1291109172 ClassXI Math TopperSamplepaper 50

7/30/2019 1291109172 ClassXI Math TopperSamplepaper 50

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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 sentence

or as per the exact requirement of the question.4. There is no overall choice. However, internal choice has been provided in

04 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 A

1. 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 2

1

x

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

251

3. Find the value of i

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 are

the 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


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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. Evaluate

1-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: 9x2

- 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 thatsec8A 1 tan8A

sec4A 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?


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3 x,x 1

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

2x, x 1

OR

1, 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 and

into 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)


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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

OR

Evaluate the limits of the following two functions of x

2 3 2x 1

x 0

x 2 1(i)lim

x x x 3x 2x

sin4x(ii)lim

sin2x

27. Find the length of the perpendicular drawn from the points

2 2 2 2 x y( a b ,0) and ( a b ,0) to the line cos sin 1a 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

OR

How 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?


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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 tabulated

values 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 i

i 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 the

money () 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 onx-axis, so the equation will be of the form2 2

2 2

x y1

a 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 y1

13 12

x y1

169 144

[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


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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 2z

6x 2x 4y 4y 6z 2z 0

4x 8z 0

x 2z 0

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

SECTION B

2

2

2 7

11. ,k, are in GP7 2

2 7k 1 [1 Mark]

7 2

k 1

k 1 [1 Mark]

2 7Whenk 1;GP : ,1,

7 2

1 7r

2 2

7

[1 Mark]

2 7Whenk 1;GP : , 1,

7 2

1 7r [1 Mark]

2 2

7

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

n 2n

[1 Mark]

Sn = nn

a [1 Mark]

=

n

i 1

2

1(n 1)

2

1 n(n 1) n

2 2 2

(n n) n

4 2

n(n 3)

4


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[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 2i

1 i 1 i

2i 2i 4i2i [1 Mark]1 ( 1) 2

1+i 1-i

1-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 + iB

Let 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]




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2

2

2

2

14. 9x - 12 x + 20 = 0

203x 4x 0 [1 Mark]

3b b 4ac

x2a

20( 4) ( 4) 4.3.

4 16 4.203x [1 Mark]2.3 6

4 16 80 4 64 4 8 1 2 4i

6 6 6 3

2 4 2 4 2 4x i x i; i

3 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 cards

in 48C4 ways [1 Mark]

The total number of ways = 4C 1 x48C4 [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 6

0 1 2 3 4

5 66 6

5 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 7

0 1 2 3 4

5 6 77 7 7

5 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]


9/21

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

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 tan2xtanx

tan 3x tan 3x tan 2x tan x = tan 2x + tan xor tan 3x tan 2x tan x = tan 3x tan 2x tan x

or tan 3x tan 2x tan x = tan 3x tan 2x tan x [2 Marks)18.


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10

2

2

2

2

2 2

2 2

118. sinx ; x Q

4

2tanWe know that sin2 =1 tan

x2tan

12sinx = [1 Mark]x 4

1 tan2

xLet tan z

2

2z 1

1 z 4

8z 1 z z 8z 1 0 [1 Mark]

b b 4ac 8 8 4.1.1 8 64 4z

2a 2.1 2

8 60 8 2 154 15

2 2

We know 16

2

1

3

4 15 4 4 15 0 and 4 15 0

Butx Q

x2n x 2n n n

2 4 2 2x x x

When n is even n= 2k 2k 2k Q tan 04 2 2 2 2

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

4 2 2

x xQ tan 0

2 2

xSo,tan 4

2

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 kbelongs 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


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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]


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

OR


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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 (n1 1 1 1Let thestatementP(n)be: ........

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

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


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Let us assume that P(k) is true

k (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 11 1 1 1 1P(k 1): ...

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

1LHS

1

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)

60

5

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

20

5

60

5

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

60! 5!15!60!C

5!55!

20.19.18.17.16 34[1Mark]

60.59.58.57.56 11977


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30

5

305

60

5

(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!C

5!55!

30.29.28.27.26. 117

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 with

number '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 4

76

x 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, 12

Median = 6 7 6 . 5 . . . . . . . . . [ 1 M a r k ]2

Forming the table, we get

xi 4 5 6 7 8 12

xi-6.5 -2.5 -1.5 -0.5 0.5 1. 5.5

6 . 5i

x 2.5 1.5 0.5 0.5 1. 5.5


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n n

i ii 1 i 1

n n

i i

i 1 i 1

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

x M x 6 . 5

M . D ( M ) =n 6

1 22 . 0 [ 1 M a r k ]

6

26 (i)sin(x+1) by the definition method

Let 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)

2

xsin( )

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

2

xsin( )

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

2

xsin( )

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

2

dy 2x+2cos( ).1

dx 2

dycos(x 1) [1Mark]

dx

2

2

2

2

2

x

26 (ii)Lety 1 tanx

d d1 tanx x x 1 tanx

dy 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


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

2 3 2 2

x 2 1 x 2 1

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

2

2

x 2 1

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

x 4x 4 1

x(x 1)(x 2)

x 4x 3

x(x 1)(x 2)

2 2

2 3 2x 1 x 1

x 1

x 1

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

(x 3)(x 1)lim

x(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 . .2

sin2x 4x sin2x

sin4x sin2x2.lim4x 2x

sin4x sin2x2. lim lim

4x 2x

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


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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) +ca b

Perpendicular distance of the point ( a b ,0) from the line

x ycos sin 1 0, is given by

a b

cos

2 2 2 2

2 2 2 2

2 2

sin cos( a b ) + (0) -1 ( a b ) -1

a b a [1Mark]

cos sin cos sin

a b a b

Perpendicular distance of the point ( a b ,0) from the line

x ycos sin 1 0, is given by

a b

cos(

a

2 2 2 2

2 2 2 2

2 2 2 2

2 2

sin cosa b ) + (0) -1 ( a b ) -1

b a ...[1Mark]

cos sin cos sin

a 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 sin

a b

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

a a=

cos sin cos sin

a b a b

2 2 22

2 2

a b cos-1

a= [1Mark]

cos sin

a b


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2 2 22

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 sin

a b

a cos b cos aa b

a=

b cos a sin

b a cos b cos a=

b cos a sin

b 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

2

1 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 x

2

1 31

2 2

[2Marks]


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

2

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 0

4x 44

1x 11 [1 Marks]

2

Inequality 2

2x + 19 6x + 47

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

-x 7

1x -7 [1 Marks]

2

Inequality 3:

(3x+11)7 11

2

14 3x+11 22

14-11 3x 22 11

3 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]

OR

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

100

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

solution


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451125

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

45112525 30100100 1125 +x 100

25 1125 45 30

100 1125 +x 100 100

1125 4525 30

1125 +x

1125 +x1 1

25 1125 45 30

1125 45 1125 451125 +x25 30

506251125 +

25

50625x

30

12025 1125 +x 1687.5 [2 Mark]

2

2025 1125 +x 1687.5

2025 1125 x 1687.5 1125

900 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 [ Ma

2

1291109172 ClassXI Math TopperSamplepaper 50

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