MATHS- (12th & 13th) Paper-2.pdf

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MATHS- (12th & 13th) Paper-2.pdf

• 1. Class : XIII Time : 3 hour Max. Marks : 240 INSTRUCTIONS 1. The question paper contain 00 pages and 2-parts. Part-B contains 9 questions of "Match the Column" type and Part-C contains 15. 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 -B (iii) Q.1 to Q.9 are "Match the Column" type which may have one or more than one matching options and carry 10 marks for each question. 2.5 marks will be awarded for each correct match within a question. There is NEGATIVE marking. 0.5 Marks will be deducted for each wrong match. Marks will be awarded only if all the correct alternatives are selected. PART -C (iv) Q.1 to Q.15 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). PAPER CODE : A 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 will be treated as incorrect. Insert leadingzero(s)ifrequiredafter rounding the result to 2 decimal places. e.g. 86 should be filled as 0086.00 . . . . . . . . . . PART-D Ensure that all columns {1 before decimal and 2 after decimal with proper sign (+) or (–)} arefilledand columns after 'E'usedfor fillingpower of 10 with proper sign (+) or (–). Answer having blank column will be treated as incorrect. e.g. – 4.19 × 1027 should be filled as – 4.19 E + 27 P A P E R - 2
• 2. PART-B MATCH THE COLUMN [3 × 10 = 30] INSTRUCTIONS: Column-Iand column-IIcontainsfour entries each. Entries ofcolumn-Iareto bematchedwith some entriesofcolumn-II.Oneormorethanoneentriesofcolumn-Imayhavethematchingwiththesameentries ofcolumn-IIandoneentryofcolumn-Imayhaveoneormorethan onematchingwithentriesofcolumn-I Q.1 Column–I Column–II (A) The expression tan 55° · tan 65º · tan 75º simplifies to cot xº (P) 10 where x  (0, 90) then x equals (B) Suppose abc < 0, a + b + c > 0 and a | a | + b | b | + c | c | = x (Q) 9 then the value x3 + 16 x – 7 equals (C) f : R  R and satisfies f (2) = – 1, f '(2) = 4. If (R) 8   3 2 dx ) x ( " ) x 3 ( f = 7, then f (3) has the value equal to (S) 5 (D) The intersection of the planes 2x – y– 3z = 8 and x + 2y– 4z = 14 is the line L. The value of 'a'for which the line Lis perpendicular to the line through (a, 2, 2) and (6, 11, –1) is [Ans. (A) S; (B) P; (C) P; (D) Q] [Sol.(A) tan (60º – 5º) · tan (60º + 5º) · tan 75º = t · 3 1 t 3   × t · 3 1 t 3   × tan 75º [Let t = tan 5º] = 2 2 t 3 1 t 3   × tan 75º = 2 3 t 3 1 t t 3   × t º 75 tan = º 5 tan º 75 tan · º 15 tan = cot 5º  x = 5 Ans. (B) Q          0 c b a 0 abc  one of these is –ve and two of them are +ve  x = 1 + 1 – 1 = 1  x3 + 16x – 7 = 10 Ans. (C)   3 2 dx ) x ( " ) x 3 ( f = 7; (3 – x) · 3 2 ) x ( ' f +  3 2 dx ) x ( ' f = 7 0 – f '(2) + f (3) – f (2) = 7 – 4 + f (3) + 1 = 7  f (3) = 10 Ans. (D) Let V  is the vector alongthelineof intersection ofthe planes 2x – y – 3z – 8 = 0 and x + 2y – 4z – 14 = 0, then V  = 4 2 1 3 1 2 k̂ ĵ î    = ) k̂ j ˆ î 2 ( 5   This is to join of (a, 2, 2) and (6, 11, –1). Vector 1 V  along this is, 1 V  = k̂ 3 j ˆ 9 î ) 6 a (    Now, 1 V · V   = 0 gives a = 9 ] XIII (STERLING) MATHEMATICS (DATE: 07-01-2007) REVIEWTEST-6
• 3. Q.2 Column–I Column–II (A) If f (x) =   ) x ( g 0 3 t 1 dt where g (x) =   x cos 0 2 dt ) t sin 1 ( (P) 3 then the value of f '   2  (B) If f (x)isa non zerodifferentiablefunction such that (Q) 2  x 0 dt ) t ( f =  2 ) x ( f for all x, then f (2) equals (R) 1 (C) If    b a 2 dx ) x x 2 ( is maximum then (a + b) is equal to (S) – 1 (D) If          2 3 0 x x b a x x 2 sin Lim = 0 then (3a + b) has the thevalueequal to [Ans. (A) S; (B) R; (C) R; (D) Q ] [Hint: (A) f ' (x) = ) x ( g 1 ) x ( ' g 3  and g ' (x) = [1 + sin(cos2x)] (– sin x) hence f ' (x) = ) x ( g 1 ) x sin )]( x sin(cos 1 [ 3 2    f '        2 =   2 g 1 0 1 3    = 0 1 1   = – 1 as g        2 = 0  f '        2 = – 1 Ans. (C) Maximum when a = – 1; b = 2  a + b = 1 (D) If 0 x b a x x 2 sin Lim 2 3 0 x      0 x bx ax x 2 sin Lim 3 3 0 x     forlimittoexist 2 + b = 0  2 b    0 x x 2 ax x 2 sin Lim 3 3 0 x     applyLHS rule, 0 x 3 2 ax 3 x 2 cos 2 Lim 2 2 0 x      a = 2 0 x x 3 ) x 2 cos 1 ( 2 Lim    3 4 x 3 x sin 4 a 2 2    3a + b = 3 · 3 4 – 2 = 2 Ans. ]
• 4. Q.3 Column–I Column–II (A) Let X be the set of all 13 digit integers where each digit can take (P) 5 values from [0, 9].An integeris chosen at random from X.The probabilitythat it is a palindrome, is n 10 1 . The valueof n is (Q) 6 (B) If the coefficient of xn is theexpansion of 2 2 ) x 1 ( ) x 1 (   is 32 (R) 7 then the value of n equals (C) Let a function is defined as f : R  R, with f (x) = x e 31 1 6  . (S) 8 Numberofdifferentintegral values which f (x)can take, is (D) Consider two lines in space as L1: ) k̂ ĵ î 3 ( k̂ 2 ĵ r1        and L2: ) k̂ 2 î ( k̂ 6 ĵ 3 î 4 r2        . If the shortest distance between these lines is d then d equals [Ans. (A) Q; (B) S; (C) P; (D) Q] [Sol. (A) n (s) = 1013 n (A) = 107  probability= 13 7 10 10 = 6 10 1  n = 6Ans. (B) Coefficient of x2 in (1 – x)–n, n  N is n + r – 1Cr now coefficient of xn in (1 + x)2 (1 – x)–2 or coefficient of xn in (1 + 2x + x2)(1 – x)–2 or coeff. of xn in (1 – x)–2 + 2 · coeff. of xn – 1 in (1 – x)–2 + coeff. of xn – 2 in (1 – x)–2 = n + 1Cn + 2 · nCn – 1 + n – 1Cn – 2 = (n + 1) + 2n + n – 1 = 4n hence 4n = 32  n = 8 Ans. ] (C) For the graph f (x) can be integer 1, 2, 3, 4, 5 as f (x) is continuous from I.V.T. (D) Vector perpendicular to both 2 0 1 1 1 3 k̂ ĵ î n     = ) 1 0 ( k̂ ) 1 6 ( j ˆ ) 2 ( î      = k̂ j ˆ 7 î 2    says k̂ j ˆ 7 î 2   now k̂ 4 ĵ 2 î 4 B A V      now S.D. = | n | n · V    = 54 4 14 8   = 54 18 = 6 3 18 = 6 6  S.D = 6  d = 6 Ans. ]
• 5. PART-C SUBJECTIVE: [5 × 10 = 50] Q.1 Let f (x)is a quadratic functionsuch that f (0)=1 and   3 2 ) 1 x ( x dx ) x ( f isarational function, find thevalue of f ' (0). [Ans. 3] [Sol. Suppose g (x) =   3 2 ) 1 x ( x dx ) x ( f ....(1) =                 dx ) 1 x ( E ) 1 x ( D 1 x C x B x A 3 2 2 = A ln x – x B + C ln (1 + x) – x 1 D  – 2 ) 1 x ( 2 E  since g (x) is a rational function hence logarithmic functions must be asent  A= C = 0 g (x) =              dx ) 1 x ( E ) 1 x ( D x B 3 2 2 ....(2) comparing Nr of (1) and (2) f (x) = B(x + 1)3 + Dx2(x + 1) + Ex2 f (x) = (B + D)x3 + (3B + D + E)x2 + 3Bx + B  f (x) is quadratic function, hence B +D = 0 also f (0) = 1 gives B = 1  D = – 1  f (x) = (2 + E)x2 + 3x + 1 f ' (x) = 2(2 + E)x + 3 f ' (0) = 3 Ans. ] Q.2 Let P be a point on the curve C1: y= 2 x 2  and Q be a point on the curve C2: xy= 9, both P and Q lie inthefirst quadrant. If'd'denotes the minimum valuebetween P andQ, find the valueof d2. [Ans. 8] [Sol. Note that C1 is a semicircle and C2 is a rectangular hyperbola. PQwillbeminimumifthenormalatPonthesemicircleisalsoa normal at Q on xy = 9 Let the normal at P be y= mx....(1) (m > 0) solvingitwith xy = 9 mx2 = 9  x = m 3 ; y = 3 m 9  Q        m 3 , m 3 differentiating xy = 9 x dx dy + y = 0  dx dy = – x y  Q dx dy = – 3 m · m 3 = – m  tangent at P and Q must be parallel  – m = – m 1  m2 = 1  m = 1  normal at P and Q is y = x solving P(1, 1) and Q(3, 3)  (PQ)2 = d2 = 4 + 4 = 8 Ans. ]
• 6. Q.3 Let u =          4 0 2 dx x cos x sin x cos and v =          4 0 2 dx x cos x cos x sin . Find the value of u v . [Ans. 4] [Sol. v =    4 0 2 dx x cos x 2 sin 1 =    4 0 2 dx ) x tan 2 x (sec = tanx + 2 ln (sec x) 4 0  = (1 + ln 2) – 0 = 1 + ln 2 again u =    4 0 2 dx x 2 sin 1 x cos ; putting sin 2x = x tan 1 x tan 2 2  u =     4 0 2 x tan 2 x tan 1 dx =    4 0 2 ) x tan 1 ( dx =        4 0 2 x ) 4 tan( 1 dx =            4 0 2 x tan 1 x tan 1 1 dx =    4 0 2 dx 4 ) x tan 1 ( u =             4 0 2 dx ) x tan x tan 2 1 ( 4 1 =               4 0 4 0 x x tan x sec n 2 4 4 1 l =            4 1 ) 2 n ( 2 2 4 4 1 l =   1 ) 2 n ( 4 1  l Hence u =   1 ) 2 n ( 4 1  l and v = (1 + ln 2); u v = 4 Ans. ] Q.4 In how many ways 13 non distinguishable objects can be distributed among 7 persons so that every person get at least one book and atleast one person gets 4 books but not more. [Sol. When atleast one person to be given 4 books category (i) 4 4 1 1 1 1 1 (ii) 4 3 2 1 1 1 1 (iii) 4 2 2 2 1 1 1 Case(i): B B B B B B B B B B B B B number of ways = ! 5 · ! 2 ! 7 = 21 Case-(ii): B B B B B B B B B B B B B number of ways = ! 4 ! 7 = 210 Case-(iii): B B B B B B B B B B B B B number of ways = ! 3 · ! 3 ! 7 = 140 Total = 21 + 210 + 140 = 371 Ans. ]
• 7. Q.5 Given the cubic equation x3 – 2kx2 – 4kx + k2 = 0. If one root of the equation is less than 1, other root is in the interval (1, 4) and the 3rd root is greater than 4, then the value of k lies in the interval   ) b a ( b , b a   where a, b  N. Find the value of (a + b)3 + (ab + 2)2. [Ans. ] [Sol. f (x) = x3 – 2kx2 – 4kx + k2 = 0 note that f (0) = k2 > 0 f (1) > 0  1 – 2k – 4k + k2 > 0 k2 – 6k + 1 > 0 [k – ( 2 2 3 )] [k – ( 2 2 3 )] > 0 ....(1) Also f (4) < 0  64 – 32k – 16k + k2 < 0 k2 – 48k + 64 < 0 (k – 24)2 < 512 (k – 24 + 2 16 )(k – 24 – 2 16 ) < 0 [k – 8( 2 2 3 )] [k – 8( 2 2 3 )] < 0 ....(2) (1)  (2)  2 2 3 < k < 8( 2 2 3 ) 8 3 < k < 8( 8 3 )  a = 3; b = 8 (a + b)3 + (ab + 2)2  (11)3 + (26)2 = 1331 + 676 = 2007 Ans. ]