TOPPER SAMPLE PAPER - 5 CLASS XI MATHEMATICS. Questions. Time Allowed : 3 Hrs Maximum Marks: 100

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1 TOPPER SAMPLE PAPER - 5 CLASS XI MATHEMATICS Questions Time Allowed : 3 Hrs Mximum Mrks: All questions re compulsory.. The question pper consist of 9 questions divided into three sections A, B nd C. Section A comprises of 10 questions of one mrk ech, section B comprises of 1 questions of four mrks ech nd section C comprises of 07 questions of six mrks ech. 3. All questions in Section A re to e nswered in one word, one sentence or s per the exct requirement of the question. 4. There is no overll choice. However, internl choice hs een provided in 04 questions of four mrks ech nd 0 questions of six mrks ech. You hve to ttempt only one of the lterntives in ll such questions. 5. Use of clcultors is not permitted. You my sk for logrithmic tles, if required. Section A 1. Let A = { x : x = n, n Z } nd B = { x : x = 3n, nz}, then find A B.. From the given tle, is y function of x. Justify your nswer? x x Find the vlue of i 4. Write the negtion of the given sttement P: Every rectngle is qudrilterl. 5. Write the given sttement in the form If- then, nd stte wht re the component sttements p nd q If I hve the money, i will uy n i-phone 6. Write the hypothesis nd the conclusion in the given impliction. If one root of qudrtic eqution is + i then the other root of the qudrtic eqution is i. 7. Find the eqution of the ellipse whose vertices re (±13, 0) nd foci re (±5, 0). 8. If (x, y) is point on the hyperol, then give three other points tht lie on it 9. The figure elow gives reltion. Write it in the roster form 1

10. Find the eqution of the set of points which re equidistnt from the points (1,, 3) nd (3,, -1) SECTION B 11. Find the vlues of' k' for which rtio/s of the GP 7,k, re in G.P. Find the common 7 (1 )(1 3) 1.

2 10. Find the eqution of the set of points which re equidistnt from the points (1,, 3) nd (3,, -1) SECTION B 11. Find the vlues of' k' for which rtio/s of the GP 7,k, re in G.P. Find the common 7 (1 )(1 3) 1. If 1... to n terms is S, then find S. 3 1+i 1-i 13. Evlute 1-i 1+i OR If ( + i ) (c + id) (e + if) (g + i h) = A + i B, then show tht ( + ) (c + d ) (e + f ) (g + h ) = A + B 14. Solve the given qudrtic eqution: 9x - 1x + 0 = Wht is the numer of wys in which set of 5 crds cn e chosen out of deck of 5 crds if ech set of 5 crds hs exctly one ce? Find the coefficient of x in the expnsion of the product (1 + x )(1 x) 17. Show tht tn 3x tn x tn x = tn 3x tn x tn x sec 8A 1 tn8a 18. Prove tht sec 4A 1 tna OR 1 x If x nd sinx =, find tn Let A = {, e, i, o, u} nd B = {, i, k, u}. Find A B nd B A. Are the two sets A B nd B A (i) equl (ii) mutully disjoint. Justify your nswer. 0. The entrnce of monument is in the form of prolic rch with verticl xis. The rch is 10 m high nd 5 m wide t the se. How wide is it m from the vertex of the prol?

3 3 x, x 1 1. Drw the grph of f(x) 1, x 1 nd find the Rnge of f. x, x 1 OR 1, x 1 Drw the grph of f(x) x 1 x 1nd find the Rnge of f 1, x 1. Let A= {,, c, d} nd B = {p, q, r}. Write n exmple of onto nd into function from A to B. Does there exists one-one function from A to B. Justify your nswer. SECTION C 3. Using mthemticl induction prove the following : n(n 3) n(n 1)(n ) 4(n 1)(n ) 4. (i)a ox contins 10 red mrles, 0 lue mrles nd 30 green mrles. 5 mrles re drwn from the ox, wht is the proility tht (i) ll will e lue? (ii) tlest one will e green? (ii) A die hs two fces ech with numer 1, three fces ech with numer nd one fce with numer 3. If die is rolled once, determine (i) P() (ii) P(1 or 3) (iii) P(not 3) 3

4 5. The men of 8, 6, 7, 5, x nd 4 is 7. Find (i) the vlue of x (ii) the men if ech oservtion ws multiplied y 3 (iii) the men devition out the medin for the originl dt 6. Find the derivtive of (i)sin(x+1) y the initio method x (ii) 1+tnx OR Evlute the limits of the following two functions of x x 1 (i)lim 3 x1 x x x 3x x sin4x (ii)lim x0 sinx 7. Find the length of the perpendiculr drwn from the points x y (,0) nd (,0) to the line cos sin 1 Show tht their product is Prove tht cos x cos(x) cos(x) Solve the inequlities nd represent the solution grphiclly (3x+11) 5(x - 7) - 3(x + 3) 0; x x + 47 nd 7 11 OR How mny litres of wter will hve to e dded to 115 litres of the 45% solution of cid so tht the resulting mixture will contin more thn 5% ut less thn 30% cid content? 4

5 TOPPER Smple Pper 5 Answers SECTION A 1. A B = {x : x = n, n Z} {x : x = 3n, n Z} = {,-,0, } { -3,0,3 } ={ -6,0,6 } = {x : x = 6n, n Z} [1 Mrk]. For every vlue of x, there is unique vlue of y, so the tulted vlues form function. [1 Mrk] i i 1 i 1 i i 1 i i i i i i i 1 1 i i [1 Mrk] 4. P: Every rectngle is not qudrilterl. [1 Mrk] 5. p: I hve the money; q: I will uy n i-phone, p q: If I hve the money () then i will uy n i-phone [1 Mrk] 6: Hypothesis: one root of qudrtic eqution is + i Conclusion: the other root of the qudrtic eqution is i. [1 Mrk] 7. The vertices re on x-xis, so the eqution will e of the form x y 1 Given tht = 13, c = 5. Therefore, from the reltion c =, we get 5 = 169 i.e., = 1 x y [1 Mrk] x y Hyperol is symmetric with respect to oth the xes So, If (x, y) is point on the hyperol, then ( x, y), (x, y) nd ( x, y) re lso points on the hyperol. [1 Mrk] 9. Reltion R from P to Q is R = {(9, 3), (9, 3), (4, ), (4, ), (5, 5), (5, 5)} [1 Mrk] 10. Let the given points e A(1,, 3) nd B( 3,, -1) Let P(x, y, z) e ny point which is equidistnt from the points A nd B. Then PA = PB 5

6 (x 1)(y )(z 3)(x 3)(y )(z 1) (x 1)(y )(z 3)(x 3)(y )(z 1) x 1 x y 4 4y z 9 6z x 9 6x y 4 4y z 1 z 6x x 4y 4y 6z z 0 4x 8z 0 x z 0 This is the required eqution of the set of points in reference. [1Mrk] SECTION B 7 11.,k, re in GP 7 7 k 1 [1 Mrk] 7 k 1 k 1 [1 Mrk] 7 Whenk 1;GP :,1, r [1 Mrk] 7 7 Whenk 1;GP :, 1, r [1 Mrk] 7 (1 3...n) n(n 1) 1. n = n n S n = n n [1 Mrk] [1 Mrk] = 1 n i1 (n 1) 1 n(n 1) n (n n) n 4 n(n 3) 4 6

7 [ Mrks] i 1 i 1 i 1 i 1( 1) 1+i 1-i 1-i 1+i 1 i 1 i 1 i i 1 i i. [1 Mrk] 1 1 i 1 1 i i i 1 i 1 i i i 4i i 1+i 1-i 1-i 1+i [1 Mrk] i 0 4 [ Mrks] OR ( + i) (c + id) (e + if) (g + ih) = A + ib Let us tke modulus on oth sides, ( + i) (c + id) (e + if) (g + ih) = A + ib [1 Mrk] We know, z z z z 1 1 ( + i) (c + id) (e + if) (g + ih)( + i)(c + id) (e + if)(g + ih) = A + ib. c d. e f. g h A B [1 Mrk]. c d. e f. g h A B [1 Mrk]. c d. e f. g h A B [1 Mrk] 7

8 14. 9x - 1 x + 0 = x 3x 4x 0 [1 Mrk] 4c 0 ( 4)( 4) x [1 Mrk] i x i x i; i [ Mrks] One ce cn e selected from 4 ces in 4 C 1. [1 Mrk] Other 4 crds which re non - ces cn e selected out of 48 crds in 48 C 4 wys [1 Mrk] The totl numer of wys = 4 C 1 x 48 C 4 = 4 x x 47 x 46 x 45 = [1 Mrk] [ 1 Mrk] To find the coefficient of x in the expnsion of the product (1 + x)(1 x) 6 7 let us find the expnsions of the inomils (1 + x ) C0 x C1 x C x C3 x C4 x C5 x C6 x x 15. x 0. x 15. x 6. x 1. x (1 x) C0 x C1 x C x C3 x C4 x C5 x C6 x C7 x 1 1x 60x 0. x 15. x 6. x 1. x [1 Mrk] x 1. x 35. x 35. x 1. x 7. x 1. x [1 Mrk] 8

9 (1 + x )(1 x) 1 1x 60x 0. x 15. x 6. x x 1 7. x 1. x 35. x 35. x 1. x 7. x x We will find only those terms tht contin x x 6. x 7. x 15. x 1. x 0. x 35. x.60x 35. x.1x 5 coefficients of x : [1 Mrk] Consider 3x = x + x Operting tn 3x = tn (x + x) [1 Mrk] tnx tny tn(x y) 1 tnx tny tnx tn x tn3x 1 tnx tnx [1Mrk] 18. tn 3x tn 3x tn x tn x = tn x + tn x or tn 3x tn x tn x = tn 3x tn x tn x or tn 3x tn x tn x = tn 3x tn x tn x [ Mrks) 9

10 1 18. sin x ; x Q 4 tn We know tht sin = 1 tn x tn 1 sin x = [1 Mrk] x 4 x Let tn z z 1 1 tn 1 z 4 8z 1 z z 8z 1 0 [1 Mrk] z.1 4c We know nd Butx Q x n x n n n 4 x x x When n is even n= k k k Q1 tn 0 4 x When n is even n= k+1 k+1 k+1 4 x x Q3 tn 0 x So, tn 4 15 [ Mrks] 19. (i) A B = {e, o}, since the elements e, o elong to A ut not to B nd B A = {k}, since the element k elongs to B ut not to A. [ Mrks] (ii)we note tht A B B A. [1 Mrk] (iii)the sets A B, nd B A re mutully disjoint sets, i.e., the intersection of these two sets is null set. [1 Mrk 10

11 0. This prol hs its xis on the y xis nd it opens downwrd. so it eqution is of the type x = - 4 y The top of the prol is its vertex pssing through the origin. The width of the se is 5 mt, therefore the coordintes of the points P nd Q re ( -.5, -10) nd (.5, 10) respectively. P nd Q lie on the prol. Sustituting the ccordintes of the point P in the eqution of the prol, we hve (-.5) = -4 (-10) [ Mrks] 6.5 = 40 = 6.5 / 40 = 5/ 3 Let w e the width of the rch t m elow the vertex. Therefore the coordintes of the points A nd B A (-w, -) nd B (w, -) A nd B lie on the prol. Sustituting the coordintes of the point A in the eqution of the prol, we hve w = - 4 (5/3) (-) w = 5/4 w = 5/ w=5 =.3m [ Mrks] 11

12 1. Rnge of f= (-, ) [1Mrk for correct rnge nd 1 mrk ech for the 3 rnches of the grph] OR 1

13 Rnge f =[-1,1] [1Mrk for correct rnge nd 1 mrk ech for the 3 rnches of the grph]. Into Function: {, p), (, q), (c,p),(d,p)} rnge must e the proper suset of set B [1 Mrk] Onto: {, p), (, q) (c, r), (d, r)} rnge must e sme s set B [1 Mrk] No one- one function cn e defined from A to B ecuse n(b) < n(a) [ Mrks] SECTION C n(n Let the sttement P(n)e: n(n 1)(n ) 4(n 1) 1 1(1 3) Consider P(1) : (1 1)(1 ) 1 1(1 3) [P(1)is true] [1Mrk] (1 1)(1 )

14 Let us ssume tht P(k) is true k(k 3) P(k) :...(1mrk) k(k 1)(k ) 4(k 1)(k ) To prove: k 1(k P(k 1) : k(k 1)(k ) k 1(k )(k 3) 4(k )( LHS k(k 1)(k ) k 1(k )(k 3) 1 k(k 3) k(k 1)(k ) k 1(k )(k 3) k(k 3) 1 (usingp(k) 4(k 1)(k ) k 1(k )(k 3) 1 k(k 3) 1 (k 1)(k ) 4(k 3) (k 1)(k ) 4(k 3) 1 k(k 9 6k) 4 (k 1)(k ) 4(k 3) 3 1 k 9k 6k 4 (k 1)(k ) 4(k 3) 1(k 1)(k 4) (k 1)(k ) 4(k 3) k 1(k 4) RHS [3Mrks] 4(k )(k 3) (i) There re totl of 60 mrles out of which 5 mrles re toe selected Numer of wys in which 5 mrles re to e selected out of 60= 0 Out of 0 lue mrles, 5 cn e selected in= C 5 P(ll 5 lue mrles) = ! C5 0!5!55! 5!15! C 60! 5 5!15!60! 5!55! [1Mrk] 60 C 5 14

15 (ii) Numer of wys in which 5 mrles re to e selected out of 60=60C5. Out of 30 non-green mrles, 5 cn e selected in P(ll 5 non-green mrles) = P(tlest one green mrle) C 5 30! C5 5!5! 30!5!55! C 60! 5 5!5!60! !55! = 1 - P(ll 5non-green mrles) 117 = = [1Mrk] (ii)on the dice two fces re with numer '1', three fces re with numer '' nd one fce is with numer '3' P(1) ;P() ;P(3) (i) P() [1Mrk] (ii) P(1or3) P(1) P(3)[The events re mutully exclusive] = 3 6 [1Mrk] 1 5 (iii) P(not3) 1 P(3) [1Mrk] 5. (i) x 4 6 x 1 [ 1 M r k ] (ii) If ech oservtion is multiplied y 3 men will lso e multiplied y 3 so the men is 1 [1 Mrk] (iii) Arrnging in scending order 4, 5, 6, 7, 8, 1 Medin = [ 1 M r k ] Forming the tle, we get x i x i x i

16 n i i 1 i 1 n n x M x [ 1 M r k ] x i M x i 6. 5 i 1 i 1 M. D( M) = n [ 1 M r k ] 6 6(i)sin(x+1) y the definition method Let y=sin(x+1) y y sin(x+ x+1) i y y y y sin(x+ x+1) - sin(x+1) x+ x+1+x+1 x+ x+1- (x+1) =cos()sin() x+ x+ x cos()sin() [1Mrk] y 1 x+ x+ x cos()sin( ) x x x sin() y x+ x+ cos() [1Mrk] x x x sin() y x+ x+ lim lim cos() x0 x x0 x x sin() y x+ x+ lim lim cos() lim x0 x x0 x0 x dy x+ cos().1 dx dy cos(x 1) [1Mrk] dx x 6(ii)Lety 1 tnx d d 1 tn x x x 1 tn x dy dx dx dx 1 tn x n 1 tn x.1 x sec x dy dx 1 tn x dy 1 tn x x sec x dx 1 tn x OR [1Mrk] [1Mrk] [1Mrk] 16

17 In ech prt, two mrks for correct simplifiction nd one for finl evlution of the limit x 1 x 1 3 x x x 3x x x(x 1) x(x 3x ) x 1 x(x 1) x(x 1)(x ) x 4x 4 1 x(x 1)(x ) x 4x 3 x(x 1)(x ) x 1 x 4x 3 lim lim x x x 3x x x(x 1)(X ) (x 3)(x 1) lim x1 x(x 1)(x ) x1 3 x1 x lim (3Mrks) x1 x(x ) 1(1 ) sin 4x sin 4x x (ii) lim lim.. x0 sinx x0 4x sinx sin 4x sinx.lim x0 4x x sin 4x sinx. lim lim 4x0 4x x0 x.1.1 (s x 0, 4x 0 nd x 0)(3Mrks) 17

18 7. Perpendiculr distnce of the point (m,n) from the line x+y+c=0, is given y (m) + (n) +c Perpendiculr distnce of the point ( x y cos sin 1 0, is given y,0) from the line cos sin cos ( ) +(0) -1( ) -1 cos sin cos sin Perpendiculr distnce of the point (,0) from the line x y cos sin 1 0, is given y cos ( sin cos ) +(0) -1( ) -1 cos sin cos sin [1Mrk]...[1Mrk] Product of the perpendiculr distnces = cos cos ( ) -1( ) -1 cos sin cos sin cos cos ( ) -1( ) +1 = cos sin cos sin cos -1 = [1Mrk] cos sin 18

19 cos -1 = cos sin = 8. cos x cos(x) cos(x) cos cos cos sin cos cos = cos sin (1 cos) cos = cos sin (sin) cos = cos sin (sin) cos = cos sin (sin) cos = cos sin = [3Mrks] cos x cos(x) 1 sin(x) [1Mrk] 1 cos x cos(x) sin(x) [1Mrk] ) cos x cos(x x)cos(x x 3 1 cos x cos(x)cos [1Mrk] 1 cos x cos(x) cos x cos x 1 [1Mrk] 1 1 cos x cos x [Mrks] 19

20 (3x+11) 9. 5 (x - 7) - 3 (x + 3) 0 ; x x + 47 nd 7 11 Let us solve the inequlities one y one nd then work out the common solution. Inequlity 1: 5 (x - 7) - 3 (x + 3) 0 10x x x x 44 1 x 11 [1 Mrks] Inequlity x x + 47 x -6x x 8 -x 7 1 x -7 [1 Mrks] Inequlity 3: (3x+11) x x x x [1 Mrks] x 11, x 7,1 x together 1 x [ Mrk] 3 3 [1 Mrk] OR The mount of cid in 115lt of the 45% solution=45% of 115= 100 Let x lt of the wter e dded to it to otin solution etween 5% nd 30% solution 0

21 % 30% [1 Mrk] 115 +x x x x x x x x [ Mrk] x x x x 900 [1 Mrk] 1 So the mount of wter to e dded must e etween 56.5 to 900 lt [ M 1

Linear Inequalities: Each of the following carries five marks each: 1. Solve the system of equations graphically.

Linear Inequalities: Each of the following carries five marks each: 1. Solve the system of equations graphically. Liner Inequlities: Ech of the following crries five mrks ech:. Solve the system of equtions grphiclly. x + 2y 8, 2x + y 8, x 0, y 0 Solution: Considerx + 2y 8.. () Drw the grph for x + 2y = 8 by line.it