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MATHS- 13th Paper-2 TEST-3.pdf

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Class : XIII PART TEST- 2 INSTRUCTIONS 1. The question paper contain pages and 3-parts. Part-A contains 6 objective question , Part-B contains 2 questions of "Match the Column" 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 each wrong answer. PART-B (iii) Q.1 to Q.2 are "Match the Column" type which may have one or more than one matching options and carry 8 marks for each question. 2 marks will be awarded for each correct match within a question. There is NO NEGATIVE marking. Marks will be awarded onlyif all the correct alternatives are selected. PART-C (iv) Q.1 to Q.4 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded only if all the correct bubbles are filled in your OMR sheet. 2. Indicate the correct answer for each question by filling appropriate bubble(s) in your answer sheet. 3. Use only HB pencil for darkening the bubble(s). 4. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed. 5. The answer(s) of the questions must be marked by shading the circles against the question by dark HB pencil only. PART-A PART-B PART-C For example if only 'B' For example if Correct Ensure that all columns (4 before decimal and 2 after decimal) are filled. Answer having blank column will be treated as incorrect. Insert leading zero(s) if required after rounding the result to 2 decimal places. e.g. 86 should be filled as 0086.00 . . . . . . . . . . choice is correct then, match for (A) is P, Q; for the correct method for (B) is P, R; for (C) is P filling the bubble is and for (D) is S then the A B C D correct method for filling the bubble is P Q R S For example if only 'B & (A) D' choices are correct then, the correct method (B) for filling the bubbles is A B C D (C) (D) The answer of the question in any other manner (such as putting , cross , or partial shading etc.) will be treated as wrong. Class - XIII MATHS Part Test - 2 PART A Select the correct alternative(s) (choose one or more than one) Q.1 You are given an unlimited supply of each of the digits 1, 2, 3 or 4. Using only these four digits, you construct n digit numbers. Such n digit numbers will be called L E G I T I M AT E if it contains the digit 1 either an even number times or not at all. Number of n digit legitimate numbers are (A) 2n + 1 (B) 2n + 1 + 2 (C) 2n + 2 + 4 (D*) 2n – 1(2n + 1) [Sol. × × × × (n places) Total numbers = 3n + nC2 · 3n – 2 + nC4 · 3n – 4 + (nC2 indicates selection of places) (3 1)n

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- Class : XIII (Sterling) Time : 3 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 the Column" type and Part-C contains 4 subjective type questions.Allquestions 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 markwillbe deducted for eachwrong answer. PART-B (iii) Q.1 to Q.2 are "Match the Column" type whichmayhave oneor more than one matching options and carry8 marks for each question. 2 marks willbe awarded for each correct matchwithin a question. ThereisNONEGATIVE marking. Markswillbeawardedonlyifallthecorrectalternativesareselected. PART-C (iv) Q.1 to Q.4 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded onlyifallthe correct bubbles are filled in your OMR sheet. 2. Indicate the correct answer for eachquestionbyfilling appropriate bubble(s) inyour answer sheet. 3. Use onlyHB pencilfor darkening the bubble(s). 4. Use ofCalculator, Log Table, Slide Rule and Mobile is not allowed. 5. The answer(s) ofthe questionsmust bemarked byshading the circlesagainst the questionbydark HBpencil only. PART TEST-2 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 the bubble is 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 be filled as 0086.00 . . . . . . . . . . PART-A For example if only 'B' choice is correct then, the correct method for filling thebubble is A B C D For example ifonly'B & D' choices are correct then, thecorrect method for fillingthe bubbles is 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.
- Class - XIII MATHS Part Test - 2 PARTA Select the correct alternative(s) (choose one or more than one) Q.1 You are given an unlimited supply ofeach ofthe digits 1, 2, 3 or 4. Using only these four digits, you construct n digit numbers. Suchn digit numbers willbe calledLE G IT I MAT E ifit contains the digit 1 either aneven number times or not at all. Number of n digit legitimate numbers are (A) 2n + 1 (B) 2n + 1 + 2 (C) 2n + 2 + 4 (D*) 2n – 1(2n + 1) [Sol. × × ×............× (n places) Totalnumbers = 3n + nC2 · 3n – 2 + nC4 · 3n – 4 + .......... (nC2 indicates selection of places) = 2 ) 1 3 ( ) 1 3 ( n n = 2 1 [4n + 2n] = 22n – 1 + 2n – 1 = 2n – 1(2n + 1) ] Q.2 Whichofthefollowing statements are correct? (A*) Number ofwords that can beformed with 6 onlyofthe letters ofthe word "CENTRIFUGAL" if each word must contain allthe vowels is 3 · 7! (B*) There are 15 balls ofwhichsome are white and the rest black. Ifthe number ofways in which the balls can be arranged in a row, is maximumthan the number ofwhite balls must be equalto 7 or 8. Assume balls ofthe same colour to be alike. (C) There are12 things, 4 alikeofone kind, 5alike ofanother kind andtherest are alldifferent. The total number ofcombinations is 240. (D*) Number ofselections that canbe made of6 letters fromthe word "COMMITTEE" is 35. [Sol. (A) 11 c 7 v 4 / 7C2 · 6! = 3 · 7 · 6! = 3 · 7! (B) No. of ways = )! r 15 ( ! r ! 15 = 15cr r W ..... W W r 15 B ...... B B B This is maximumif r = 7 or 8 (C) Totalno. ofcombinations = 5 · 6 · 23 – 1 = 240 –1 = 239 (D) 2 alike + 2 other alike + 2 other alike = 1 2 alike + 2 other alike + 2 different = 3C2 · 4C2 = 18 2 alike + 4 different =3C1 · 5C4 = 15 All6 different = 1 —— = 35 Ans. ] Q.3 Given three points P1, P2 and P3 with position vectors k̂ j ˆ î 2 r1 ; ĵ 3 î r2 and k̂ 3 ĵ î r3 respectively. Equationofa line orthogonalto the plane containing these points and passing through the centroid P0 ofthe triangle P1P2P3, is (A) k̂ 2 j ˆ 3 î 4 3 1 r + ) k̂ 2 j ˆ 3 î 14 ( (B*) k̂ 3 2 j ˆ î 3 4 r + ) k̂ 2 j ˆ 3 î 14 ( (C) k̂ 2 j ˆ 3 î 4 r + ) k̂ 2 j ˆ 3 î 14 ( (D) k̂ 2 j ˆ 3 î 4 3 1 r + ) k̂ 2 j ˆ 3 î 10 (
- [Hint: P0 = 3 2 , 1 , 3 4 vector normalto the plane b a n = 1 4 1 3 2 0 k̂ ĵ î = ) 2 0 ( k̂ ) 3 0 ( j ˆ ) 12 2 ( î = – ) k̂ 2 ĵ 3 î 14 ( equationofthe line is k̂ 3 2 j ˆ î 3 4 r + ) k̂ 2 j ˆ 3 î 14 ( Ans. ] Q.4 Evaluate 3 0 2 1 2 dx x 1 x 2 ·sin x 1 1 = (A*) 72 7 2 (B) 16 2 (C) 9 2 (D) none [Sol: I = 3 0 2 1 2 dx x 1 x 2 ·sin x 1 1 sin–1 2 x 1 x 2 = 1 x if , x tan 2 1 x 1 if , x tan 2 1 1 I = 1 0 3 1 2 1 2 2 1 2 dx x 1 x 2 ·sin x 1 1 dx x 1 x 2 sin x 1 1 = 1 0 3 1 2 1 2 1 dx x 1 x tan 2 dx x 1 x tan 2 = 3 1 2 1 1 0 3 1 2 2 1 dx x 1 x tan 2 dx x 1 1 dx x 1 x tan 2 = 3 / 4 / 3 1 1 4 / 0 , dt t 2 x tan dt t 2 Put, tan–1x = t = 3 / 4 / 2 1 1 4 / 0 2 2 t 2 1 tan 3 tan 2 t 2 = 16 9 4 3 16 2 2 2 = 2 72 7 ]
- Q.5 Which ofthe following is always true about a function f(x) on the interval[a, b] = {x |a < x <b|}? (A) If f (x) > 0 on [a, b], then b a b a dx ) x ( dx ) x ( 2 f f (B) Iff (x) is increasing on[a, b] then f 2(x) is increasing on [a, b] (C) If f (x) attains a minimumat c where a < c < b, then f ' (c) = 0 (D*) none [Sol: (A) may be false if 0 < f (x) < 1, f 2 (x) < f (x) and b a b a dx ) x ( dx ) x ( 2 f f (B) may be false since if f (x) < 0, dx d f 2 (x) = 2 f (x) f (x) < 0 when f (x) > 0 and so f (x) is decreasing while f (x) is increasing. (C) maybefalse since a functionmaynot be differentiableat x= cfor whichit attainsits minimum. Since none of the statement are always true, the correct answer is (D) ] Q6. Let dx e 2 x = and f (t) = dx e 2 x t , (t > 0) (A) f (x) is equal to x (B*) f(x)is monotonic decreasing (C*) if area bounded by y = f (x), x = 2, x = 16 and y = c is minimum then c = 3 (D*) equation of normal to curve y = f (x) at x = is y – 1 = 2(x – ) [Sol. dx e 2 x = ; put x = t u dx = t du = t du e 2 tu = t dx e 2 tx f (t) = t f (x) = x f ' (x) = – 2 3 x 2 f (x) is monotonic decreasing. if area bounded by y = f (x), x = 2, x = 16 and y = c is minimum then c = f 2 16 2 = f (9) = 3 x N m = 2 = 2 equation of normal is (y– 1) = 2(x – ) ]
- PART B Match the column Q1. ColumnI ColumnII (A) P(m, n) (where m, n N) is anypoint inthe interior of (P) 20 the quadrilateral formed by the pair of lines xy= 0 & 8x2 + 5y2 + 14xy – 48x – 30y + 40 = 0 the possible number ofpositions of the point P is (B) The graph of y= x + k intersect the graph of (x–3)2 + (y + 2)2 (Q) 4 = 50 in one or more points if and only if a < k < b then b–a is (C) If a , b and c are three mutuallyperpendicular unit vectors and (R) 12 d is unit vector whichmakes equalangle with c and b , a then 2 d c b a = p + q r where q, r and p are three consecutive natural numbers then find the value ofp + q + r (D)Abscissa and ordinates ofngivenpoints are inAPwithfirst terma (S) 9 and common difference 1 and 2 respectively. Ifalgebraic sumof perpendiculars fromthese givenpoints on a variable line which always passes through the point 11 , 2 13 is zero, then a + n = [Ans: A Q; B P; C S; D R] [Hint: OP < OM 2 2 n m < 25 16 20 m2 + n2 < 41 400 m2 + n2 < 9 ..............(1) ( m, n N) & OP > OL 2 2 n m > 5 2 m2 + n2 > 5 4 m2 + n2 > 2 ..............(2) From (1) & (2) 2 < m2 + n2 < 9 Possibilities If m2 + n2 = 2 m = n = 1 P(1, 1) If m2 + n2 = 5 m = 2, n = 1 or m = 1, n = 2 P(2, 1) & P(1, 2) If m2 + n2 = 8 m = n = 2 P(2, 2) Totalno. ofpossibilities are 4 ]
- The value ofk can be obtained ifperpendicular distance from(3, – 2) onthe line x– y+ k =0 is 50 , which gives | 5 + k| 10 – 10 5 + k 10 – 15 k 5 a = –15; b = 5 (C) 2 d c b a = b · a 2 a 2 = 4 + c b a · d 2 Let c b a d , then c · d b · d a · d = cos = = = cos also 2 + 2 + 2 = 1 3cos2 = 1 cos = 3 1 2 d c b a = 3 3 . 2 4 = 4 + 3 2 q = 2, r = 3, p = 4 (D) a = 2, n = 10 ] Q.2 ColumnI ColumnII (A) The value of n 4 r 1 r 2 n n 4 r 3 r n Lim = q p (P) 2 in its lowest form where p + q = (B) No. ofintegralvalues ofa for which the cubic (Q) 0 f(x) = x3 + ax + 2 is non monotonic and has exactly one real root. (C) The radicalcentre ofthree circles is at the origin. (R) 1 The equations of two of the circles are x2 + y2 = 1 and x2 + y2 + 4x + 4y – 1 = 0. If the equation of third circle passes through the point (1, 1) and (–2, 1) is x2 + y2 + 2gx + 2fy + c = 0 then f – c = (D) If x2 + y2 + z2 – 2xyz = 1 , then the value of (S) 11 2 2 2 z 1 dz y 1 dy x 1 dx [Ans: A S; B P; C Q; D Q] [Sol (A) Tr = 2 4 n r 3 n · n r 1 S = n 4 1 2 n r · 4 n r 3 1 n 1 = 4 0 2 ) 4 x 3 ( x dx put 4 x 3 = t dt dx x 1 2 3
- = 10 4 2 t dt 3 2 = 10 1 4 1 3 2 t 1 3 2 4 10 = 10 1 40 6 · 3 2 (C) Ans: x2 + y2 + x – 2y – 1 = 0 Let the equation ofthe 3rd circle be x2 + y2 + 2gx + 2fy + c = 0 radicalaxis between 1st and 3rd is 2gx + 2fy + c + 1 = 0 It passes through (0, 0) c = – 1 Hence equation of the 3rd circle is x2 + y2 + 2gx + 2fy – 1 = 0 since the circle x2 + y2 + 2gx + 2fy – 1 = 0 passes through (1, 1) and (–2, 1), hence 2 + 2g + 2f – 1 = 0 2g + 2f + 1 = 0 ....(1) or 4 + 1 – 4g + 2f – 1 = 0 2g – f = 2 ....(2) Solving (1) and (2) f = – 1 ; g = 2 1 Hence the equation ofthe third circle is x2 + y2 + x – 2y – 1 = 0 Alternatively : Equation of the third circle canbe written as (x – 1) (x + 2) + (y – 1)2 + (y – 1) = 0 [ familyofcircle through (1, 1) and (– 2, 1) ] x2 + y2 + x + ( – 2) y – (1 + ) = 0 Radical axis of S2 and S3 is S2 – S3 = 0 i.e. 3x + (6 – )y + = 0 point (0, 0) willsatisfythis equation = 0 hence S3 = x2 + y2 + x – 2y – 1 = 0 (D) Diff. thegivenrelation 2x dx + 2y dy + 2z dz – 2 (xy dz + yz dx + zx dy ) = 0 ( x – yz ) dx + ( y – zx ) dy + ( z – xy ) dz = 0 dx y zx z xy dy x yz z xy dz x yz y zx ( )( ) ( )( ) ( )( ) 0 dx P dy P dz P 1 2 3 0 Now consider P y zx z xy 1 2 2 2 ( ) ( ) ( ) ( ) y xyz z x z xyz x y 2 2 2 2 2 2 2 2 ) y x y x 1 ( ) x z z x 1 ( 2 2 2 2 2 2 2 2 = ( ) ( ) ( ) ( ) 1 1 1 1 2 2 2 2 z x x y ) x 1 ( · ) z 1 )( y 1 ( ) x 1 ( P 2 2 2 2 1 |||ar expressions for P and P 2 3 Hence the result . ]
- PART C Q.1 Number of eight digit numbers which can be formed using the digits 1, 3, 4, 5, 6, 7, 8, 9 (without repetition) ifeachnumber has to be divisible by275 is [Hint: Anumber willbe divisible by275 ifit is divisible by 25 and 11 . For divisible by x + y = 1 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 43 For divisible by25 last two digits must be 7 5 Hence Now we have 1, 3, 4, 6, 8, 9 to be filled in six places, such that sum of even places = 27 5 = 22 or 16 5 = 11 sum of odd places = 27 7 = 20 or 16 7 = 9 36 is possible when we have 1, 4, 6 at even places and 3, 8, 9 at odd places ] Q.2 The least integral value of a for which the function f(x) = 3 x 3 a + (a + 2)x2 + (a – 1)x + 2 possess a negativepointofminimum. [Ans: a = 2] Q.3 Let c and b , a be threenon-coplanar unit vectors equallyinclinedto one another at anacute angle. If c b b a = c r b q a p , then p2 + q2/cos + r2 is equal to. [Ans: 2] [Sol: 1 c · c b · b a · a and cos c · b c · a b · a consider c r b q a p c b b a . Taking dot product with b we get 0 = q + (p + r) cos .....(1) Also c r b q a p c b b a ) c · a pr c · b qr b · a pq ( 2 c · c r b · b q a · a p c b · b a 2 c b b a 2 2 2 2 2 pr qr pq cos 2 r q p c · a b · b c · b b · a 2 sin 2 2 2 2 2 pr qr pq cos 2 r q p cos cos sin 2 2 2 2 2 2 cos 1 2 pr qr pq cos 2 r q p 2 2 2 p2 + r2 + q2 + 2q(p + r)cos + 2prcos = 2(1 – cos) From (1) we get 2q (p + r) cos = –2q2 and p2 + r2 + 2pr = 2 2 cos q p2 + r2 – q2 + 2pr cos = 2(1 – cos) p2 + r2 – q2 + 2 2 2 2 r p cos q cos = 2(1 – cos) (p2 + r2) ( 1 –cos) + cos q2 (1 – cos) = 2(1 – cos) p2 + r2 + cos q2 = 2 ]
- Q.4 Equation ofstraight linemeeting the circle x2 + y2 = 9intwo points at equaldistances of5 froma point (x1, y1) on the circumference ofthe circle is given byxx1 + yy1 + q p = 0. Where p + q = (no. q p isinits lowest form) [Ans: q p = 2 7 ; p + q = 9]

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