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REVIEW TEST-8 REVIEW TEST-8 INSTRUCTIONS 1. The question paper contain 00 pages and 4-parts. Part-A contains 9 objective question, Part-B contains 6 questions of "Match the Column" type and Part-C contains 14 and Part-D contains 3 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.14 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.3 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 RuPleAaRnTd-BMobile is not allowed. 5 PART-C 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 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) . . . . . . . . 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 Let f (x) = x 1

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REVIEW TEST-8 REVIEW TEST-8 INSTRUCTIONS 1. The question paper contain 00 pages and 4-parts. Part-A contains 9 objective question, Part-B contains 6 questions of "Match the Column" type and Part-C contains 14 and Part-D contains 3 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.14 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.3 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 RuPleAaRnTd-BMobile is not allowed. 5 PART-C 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 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) . . . . . . . . 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 Let f (x) = x 1

- 1. REVIEW TEST-8 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 : XII (ABCD) Time : 3 hour Max. Marks : 243 INSTRUCTIONS 1. The question paper contain 00 pages and 4-parts. Part-A contains 9 objective question, Part-B contains 6 questions of "Match the Column" type and Part-C contains 14 and Part-D contains 3 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 onlyif all the correct alternatives are selected. PART-C (iv) Q.1 to Q.14 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded onlyifall the correct bubbles are filled in your OMR sheet. PART-D (iv) Q.1 to Q.3 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 ] Only for12th (Kota andAjmer) Q.9 Let P andQ are two points denoting thecomplex numbers and respectivelyon the complex plane. Which of thefollowingequations can representthe equationofthecirclepassingthrough Pand Qwith least possible area? (A)Arg z z = 2 (B*) Re (z – )( z ) = 0 (C*) 2 2 2 | | | Z | | Z | (D) z 2 z 2 z z = 0
- 3. [Sol. (A) is not correct as it should be ± 2 (B)equationofcircle is, z z ispurelyimaginary Re z z = 0 or Re (z – )( z ) = 0 (B) is correct ] (C) is obviouslycorrect. (D) can not be correct as centre of required circle is 2 but from this given equation centre of the circle is – coefficient of z = – 2 . Only for 12th (Jaipur) Q.3 Takingyas thedependentandxastheindependentvariable,whichofthefollowingordinarydifferential equations arenot linear? (A*) y'y'' + y2 = x2 (B) x2y'' - xy' + 6y = lnx (C*) [1 + (y')2]1/2 = 5y (D*) y' = y x Only for13th (Kota, Jaipur &Ajmer) 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] ]
- 4. 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]
- 5. PART-C SUBJECTIVE: Q.10vector 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. ] For Both 12th & 13th (Kota / Jaipur /Ajmer) Q.11217/3 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] [Ans. k = 8] [Sol. Considering tan–1x as 1st and x 1 1 as 2nd and applying IBP 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. ]
- 6. For Both 12th & 13th (Kota / Jaipur /Ajmer) Q.12249/3 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. ]
- 7. Only for 12th (Kota / Jaipur /Ajmer) Q.1318/5 Withrespecttoaparticularquestiononamultiplechoicetest(having4 alternatives withonly1correct) a student knows the answer and therefore can eliminate 3 of the 4 choices from consideration with probability2/3,caneliminate2ofthe4choicesfromconsideration withprobability1/6,caneliminate1 choice from consideration with probability1/9, and can eliminate none with probability 1/18. If the student knows the answer, he answers correctly, otherwise he guesses from among the choices not eliminated. If theanswergivenbythestudent was found correct,then the probabilitythathe knewthe answeris b a where a and b are relativelyprime. Find the value of (a + b). [10] [Ans. p = 144 + 173 = 317] [Sol. A: his answer was correct B1 : hecan eliminate3 out of 4 alternatives B2 : hecan eliminate2 out of 4 alternatives B3 : hecan eliminate1 out of 4 alternatives B4 : hecan eliminatenoneout of4alternatives P(B1) = 3 2 ; P(B2) = 6 1 ; P(B3) = 9 1 ; P(B4) = 18 1 P(A/B1) = 1; P(A/B2) = 2 1 ; P(A/B3) = 3 1 ; P(A/B4) = 4 1 P(B1 / A) = ) A ( P ) A B ( P 1 = ) B A ( P ) B A ( P ) B A ( P ) B A ( P ) A B ( P 4 3 2 1 1 = 4 1 i i i 1 1 ) B / A ( P · ) B ( P ) B / A ( P · ) B ( P = 4 1 · 18 1 9 1 · 3 1 2 1 · 6 1 1 · 3 2 1 · 3 2 = 173 216 · 3 2 = 173 144 = b a a + b = 144 + 173 = 317 Ans. ] Only for 13th (Kota / Jaipur /Ajmer) Q.13 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. [12th & 13th 03-12-2006] Let C = A = 2 [T/S, Q.25, Ex-2, Ph-3] also a = n + 2, b = n + 1, c = n 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)
- 8. 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. ] (Only in 12th and only forKota andAjmer) Q.14 If the expression z5 – 32 canbe factorised intolinear and quadratic factors over real coefficients as (z5 – 32) = (z – 2)(z2 – pz + 4)(z2 – qz + 4) then find the value of (p2 + 2p). [10] [Ans. 4] [Sol. z5 – 32 = (z – z0)(z – z1)(z – z2)(z – z3)(z – z4) [T/S, Q.24, Ex-1, Complex] Where zi's , i = 0, 1, 2, 3, 4 are given by [12th 3-12-2006] zi = 2 5 m 2 sin 5 m 2 cos i (usingDemoivre'sTheorem) with m = 0, 1, 2, 3, 4, we get z0 = 2, amd z1 = 5 2 sin 5 2 cos 2 i ; z2 = 5 4 sin 5 4 cos 2 i ; z3 = 5 6 sin 5 6 cos 2 i z4 = 5 8 sin 5 8 cos 2 i hence, z5 – 32 = (z – 2) 4 z 5 2 cos 4 z2 4 z 5 4 cos 4 z2 z z ) ( z ) z )( z ( using 2 p = 5 2 cos 4 = 4 cos 72° = 4 sin 18° p2 + 2p = [16sin218° + 8 sin18°] = 8[1 – cos 36° + sin 18°] = 8 4 1 5 4 1 5 1 = 4 Ans.]
- 9. (Only in 13th and only for Kota andAjmer) and (Both in 12th & 13th ForJAIPUR) Q.1433/3 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 ]