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Class : XIII PAPER CODE : A 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). 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) . . . . . . . . XIII MATHEMATICS REVIEW TEST-6 PART-B MATCH THE COLUMN [3 × 10 = 30] INSTRUCTIONS: Column-I and column-II contains four entries each. Entries of column-I are to be matched with some entries of column-II. One or more than one entries of column-I mayhave the matching with the same entries of column-II and one entry of column-I may have one or more than one matching with entries of column-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 | + | b | + | c | = x (Q) 9 a b c then the value x3 + 16 x – 7 equals (C) f : R R and satisfies f (2) = – 1, f '(2) = 4. If (R) 8 3 (3 x) f "(x)dx = 7, then f (3) has the value equal to (S) 5 2 (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 L is 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 = × × tan 75º [Let t = tan 5º] 3 t2 = 1 3t2 3t t3 × tan 75º = 1 3t 2 × tan 75º t tan15º·tan 75º = tan 5º = cot 5º x = 5 Ans. (B) Q abc < 0 one of these is –ve and two of them are +ve a + b + c > 0 x = 1 + 1 – 1 = 1 x3 + 16x – 7 = 10 Ans. 3 (C) (3 x) f "(x)dx 2 3 = 7; (3 – x) · f ' (x) 2 3 + f ' (x) dx = 7 2 0 – f '(2) + f (3) – f (2) = 7 – 4 + f (3) + 1 = 7 f (3) = 10 Ans. (D) Let V is the vector along the line of intersection of the planes 2x – y – 3z – 8 = 0 and x + 2y – 4z – 14 = 0, then ˆi V = 2 1 ˆj kˆ 1 3 2

- 1. REVIEW TEST- 10 / 7 PAPER CODE : A P A P E R - 1 Class : XII & XIII Time : 3 hour Max. Marks : 240 INSTRUCTIONS 1. The question paper contains 60 questions and 10 pages. Each question carry 4 marks and all ofthem are compulsory. There is NEGATIVE marking. 1 markwillbe deducted for eachwrong answer. 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. 2. Indicate the correct answer(s) for eachquestionbyfilling appropriate bubble(s) inyour OMR sheet. 3. Use onlyHB pencilfor darkening the bubble(s). 4. Use ofCalculator, Log Table, Slide Rule and Mobile is not allowed. 5. For example ifonly'B' choice is correct then, thecorrect 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 ofthe question in anyother manner (such as putting , cross , or partialshading etc.) willbe treated as wrong. USEFUL DATA Atomic Mass:Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31, Ag = 108, N = 14, Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.4, Ba = 137, Co = 59, Hg = 200, Pb = 207, He = 4, F=19. Radius of nucleus =10–14 m; h = 6.626 ×10–34 Js; me = 9.1 ×10–31 kg, R = 109637 cm–1.
- 2. XII & XIII MATHEMATICS REVIEW TEST-10 / 7 Select the correct alternative(s). (One or More than one is/are correct) [15 × 5 = 75] There is NEGATIVE marking. 1 markwillbe deducted for eachwrong answer. Q.41 PointAliesonthe line y= 2xand thesumofits abscissaand ordinate is 12. Point B lies onthe x-axis and the lineAB is perpendicular to the line y= 2x. Let 'O' be the origin. The area ofthe triangleAOB is (A) 20 (B) 40 (C) 60 (D*) 80 [Sol. Let A (x1, y1) x1 + y1 = 12 y1 = 12 – x1 A lies on y= 2x y1 = 2x1 12 – x1 = 2x1 x1 = 4. Hence A (4, 8) equation ofAB is x + 2y = ....(1) (1) passes through (4, 8) = 20 x + 2y = 20 B is (20, 0) Area of AOB = 2 8 · 20 = 80 sq. units.] Q.42 Suppose f (x) = sin x and g (x) = 1 – x . Whichofthe following composite functions have the same range? (A)fog (B*) gof (C) fof (D*) gog [Sol. (A) f [g(x)] = f (1 – x ) = sin(1 – x ) domain is x 0 ; range [–1, 1] (B) g[f (x)] = 1 – ) x ( f = 1 – x sin domain 2k x 2k + ; range [0, 1] (C) (fof)(x) = f[f(x)] = f(sin x) = sin(sin x) Domain x R; range [– sin 1, sin 1] (D) (gog)(x) = g [g (x)] = 1 – ) x ( g = 1 – x 1 domain is 0 x 1; range is [0, 1] hence (B) and (D) have the same range ] Q.43 Giventhe vectors k̂ j ˆ î 2 u k̂ 2 ĵ î v k̂ î w Ifthe volume of the parallelopiped having – u c , v and w c as concurrent edges, is 8 then 'c' can be equalto (A*) 2 (B*) – 2 (C) 4 (D) cannot be determined [Sol. V = – c2 ] w v u [ = – c2 1 0 1 2 1 1 1 1 2 = – c2[2(1 – 0) – 1(1) + (–2 – 1)] = – c2[2 – 1 – 3] = 8 2c2 = 8 c = 2 or – 2 Ans. ]
- 3. Q.44 Let A = the number ofways ofselecting a committee of5 froma group of9 persons. B = the number ofpermutations ofthe word PRELEPCO takenallat a time. C =the number ofwaysinwhich8 peoplecanbe arranged ina line iftwo particular people must stand next to each other. The value of C AB equals (A) 63 (B*) 126 (C) 360 (D) none [Hint: A = 9C5; B = ! 2 · ! 2 ! 8 ; C = 7! · 2!] Q.45 If 349(x + iy) = 100 3 2 3 2 i and x = ky then the value of k equals (x, y R) (A) 3 (B) – 3 (C) 3 1 (D*) – 3 1 [Sol. Let r cos = 2 3 and r sin = 2 3 r2 = 4 9 + 4 3 = 3 r = 3 also tan = 2 3 · 3 2 = 3 1 ; = 6 349(x + iy) = 100 6 sin i 6 cos 3 = 350 3 50 sin i 3 50 cos equating realand imaginaryparts 349 x = 350 cos 3 50 and 349y = 350 sin 3 50 y x = cot 3 50 = cot 3 2 x = – 3 1 y ; hence K = – 3 1 Ans. ] Q.46 Let x, ybe reals satisfying sin x+ cos y=1 and siny+ cos x= – 1. Then which ofthe following must be correct? (A*) sin(x + y) = 0 (B) cos(x – y) = 0 (C) cos(x + y) = 1 (D*) cos 2x = cos 2y [Sol. Given sin x + cos y = 1 ....(1) and sin y + cos x = – 1 ....(2) squaring equation(1), sin2x + cos2y + 2sin x · cos y = 1 ....(3) squaring equation(2), cos2x + sin2y + 2 sin y · cos x = 1 ....(4) adding equation (3) and (4), we get 2 + 2sin(x + y) = 2 sin(x + y) = 0 (A) is correct hence cos (x + y) = 1 or – 1 (C) need not be correct hence x + y = k y = k – x 2y = 2k – 2x cos 2y = cos(2k – 2x) cos 2y = cos 2x (D) is correct ]
- 4. Q.47 Two pointsA(x1, y1) and B(x2, y2) are chosen on the graphof f (x) = ln x with0 < x1 < x2. The points C and Dtrisect line segmentABwithAC < CB.ThroughC a horizontallineis drawn to cut the curveat E(x3, y3). If x1 = 1 and x2 = 1000 then the value of x3 equals (A*) 10 (B) 10 (C) (10)2/3 (D) (10)1/3 [Sol. Usingsectionformula a = 3 1000 · 1 1 · 2 = 334 b = 3 1000 n · 1 0 · 2 l b = 3 1000 n l ....(1) now line y= b intersects the curve y= lnx b = ln x ....(2) from(1) and (2) 3 1000 n l = ln x ln (1000)1/3 = ln x x = (1000)1/3 = 10 Ans. ] Q.48199/4 Suppose f is a differentiable realfunctionsuch that f (x) + f '(x) 1 for all x, and f (0) = 0, then the largest possible value of f (1), is (A) e–2 (B) e–1 (C*) 1 – e–1 (D) 1 – e–2 [Sol. Given f ' (x) + f (x) 1 multiplying byex f ' (x) ex + f (x) ex ex ) x ( f · e dx d x ex integrating between0 and 1 1 0 x dx ) x ( f e dx d 1 0 x dx e 1 0 x ) x ( f e 1 0 x e e · f (1) – e0 · f (0) e – 1 f (1) e 1 e ] Q.49 Let S be a set consisting offirst five prime numbers. SupposeAand B are two matrices oforder 2 each with distinct entries S. The chance that the matrixAB has atleast one odd entry, is (A) 0 (B) 20% (C) 80% (D*) 96% [Sol. S = {2, 3, 5, 7, 11} Total ways in whichAand B can be chosen = (5C4 · 4!)2 = (5!)2 P(E) = 1 – P(A and B does not contain 2) 1 – 2 2 ) ! 5 ( ) ! 4 ( = 1 – 25 1 = 25 24 = 96% Ans. ]
- 5. Q.50 A curve is represented parametrically by the equations x = t + eat and y = – t + eat when t R and a > 0. Ifthe curve touches the axis ofx at the pointA, thenthe coordinates of the pointAare (A) (1, 0) (B) (1/e, 0) (C) (e, 0) (D*) (2e, 0) [Sol. x = t + eat ; y = – t + eat dt dx = 1 + aeat ; dt dy = – 1 + aeat; dx dy = at at ae 1 ae 1 at the pointA, y = 0 and dx dy = 0 for some t = t1 1 at ae = 1 ....(1) ; also 0 = – t1 + 1 at e ; 1 at e = t1 .....(2), putting thisvalue in(1) we get, at1 = 1 t1 = a 1 ; now from(1) ae = 1 a = e 1 hence xA = t1 + 1 at e = e + e = 2e A (2e, 0) Ans. ] Q.51 Suppose g (x) = 2x + 1 and h (x) = 4x2 + 4x + 5 and h (x) = (fog)(x). The area enclosed bythe graph ofthe functiony= f (x) andthe pair oftangents drawn to it fromthe origin, is (A) 8/3 (B*) 16/3 (C) 32/3 (D) none [Sol. Given g (x) = 2x + 1; h (x) = (2x + 1)2 + 4 now h (x) = f[ g (x) ] (2x + 1)2 + 4 = f (2x + 1) let 2x + 1 = t f (t) = t2 + 4 f (x) = x2 + 4 ....(1) solving y = mx and y = x2 + 4 x2 – mx + 4 = 0 put D = 0 m2 = 16 m = ± 4 tangents are y = 4x and y = – 4x A= 2 0 2 dx ] x 4 ) 4 x [( 2 = 2 0 2 dx ) 2 x [( 2 = 2 0 3 ) 2 x ( 3 2 = 3 16 sq. units Ans. ] Q.52vector Suppose a vector V satisfies V ) V · V ( = V 169 and 4 ) î · V ( = 3 ) ĵ · V ( = ) k̂ · V ( where ) î · V ( >0 then ) î · V ( + ) ĵ · V ( + ) k̂ · V ( equals (A) 16 (B*) 19 (C) 21 (D) 23 [Sol. Given, V | V | 2 = 169 V | V | = 13 [To be put in teaching notes] let 4 ) î · V ( = 3 ) ĵ · V ( = ) k̂ · V ( = (say); Also k̂ ) k̂ · V ( ĵ ) j ˆ · V ( î ) î · V ( V hence, ) î · V ( = 4 ; ) ĵ · V ( = 3 ; ) k̂ · V ( = 2 | V | = 16 2 + 9 2 + 2 = 169 144 ) 144 16 9 ( 2 = 169
- 6. = 12 or – 12 (since î · V > 0, hence rejected) hence î · V = 3; ĵ · V = 4 and k̂ · V = 12 k̂ ) k̂ · V ( ĵ ) j ˆ · V ( î ) î · V ( V ) k̂ 12 j ˆ 4 î 3 ( V sum = 19 Ans. ] Q.53 Suppose the function gn(x) = x2n + 1 + anx + bn (n N) satisfies the equation 1 1 n dx ) x ( g ) q px ( = 0 for alllinear functions (px+ q) then (A) an = bn = 0 (B*) bn = 0; an = – 3 n 2 3 (C) an = 0; bn = – 3 n 2 3 (D) an = 3 n 2 3 ; bn = – 3 n 2 3 [Sol. We have 1 1 n n 1 n 2 dx ) b x a x )( q px ( = 0 equating the oddcomponent to be zero and integrating we get 3 n 2 p 2 + 3 p a 2 n + 2bnq = 0 for all p, q hence bn = 0 and an = – 3 n 2 3 ] Q.54 Suppose f: R R is a differentiable function satisfying f (x + y) = f(x) + f(y) + xy(x + y) for every x, y R. If f ' (0) = 0 then which of the following hold(s) good? (A*) fis an odd function (B*) fis a bijective mapping (C) fhas a minima but no maxima (D*) fhas aninflectionpoint [Sol. Given f (x + y) = f (x) + f (y) + xy(x + y) ....(1) f ' (x) = h ) x ( f ) h x ( f Lim 0 h = h ) h x ( xh ) h ( f Lim 0 h [ using (1) ] = 0 h Lim x(x + h) + 0 h Lim h ) h ( f = x2 + f ' (0) ( as f (0) = 0 ) f ' (x) = x2 f (x) = 3 x3 + C but f (0) = 0 C = 0 Hence f(x) = 3 x3 A, B, D ]
- 7. Q.55ellipse Consider the parabola y2 = 4x and the ellipse 2x2 + y2 = 6, intersecting at P and Q. Which of the following hold(s) good? (A*) The two curves are orthogonal. (B) The area enclosed bythe parabola and the double ordinate, commonto ellipse and parabola is 4/3. (C*) The commondouble ordinate is the latus rectumofthe parabola. (D*) Iftangent and normalat the point P onthe ellipse intersect the x-axis at T and G respectivelythen area ofthe triangle PTG is 4. [Sol. C1 : 2x2 + y2 = 6 [T/S, Q.9, Ex-2, Ellipse as subjective] and C2 : y2 = 4x solving (1) and (2) P = (1, 2) and Q = (1, – 2) Hence commonchord passes focus (C) is correct. also A = 2 1 0 dx ) x 2 ( = 1 0 2 3 x 3 2 · 4 = 3 8 (B)is not correct Also 1 C P dx dy · 2 C P dx dy = – 1 (A) is correct now tangent at P on C1 2(x · x1) + yy1 = 6 2x + 2y = 6 x + y = 3 normalis x – y = – l it passes through (1, 2) ] Q.56 TriangleABC has BC = 1 andAC = 2. The maximumpossible value ofthe angleAis (A*) 6 (B) 4 (C) 3 (D) 2 [Sol. Using cosinerule cos = x 4 1 4 x2 = x 4 3 x2 = x 3 x 4 1 = 3 2 x 3 x 4 1 2 hence cos is minimumif x= 3 minimumvalue of cos = 2 · 4 3 = 2 3 maximumvalue of = 6 Ans. ] Q.57complex Let the equation of a line in the Cartesian plane is Ax + By + C = 0 where A, B, C R and A2 + B2 0. If the corresponding equation on the complex plane is 0 z z where is complex and is real, then the correct statement(s) is/are (A*) = 2 B A i and = C (B*) If the line is parallelto imaginaryaxis. (C*) If = 0 line is parallelto realaxis. (D*) slope m of the line is i . [Sol. Let z = x + iy; y x z i [Illustration, Complex No.] now Ax + By + C = 0 2 ) z z ( A + i 2 ) z z ( B + C = 0
- 8. 2 ) z z ( A – 2 ) z z ( B i + C = 0 C z 2 B A z B A i 2 i = 0 comparing this with 0 z z we have = 2 B A i and = C (A) is correct If then 2 i i B A 2 B A B = 0 hence line is x = – A C | | to y-axis (B) is correct If = 0 then A– Bi + A+ Bi = 0 A = 0 hence line is y = – B C || to x-axis. (C) is correct Again m = – B A ....(1) we have A – Bi = 2 A + Bi = 2 A = and – 2Bi = 2( – ) B = – i m = – B A = i (D) is correct hence (A), (B), (C), (D) are correct.] Q.58misc on def & inde Let f (x) = x 1 t dt e 2 and h (x) = ) x ( g 1 f , where g (x) is defined for allx, g'(x) exists for allx, and g (x) 0 for x > 0. If h'(1) = e and g'(1) = 1, then the possible values which g(1) can take (A*) 0 (B) – 1 (C*) – 2 (D) – 4 [Sol. Given f (x) = x 1 t dt e 2 ; h (x) = ) x ( g 1 f ; g (x) 0 for x > 0 now h (x) = ) x ( g 1 1 t dt e 2 differentiating h'(x) = ) x ( ' g · e 2 ) x ( g 1 h'(1) = e (given) ) 1 ( ' g · e 2 ) 1 ( g 1 = e 2 ) 1 ( g 1 = 1 1 + g (1) = ± 1 g (1) = 0 or g(1) = – 2 (A), (C) ]
- 9. Q.59prob P and Q are two points on the upper halfofthe ellipse 2 2 2 2 b y a x = 1. The centre ofthe ellipse is at the origin 'O'and PQ is parallelto the x-axis suchthat the triangle OPQ has the maximumpossible area.A point is randomlyselected frominside ofthe upper halfoftheellipse. The probabilitythat it lies outside thetriangle, is (A*) 1 (B) 2 1 2 (C) 2 1 (D) 4 1 [Sol. A = 2 sin b cos a 2 = ab sin cos = 2 2 sin ab Hence Ais maximum if = /4 and Amax = 2 ab ; Also sample space 2 ab Favourable and come = 2 ab – 2 ab = 2 ab ( – 1) Probability = 2 ab · ab 2 ) 1 ( = 1 Ans. ] Q.60 Asquare OABC is formed by line pairs xy= 0 and xy+ 1 = x + y where 'O' is the origin.Acircle with centre C1 inside the square is drawn to touchthe line pair xy= 0 and another circle with centre C2 and radius twice that of C1, is drawn to touch the circle C1 and the other line pair. The radius ofthe circle with centre C1 is (A) 1 2 3 2 (B) 1 2 3 2 2 (C*) 1 2 3 2 (D) 2 3 1 2 [Sol. diagonalofthe square = 2 also d = 2 r + 3r + r 2 2 2 = r 2 3 + 3r r = 1 2 3 2 Ans. ]