1

TOPPER SAMPLE PAPER - 5

CLASS XI – MATHEMATICS

Questions

Time Allowed : 3 Hrs Maximum Marks: 100_____________________________________________________________1. All questions are compulsory.2. The question paper consist of 29 questions divided into three sections A,

B and C. Section A comprises of 10 questions of one mark each, section Bcomprises of 12 questions of four marks each and section C comprises of07 questions of six marks each.

3. All questions in Section A are to be answered in one word, one sentenceor as per the exact requirement of the question.

4. There is no overall choice. However, internal choice has been provided in04 questions of four marks each and 02 questions of six marks each. Youhave to attempt only one of the alternatives in all such questions.

5. Use of calculators is not permitted. You may ask for logarithmic tables, ifrequired.

Section A1. Let A = { x : x = 2n , n Z } and B = { x : x = 3n , nZ} , then find

A B.

2. From the given table, is y a function of x. Justify your answer?

x -2 -1.5 -1 -0.5 .25 .5 1 1.5 21x

-0.5 -0.67 -1 -2 4 2 1 .67 .5

2513. Find the value ofi

4. Write the negation of the given statementP: Every rectangle is a quadrilateral.

5. Write the given statement in the form “If- then”, and state what arethe component statements p and qIf I have the money, i will buy an i-phone

6. Write the hypothesis and the conclusion in the given implication. If oneroot of a quadratic equation is a + ib then the other root of thequadratic equation is a – ib.

7. Find the equation of the ellipse whose vertices are (±13, 0) and fociare (±5, 0).

8. If (x, y) is a point on the hyperbola, then give three other points thatlie on it

9. The figure below gives a relation. Write it in the roster form

2

10. Find the equation of the set of points which are equidistant from thepoints (1, 2, 3) and (3, 2, -1)

SECTION B

11. Find the values of' k' for which2 7,k,7 2

are in G.P. Find the common

ratio/s of the GP

12. If (1 2) (1 2 3)1 ...2 3

to n terms is S, then find S.

1+i 1-i13. Evaluate1-i 1+i

OR

If (a + i b) (c + id) (e + if) (g + i h) = A + i B, then show that (a2+b2)(c2 + d2) (e2 + f2 ) (g2 + h2) = A2 + B2

14. Solve the given quadratic equation: 9x 2 - 12x + 20 = 0

15. What is the number of ways in which a set of 5 cards can be chosenout of a deck of 52 cards if each set of 5 cards has exactly one ace?

5 6 716. Find the coefficient of x inthe expansion of the product (1 + 2x ) (1 x)

17. Show that tan 3x tan 2x tan x = tan 3x – tan 2 x – tan x

18. Prove that sec8A 1 tan8Asec4A 1 tan2A

OR

1 xIf x and sinx = , find tan2 4 2

19. Let A = {a, e, i, o, u} and B = {a, i, k, u}. Find A – B and B – A. Arethe two sets A – B and B – A (i) equal (ii) mutually disjoint. Justifyyour answer.

20. The entrance of a monument is in the form of a parabolic arch with avertical axis. The arch is 10 m high and 5 m wide at the base. Howwide is it 2 m from the vertex of the parabola?

3

3 x,x 121. Draw the graph of f(x) 1, x 1 and find the Range of f .

2x, x 1

OR1, x 1

Draw the graph of f(x) x 1 x 1and find the Range of f1, x 1

22. Let A= {a, b, c, d} and B = {p, q, r}. Write an example of onto andinto function from A to B. Does there exists a one-one function from Ato B. Justify your answer.

SECTION C

23. Using mathematical induction prove the following :1 1 1 1 n(n 3)........

1.2.3 2.3.4 3.4.5 n(n 1)(n 2) 4(n 1)(n 2)

24. (i)A box contains 10 red marbles, 20 blue marbles and 30 greenmarbles. 5 marbles are drawn from the box, what is the probabilitythat(i) all will be blue? (ii) atleast one will be green?

(ii) A die has two faces each with number ‘1’, three faces each withnumber ‘2’ and one face with number ‘3’. If die is rolled once,determine(i) P(2) (ii) P(1 or 3) (iii) P(not 3)

4

25. The mean of 8, 6, 7, 5, x and 4 is 7. Find (i) the value of x (ii) themean if each observation was multiplied by 3 (iii) the mean deviationabout the median for the original data

26. Find the derivative of(i)sin(x+1) by the abinitio method

x(ii)1+tanx

OREvaluate the limits of the following two functions of x

2 3 2x 1

x 0

x 2 1(i)limx x x 3x 2xsin4x(ii)limsin2x

27. Find the length of the perpendicular drawn from the points2 2 2 2 x y( a b ,0) and ( a b ,0) to the line cos sin 1

a b

Show that their product is b2.

2 2 2 328. Prove that cos x cos (x ) cos (x )3 3 2

29. Solve the inequalities and represent the solution graphically(3x+11)5(2x - 7) - 3(2x + 3) 0; 2x + 19 6x + 47 and 7 11

2

ORHow many litres of water will have to be added to 1125 litres of the45% solution of acid so that the resulting mixture will contain morethan 25% but less than 30% acid content?

5

TOPPER Sample Paper 5

Answers

SECTION A

1. A B = {x : x = 2n, n Z} {x : x = 3n, n Z} = {…,-2,0,2 …}{…-3,0,3…} ={… -6,0,6…} = {x : x = 6n, n Z} [1 Mark]

2. For every value of x, there is a unique value of y, so the tabulatedvalues form a function. [1 Mark]

25 25 251225 25 24 2

2

12

1 1 i 13. i 1 i 1 i i 1 i ii i i i

1 1 i 1 1 i i [1 Mark]

4. P: Every rectangle is not a quadrilateral. [1 Mark]

5. p: I have the money; q: I will buy an i-phone, p q: If I have themoney () then i will buy an i-phone

[1 Mark]6: Hypothesis: one root of a quadratic equation is a + ib Conclusion: the

other root of the quadratic equation is a – ib. [1 Mark]

7. The vertices are on x-axis, so the equation will be of the form2 2

2 2

x y 1a b

Given that a = 13, c = 5.Therefore, from the relation c2 = a2 – b2, we get25 = 169 – b2 i.e., b = 12

2 2

2 2

2 2

x y 113 12

x y 1169 144

[1 Mark]

8. Hyperbola is symmetric with respect to both the axes So, If (x, y) is apoint on the hyperbola, then (–x, y), (x, –y) and (– x, –y) are alsopoints on the hyperbola. [1 Mark]

9. Relation R from P to Q is R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5),(25, –5)} [1 Mark]

10. Let the given points be A(1, 2, 3) and B( 3, 2, -1)Let P(x, y, z) be any point which is equidistant from the points A andB.Then PA = PB

6

2 2 2 2 2 2

2 2 2 2 2 2

2 2 2

2 2 2

(x 1) (y 2) (z 3) (x 3) (y 2) (z 1)

(x 1) (y 2) (z 3) (x 3) (y 2) (z 1)x 1 2x y 4 4y z 9 6z

x 9 6x y 4 4y z 1 2z6x 2x 4y 4y 6z 2z 04x 8z 0x 2z 0

This is the required equation of the set of points in reference. [1Mark]

SECTION B

2

2

2 711. ,k, are in GP7 2

2 7k 1 [1 Mark]7 2

k 1k 1 [1 Mark]

2 7Whenk 1;GP : ,1,7 2

1 7r2 27

[1 Mark]

2 7Whenk 1;GP : , 1,7 2

1 7r [1 Mark]2 27

12. an = (1 2 3 ....n) n(n 1)n 2n

[1 Mark]Sn = n

na [1 Mark]

=

n

i 1

2

1 (n 1)2

1 n(n 1) n2 2 2(n n) n

4 2n(n 3)

4

7

[2 Marks]

2 2

2 2

2 2

1 i 1 i1+i 1-i13.1-i 1+i 1 i 1 i

1 i 2i 1 i 2i. …[1 Mark]

1 i 1 i

1 1 2i 1 1 2i 2i 2i1 i 1 i

2i 2i 4i 2i …[1 Mark]1 ( 1) 2

1+i 1-i1-i 1+i

2 22i 0 2 4 2 …[2 Marks]

OR

1 2 1 2

(a + ib) (c + id) (e + if) (g + ih) = A + iBLet us take modulus on both sides ,(a + ib) (c + id) (e + if) (g + ih) = A + iB …[1 Mark]

We know, z z z z

(a + ib) (c + id) (e + if) (g + ih) (a + ib) (c +

2 2 2 2 2 2 2 2 2 2

2 22 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2

id)

(e + if) (g + ih) = A + iB

a b . c d . e f . g h A B …[1 Mark]

a b . c d . e f . g h A B …[1 Mark]

a b . c d . e f . g h A B …[1 Mark]

8

2

2

2

2

14. 9x - 12 x + 20 = 0203x 4x 0 …[1 Mark]3

b b 4acx2a

20( 4) ( 4) 4.3. 4 16 4.203x …[1 Mark]2.3 6

4 16 80 4 64 4 8 1 2 4i6 6 6 3

2 4 2 4 2 4x i x i; i3 3 3 3 3 3

…[2 Marks]

15. One ace can be selected from 4 aces in 4C1. [1 Mark]Other 4 cards which are non - aces can be selected out of 48 cardsin 48C4 ways [1 Mark]

The total number of ways = 4C 1 x 48C4 [1 Mark]= 4 x 2 x 47 x 46 x 45 = 778320

[ 1 Mark]

5

6 7

0 1 2 3 46 6 6 6 6 60 1 2 3 4

5 66 65 6

2

16. To find the coefficient of x inthe expansion of the product(1 + 2x) (1 x)let us find the expansions of the 2 binomials.

(1 + 2x ) C 2x C 2x C 2x C 2x C 2x

C 2x C 2x

1.1 6. 2x 15. 2x

3 4 5 6

3 4 5 62

0 1 2 3 47 7 7 7 7 70 1 2 3 4

5 6 77 7 75 6 7

2 3 4 5 6 7

20. 2x 15. 2x 6. 2x 1. 2x

1 12x 60x 20. 2x 15. 2x 6. 2x 1. 2x [1 Mark]

(1 x) C x C x C x C x C x

C x C x C x

1.1 7. x 21. x 35. x 35. x 21. x 7. x 1. x [1 Mark]

9

3 4 5 66 7 2

2 3 4 5 6 7

5

5 4 2 3 3 42

(1 + 2x ) (1 x) 1 12x 60x 20. 2x 15. 2x 6. 2x 2x

1 7. x 21. x 35. x 35. x 21. x 7. x x

We will find only those terms that contain x …[1 Mark]

6. 2x 7. x 15. 2x 21. x 20. 2x 35. x .60x 35. x .12x

5

5

5 4 3

21. x

coefficients of x :

6. 2 7.15. 2 21.20. 2 35.60 35.12 21.

21 420 2100 3360 1680 192 171 [1 Mark]

17. Consider 3x = 2x + xOperating tan 3x = tan (2x + x) [1Mark]

tanx tanytan(x y)1 tanx tany

tan2x tanxtan3x [1Mark]1 tan2x tanx

tan 3x – tan 3x tan 2x tan x = tan 2x + tan xor tan 3x – tan 2x – tan x = tan 3x tan 2x tan xor tan 3x tan 2x tan x = tan 3x – tan 2x – tan x [2 Marks)

18.

10

2

2

2

2

2 2

2 2

118. sinx ; x Q4

2tanWe know that sin2 =1 tan

x2tan 12sinx = …[1 Mark]x 41 tan2

xLet tan z2

2z 11 z 48z 1 z z 8z 1 0 [1 Mark]

b b 4ac 8 8 4.1.1 8 64 4z2a 2.1 2

8 60 8 2 15 4 152 2

We know 16

2

1

3

4 15 4 4 15 0 and 4 15 0Butx Q

x2n x 2n n n2 4 2 2

x x xWhen n is even n= 2k 2k 2k Q tan 04 2 2 2 2

xWhen n is even n= 2k+1 2k+1 2k+14 2 2

x xQ tan 02 2

xSo, tan 42

15 …[2 Marks]

19. (i) A – B = {e, o}, since the elements e, o belong to A but not to Band B – A = {k}, since the element k belongs to B but not to A.

[2 Marks](ii)We note that A – B ≠B – A. [1 Mark](iii)The sets A – B, and B – A are mutually disjoint sets, i.e., theintersection of these two sets is a null set. [1 Mark

11

20.

This parabola has its axis on the y axis and it opens downward. so itequation is of the type

x 2 = - 4 a yThe top of the parabola is its vertex passing through the origin. Thewidth of the base is 5 mt, therefore the coordinates of the points P andQ are (-2.5, -10) and (2.5, 10) respectively. P and Q lie on theparabola.Substituting the ccordinates of the point P in the equation of theparabola, we have(-2.5) 2 = -4a (-10) …[2 Marks]6.25 = 40 aa = 6.25 / 40 = 5/ 32Let 2w be the width of the arch at 2 m below the vertex. Therefore thecoordinates of the points A and B A (-w, -2) and B (w, -2)A and B lie on the parabola.Substituting the coordinates of the point A in the equation of theparabola, we havew 2 = - 4 (5/32) (-2)w 2 = 5/4w = 5/22w=5 = 2.23m …[2 Marks]

12

21. Range of f= (-∞, 2) [1Mark for correct range and 1mark each for the 3 branches of the graph]

OR

13

Range f =[-1,1] [1Mark for correct range and 1 markeach for the 3 branches of the graph]

22. Into Function: {a, p), (b, q), (c,p),(d,p)} range must be the propersubset of set B [1 Mark]Onto: {a, p), (b, q) (c, r), (d, r)} range must be same as set B

[1 Mark]No one- one function can be defined from A to B because n(B) < n(A)

[2 Marks]

SECTION C

23.

n (n 3)1 1 1 1Let thestatementP(n)be: ........1.2.3 2.3.4 3.4.5 n (n 1)(n 2) 4(n 1)(n 2)

1 (1 3)1ConsiderP(1):1.2.3 4(1 1)(1 2)

1 (1 3)1 1.4 1 [P(1)is true] [1Mark]1.2.3 4(1 1)(1 2) 4.2.3 1.2.3

14

Let us assume that P(k) is truek (k 3)1 1 1 1P(k): ... (1mark)

1.2.3 2.3.4 3.4.5 k (k 1)(k 2) 4(k 1)(k 2)

To prove:k 1 (k 4)1 1 1 1 1P(k 1): ... [1Mark]

1.2.3 2.3.4 3.4.5 k (k 1)(k 2) k 1 (k 2)(k 3) 4(k 2)(k 3)

1LHS1

2

1 1 1 1....2.3 2.3.4 3.4.5 k (k 1)(k 2) k 1 (k 2)(k 3)

1 1 1 1 1...1.2.3 2.3.4 3.4.5 k(k 1)(k 2) k 1 (k 2)(k 3)

k(k 3) 1 (usingP(k)4(k 1)(k 2) k 1 (k 2)(k 3)

1 k(k 3) 1(k 1)(k 2) 4 (k 3)

1 k(k 3) 4(k 1)(k 2) 4

2

3 2

2

(k 3)

1 k(k 9 6k) 4(k 1)(k 2) 4(k 3)

1 k 9k 6k 4(k 1)(k 2) 4(k 3)

1 (k 1) (k 4)(k 1)(k 2) 4(k 3)

k 1 (k 4)RHS [3Marks]

4(k 2)(k 3)

605

20

(i) There are a total of 60 marbles out of which 5 marbles are tobe selectedNumber of ways in which 5 marbles are to be selected out of 60= C

Out of 20 blue marbles , 5 can be selected in= C5

205

605

20!C 20!5!55!5!15! P(all 5 blue marbles) =

60! 5!15!60!C5!55!

20.19.18.17.16 34 [1Mark]60.59.58.57.56 11977

15

305

305

605

(ii) Number of ways in which 5 marbles are to be selected out of 60=60C5.Out of 30 non-green marbles , 5 can be selected in C

30!C 30!5!55!5!25!P(all 5 non-green marbles) =

60! 5!25!60!C5!55!

30.29.28.27.26. 117 [1Mark]60.59.58.57.56 4484

P(atleast one green marble) = 1 - P(all 5non-green marbles)117= 1 -4484

4484 117 4367 = [1Mark]4484 4484

(ii)On the dice two faces are with number '1', three faces are withnumber '2' and one face is with number '3'

2 1 3 1 1P(1) ;P(2) ;P(3)6 3 6 2 6

1(i) P(2) [1Mark]2

(ii) P(1or3) P(1) P(3)[The events are mutua

lly exclusive]1 1 1= [1Mark]3 6 2

1 5(iii) P(not3) 1 P(3) 1 [1Mark]6 6

25. (i)8 6 7 5 x 47

6x 1 2 [ 1 M a r k ]

(ii) If each observation is multiplied by 3 mean will also bemultiplied by 3 so the mean is 21 [1 Mark]

(iii) Arranging in ascending order 4, 5, 6, 7, 8, 12Median = 6 7 6 . 5 . . . . . . . . . [ 1 M a r k ]

2

Forming the table, we getxi 4 5 6 7 8 12xi-6.5 -2.5 -1.5 -0.5 0.5 1.

55.5

6 . 5ix 2.5 1.5 0.5 0.5 1.5

5.5

16

n n

i ii 1 i 1

n n

i ii 1 i 1

x M x 6 . 5 1 2 [ 1 M a r k ]

x M x 6 . 5M . D ( M ) =

n 61 2 2 . 0 [ 1 M a r k ]6

26 (i)sin(x+1) by the definition methodLet y=sin(x+1)

y y sin(x+ x+1)y y y y sin(x+ x+1) - sin(x+1)

x+ x+1+x+1 x+ x+1- (x+1)=2cos( )sin( )2 2

2x+ x+2 x2cos( )sin( ) [1Mark]2 2

y 1 2x+ x+22cos( )sin(x x 2

x 0 x 0

x 0 x 0 x 0

x )2xsin( )y 2x+ x+2 2cos( ) [1Mark]

xx 22

xsin( )y 2x+ x+2 2lim lim cos( )xx 22

xsin( )y 2x+ x+2 2lim lim cos( ) limxx 22

dy 2x+2cos( ).1dx 2dy cos(x 1) [1Mark]dx

2

2

2

2

2

x26 (ii)Lety1 tanx

d d1 tanx x x 1 tanxdy dx dx [1Mark]dx 1 tanx

1 tanx .1 x sec xdy [1Mark]dx 1 tanx

dy 1 tanx x sec x [1Mark]dx 1 tanx

OR

17

In each part, two marks for correct simplification and one for finalevaluation of the limit

2 3 2 2

x 2 1 x 2 1x(x 1)x x x 3x 2x x(x 3x 2)

2

2

x 2 1x(x 1) x(x 1)(x 2)

x 4x 4 1x(x 1)(x 2)

x 4x 3x(x 1)(x 2)

2 2

2 3 2x 1 x 1

x 1

x 1

x 2 1 x 4x 3lim limx(x 1)(X 2)x x x 3x 2x

(x 3)(x 1)limx(x 1)(x 2)x 3 1 3lim 2 (3Marks)

x(x 2) 1(1 2)

x 0 x 0

x 0

4x 0 2x 0

sin4x sin4x 2x(ii) lim lim . .2sin2x 4x sin2x

sin4x sin2x2.lim4x 2x

sin4x sin2x2. lim lim4x 2x

2.1.1 2(as x 0,4x 0 and 2x 0) (3Marks)

18

2 2

2 2

27. Perpendicular distance of the point (m,n) from the line ax+by+c=0,is given by

a(m) + b(n) +c

a b

Perpendicular distance of the point ( a b ,0) from the linex ycos sin 1 0, is given bya b

cos

2 2 2 2

2 2 2 2

2 2

sin cos( a b ) + (0) -1 ( a b ) -1a b a [1Mark]

cos sin cos sina b a b

Perpendicular distance of the point ( a b ,0) from the linex ycos sin 1 0, is given bya b

cos (a

2 2 2 2

2 2 2 2

2 2 2 2

2 2

sin cosa b ) + (0) -1 ( a b ) -1b a ...[1Mark]

cos sin cos sina b a b

cos cos( a b ) -1 ( a b ) -1a aProduct of the perpendicular distances =cos sin co

a b

2 2

2 2 2 2

2 2 2 2

s sina b

cos cos( a b ) -1 ( a b ) +1a a =cos sin cos sin

a b a b

2 2 2

2

2 2

a b cos -1

a = [1Mark]

cos sina b

19

2 2 2

2

2 2 2 2

2 2

2 2 2 2 22 2

2

2 2 2 2

2 2 2 2 2 2

2 2 2 2

2 2 2 2 2

2 2 2 2

2 2 2 2 2

2 2 2

a b cos -1

a=

b cos a sina b

a cos b cos aa ba

=b cos a sin

b a cos b cos a=

b cos a sinb a (1 cos ) b cos

=b cos a sin

b a (sin ) b cos=

b cos a sin

2

2 2 2 2 2

2 2 2 2

2 2 2 2 2

2 2 2 2

2

b a (sin ) b cos=

b cos a sin

b a (sin ) b cos=

b cos a sin=b [3Marks]

2 2 2

2 2 2

2 2 2

2

28. cos x cos (x ) cos (x )3 3

cos x cos (x ) 1 sin (x ) [1Mark]3 3

1 cos x cos (x ) sin (x ) [1Mark]3 3

1 cos x cos(x x )cos(x x3 3 3

2

2

2 2

2 2

)3

21 cos x cos(2x)cos [1Mark]3

11 cos x cos(2x)2

11 cos x 2cos x 1 [1Mark]2

11 cos x cos x2

1 312 2

[2Marks]

20

(3x+11)29. 5 (2x - 7) - 3 (2x + 3) 0 ; 2x + 19 6x + 47 and 7 112

Let us solve the inequalities one by one and then work out the common solution.Inequality 1:5 (2x - 7) - 3 (2x + 3) 0

10

x - 35 -6x - 9 0

4x - 44 04x 44

1x 11 [1 Marks]2

Inequality 22x + 19 6x + 47

2x -6x -19 + 47-4x 28-x 7

1x -7 [1 Marks]2

Inequality 3:(3x+11)7 11

214 3x+11 2214-11 3x 22 113 3x 11

11 11 x [1 Marks]3 2

11 11 1x 11,x 7,1 x together 1 x [ Mark]3 3 2

[1 Mark]

OR45 1125The amount of acid in 1125lt of the 45% solution=45% of 1125=

100Let x lt of the water be added to it to obtain a solution between 25% and 30%solution

21

451125 110025% 30% [1 Mark]1125 +x 2

45112525 30100100 1125 +x 10025 1125 45 30100 1125 +x 100 100

1125 4525 301125 +x

1125 +x1 125 1125 45 301125 45 1125 451125 +x

25 3050625 1125 +

25

50625x30

12025 1125 +x 1687.5 [2 Mark]2

2025 1125 +x 1687.5

2025 1125 x 1687.5 1125900 x 562.5

1562.5 x 900 [1 Mark]2

1So the amount of water to be added must be between 562.5 to900 lt [ Mark]2