Anzeige
Anzeige
Nächste SlideShare
MATHEMATIC PART TEST-3 12th.pdf
Wird geladen in ... 3
1 von 8
Anzeige

### MATHEMATIC PART TEST-3 13th.pdf

1. Class : XIII (Sterling) Time : 1 hour Max. Marks : 60 INSTRUCTIONS 1. The question paper contain pages and 3-parts. Part-A contains 6 objective question , Part-B contains 2 questions of "Match theColumn" type and Part-C contains 4 subjective type questions.All questions are compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. PART-A (i) Q.1 to Q.6 have One or More than one is / are correct alternative(s) and carry 4 marks each. There is NEGATIVE marking. 1 mark will be deducted for eachwrong answer. PART-B (iii) Q.1 to Q.2 are"Match the Column" typewhich mayhave oneor more than onematching options and carry8 marks foreach question. 2 marks will be awarded for each correct match within a question. ThereisNONEGATIVEmarking.Markswillbeawardedonlyifallthecorrectalternativesareselected. PART-C (iv) Q.1 to Q.4 are "Subjective"questions. There is NO NEGATIVE marking. 5 Marks will be awarded onlyifall the correct bubbles are filled in yourOMR sheet. 2. Indicate the correctanswer for each question byfillingappropriate bubble(s) inyour answer sheet. 3. Use onlyHB pencil fordarkeningthe bubble(s). 4. Use ofCalculator, LogTable, SlideRule and Mobile is not allowed. 5. Theanswer(s)ofthequestionsmustbemarked byshadingthecirclesagainst thequestionbydarkHBpencil only. PART TEST-3 PART-B For example if Correct match for (A) is P, Q; for (B) is P, R; for (C) is P and for (D) is S then the correct method for filling thebubbleis P Q R S (A) (B) (C) (D) PART-C Ensure that all columns (4 before decimal and 2 after decimal) are filled. Answer having blank column willbe treated as incorrect. Insert leading zero(s) if required after rounding the result to 2 decimalplaces. e.g. 86 should befilled as 0086.00 . . . . . . . . . . PART-A For example if only 'B' choice is correct then, the correct method for fillingthebubbleis A B C D Forexampleifonly'B& D' choices are correct then, the correct method forfillingthebubblesis A B C D The answer of the question in any other manner (such as putting , cross , or partial shading etc.) will be treated as wrong. MATHEMATICS
2. PARTA Select the correct alternative(s) (choose one or more than one) Direction for Q.1 & Q.2 (2 questions) Cards aredrawnonebyonefromawell shuffledpack of52 playingcardswithoutreplacementuntilthe queen of hearts appears. Let the chance that 8th card drawn the queen ofhearts is s r where r and sare relativelyprime.Acircle of radius r obtained as above is tangent to two congruent circles which are tangent toeachother.Allthethreecircles aredisjoint andthesethreecircles havecommon tangent line. Let xbetheradii ofthetwo congruent circles.Suppose threenumbers a2, b2 and c2 forman arithmetical progression. If (a + b) = – 12 and (c – b) = x and a b a c   has the value equal to q. If now the roots of the equation x2 + Bx + D = 0 are r1 and r2 where r1 + r2 = 2 and 2 2 2 1 r r  = q then Q.1 The value of 'x'is equal to (A) 2 (B) 4 (C) 6 (D) 8 Q.2 The value of 'D' is equal to (A) 2 1 (B) 4 1 (C) 6 1 (D) 8 1 Q.3 Refering to the parabola y2 = 4ax (a > 0) four statements are given I Chord of contact of the pair of tangent from (–a, 0) passes through the focus of the parabola. II All chords subtendinga right angle at the vertex passesthrough focus. III Tangent totheparabolaat anypoint Pmeets thetangent at thevertexinY.Alinedrawn through Y perpendicularto the tangent passes through the focus. IV Acirclecircumscribinganytriangleformedby3co-normalpointsontheparabolapassesthrough itsfocus. (A) exactlyonestatement is correct (B) exactlytwo statements are correct. (C) exactlythreestatements are correct (D) all thestatements are correct. Q.4 The area of region log2(logyx) > 0  x        2 , 2 1 is (A) 8 5 sq. units (B) 8 7 sq. units (C) 8 13 sq. units (D) 8 15 sq. units Q.5 The vertices of a triangle in the argand plane are 3 + 4i, 4 + 3i and 2 6 + i then distance between orthocentreandcircumcentreofthetriangleis (A) 6 28 137  (B) 6 28 137  (C) 6 28 137 2 1  (D) 6 28 137 3 1  Q6. A purse contains 3 cards. Oneof them is red on both the sides, one is blue on both the sides and one is red on oneside and blue onthe other side.Acard is randomlychosen andput on the table. It shows red colour on theupper side. Then the probabilitythat the other side of the card is also red is (A) 3 2 (B) 2 1 (C) 3 1 (D) none
3. PART B MATCH THE COLUMN: INSTRUCTIONS: Column-Iand column-IIcontainsfour entries each. Entries ofcolumn-Iareto bematchedwith some entriesofcolumn-II.Oneormorethanoneentriesofcolumn-Imayhavethematchingwiththesameentries ofcolumn-IIandoneentryofcolumn-Imayhaveoneormorethanonematchingwithentriesofcolumn-II. Q1. ColumnI ColumnII (A) The fractional part of 15 2 n 4 is 15 k where k is (P) 0 (B) If C0, C1, C2, ..........Cn denote the binonial (Q) 1 coefficients in the expansion of (1 + x)n, then        n 0 r r n e e r n r 10 log 1 10 log r 1 C ) 1 ( is equal to (C) The remainder obtained when 1! + 2! + 3! + .... (R) 2 .............. 95! is divided by 15 is (D) The possible values of b > 0, so that the area of (S) 3 the bounded region enclosed between the parabola y = x – bx2 and y = b x2 is maximum is Q.2 ColumnI ColumnII (A) The area bounded by y = x2 + 2 and y = 2| x | – cosx (P) 2 is of the form of q p where p & q are relatively prime then p – q = (B) Tangent drawn to circle | z | = 2 atA(z1) and normal at (Q) 8 B(z2) meet at the point P(zp) then zp = 2 1 2 1 2 2 1 z z z z z | z | k  where k = (C)Acircle cuts two perpendicularlines so that each intercept is (R) 0 of given length a and b, the locus of the centre of the circle is given by ), b a ( m y x 2 2 2 2     where l, m, are relatively prime numbers then the value of l + m is (D)Theorderofdifferentialequationoffamilyof circleswithone (S) 5 diameter along y = x
4. PART C Q.1 If a function f(x) is defined as f(x) = max {4 – x2, |x – 2|, (x – 2)1/3} for x  [–2, 4]. Then find the area bounded bythe curve and x-axis is q p where p & q are co-prime then p+q is equal to. Q.2 P is anypoint, O being the origin. The circle on OP as diameter is drawn. Points Q and R are taken on the cirlceto lies on thesame side of OPsuch that POQ= QOR = . If P,Q, R are z1, z2, z3 such that 3 2 z2 2 = (2 + 3 ) z1z3. Then find angle . Q.3 A student can solve 2 out of 4 problems of mathematics, 3 out of 5 problems of physics and 4 out of 5 problemsofchemistry.Thereareequalnumberofbooksofmaths,physicsandchemistryinhisshelf.He selects one book randomlyand attempts 10 problems from it. If he solves the 1st problem, find the probabilitycorrectly upto 2 places ofdecimal that he will be able to solvethe second problem. Q.4 From an external point P(, 2) a variable line is drawn to meet the ellipse 4 y 9 x 2 2  = 1 at the pointA A and D. Same line meets thex-axis and y-axis at the points B and C respectively. Find theleast positive value of '' such that PA·PD = PB·PC.
5. SOLUTIONS PARTA Q.1 B Q.2 A [Q1 & Q2 Sol. Prove that queen of hearts is obtained and the 8th draws = 52 1 Hence r = 1 Now radius of congurrent circle is x as shown (x + 1)2 = (x – 1)2 + x2 x2 + 2x + 1 = 2x2 – 2x + 1 x2 – 4x = 0 x = 0 (not possible)  x = 4 now b2 – a2 = d c2 – a2 = 2d 8 c a — — — — — 4 b c x 12 b a given          2 2 2 2 a b a c   = 2 a b a c   · a b a c   = 2 q = c a ) b a ( 2   = 8 ) 12 ( 2   = 3 now r1 + r2 = 2; 2 2 2 1 r r  + 2r1r2 = 4  q + 2D = 4  D = 1/2] Q.3 (B) [Sol I & III correct ; II – passes through (4a, 0)  False ; IV – passes through vertex hence false] Q.4 (A) [Hint: logy(x) > 1 Case-I: y  (0, 1), x < y Case-II: y  (1, ), x > y Area = 8 5 2 1 8 1 ) 1 )( 1 ( 2 1 2 1 2 1 2 1                 Ans. ] Q.5 (B) [Sol: Clearly |Z1 | = |Z2 | = |Z3 | = 5,  points would lie on the circle centred at origin O.  Circumcenter C is origin Centroid G =           3 8 3 6 2 7 i OC = 3CG = 3 x 3 1   64 6 2 7 2   = 6 28 137  Ans. ] Q6. (A) [Hint: Bayes theorem  A : card randomly chosen shows up red B1 : card with both side red B2 : card which both side blue B3 : one side red & one side blue P(B1) = P(B2) = P(B3) = 3 1 A(A/B1) = 1 P(A/B2) = 2 1 P(A/B3) = 0 P(B1/A) = 3 2 ]
6. PART B MATCH THE COLUMN: Q1. [Ans: A  Q; B  P; C  S; D  Q] [Sol:(A) 15 2 n 4 = 15 16n = 15 ) 15 1 ( n  = 15 15 C ....... 15 C 15 C 1 n n n 2 2 n 1 n     = 15 k 15 1 , where k  N, = 15 1 + k          15 2n =        k 15 1 = 15 1 Ans: Q (B) Let loge10 = x, then        n 0 r r n e e r n r 10 log 1 10 log r 1 C ) 1 ( =        n 0 r r r n r nx 1 rx 1 C ) 1 ( =                    n 0 r r 1 r 1 n r n 0 r r r n r nx 1 rx C r n ) 1 ( nx 1 1 C ) 1 ( =                           n 0 r 1 r 1 r 1 n 1 r n 0 r r r n r nx 1 1 C ) 1 ( · nx 1 nx nx 1 1 C ) 1 ( = 1 n n nx 1 1 1 nx 1 nx nx 1 1 1                    = n n nx 1 nx nx 1 nx                = 0 Ans: P (C) 1! + 2! + 3!+ 4! = 33  remainder is 3 Ans: S (D) The given parabolas are y = x – bx2 ..........(1) and y = b x2 ..........(2) The abscissae of their points of intersection are given by b x2 = x – bx2  x2 (1 + b2) – bx = 0  x = 0, x = 2 b 1 b  Thus, the two parabolas intersect at O (0, 0) and A             2 2 2 b 1 b , b 1 b The region enclosed by the two parabolas is shaded in figure. To find the area of this region, let us slice it into vertical strips. The approximating rectangle shown in figure has length = y2 – y1, width = x and it can move between x = 0 and x = 2 b 1 b  . So, the area A enclosed by the two parabolas is given by A =      2 b 1 b 0 1 2 dx y y =             2 b 1 b 0 2 2 dx b x bx x = 2 b 1 b 0 3 3 2 b 3 x 3 bx 2 x          =   3 2 2 2 2 b 1 b b 1 b 3 1 b 1 b 2 1                 =  2 2 2 2 2 2 2 b 1 b 6 1 b 1 b 3 1 b 1 b 2 1                  Now, A =  2 2 2 b 1 b 6 1                      4 2 2 2 2 2 b 1 b 2 · b 1 2 · b b 2 · b 1 6 1 db dA      2 2 3 2 b 2 b 1 b 1 3 b db dA    
7.      3 2 b 1 3 b 1 b 1 b db dA     For maximum value of A, we must have db dA = 0  b(1– b) (1 + b) = 0  1 – b = 0  b = 1   0 b   Since b > 0. Therefore, 1 + b > 0. Thus, db dA changes its sign from positive to negative in the neighbourhood of b = 1. Ans: Q] Q.2 [Ans: A  S; B  P; C  S; D  P] [Sol:(A) x2 + 2 = 2| x | – cosx (|x| – 1)2 + 1 = –cosx  x = ± 1 Area =         1 1 2 dx x cos | x | 2 2 x = 3 8 (B) 2 z z z 0 arg 1 p 1              p 1 1 p 1 1 z z z z z z    = 0 ..................(1) Also B, O and P are collinear  arg p 2 z z =  or p 2 z z = p 2 z z ..................(2) from (1) & (2) zp = 2 1 2 1 2 2 1 z z z z z | z | 2  ; p 2 z z = p 2 z z , 2 2 p p z z z z  (C) Let the perpendicular lines are x-axis and y-axis and let the equation of circle is x2 + y2 + 2gx + 2fy + c = 0 xint = a c g 2 2   and b c f 2 y 2 int                4 b a f g 2 2 2 2  locus of centre is x2 – y2 = ) b a ( 4 1 2 2  5 m , 4 m , 1         (D) (x – a)2 + (y – a)2 = r2 ] PART C Q.1 [Ans: 63] [Sol: y = 4 – x2 is a parabola EMA. y = |x – 2| is a pair of straight lines ABF and ACD y = (x – 2)1/3 is the curve IHABG Thus frm the graph f(x) = max {4 – x2, |x – 2|, (x – 2)1/3} is f(x) =                         4 x 3 if 2 x 3 x 2 if 2 x 2 x 1 if x 4 1 x 2 if x 2 3 / 1 2 Now, area bounded by the curve and x-axis =               3 2 4 3 3 / 1 2 1 2 1 2 dx ) 2 x ( dx ) 2 x ( dx ) x 4 ( dx ) x 2 ( =   4 3 2 3 2 3 / 4 2 1 3 1 2 2 x 2 2 x 2 x 4 3 3 x x 4 2 x x 2                                    = 4 59 2 3 4 3 9 2 7     sq. units ]
8. Q.2 [Ans: 15o] [Sol: By rotation formula we have z2 =  i 1e z OP OQ = z1cos ei ...........(1) (considering  OQP) z3 =  i 2 1e z OP OR = z1cos2 e2i ...........(2) (considering  ORP) Now z2 2 = z1 2 cos2 e2i z2 2 =    cos z cos z z 1 2 3 2 1 (from 2)  z2 2 cos2 z1z3 cos  3 2 z2 2 cos2 3 2 z1z3 cos .............(3) It is given that 3 2 z2 2 =   3 2  z1z3 Using this in equation (3)   3 2  z1z3cos2 = 3 2 z1z3cos2    3 2  (2cos2 – 1) = 3 2 cos2  cos2  = 4 3 2  = 2 2 2 1 3           cos = 2 2 1 3  (where  is acute and hence cos > 0)   = 15°. Q.3 [Ans: 0.65] [Sol: Let P(m), P(p), P(c) be the probability of selecting book of maths, physics and chemistry respectively. Clearly P(m) = P(p) = P(c) = 3 1 Again let P(s1) and P(s2) be the probability that he solves the 1st and 1st as well as 2nd problem respectively. Then P (s1) = P(m) × P       m s1 + P(p) × P         p s1 + P(c) × P       c s1  P(s1) = 30 19 5 4 3 1 5 3 3 1 2 1 3 1       Similarly P(s2) = 300 125 5 4 3 1 5 3 3 1 2 1 3 1 2 2 2                          ) s ( P ) s ( P 1 2 = 30 19 300 125 = 38 25 = 0.65 ] Q.4 [Ans:  = 6] [Sol: We have been give PA·PD = PB·PC Equation of any line through point 'P' is : =       sin 2 y cos x = r or, x =  + r cos, y = 2 + r sin Putting this point in the equation of given ellipse we get: 4 (r cos + )2 + 9(rsin + 2)2 = 36 or r2 (4cos2 + 9sin2) + 4r (9cos + 2cos) + 42 = 0 Since PA and PD are the roots of this quadratic in r, we get PA · PD =       2 2 2 sin 9 cos 4 4 Similarly, putting x = r cos + , y = r sin + 2 in the equation of the coordinate axes i.e. xy = 0 we get; (r cos + ) (r sin + 2) = 0 or r2 sin cos + r(2cos + sin) + 2 = 0 Since PB and PC are the roots of this quadratic in 'r', we get PB · PC =    cos sin 2 =   2 sin 4 Thus, we finally get:   2 sin 4 =     2 2 2 sin 9 cos 4 4 =           2 cos 1 9 2 cos 1 4 8 2   2 sin 1 =    2 cos 5 13 2  13 = 5cos2 + 2sin2 for existence of          4 25 13 4 25 2 R   is the solution of L.H.S. inequality ....(1) and ) , 6 [ U ] 6 , (      is the solution of R.H.S. inequality ....(2)  (1)  (2) is  (–, –6] U [6, )
Anzeige