Class Xi Math Sample Paper

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TOPPER SAMPLE PAPER - 2

CLASS XI – M ATH EMA TI CS

(Ques t i ons )

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 B comprises of 12questions of four marks each and section C comprises of 07 questions of six markseach.

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

4. Use of calculators is not permitted. You may ask for logarithmic tables, if required. ______________________________________________________________________

1. Is the given sentence a mathematical statement?P: 15 is not a multiple of 3

2. Write the truth value of the statement p: Intersection of two disjoint setsis an empty set.

3. Find cos cos sin sin4 4 4 4

1

4. Find the argument of 1- i

5. Find the roots of the equation 2x2 +10 x+ 20 = 06. Find the sum of the series 1+2+22+23+…. +2n

7. Write the connective used in the statementP: All whole numbers are either prime or composite.

8. What is the eccentricity of the curve 4 x 2 + y2 = 100?

9. What is the probability that two friends will have the same birthday.

10. All cards of ace, jack and queen are removed from a deck of playingcards. One card is drawn at random from the remaining cards, find theprobability that the card drawn is a face card.



Sect ion B

11. Divide 20 into 4 parts which form an AP such that ratio of the product

of Ist

and 4th

term to the product of 2nd

and 3rd

is 2: 3ORIf the sum of n terms of an A.P is (pn+qn2) where p,q are constants, findcommon difference.

12. A school gave medals on its sports day 38 medals were given for soccer,15 for basketball, 20 for cricket. These medals were given to 58 studentsin all. Only three students got medals in all three sports. How manystudents received medals in exactly two of the three sports?

13. Show that 2 cos6θ = 64 cos6θ – 96 cos4θ + 36 cos²θ

OR

sin3 cos32

sin cos

14. In how many ways can 5 children be arranged in a row such that (i) 2boys x and y are always together (ii) are never sit together?

OR

In how many ways 5 men and 4 women can be seated in a row so thatthe women occupy even places only?

15. Find the equation of the set of points P, the sum of whose distances fromA (4, 0, 0) and B (-4, 0, 0) is equal to 10 ?

9 3 516. Prove that 2 cos cos cos cos 0

13 13 13 13

17. Find the domain and range of f(x) = x 5

OR

Find the domain and range of f(x) =2

3

(2 x )

18. A ladder 12 m long leaning against the wall begins to slide down. Its oneend always remains on the wall and the other on the floor. Find theequation of the locus of a point P which is 3 m from the end in contactwith the floor. Identify the conic section represented by the equation.

19. Show, using Binomial theorem that an - bn is a multiple of (a – b) if a, b, n N

20. Find the equation of the line perpendicular to the line whose equation is6x – 7y + 8 = 0 and that passes through the point of intersection of thetwo lines whose equations are 2x – 3y – 4 = 0 and 3x + 4y – 5 = 0.

21. An administration assistant is given three letters written for three different

people. He is also given three addressed envelopes in which to put them



and mail them to three people X, Y and Z. What is the probability thatatleast one person out of X, Y and Z got the letter written to him.

22. A student appears for an entrance examination. Section A of the paper is

Mathematics having 5 questions and Section B is General studies having 7questions. The student is required to answer 8 questions in all but heshould have attempted at least 3 from mathematics and the same forGeneral studies .In how many ways can he do so?

Sect ion C

23. The sum of n terms of two APs are in the ratio (7n + 1):(4n + 27).Findthe ratio of their 13th term.

cos 7x+ cos5x cos 9x+ cos3x24. Prove that cot6x

sin 7x+ sin5x sin 9x+ sin3x

25. Show by mathematical induction that the sum to n terms of the series

2 2 2 2 2 2

2

n 2

1 2 2 3 2 4 5 2 6 ....is

n n 1, when n is even

2Sn n 1

, when n is odd2

26. Graph the given inequalities and shade the common solution region.x + y ≥35

27. Let R be a relation from N to N defined by

R = {(a, b):a, bN and a = b2 }.Which of the following are true ?(i) (a, a) R, for all a N(ii) (a, b) R, implies (b, a)R(iii) (a, b) R, (b, c) R implies (a, c) R

28. Given below is the frequency distribution of weekly study hours of a groupof class 11 students. Find the mean, variance and standard deviation of the distribution using the short cut method.

Classes Frequency0 - 10 5

10 - 20 820 - 30 1530 - 40 1640 - 50 6

29. Find the derivative of the given function1

(i)f(x) , using first principlex

(ii) y = (x +sec x)( x -tanx)

2x+y≥40, x+2y≥ 50,



TOPPER SAMPLE PAPER - 2

CLASS XI – M ATH EMA TI CS

So lu t i ons

1. Since the statement has definite truth value so it is a mathematical

statement. [1 mark]

2. Since A and B are disjoint sets, so their intersection is empty or a null

set, so the statement is true.

[1 Mark]

3. cos cos sin sin cos cos ( )4 4 4 4 4 4 2

sin( )

[1 Mark]

2 2

1 1 1

1 1 1+ i 1+ i 1+ i 1 14. i

1- i 1- i 1+ i 1 1 2 21 - i

1 1Comparing with x +iy, x= ,y

2 2

1y 2Argument = tan tan tan 1 [1 Mark]

1x 42

25. 2x 10x 20 0

-10 100 160 -5 i 15x= =

4 4

-5+i 15 -5 i 15Sorootsare , [1 Mark]

4 4

6. 1+2+22+23+…. +2n = 20 + 2 + 22 +......+ 2n

This is a GP with first term a = 1 and common ratio r = 2

Sn =

n nna(r 1) 2 1

(2 1)(r 1) 2 1

[1 mark]

7. The word ‘Or’ connects the two statements p: All whole numbers areprime, q: All whole numbers are composite, so the connective is ‘Or’

[1 Mark]



2 2

2 2

2 2 2 2

8. 4x + y = 100

x y1

25 100

a 100,b 25;c a b 100 25 75 5 3

c 5 3 3eccentricity,e [1 Mark]

a 10 2

9.Probability of both the friends not having same birthday is 365 364

365 365

So probability of two friends having the same birthday is365 364 364 1

= 1- 1 1 [1 Mark]365 365 365 365

10. Total cards in the deck =52

There are 4 cards each of ace (one card per suit), jack and queen.

So if all cards of ace, jack and queen are removed from a deck of playing cardsthen 40 cards are left.

In the remaining cards there are kings as 4 face cards

So probability of drawing a face card =4 1

40 10 [1 mark]

Sect ion B

11. Let the four parts be a - 2d, a - d, a + d, a + 2d [1 Mark]So a - 2d + a - d + a + d + a + 2d = 20a = 5 [1 Mark]

2 2

2 2

[(a 2d)(a 2d)] 2[(a d)(a d)] 3

(a 4d ) 2

3(a d )

3a2 - 12d2 = 2a2 - 2d2

a2 - 10 d2 = 0So 25=10 d2

5d

10

[1 Mark]



So AP is 5 - √10, 5 -5

10, 5 +

5

10, 5 + √10 or 5 - √10, 5 -

5

10 , 5 +

5

10 , 5 + √10

[1 Mark]

OR(Sn = (pn + qn2)S1 = (p.1 + q.12) = p + q = a1 [1 Mark]S2 = (p.2 + q.22) = 2p + 4q = a1 + a2 [1 Mark]a2 = S2 - S1 = p + 3q [1 Mark]d = a2 - a1 = (p + 3 q) - (p + q) = 2q [1 Mark]

12.Let A, B and C represents the set of students who got medals for Soccer,Basketball and Cricket.

n(A)=38 n(B)=15,n(C)=20 n(A B C)=58 n(A B C) = 3Using counting theoremsn(A B C) n(A) n(B) n(C) n(A B) n(B C) n(A C) n(A B C) [1Mark]

Substituting the values weget

58 38 15 20 n(A B) n(B C) n(A C) 3 [1Mark]

n(A B) n(B C) n(A C) 76 58 18 [1Mark]

Now,each of n(

A B),n(B C),n(A C) include the 3 students who recieved

medal for all three sports

Number of students who recieved medals in exactly two sports

n(A B) 3 n(B C) 3 n(A C) 3 18 3 3 3 9 [1Mark]

3 3

3

32 2 2

26 4 2

13. cos6 cos3(2 )

4cos (2 ) 3cos(2 ) Using,cos3 4cos 3cos [1Mark]

4 cos(2 ) 3cos(2 )

4 2cos 1 3 2cos 1 Using cos2 =2cos 1 [1Mark]

4 8cos 3 4cos 1 3 2cos 1 1

2

6 4 2 2

6 4 2

6 4 2

6 4 2

3 2cos 1

32cos 48cos 24cos 4 6cos 3

32cos 48cos 18cos 1

2cos6 2 32cos 48cos 18cos 1

64cos 96cos 36cos 2 [2Marks]

OR



Givensin3 cos3

2sin cos

Using sin 3θ = 3 sin θ – 4 sin³θ and cos 3θ = 4 cos³θ – 3 cosθ in LHS

L.H.S. =

3 33sin 4sin 4cos 3cos

sin cos

[1Mark]

=sin (3 4sin² ) cos (4cos² 3)

sin cos

= 3 – 4 sin²θ – 4 cos²θ + 3 [1Mark]

= 6 – 4 sin²θ – 4 cos²θ [1Mark]= 6 – 4(sin2θ + cos²θ)= 6 – 4= 2

= R.H.S. [1Mark]

14. Five children could be arranged in 5! ways.(i) If x and y have to sit together, then taking x and y as 1 unit there are4 ways of arranging them

and the two can interchange places , so 2! x 4! = 48 ways[2Marks](ii) Number of ways in which two children x and y are

Total ways - ways in which x and y are together = 5 ! - 2! x 4! =

72Ways [2 Marks]

OR

There are 9 places in all i.e 1,2,3,4,5,6,7,8,9.Out of these 5 are odd and 4 are even.The 5 odd places i.e 1, 3, 5, 7, 9 have to be occupied by the 5 men as thewomen do not have to sit on theseThis they can do in 5P5 ways = 5! [1 Mark]Now the 4 even places i.e., 2, 4, 6, 8 have to be occupied by the 4 womenThis they can do in 4P4 ways = 4! [1 Mark]Together the men and women can be seated in 5! x 4! = 120 x 24 =

2880 ways [2 Marks]

15. Let P(x, y, z) be the required point.Given PA + PB =10i.e PA = 10 – PBSquaring both sides,PA2 = (10 - PB)2 = 100 + PB2 - 20PB

(x-4)2 + y 2 + z 2 = 100 + (x +4)2 + y2 + z 2 -20 2 2 2((x +4) + y + z )

simplifying, we get

-16x - 100 = -20 2 2 2((x +4) + y + z ) [1 Mark]



2 2 2

2 2 2 2

2 2 2 2

2

(16x 100) 20 (x +4) + y + z

Squaring both sides again and simplifying

16x 100 = 400 (x +4) + y + z ) [ 1 Mark]

256x 10000 3200x 400 x +16+ 8x + y + z

256x 10000

2 2 2

2 2 2 2

2 2 2

2 2 2

2 2 2

3200x 400x +6400+ 3200x + 400y + 400z

256x 100 32x 4x +64+ 32x + 4y + 4z

100

144x 4y - 4z 36 0

100

144x + 400y + 400z 3600 0

18x + 50y + 50z 450 0 [ 2 Marks

]

9 3 516. 2 cos cos cos cos13 13 13 13

9 9 3 5cos +cos cos cos [1Mark]

13 13 13 13 13 13

10 8 3 5cos +cos cos cos

13 13 13 13

10 8 3cos +cos cos cos

13 13 13

5[1Mark]

13

3 5 3 5

= cos +cos cos cos [1Mark]13 13 13 133 5 3 5

cos - cos cos cos 0 [1Mark]13 13 13 13

2

2

17. Let y f(x) x 5

Square root is defined only for non-negative real numbers so

Domain of f is the set of numbers for which x 5 0

i.ex 5

Domain of f=[5, ) [2Marks]

Now, x 5 y

(x 5) yx y

5

since x is taking values greater then or equal to 5 so

Range of f = [0, ) [2Marks]

OR

f(x) =2

3

2 xFor f to be defined 2 - x2 0, i.e. 2 x 2

So Domain = {x: x is a real number and x ≠ ± √2}

[ 2 Marks]



For range

2

3y

2 x

2y - x2y = 3x2 =

2y 3

y

x =2y 3

y

For x to be defined2y 3

y

must be positive, i.e either both Numerator and

Denominator should be positive, in which case y > 3/2 Or else both shouldbe negative, in which case y < 0.

Range =

3

( ,0) ( , )2 [2 Marks]18. Let the Ladder be AB, where A is its end on the horizontal axis and B is the

end on the vertical axis.

A right angled triangle is formed here: (12)2 = a2 + b2

The distance of the point P from the horizontal axis is 3 m, so from theend where it touches the vertical axis is 12 – 3 = 9 mi.e. P divides AB in the ratio 3: 9 = 1: 3 [1 Mark]Therefore the coordinates of P, using the section formula are:

2

2

2 2

2 2

1 0 3 a 1 b 3 0 3a b(x,y) , ,

1 3 1 3 4 44x

a ,b 4y [1Mark]3

4x144 4y [1Mark]

3

x y1

9 9 9

x y1 [1Mark]

81 9



n nn n n n

n n 1 n 2 n (n 1) n nn 0 n 1 n 2 n n 1 n n0 1 2 n 1 n

n n 1 n 2 1n 0 n 1 n 2 n n 10 1 2 n 1

0n n n0

n

9. a - b = a - b + b - b = (a-b) + b - b

C a - b b C a - b b C a - b b ... C a - b b C a - b b b

C a - b b C a - b b C a - b b ... C a - b b

C a - b b b

n n 1 n 20 n 1 n 2 n n 10 1 2 n 1

n n

n n 1 n 2n 0 n 1 n 2 n n 10 1 2 n 1

n 1 n 2 n 3n 0 n 1 n 2 n n 10 1 2 n 1


b b


[2Mark]

a - b C a - b b C a - b b C a - b b ... C b

1[ M2

n n n0 1 n 1

n 1 n 2 0 1 2 n 1

n n

n n

ark]

Now, C , C ,.., C Z; Since a and b are naturalnumbers

a - b , a - b ,... Z;b ,b ,b ,b Z

1[ Mark]2

1a - b = a - b an integer [ Mark]

2

1a - b is a multiple of (a - b) [ Mark]2

20. For finding the point of intersection of lines 2x-3y-4 =0 and 3x+4y-5 =0



Multiplying (i) by 4 and (ii) by 3 we get

8x 12y 16 0

9x 12y 15 0

17x 31

31 31 2x 2 - 3y - 4 =0 y [2 Marks]

17 17 17

6Now the slope of the given line 6x - 7y + 8 = 0 is

7

Sothe slope

1 1

7of the required line = - [1Mark]

6

The equation of the required line is:

y - y m(x - x )

2 7 31y - - x -

17 6 17

2 7x 217y + - +

17 6 102

119x 102y 205 [1Mark]

21. We need to find the probability that atleast one letter was put in the correct envelope.

The total number of ways of putting three letters in 3 envelopes = 3x2x1=6 ...[1 Mark]

The number of ways of putting all three letters in incorrect envelopes is

Letter for X goes into envelope for Y and Letter for Y goes into envelope for Z, Letter for

Z goes into envelope for X or Letter for X goes into envelope for Z and Letter for Y goes

into envelope for X and Letter for Z goes into envelope for Y

So there are 2 ways

2 1P(None of the letters is put in the right envelope) = ...[2 Marks]

6 3P(Atleast one l

etter is put in the right envelope)= 1 - P(None of the letters is put in the

1 2right envelope) = 1 ...[1 Mark]

3 3

22. There are 3 ways in which he can answer 8 questions and taking atleast 3from each sectionCase I: 3 from the first section and 5 from the secondCase II: 4 from the first section and 4 from the secondCase III: 5 from the first section and 3 from the second [1 Mark]The number of ways for doing this isCase I: 5C3 x 7C5 ; Case II: 5C4 x 7C4; Case III: 5C5 x 7C3 [2 Marks]Total number of ways= 5C3 x 7C5 + 5C4 x 7C4 + 5C5 x 7C3 = 210 + 175 + 35 = 420

[1 Mark]



Sect ion C

23.The sum of n terms of two APs are in the ratio (7n+1):(4n+27)

1 1

2 2

1 1

2 2

1 1

2 2

n2a (n 1)d

7n 12n 4n 272a (n 1)d2

2a (n 1)d 7n 1

2a (n 1)d 4n 27

(n 1)2a d

7n 12

(n 1) 4n 272a d

2

[3 Marks]th 1 1

2 2

1 1

2 2

a 12dTo find the ratio of the 13 terms , we need

a 12d

n 112 n 1 24 n 25.

2a 12d 7n 1 7 25 1 176

[3 Marks]a 12d 4n 27 4 25 27 127

24. Numerator cos 7x+ cos5x cos 9x+ cos3x

7x 5x 7x 5x 9x 3x 9x 3x2cos cos 2cos cos

2 2 2 22cos6xcosx 2cos6xcos3x [1Mark]

Denominator sin 7x+ sin5x sin

9x+ sin3x

7x 5x 7x 5x 9x 3x 9x 3x2sin cos 2sin cos

2 2 2 2

2sin6xcosx 2cos6xcos3x [1Mark]

cos 7x+ cos5x cos 9x+ cos3x 2co

sin 7x+ sin5x sin 9x+ sin3x

s6xcosx 2cos6xcos3x

2sin6xcosx 2cos6xcos3x

2cos6x cosx cos3x[2Marks]2sin6x cosx cos3x

2cos6x

2sin6xcot6x

[2Marks]



2 2 2 2 2 2

n

2 2

2

n 2

2 2

1

25. LetS 1 2 2 3 2 4 5 2 6 ..........

Every odd term is n and every even term is 2 n

n n 1 , when n is even2Let P(n) :S

n n 1, when n is odd

2

n n 1 1 1 1whenn 1,S 1,which is true [1

2 2

Mark]

Let the result holds for k i.e P(k) be true to prove P(k+1) to be true.There are two cases

22

k+1

2 2th 2

k

2 22

Case I : If k is odd then k+1 is even

k 1 k 1 1(k 1)(k 2)To prove :P(k 1): S 2 2

k k 1 k k 1P(k 1): P(k) + (k + 1) term = 2(k + 1) (since P(k) = S )

2 2

k k 1 k (k 1)2(k + 1) (k 1) 2(k + 1)

2 2

2

22

k+1

k 4(k + 1)2

(k 1) (k 1)(k 2)k 4k + 4

2 2Sp P(k + 1) is true [2 Marks]

Case II : k is eventhen (k+1) is even

k 1To prove :P(k 1): S

2 2

2 22th

k

22

k 1 1 k 1 k 2=

2 2

k k 1 k k 1P(k 1):P(k)+ (k+1) term = k 1 (since,P(k) = S )

2 2

k 1 k 2kk 1 1

2 2

P(k+1) is true [2 Marks]



26. The In equations are 2x+y 40 x+2y50 and x + y 35Converting the inequations to equations and plotting the corresponding lines

2x+y = 40 or y = 40-2x

x 0 20 10y 40 0 20

x+2y= 50 or x =50-2y

x 0 50 10y 25 0 20

x + y = 35 or y =35-x

x 0 35 10y 35 0 25

(1/2 mark for each table)

( 1 mark for plotting each line +1/2 mark for shading the region of each line)

27. If we need to prove something false one counter example is sufficient.

(i) (a, a) R, for all a N is not true, for example take 2 N. we have 2 ≠ 22, therefore (2, 2) R. (2 marks)



(ii) (a, b) R, implies (b, a)R isalso not true for example take a = 9, b = 3.

As 9 = 32

, we have (9, 3) R, but 3 ≠ 92

, therefore (3, 9) R. (2 marks)(iii) (a, b) R, (b, c) R implies (a, c) R

. For example take a = 16, b = 4 and c = 2.we have 16 = 42and 4 = 22.Therefore (16, 4) R and (4, 2) RBut 16 ≠ 22, so (16, 2) R.

(2 marks)

So for the given data Mean = 27, Standard Deviation = 11.49 andVariance = 132.02

2 8 . Let assumed mean be a =25Classes f i xi yi=

(x-a)/10yi

2 f i yi f i yi2

0 - 10 5 5 -2 4 -10 2010 - 20 8 15 -1 1 -8 820 - 30 15 25 0 0 0 030 - 40 16 35 1 1 16 1640 - 50 6 45 2 4 12 24

50 10 68

f i = 50 , fiyi = 10 , f iyi2

= 68 …[3Marks]

a = 25 ii

x 2 5y =

1 0

n n

i i ii 1 i 1

f y = 1 0 , f 5 0 , h

n

i ii 1

n

ii 1

f y

x a

f

We get, x= 1 0 1 02 5

5 0

…[1Mark]

2n n2

x i i i ii 1 i 1

hN fy fy

N

2

X

2

X

1 0= 5 0 6 8 1 0

5 0

11 0 3 3 1 1 . 4 9 . . . . . . . [1 M a r

5

1 3 2 . 0 2 . . . . . . . [1 M a r k ]



129(i) Derivativeof f(x) , using first principle

x1

f(x) x1

f(x x)x x

1 1 1 1f(x x) f(x)

x x x x x x

x x x xf(x x) f(x) [1Mark]

x x x x x x

f(x x) f(x) 1 x.

x x x x

'

x 0 x 0

'2

1[1Mark]

x x x x

f(x x) f(x) 1 1f (x) lim limx x x x x(x)

1f (x) [1Mark]

x

2

2

(ii) y = (x + secx)(x - tanx)

dy d d(x +secx) (x - tanx) +(x - tanx) (x + secx) [1Mark]

dx dx dxdy

(x + sec x) 1 - sec x +(x - tanx) 1 + secx tanxdxdy

(x + secx).1 - sec x(x +sec x)dx

2 3 2

2 3 2

+( x -tanx) + sec x tanx(x - tanx)[1Mark]

dy(x + secx) - xsec x sec x + (x - tanx) + xsecx tanx - xtan x sec x

dxdy

2x + secx - xsec x sec x - tanx + xsecx tanx - xtan x secxdx

[1Mark]

Class Xi Math Sample Paper

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