GLOBAL TALENT SEARCH EXAMINATIONS (GTSE) CLASS -XI

Transcription

1 GLOBAL TALENT SEARCH EXAMINATIONS (GTSE) Date: rd November 008 CLASS -XI MATHEMATICS Max Marks: 80 Time: :0 to :5 a.m. General Instructions: (Read Instructions carefully). All questions are compulsory. First 5 minutes for reading instructions.. This paper contains 0 objective type questions. Each question or incomplete sentence is followed by four suggested answers or completions. Select the one that is the most appropriate in each case and darken the correct alternative on the given answer-column, with a pencil or pen.. For each correct answer marks will be awarded and mark will be deducted for each incorrect answer.. No extra sheet will be provided. 5. Use of calculators & mobile is not permitted in examination hall. 6. Use of unfair means shall invite cancellation of the test Name of the Student : Roll No. : Centre : Invigilator s Signature : AMITY INSTITUTE FOR COMPETITIVE EXAMINATIONS E-5, Defence Colony, New Delhi Ph.: 6, 6. B-/6, Main Najafgarh Road, Janakpuri, New Delhi Ph.: 557 / / /. Amity International School, Mayur Vihar, Phase-I Ext., Delhi. Ph.: Amity Campus, Sector-, Noida Ph.: 950-8, 89. Amity International School, Sector-6, Vasundhara Yojna, Ghaziabad. Ph.: Amity International School, Sector-6, Gurgaon, Haryana. Ph.:

2 Mathematics-XI GTSE. The centre of circle arg z z is MATHEMATICS (a) i (b) i (c) i (d) + i. If ab <, then for equation (x a)(x b) 0 (a) both roots are positive (b) one root is positive, one is negative (c) both roots are negative (d) roots are imaginary. A line through O meets the lines x + y and x + y at the points P & Q, then OP : PQ is (a) : (b) : (c) : (d) :. PQ and RS are two perpendicular chords of the rectangular hyperbola xy c. If O is the centre of the rectangular hyperbola, then the product of the slopes of OP, OQ, OR and OS is equal to (a) (b) (c) (d) 0 5. The value of i 0 0 C C is equal to i 0 i (a) C 0 (b) C 0 (c) C (d) none of these 6. The maximum value of t t, where t t ( t ) ( t ) ( ) & 0 t i, is (a) 8 (b) (c) (d) - : Rough Space : - Amity Institute For Competitive Examinations []

3 Mathematics-XI GTSE 7. If a circle passes through the point (, ) and cuts the circle x +y orthogonally, then the locus of its centre is (a) x + y 9 0 (b) x + y 0 (c) x + y x 8y + 0 (d) x + y x 6y If,,, R, satisfy + ( ) + + 0, + ( ) + + 0, then (a) min ( ) (b) min () 8 (c) max. () 8 5 (d) none of these 9. The reflection of the curve xy in the line y x is the curve x + rxy + sy + t 0, then the value of r is (a) 7 (b) 5 (c) 75 (d) none of these 0. If normal to the parabola y x at P(t ) meets the curve again at Q(t ) such that area of the triangle OPQ (O is the origin) ; then t (a) t (b) t (c) P lies on the right of the latus rectum (d) all of these. If P 00 0, P 0 00, P 99 00, P 00 99, then which of the following is true (a) P > P > P > P (b) P > P > P > P (c) P > P > P > P (d) none of these. The circumcentre of the triangle formed by the tangents and the chord of contact from the point (, ) to the parabola y x is (a) (9, ) (b) (, 9) (c) (, 6) (d) (6, ) - : Rough Space : - Amity Institute For Competitive Examinations []

4 Mathematics-XI GTSE x y. From a point (h, k), tangents are drawn to b such that the chord of contact passes through a the point (0, 0), then (h, k) can be (a) (5, ) (b) (, 5) (c) (, 5) (d) no such point exists. If y, x, z are in A.P., then x+y, y+z, x+z are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 5. The point A on the parabola y ax for which AC AB is maximum, where B (0, a) and C ( a, 0) is (a) (a, a) (b) (a, a) (c) (a, a) (d) none of these 6. The sum of all the even divisors of 0 is (a) 860 (b) 9 (c) (d) 5 7. For the curve z i, which one is true (a) the maximum value of arg z (c) the maximum value of arg z (b) the maximum value of z (d) none of these 8. If in a triangle ABC, CD is the angular bisector of the angle ACB, then CD is equal to (a) a b C cos ab (b) a b cos C ab (c) ab C cos a b (d) none of these 9. India and Sri Lanka play one day international series until one team wins matches. No match ends in a draw. The number of ways in which India can win the series is (a) 5 (b) 70 (c) 0 (d) none of these 0. In a triangle ABC, the minimum value of cot A + cot B + cot C is equal to (a) 0 (b) (c) (d) none of these Amity Institute For Competitive Examinations []

5 Sol.: (c): Centre will lie on perpendicular bisector of and i.e. x 0. Z since chord joining and subtends an angle of at circumference, it will subtend an angle of at the centre. 5 so centre of circle i Sol.:. (b): (x a) (x b) 0 x x(a + b) + ab 0 Product of roots ab < 0 so one root is positive, other is negative. Sol.:. (a): same sign. The two given lines are parallel. Also origin does not lie between the lines as S and S have Now OPD ~ OQD Q OP PQ OD D D D 0 0 Now OD 5 P D x + y 0 0 OD 5 O x + y D D OD OD 5 So OP PQ OD D D c Sol.:. (b): Parametric point of xy c is ct, t Let P, Q, R, S be ct c c c,, ct,, ct, t t t and, c ct t respectively. slope of PQ c c t t c( t t ) t t similarly slope of RS t t

6 As PQ and RS are perpendicular,. t t t t t t t slope of OP c t t ct similarly slopes of OQ, OR and OS are,, t t t respectively. so product of slopes of OP, OQ, OR and OS... t t t t Sol.: 5. (a): ( + x) 0 0 C C x + 0 C x C 0 x 0 ( + x) C 0 + C x + C x C x ( t ttt) 0 0 C i C0 i 0 C 0 C C C C C C 0 C 0 i 0 coefficient of x 0 in ( + x) 0. ( + x) coefficient of x 0 in ( + x) C 0 Sol.: 6. (a): (t t ) / [( t ) ( t ) ( )]/ ( t) ( t ) ( t) (t t ) / (t t ) / (t t ) / t t 8 ( t t ) t Max (t t ) 8 Sol.: 7. (a): Let equation of circle is x + y + gx + fy + c 0 as this circle cuts the circle x + y 0 orthogonally (+g) 0 + (+f ) 0 c c x + y + gx + fy + 0 circle passing from (, ) g + f + 9 0

7 for locus of centre g x, f y x + y 9 0 Sol.: 8. (b): Obviously and are the roots of r + (x )r + x x + 0 i.e., x + x(r ) + r r + 0 r r + r So min () 8 Sol.: 9. (a): Let (h, k) be the general point of deflected curve. k h h k Mirror image of (h, k) in the line y x is, 5 5 k h h k 5 5 x y 7xy r 7 Sol.: 0. (d): Obviously t t t t t t Area of OPQ t t 0 t t 0 t t (t t ) (t + ) (t + /t ) t (t + ) (t + ) 6 t + t 0 (t ) (t + ) 0 t t So P lies on the right of latus rectum. Sol.:. (a): P 0 00 (00 + )

8 The expansion contains 0 terms with sum of last two terms < and all other terms So P < P Similarly, P (99 + ) The expansion contains 00 terms with sum of last two terms < and all other terms So P < P Obviously P > P So, P > P > P > P Sol.:. (a): The point (, ) lies on the directrix of y x, so the two tangents from it will be perpendicular. So chord of contact will be the diameter of the circumcircle. So circumcentre will be the midpoint of chord of contact. Let circumcentre be (h, k). The equation of chord of contact is T S ky (x + h) k h ky x + k h...(i) Also equation of chord of contact of (, ) is T 0 y (x )...(ii) (i) & (ii) represent same line, so k k, h 9 k h So circumcentre (9, ) Sol.:. (d): Chord of contact of (h, k) is T 0 xh yk b a This line can never pass through (0, 0) So no such point (h, k) exists. Sol.:. (b): Sol.: 5. (a): y, x, z are in A.P. y, x, z are in A.P. x + z, y + z, x + y are in A.P. x+z, y+z, x+y are in G.P. AB + BC AC

9 AC AB BC So maximum value is BC, which will be achieved when A, B, C are collinear. So A is point of intersection of BC and y ax. ( a, 0) C (0, a) B A x y equation of BC is a a If A (at, at) at at a t A (a, a) y x a Sol.: 6. (d): Sum of even divisors ( + )( + )( + 5)( + 7) 5 Sol.: 7. (a): z i represents a circle with centre (0, ) and radius P is the point of circle with maximum argument. sin / 0 So maximum value of arg Z / + /6 / P O i Sol.: 8. (c): AD : DB b : a Let AD bk, DB ak C Then bk + ak c k b So AD c a bc b a A b C/ D b : a a B Now in ACD, sin C / AD sin A CD sin C / bc /( b a) sin A CD CD bc sin A ( b a) sin( C / ) ba sin C ( b a) sin( C / ) ab C cos a b Sol.: 9. (a): Obviously maximum number of matches required is 7. Let us assume India wins th time in i th match, i 7. If we assume that India gets defeated in remaining (7 i) matches, then India will win exactly out of the 7 matches and there is a one-one correspondence between the given question and India winning out of 7 matches. So no. of ways 7 C 5

10 Sol.: 0. (b): (cot A cot B) + (cot B cot C) + (cot C cot A) 0 cot A + cot B + cot C cot A cot B + cot B cot C + cot C cot A But in any triangle, tan A + tan B + tan C tan A tan B tan C cot B cot C + cot C cot A + cot A cot B So cot A + cot B + cot C cot A cot B + cot B cot C + cot C cot A Mathematics Class-XI Answers (c). (b). (a). (b) 5. (a) 6. (a) 7. (a) 8. (b) 9. (a) 0. (d). (a). (a). (d). (b) 5. (a) 6. (d) 7. (a) 8. (c) 9. (a) 0. (b)