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REVIEW TEST-5 PAPER CODE : A INSTRUCTIONS 1. The question paper contain 24 pages and 4-parts. Part-A contains 9 objective question, Part-B contains 6 questions of "Match the Column" type and Part-C contains 15 and Part-D contains 2 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.9 have One or More than one is / are correct alternative(s) and carry 5 marks each. There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. PART-B (iii) Q.1 to Q.6 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 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. PART-D (iv) Q.1 to Q.2 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. Th 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 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 P Q R S leading zero(s) if required after (A) rounding the result to 2 decimal places. (B) e.g. 86 should be filled as 0086.00 (C) . . (D) . . . . . . . . Class - XIII Mathematics Paper - 2 Select the correct alternative. (More than one are correct) [3 × 5 = 15] Q.7 Let a1, a2, a3 ....... and b1, b2, b3 be arithmetic progressions such that a1 = 25, b1 = 75 and a100 + b100 = 100. Then (A*) the difference between successive terms in progression 'a' is opposite of the difference in progression 'b'. (B*) an + bn = 100 for any n. (C*) (a1 + b1), (a2 + b2), (a3 + b3), are in A.P. 100 (D*) (ar + br ) = 10000 r=1 [Sol. a1 + (a1 + d), (a1 + 2d), ......... b1 + (b1 + d1), (b1 + 2d1), ......... hence a100 = a1 + 99d b100 = b1 + 99d1 add ——————— a100 + b100 = 100 + 99(d + d1) hence d + d1 = 0 d = – d1 (A) (B) and (C) are obviously true. 100 100 100 200 n (ar + br ) = r=1 2 [(a1 + b1) + (a100 + b100)] = 2 = 104 (D) (using Sn = (a + d) ) ] 2 Q.8 L

- 1. REVIEW TEST-5 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-A For example if only'B'choice is correct then, the correct method for filling the bubble is A B C D For example if only 'B & D' choices are correct then, the correct method for filling the 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. 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 Class : XIII (XYZ) Time : 3 hour Max. Marks : 243 INSTRUCTIONS 1. The question paper contain 24 pages and 4-parts. Part-A contains 9 objective question, Part-B contains 6 questions of "Match the Column" type and Part-C contains 15 and Part-D contains 2 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.9 have One or More than one is / are correct alternative(s) and carry 5 marks each. There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. PART-B (iii) Q.1 to Q.6 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 NEGATIVE marking. 0.5 Marks will be deducted for eachwrong match. Marks will be awarded onlyifall 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. PART-D (iv) Q.1 to Q.2 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.
- 2. Select the correct alternative. (More than one are correct) [3 × 5 = 15] Q.7 Let a1, a2, a3 ....... and b1, b2, b3 ...... be arithmetic progressions such that a1 = 25, b1 = 75 and a100 + b100 = 100. Then (A*)thedifferencebetweensuccessivetermsinprogression'a'isoppositeofthedifferenceinprogression'b'. (B*) an + bn = 100 for any n. (C*) (a1 + b1), (a2 + b2), (a3 + b3), ....... are in A.P. (D*) 100 1 r r r ) b a ( = 10000 [Sol. a1 + (a1 + d), (a1 + 2d), ......... b1 + (b1 + d1), (b1 + 2d1), ......... hence a100 = a1 + 99d b100 = b1 + 99d1 add ——————— a100 + b100 = 100 + 99(d + d1) hence d + d1 = 0 d = – d1 (A) (B) and (C) areobviouslytrue. 100 1 r r r ) b a ( = 2 100 [(a1 + b1) + (a100 + b100)] = 2 200 100 = 104 (D) (using Sn = ) d a ( 2 n ) ] Q.8 Let f (x)= 0 x if 0 0 x if ] x [ x x 1 x where [x] denotes the greatest integer function, then thecorrect statements are (A*) Limit exists for x = – 1. (B*) f (x) has a removable discontinuityat x = 1. (C*) f (x) has a non removable discontinuityat x = 2. (D*) f (x)is discontinuous at allpositiveintegers. [Hint: f (1+) = 1 = f (1–) but f (1) = 2; f (–1+) = 3 = f(–1–) f (2+) = 4; f (2–) = 2; f (2) = 4 ] Q.9 f ''(x) > 0 for all x [–3, 4], then which of the following are always true? (A) f (x)has a relative minimum on(–3, 4) (B*) f (x) has a minimum on [–3, 4] (C*) f (x) is concave upwards on [–3, 4] (D*) if f (3) = f (4) then f (x) has a critical point on [–3, 4] [Hint: (A) f(x)has norelative minimumon (–3, 4) (B) f(x)is continuous function on[–3, 4] f (x) has min. and max. on [–3, 4] by IVT (C) f '' (x) > 0 f (x) is concave upwards on [–3, 4] (D) f (3) = f (4) ByRolle'stheorem c (3, 4), where f ' (c) = 0 critical point on [– 3, 4] ] Class - XIII Mathematics Paper - 2
- 3. PART-B MATCH THE COLUMN [2 × 8 = 16] INSTRUCTIONS: Column-Iand column-IIcontainsfour entries each. Entries ofcolumn-Iareto bematchedwith some entriesofcolumn-II.Oneormorethanoneentriesofcolumn-Imayhavethematchingwiththesameentries ofcolumn-IIandoneentryofcolumn-Imayhaveoneormorethanonematchingwithentriesofcolumn-II. Q.52 Column–I Column–II (A) Let Z = (cos 12° + i sin 12° + cos 48° + i sin 48°)6 (P) – 1 then Im(z) is equal to (B) ) ex x ( n x sin n Lim 3 4 3 0 x l l equals (Q) 0 (C) If twice thesquare on the diameter of a circle is equal to the (R) 1 sum of the squares on thesides of the inscribed triangleABC then sin2A + sin2B + sin2C equals (D) Let g (x) be a function such that g(a + b) = g(a) · g(b) (S) 2 for all real numbers a and b. If 0 is not an element of the range ofgthen g(x) · g(–x) equals [Ans. (A) Q; (B) P; (C) S; (D) R] [Hint: (A) Z = cos 12° + cos 48° + i (sin 48° + sin 12°) = 2 cos 30° · cos 18° + 2i sin 30° · cos 18° = 2 cos 18° (cos 30° + i sin 30°) = 2 cos 18° [cos(/6) + i sin(/6)] Z6 = 26 · cos618°(cos + i sin ) = – 26 (cos618°)6 + 0i Im Z6 = 0 Ans. (B) ) ex x ( n x sin n Lim 3 4 3 0 x l l = ) e x ( x x sin n Lim 3 3 0 x l = ln e 1 = – 1 Ans. (C) Given 2d2 = a2 + b2 + c2; hence 8R2 = 4R2 A sin2 A sin2 = 2 (D) g(0) = g2(0) g(0) = 1 (as g (0) 0) again put a = x, b = – x g (0) = g (x) · g(–x) = 1 Ans. ] Q.62 ColumnI ColumnII (A) ABC is a triangle. If P is a point inside the ABC such that (P) centroid areas of the triangles PBC, PCAand PAB are equal, then w.r.t the ABC, Pis its (Q) orthocentre (B) If c , b , a are the position vectors of thethree non collinear pointsA,B and C respectivelysuch that the vector (R) Incentre C P B P A P V is anull vector then w.r.t. the ABC, P is its (C) If Pis a point inside the ABC such that the vector (S) circumcentre ) C P )( AB ( ) B P )( CA ( ) A P )( BC ( R is anull vectorthen w.r.t. the ABC, Pis its (D) If Pis a point in theplane of the triangleABC such that the scalar product B C · A P and C A · B P , vanishes then w.r.t. the ABC, P is its [Ans. (A) P; (B) P; (C) R; (D) Q]
- 4. PART-C SUBJECTIVE: Q.11vector ABCD is a tetrahedron with pv's of its angular points as A(–5, 22, 5); B(1, 2, 3); C(4, 3, 2) and D(–1,2,–3).IftheareaofthetriangleAEFwherethequadrilateralsABDEandABCFareparallelograms is S thenfindthe valueof S. [10] [Ans. 110] [Sol. pv of M = 2 d a = k̂ j ˆ 12 î 3 |||ly pv of N = 2 c a = k̂ 2 7 j ˆ 2 25 î 2 1 Now the AEF is as shown 5 1 5 6 0 2 k̂ ĵ î 2 1 S k̂ ĵ 10 î 3 S = 110 S = 110 Ans. ] Q.12217/3 If the value of the definite integral 1 0 1 dx x 1 x tan is equal to k 2 n l then find the value of k. [10] [Sol. Considering tan–1x as 1st and x 1 1 as 2nd and applying IBP [Ans. k = 8] I = 1 0 1 ) x 1 ( n · x tan l – 1 0 2 dx x 1 ) x 1 ( n l I = 4 ln 2 – 1 I 1 0 2 dx x 1 ) x 1 ( n l ....(1); put x = tan in I1 I1 = 4 0 d ) tan 1 ( n l (applying King) ....(2); I1 = 4 0 d tan 1 tan 1 1 n l = 4 0 d tan 1 2 n l I1 = 4 0 d 2 n l – 4 0 d ) tan 1 ( n l ....(3) 2I1 = 4 0 d 2 n l = 4 ln 2 I1 = 8 ln 2 I = 4 ln 2 – 8 ln 2 I = 8 ln 2 k = 8 Ans. ]
- 5. Q.13249/3 Suppose f : R R+ be a differentiable function and satisfies 3 f (x + y) = f (x) · f (y) for all x, yR with f (1) = 6. If U = ) 1 ( n 1 1 n Lim n f f and V = 3 0 dx ) x ( f . Find the value of (U·V). [10] [Ans. 126] [Sol. f ' (x) = h ) x ( f ) h x ( f Lim 0 h = h ) x ( f 3 ) h ( f · ) x ( f Lim 0 h (using f rule) = h 3 ) x ( f 3 ) h ( f · ) x ( f Lim 0 h f ' (x) = h ] 3 ) h ( f [ · 3 ) x ( f Lim 0 h = h ) 0 ( f ) h ( f Lim 3 ) x ( f 0 h (from f rule x = 0; y = 0; 3 f (0) = f 2(0); f (0) 0 hence f (0) = 3) f ' (x) = 3 ) x ( f · f ' (0); 3 · ) x ( f ) x ( ' f = f ' (0) = k (say) integrating 3 ln ) x ( f = kx + C; put x = 0; f (0) = 3 3 ln 3 = C hence, 3 ln ) x ( f = kx + 3 ln 3 3 ln 3 ) x ( f = kx; put x = 1; f (1) = 6 3 ln 2 = k hence 3 ln 3 ) x ( f = (3 ln 2) x ln 3 ) x ( f = x ln 2 3 ) x ( f = e(ln 2)x = 2x f (x) = 3 · 2x .....(1) now U = h ) 1 ( f ) h 1 ( f Lim 0 h where n = h 1 U = f ' (1) = 1 x x 2 n 2 · 3 l ; hence U = 6 ln 2 now, V = 3 0 x dx 2 3 = 3 0 x 2 2 n 3 l = 2 n 21 l , hence V = 2 n 21 l U · V = (6 ln 2) 2 n 21 l = 126 Ans. ] Q.14 The sides of a triangle are consecutive integers n, n + 1 and n + 2 and the largest angle is twice the smallestangle.Findn. [10] [Ans. 4] [Sol. Let A> B > C i.e. Ais the largest and C is the smallest. Let C = A = 2 also a = n + 2, b = n + 1, c = n
- 6. 2 sin a = sin c cos sin 2 2 n = sin n 2 cos = n 2 n cos C = n 2 2 n ....(1) also cos C = ab 2 c b a 2 2 2 = ) 1 n )( 2 n ( 2 n ) 1 n ( ) 2 n ( 2 2 2 cos C = ) 2 n 3 n ( 2 5 n 6 n 2 2 .....(2) from (1) and (2) ) 2 n 3 n ( 2 5 n 6 n 2 2 = n 2 2 n = 2 1 + n 1 2 n 3 n 5 n 6 n 2 2 – 1 = n 2 2 n 3 n 3 n 3 2 = n 2 2n2 + 6n + 4 = 3n2 + 3n n2 – 3n – 4 = 0 (n – 4)(n + 1) = 0 n = 4 Ans. ] Q.1533/3 f : R R, f(x) = 1 x n mx x 3 2 2 . Iftherangeof this function is[– 4, 3)then find thevalueof (m2 +n2). [10] [Ans. m = 0; n = – 4 and (m2 + n2) = 16] [Sol. f (x) = 2 2 x 1 3 n mx ) 1 x ( 3 ; f (x) = 3 + 2 x 1 3 n mx y = 3 + 2 x 1 3 n mx for y to lie in [– 4, 3) mx + n – 3 < 0 x R this ispossible onlyif m = 0 when, m = 0 then y = 3 + 2 x 1 3 n note that n – 3 < 0 (think !) n < 3 if x , ymax 3– now ymin occurs at x = 0 (as1 +x2 is minimum) ymin = 3 + n – 3 = n n = – 4 ]