 0 (3) x2 + y2 – 3ax – 4by + (a2 + b2 – p2) = 0 (4) 2ax + 2by – (a2 – b2 + p2) = 0 Ans. (2) Sol.: Let the centre be (h, k) and suppose the equation of the circle is x2 + y2 – 2hx – 2ky + c = 0. The orthogonality condition gives c = p2. Substituting this value of c and putting x = a, y = b in the equation of the circle gives the locus 2ax + 2by – (a2 + b2 + p2) = 0. 48. 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) 14 (2) 13 (3) 12 (4) 12 Ans. (3) Sol.: ( , 0) ae ( 0) -ae, (0, ) b By Pythagoras theorem 2(b2 + a2e2) = 4a2e2. Hence b2 /a2 = e2. For an ellipse b2/a2 = 1 – e2 which gives 1/2 e = .Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -13 -AIEEE-2005 Mathematics 49. 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) a hyperbola (2) a parabola (3) an ellipse (4) a circle Ans. (2) Sol.: A circle which touches the x axis has equation x2 + y2 + 2gx + 2fy + g2 = 0. This touches the given circle if g2 + f2 + 6f + 9 = ( )2 2 f ± . Hence the locus is parabola. 50. The angle between the lines 2x = 3y = –z and 6x = –y = – 4z is (1) 45º (2) 30º (3) 0º (4) 90º Ans. (4) Sol.: The direction ratios of the two lines are 1 1 , , 1 2 3 æ ö - ç ÷ è ø and 1 1 , 1, 6 4 æ ö - - ç ÷ è ø. The dot product of these two vectors is zero. Hence the angle is 90º. 51. If the angle q between the line 1 1 2 1 2 2 x y z + - - = = and the plane 2 4 0 x y z - + l + = is such that 1 sin 3 q= the value of l is (1) 34 (2) 4 3 - (3) 53 (4) 3 5 - Ans. (3) Sol.: The directin ratio of the line is (1, 2,2) e= . The direction ratio of the normal to the plane is ( ) 2, 1, - l 3 5 sin × = +l q n e gives 2 5 l= + l . So 53 l = 52. The locus of a point P(a, b) moving under the condition that the line y = ax + b is a tangent to the hyperbola 2 2 2 2 1 x y a b - = is (1) a parabola (2) a hyperbola (3) an ellipse (4) a circle Ans. (2) Sol.: Condition for tangency gives b2 = a2 a2 – b2. Hence the locus is the hyperbola. 53. The distance between the line r = 2i – 2j + 3k + l(i – j + 4k) and the plane r.(i + 5j + k) = 5 is (1) 3 10 (2) 10 3 (3) 10 9 (4) 10 3 3 Ans. (4) Sol.: The distance can be obtained by taking any point on the line and any point on the plane and projecting the vector joining these two points along the unit normal to the plane. A point on the line is 2i – 2j + 3k and the point on the plane is j. The unit normal is 53 3 + + i j k. The required distance is 10 3 3 .Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -14 -Mathematics AIEEE-2005 54. For any vector a, the value of (a × i)2 + (a × j)2 + (a × k)2 is equal to (1) 2a2 (2) 4a2 (3) 3a2 (4) a2 Ans. (1) Sol.: Use 2 2 2 | | | | ( ) ´ = - × a i a a i etc. gives 2|a|2. 55. 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) –2 (2) 2 (3) –1 (4) 1 Ans. (1) Sol.: The centres of the two spheres are (–3, 4, 1) and (5, –2, 1) whose mid point is (1, 1, 1). This lies on the plane. We get a = –2. 56. Let a, b and c be distinct non-negative numbers. If the vectors ai + aj + ck, i + k and ci + cj + bk lie in a plane, then c is (1) equal to zero (2) the Harmonic Mean of a and b (3) the Geometric Mean a and b (4) the Arithmetic Mean of a and b Ans. (3) Sol.: The scalar triple product of the three vectors must be zero which gives 1 0 1 0 a a c c c b = implying ab = c2 57. If a, b, c are non-coplanar vectors and l is a real number then [ ] 2 ( ) é ù l + l l = + ë û a b b c ab c b for (1) exactly three values of l (2) exactly two values of l (3) exactly one value of l (4) no value of l Ans. (4) Sol.: 2 4 ( ) [ ] é ù l + l l = l ë û a b bc a b c , [ ] [ ] + = - a b cb a b c . Hence the equation has no solution for l. 58. Let a = i – k, b = xi + j + (1 – x)k and c = yi + xj + (1 + x – y)k. Then [a, b, c] depends on (1) both x and y (2) neither x nor y (3) only y (4) only x Ans. (4) Sol.: 1 0 1 [ ] 1 1 1 2 1 x x x y x x y - = - = + + - a b cAmity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -15 -AIEEE-2005 Mathematics 59. 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 (1) 89 (2) 79 (3) 29 (4) 19 Ans. (4) Sol.: Number of points in the sample space is 27 and the number of points favourable to the event is 3. Hence, the required probability is 19 . 60. A random variable X has Poisson distribution with mean 2. Then P(X > 1.5) equals (1) 2 3 1 e - (2) 2 3 e (3) 2 2 e (4) 0 Ans. (1) Sol.: ( 1.5) 1 ( 0) ( 1) P X PX P X > = - = - = = 2 3 1 e - 61. Let A and B be two events such that ( ) 1 , 6 P A B È = ( ) 14 P A B Ç = and ( ) 1 , 4 P A = whree A stands for component of event A. Then events A and B are (1) independent but not equally likely (2) mutually exclusive and independent (3) equally likely and mutually exclusive (4) equally likely but not independent Ans. (1) Sol.: 3 5 1 ( ) , ( ) , ( ) 4 6 4 PA P A B P A B = È = Ç = Þ 1 ( ) 3 P B = . So, ( ) ( ) ( ) P A B P A P B Ç = . 62. A lizard, at 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 catch the insect after (1) 21 s (2) 24 s (3) 20 s (4) 1 s Ans. (1) Sol.: 21 21 + 20 t t2 t s 2 21 20 t t = + gives t = 21 s. 63. 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) 3 : 2 (2) 3:2 2 (3) 2 : 1 (4) 3: 2 Ans. (2) Sol.: 3x x y By Pythagoras theorem, 9x2 = x2 + y2. So, 3 38 x y =Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -16 -Mathematics AIEEE-2005 64. 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) 2 1( ) 2 f f m f f n ¢ ¢ + = (2) 2 1 ( ) 2 f fn ffm ¢ ¢ - = (3) 2 ( ) f f m f f n ¢ ¢ - = (4) 2 ( ) f f m f f n ¢ ¢ + = Ans. (2) Sol.: Area of triangle OAD – Area of triangle OCB = n gives 12 nf T¢ = O AD B T C t v m ( ) f m T f T¢ + = Þ ( ) f f T fm ¢- = . Using the first equation, we get, 2 2( ) n f f ffm ¢ ¢ - = 65. 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 the. The resultant of A and B after combining is displaced through a distance (1) 2( ) H A B + (2) H A B - (3) 2H A B - (4) H A B + Ans. (4) Sol.: When the forces A and B are acting on the plane, the distance of resultant from the force A is equal to dB A B + where d is the distance between the points of action of the forces A and B. When the couple of moment H is also acting on the plane, then distance of the resultant force from the force dB H A A B A B = - + + . Thus the position of resultant is shifted by H A B + . 66. The sum of the series : 1 1 1 1 4.2! 16.4! 64.4! + + + +K ad inf. is (1) 1 2e e - (2) 1 2e e + (3) 1 e e - (4) 1 e e + Ans. (2) Sol.: General term = 1 , 0,2,4 2 ! n n n = even ! 2 n x x n x e e n - + = åAmity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -17 -AIEEE-2005 Mathematics 67. If a1, a2, a3, ...., an are in G.P., then the determinant 1 2 3 4 5 6 7 8 log log log log log log log log log n n n n n n n n n a a a a a a a a a + + + + + + + + D = is equal to (1) 4 (2) 2 (3) 1 (4) 0 Ans. (4) Sol.: 1 2 log ,log ,log n n n a a a + + are in A.P.. 68. 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) (–¥, 4) (2) [4, 5] (3) (5, 6] (4) (6, ¥) Ans. (1) Sol.: Discreminent equals –4(k – 5) ³ 0 Þ k £ 5. The quadratic equation at x = 5 must be +ve and sum of the roots must be less than 10. These conditions implie k2 – 9k + 20 > 0. So, k < 4. 69. If the equation anxn + an – 1xn – 1 + ... + a1x = 0, a1 ¹ 0, n ³ 2, has a positive root x = a, then the equation nanxn–1 + (n – 1)an–1xn–2 + ... + a1 = 0 has a positive root, which is (1) greater than or equal to a (2) equal to a (3) greater than a (4) smaller than a Ans. (4) Sol.: The expression nanxn–1 + (n – 1)an–1xn–2 + ... + a1 is the derivative of anxn + an – 1xn – 1 + ... + a1x. So, by Rolle’s theorem, the derivative is 0 at some point between 0 and a. 70. 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(a) + f(a – x) (2) f(–x) (3) –f(x) (4) f(x) Ans. (3) Sol.: Put y = 0 in the given functional equation to get f (a) = 0. Next put x = 0 to get f(–y) = f(y). Next, put x = y = a to get f(2a) = –1. Finally, put x = 2a and replace y by x to get f(2a – x) = –f(x). 71. The plane x + 2y – z = 4 cuts the sphere x2 + y2 + z2 – x + z – 2 = 0 in a circle of radius (1) 2 (2) 2 (3) 3 (4) 1 Ans. (4) Sol.: C(1/2, 0, 1/2) - d r R 2 2 2 r R d = - = 5 9 1 2 6 - =Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42 -18 -Mathematics AIEEE-2005 72. 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 (3) 3a2 – 10ab + 3b2 = 0 (4) 3a2 – 2ab + 3b2 = 0 Ans. (2) Sol.: q 3q From the diagram it is clear that q equals 45º, ( ) 1/2 2 2 2 2 tan a b ab h ab a b a b é ù + - - ë û q= = + + gives 2 2 3 2 3 0 a ab b + + = 73. The value of 2 cos , 0, 1 xx dx a a p -p > + ò is (1) ap (2) 2p (3) ap (4) 2p Ans. (4) Sol.: 2 2 cos cos( ) 1 1 x x x x dx dx a a p p p - p - -p -p p - p - = + + ò ò gives 2 2 cos 1 cos 1 2 2 xx dx xdx a p p -p -p p = = + ò ò . 74. 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 (1) 23u (2) 3 u (3) 3u (4) 2u Ans. (2) Sol.: t = + v u g 0 × = v u gives 2| | 3 t g = u . This implies | | | | 3 = u v 75. Let x1, x2, ..., xn be n observations such that 2 400 i x = å and 80 i x = å . Then a possible value of n among the following is (1) 9 (2) 12 (3) 15 (4) 18 Ans. (4) Sol.: By Cauchy-Schwarz inequality, ( )1 2 2 1 n i i i x n x = £ å å we get n ³ 16. bacd