1291109075 Classxi Math Toppersamplepaper 30

7/30/2019 1291109075 Classxi Math Toppersamplepaper 30

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Topper Sample Paper- 3CLASS XI MATHEMATICS

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 12 questions of four marks each and section Ccomprises of 07 questions of six marks each.

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

4. There is no overall choice. However, internal choice has been providedin some questions

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

if required.Section A

1. Write the negation of the given statement, p: Intersection of twodisjoint sets is not an empty set.

2 f(1.1) - f(1)2. If f(x) = x , the evaluate .(1.1) - (1)

2

2

x 43. Find the domain of the function f(x) =

x 8x 12

4. If f(x) is a linear function of x.f: Z Z, f(x) = a x + b. Find a and b

if { (1,3) , (-1, -7 ) , (2, 8) (-2 , -12 )} f.5. With p: It is cloudy and q: Sun is shining and the usual meanings of

the symbols: , , , , write the given statement

symbolically.It is not true that it is cloudy if and only if sun is not shining.

6. Identify the quantifier in the given statement: P: There exists awhole number which is not natural. Write its truth value.

7. Show that the points (-2, 3 , 5 ) , ( 1, 2 , 3 ) and ( 7, 0 , -1 ) arecollinear.

8. What are the real numbers 'x' and 'y', if (x - iy) (3 - 5i) is the

conjugate of (-1 - 3i)9. Find the focus of a parabola whose axis lies along the X- axis and

the parabola passes through the point (5,10)10. Write the equation, coordinates of foci, vertices and the eccentricity

of the conic section given below.

Section B

11. If the general term of a sequence is a linear polynomial, show thatit is an AP.


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OR

Find the sum of the given sequence uptill the nth term:1.2 + 2. 3 + 3. 4 +

12. In a group of 400 people, 250 can speak Hindi and 200 can speakEnglish. Everyone can speak atleast one language. How manypeople can speak both Hindi and English?

13. If n = 210, then find n2 .14. An equilateral triangle is inscribed in the parabola y2 = 4ax, where

one vertex is at the vertex of the parabola. Find the length of theside of the triangle.

15. Prove that (cos 3 x cos x) cos x + (sin 3x + sin x) sin x =0

OR

Simplify the expression: sin7x + sinx + sin3x + sin5x16. A pendulum 36 cm long oscillates through an angle of 10 degrees.

Find the length of the path described by its extremity.17. Let A = {a,b,c}, B = {c,d} and C = {d,e,f}. Find

(i) A x (B C) (ii) (A x B) (A x C)(iii) A x (B C) (iv) (A x B) (A x C)

2

2

x18. If :R R;f(x) .What is the range of f?

x 1

19. What is the number of ways of choosing 4 cards from a pack of52playing cards? In how many of these

(i) four cards are of the same suit,(ii) four cards belong to four different suits,(iii) are face cards,(iv) two are red cards and two are black cards,

20. Evaluate (99)5 Using Binomial theorem

OR

Find the ratio of the coefficient of x2 and x3 in the binomialexpansion (3 + a x)9

2

2 2a-ib21. If x - iy = ,find x y .c-id

ORLet z1 = 2 i and z2 = -2 + i , then find

1 2

1 1 2

z z 1i Re (ii)Im

z z z

2 1022. Solve the equation 3 x 4x 07


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Section C

23. Plot the given linear inequations and shade the region that iscommon to the solution of all inequations x 0, y 0, 5x + 3y 500; x 70 and y 125.

ORSolve the inequality given below and represent the solution on thenumber line.

1 3x 20 1

x 62 5 3

24. The scores of two batsmen A and B, in ten innings during a certainseason are given below:

A B

32 19

28 3147 48

63 53

71 67

39 90

10 10

60 62

96 40

14 80

Find which batsman is more consistent in scoring

25. From the digits 0, 1, 3, 5 and 7, number greater than 5000 but of 4digits are formed. What is the probability that the number formed isdivisible by 5, if(i) the digits are repeated(ii) the digits are not repeated

3

1 x 226. If x Q and cosx =- thenshow that sin .

3 2 3

27. (i) Find the derivative of the given function using first principle:

f(x)=cos x-

16

(ii) Find the derivative of

2x cos4

ysinx

28. If three lines whose equations are y = m1x + c1, y = m2x + c2 and y= m3x + c3 are concurrent, then find (i) the condition ofconcurrence of the three lines(ii) the point of concurrence

ORA beam is supported at its ends by supports which are 14 cm apart.

Since the load is concentrated at its centre, there is a deflection of

5 cm at the centre and the deflected beam is in the shape of aparabola. How far from the centre is the deflection 2 cm?


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29. Prove by using the principle of mathematical induction that

x2n

y2n

is divisible by (x + y).


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TOPPER SAMPLE PAPER 3CLASS XI MATHEMATICS

(Solutions)

1. p: Intersection of two disjoint sets is not an empty set.p: Intersection of two disjoint sets is an empty set [ 1 Mark]

2

22

2. f(x) x

f(1) 1 1;f(1.1) 1.1 1.21

f(1.1) f(1) 1.21 1 0.212.1 [ 1Mark]

(1.1 1) 0.1 0.1

3. f(x) =2

2

x 4

x 8x 12

For f(x) to be defined x2

-8x+12 must be non zero i.e x2

-8x+120( x-2)(x-6) 0ie x 2 and y6

So domain of f = R-{2,6} [1 mark]

4. f(x) ax b

(1,3) f f(1) a.1 b 3 a b 3

(2,8) f f(2) a.2 b 8 2a b 8

solving the two equations , we get a = 5, b=-2

a = 5, b=-2 also satisfy the other two ordered pairs

f( 2) 5. 2 2 12 ( 2, 12)

f( 1) 5. 1 2

7 ( 1, 7) ...[ 1Mark]

5. (p q) [1Mark]6. P: There exists a whole number which is not natural.

Quantifier in the given statement is There existsStatement P is true since 0 is a whole number which is not

natural.[1Mark]

7. Let the given points be A , B and C , such that A(-2, 3 , 5 ), B ( 1,2 , 3 ) and C( 7, 0 , -1 ).


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2 2 2

2 2 2

2 2 2

AB (1 2) (2 3) (3 5) 14

BC (7 1) (0 2) ( 1 3) 56 2 14

AC ( 2 7) (3 0) (5 1) 126 3 14

AB BC AC

Therefore, the points A, B , C are collinear points. [1Mark]

22

8. (x iy)(3 5i) 1 3i 1 3i

1 3i (3 5i)1 3i 3 5i 9i 15i(x iy)

(3 5i) (3 5i)(3 5i) (9 25i )

3 5i 9i 15 12 14i 6 7i 6 7i

9 25 34 17 17 17

6 7x ;y [ 1Mark]

17 17

9 The parabola is y = 4axThe point (5, 10) lies on it (10) = 4a (5)

100a 5 [ 1Mark]

20

2 2

2 2

10. Given conic sec tion represents a hyperbola

x y1

a b

cFoci ( c,0);Vertices ( a,0);eccentricity [ 1Mark]a

SECTION B

11. Since, the general term of the AP is a linear polynomialSo let an be kn+ pThen a1= k(1)+pa2= k(2) +pa3= k(3)+p..

.an-1 = (n-1)k+p [1Mark]Then a3 - a2 = k(3)+p - [k(2) +p]= 3k - 2k =ka2 - a1= k(2) +p - [k(1)+p] = 2k - k = kan an-1 = kn+ p - (n-1)k- p = k

Hence a3 - a2 = a2 - a1 = an an-1 =k [2Marks]

Since the difference between two consecutive terms is same so thesequence is an A.P

[1Mark]

OR


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Points Q and R will have the same x - coordinate = k(say)[1Mark]

Now in the right triangle PRT, right angled at T.

2

2

k 1 k ktan30 RT

RT RT3 3

kR(k, ) [1Mark]

3

Now R lies on the parabola : y = 4 ax

k4 a(k)

3

k4a

3k 12a

[2Marks]

12akLength of side of the triangle = 2(RT)=2. 2. 8 3a [1Mark]

3 3

15. ConsiderLHS (cos 3 x - cos x) cos x + (sin 3x + sin x) sin x

3x+x 3x-x 3x+x 3x-x 1= 2sin sin cosx 2sin cos sinx [1 Mark]

2 2 2 2 2

2sin2xcosx sinx 2sin2xsinx cosx

1

[1 Mark]2

2sinxcosxsin 2x 2sinxcosxsin2x=0 [1Mark] OR

sin 7x + sin x + sin3x + sin 5x = sin 7x + sin x + sin3x + sin 5x

7x x 7x x 8x 6xsin 7x + sin x 2sin cos 2sin cos

2 2 2 2

2sin4xcos3x ......(i) [1Mark]

3x 5x 3x 5xsin3x + sin 5x 2sin cos 2

2 2

8x 2xsin cos

2 2

2sin4xcos( x) 2sin4xcosx ....(ii) [1Mark]

From(i)and (ii)

sin7x + sinx + sin3x + sin5x 2sin4xcos3x 2sin4xcosx

2sin4x cos3x cosx

3x x 3x x 4x2sin4x 2cos cos 2sin4x 2cos

2 2 2

2xcos

2

4sin4xcos2xcosx [2Marks]

16.Length of pendulum 36 cm long


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Angle of oscillation = 10 degrees180 degrees = radians

so, 10 degrees= radians18

= radians (1 Mark)18

so using this formula =rand substituting the values of r=36,

= radians18

(1 Mark)

we get,

=36 x

18=2 x (3.14) = 6.28 cm (2 Marks)

17.A = {a,b,c} B = {c,d} C ={d,e,f}(i) (B C) = {d}

A x (B C) ={(a,d), (b,d), (c,d)}. [1Mark](ii) A x B ={(a,c), (a,d), (b,c), (b,d), (c,c), (c,d)}

A x C={(a,d), (a,e), (a,f), (b,d), (b,e), (b,f), (c,d), (c,e), (c,f)}(A x B ) ( A x C) = {(a,d),(b,d),(c,d)} [1Mark]

(iii) (B C) = {c, d, e ,f}A x (B C) = {(a,c), (a,d), (a,e), (a,f), (b,c), (b,d), (b,e), (b,f),(c,c), (c,d), (c,e), (c,f)}. [1Mark]

(iv) (A x B ) ( A x C) ={(a,c), (a,d), (a,e), (a,f), (b,c), (b,d), (b,e),(b,f), (c,c), (c,d), (c,e), (c,f)}.[1Mark]


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10

2

2

2

2

2 2

22 2 2 2 2

2

2

2

x18. f(x)

x 1

xLet y = f(x)

x 1

x 0 x 1 1 Denominator Numerator y 1

xNow,y y x 1 x yx y x x (y 1) y [1Mark]

x 1

yx

1 y

1 y 0

y 1

yNow,x 0 [1Mark]1 y

Case1: y

0;1 y 0 y 0;1 y or y 1,but y 1

i.e y [0,1) [1Mark]

Case2 : y 0;1 y 0 y 0;1 y or y 1

Not possible [1 Mark]

Range of f(x) = [0,1)


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524

134

19. The number of ways of choosing 4 cards from a pack of 52 playing cards

52! 52.51.50.49= C

4!48! 1.2.3.4

270725(i) The number of ways of choosing four cards of any one suit

13! 13.12.11.10= C

4!9!

7151.2.3.4

Now, there are 4 suits to choose from, so

The number of ways of choosing four cards of one suit=4 715=2860 [1Mark]

(ii) four cards belong to four different suits,

ie, one card from each sui

13 13 13 13 41 1 1 1

t

The number of ways of choosing one card from each suit

= C C C C 13 13 13 13 13 [1Mark]

(iii) are face cards

There are 12 face cards

The number of ways of

124

choosing four face cards from 12

= C

=495 [1Mark]

(iv) two are red cards and two are black cards,

There are 26

26 262 2

red cards and 26 black cards,

The number of ways of choosing 2 red and 2 black from 26 red and 26 black

26! 26! 26.25 26.25= C C 13 25 13 25

2!24! 2!24! 2 2

=105625

[1Mark]

5

5 5 5 4 3 20 1 2 35 5 5 510 2 3

1 04 55 54 5

5 4 3 2

20. 99

To be able to use binomial theorem , let us express 99 as a binomial : 99= 100-1

99 100-1 C 100 1 C 100 1 C 100 1 C 100 1

C 100 1 C 100 1

1. 100 5 100 10. 100 10 100 5. 100 1

[1Marks]

10000000000 5 100000000 10. 1000000 10 10000 5. 100 1 [1Mark]

10000000000 500000000 10000000 100000 500 1

10000000000+10000000+500 500000000+100000+1

10010000500 500100001

95099

00499 [2 Marks]

OR


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9

9

r 9 r9

r 1 r

r 9 rr 9r

2 9 22 9 9 2 72 2

3 9

Given:(3 + ax)

Generalterm in theexpansionof (3 + ax)

t C ax 3 [1Mark]Coefficient of x C a 3

Coefficient of x C a 3 C a 3 [1Mark]

Coefficient of x C

3 9 3 9 3 6

3 3

9 2 7 922 2

3 9 3 6 93 3

a 3 C a 3 [1Mark]

C a 3 3. CCoefficient of x 3.9!3!6! 3.3 9

2!7!9!a 7a 7aCoefficient of x C a 3 a. C

[1Mark]

2

2

22

2

22

2 2

22 2 2 2 2

a-ib21. x - iy =

c-id

a-ibx - iy =

c-id

Now, x - iy x - iy

a-ibx - iy = x - iy = [1Mark]

c-id

But x - iy x y

x - iy = x y x y ...(i)

2 22 2 2

2 2 22

2 22 2

2 2

22 2 2 2

22 2

2 22 2

[1Mark]

a ba-iba-ib a-ib a b...(ii) [1Mark]

c-id c-id c id c dc d

From (i) and (ii) , we have

a bx y

c d

a b a bx y [1Mark]

c dc d

OR

We have,z1 = 2 i and z2 = -2 + i

1 2

1

z zi To find Re

z


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2

1 2

1

2

4 i 4i2 i 2 iz z= [1Mark]

2 i 2 iz

3 4i 3 4i 2 i=- = -2 i 2 i 2 i

6 3i 8i 4 i

4 1

6 11i 4 2 11i 2 11i 1[ Mark]

5 5 5 2

1 2

1

z z 2 11i 2 11i 2 1Re =Re Re [ Mark]

5 5 5 5 2z

1 2

1(ii)Im

z z

2

1 2

1 1 1

2 i 2 i 4 2i 2i iz z

1 1.. [1Mark]

4 1 5

1 2

1 1Im Im 0 .. [1Mark]

5

z z

2

2

2

2

1022. 3 x 4x 0

7

21x 28x 10 0

1D 28 4 21 10 784 840 56 0 [ Mark]

2The equation has complex roots

28 56b b 4ac b Dx=

2a 2a 2 21

28 56i 28 2 14i 1[1 Mark]

42 42 2

28 2 14i 28 2 14i,

42 42

14 14

i 14 14i, [2Marks]

21 21


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Section C

23. System of inequationsx 0, y 0, 5x + 3y 500; x 70 and y 125.Converting inequations to equations

5x + 3y = 500 y =500 5x

3

x 100 40 -80y 0 100 300

x 70 is x =70 and y 125 is y =125.

Plotting these lines and determining the area of each line we get

[3 Marks for correct plotting of lines, three for correct shading]

OR

1 3x 20 1x 6

2 5 3

1 13x 20 x 610 3

Multiplying both sides by LCM of 10 and 3, i.e 30, we get

30 303x 20 x 6

10 3

3(3x 20) 10 x 6

9x 60 10x 60

60 60 10x 9x

120 x

x 120

x ( ,120]


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Thus, all real numbers less than or equal to 120 are the solution ofthe given inequality. The solution set can be graphed on real line asshown

[4 marks for correct solution, 2 marks for correct plotting on realnumber line]

24. Coefficient of variation is used for measuring dispersion. In that casethe batsman with the smaller dispersion will be more consistent.

(1/2 mark)Cricketer A Cricketer B

x di= x -50

di2 y fi = x -50

fi2

32 -18 324 19 -31 961

28 -22 484 31 -19 361

47 -3 9 48 -2 4

63 13 169 53 3 9

71 21 441 67 17 289

39 -11 121 90 40 1600

10 -40 1600 10 -40 1600

60 10 100 62 12 144

96 46 2116 40 -10 100

14 -36 1296 80 30 900

Total -40 6660 Total

0 5968

(2 marks)For cricketer A:

Mean = 50 + = 46

S.D =

(1 Mark)

Therefore C.V = (1/2 mark)

For Cricketer B:

Mean = 50 + = 50

S.D =


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(1 Mark)

Therefore C.V = = 49 (1/2 mark)

Since the C.V for cricketer B is smaller, he is more consistent inscoring.

(1/2 mark)

3 14 2

25. There are 4 places to be filled

The number has to be > 5000 , so in place 4 only 5 or 7 out of

0,1,3,5,7 can come

(i)When repetition of digits is allowed

The number of choices for place 4 =

Th T UH

2

The number of choices for place 3 = 5



Total number of choices = 2 x 5 x 5 x 5 = 250 [1Mark]

Now, for the number to be divisible by 5 , there should be 0 or 5 in

the units place






100P(number divisible by 5 is formed when digits are repeated ) =

250

2[1Mark]

5


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(ii) When repetition of digits is not allowed


The number of choices for next place = 4The number of choices for next place = 3

The number of choices for next place = 2


For the number to be divisible by 5 there should be either 0 or 5 in the units

place , giving rise to 2 cases.

Case I : there is 0 in the units place


Total number of choices for remaining places = 3 x 2 x 1=6

Total number of choices = 2 x3 x 2 x 1= 12

Case II : there is 5 in the units place

Then there is 7 in place 4


Total number of choices for remaining places = 3 x 2 x 1=6

Total number of choices=6

From case I and case II :

Total number of choices = 6 + 12 = 18 [1Mark]

P( a number divisible by 5 is formed when repetition of digits is not

18 3allowed)= [1Mark]

48 8


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3

2

2

2 2

3

126. x Q III quadrant and cos x =-

3

cos2 2cos 1

xcosx 2cos 121 x 2 x

- 1 2cos cos3 2 3 2 2

x 1cos [1Mark]

2 3x

sec 32

Now,x Q

32n x 2n

23

2n2n x 22 2 2

x 3n n

2 2 4Case I:W

2

4

hen n is even 2k(say)

x 32k 2k

2 2 4x

Q [1Mark]2

Case I:When n is odd 2k 1(say)

x 32k 1 2k 1

2 2 4x 3

2k 2k2 2 4

3 x 72k 2k

2 2 4x

Q [1Mark]2

x xsin 1 cos

2 2

22

2

4

1 1 21 1 [1Mark]

3 3 3

x 2InQ sin [1Mark]

2 3

x 2InQ sin [1Mark]

2 3

x 2So sin2 3


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27. (i) f(x)=cos x-16

f(x+ x)=cos x+ x - 16

f(x+ x)-f(x)=cos x+ x - cos x -16 16

x+ x - x -x+ x - x -16 1616 16

2sin sin [1Mark]2 2

2x+

2sin

x 0 x 0

x -x8

sin2 2

2x+ x - 2x+ x -x x8 8

2sin sin sin sinf(x+ x)-f(x) 2 2 2 2 [1Mark]

xx x2

xsin

x 2lim sin x+ - limx2 162

sin x - [1Mark]

16


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2

2

2 2

2

2

2

x2

x cosx4

(ii)y cossinx 4 sinx

d dsinx (x ) x (sinx)

dy dx dxcos [1Mark]dx 4 sin x

dy 2x.sinx x .cosxcos

dx 4 sin x

dy 2x.sinx xcos

dx 4

2

2

x2

2

2x2

2

2

x2

.cosx[1Mark]

sin x

2 .sin .cosdy 2 2 2 2

cosdx 4

sin2

2 .1 .02 4dy 1

cosdx 4 1 2

[1Mark]2


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1 1 2 2

3 3

1 1 2

28. The three lines whose equations are y = m x + c (1), y = m x + c (2)

and y = m x + c ....(3)aregiven

The point of intersection of (1) and ( 2) can be obtained by solving

y = m x + c , y = m x +

2

1 1 2 2

1 1 2 2

1 2 2 1

2 1

1 2

2 1 1 2 2 11 1

1 2 1 2

2 1 1 2 2 1

1 2 1

c

m x + c = m x + c

m x + c - m x - c 0

m - m x c - c

c - cx [1Mark]

m - m

c - c m c - m cy = m + c

m - m m - m

c - c m c - m c

The point of intersection = ,m - m m -

2

2 1 1 2 2 1

1 2 1 2

[1Mark]m

If this point also lies on line (3), then the three lines are concurrent and it

is the point of concurrence [1Mark]

c - c m c - m cWe substitute , into y =

m - m m - m

3 3

1 2 2 1 2 13 3

1 2 1 2

1 2 2 1 3 2 1 3 1 2

1 2 2 1 3 2 3 1 3 1 3 2

m x + c (3), to verify

m c - m c c - cm c

m - m m - m

Som c - m c = m c - c + c m - m

m c - m c m c - m c + c m -c m

` m1 (c2 c3) + m2 (c3 c1) + m3 (c1 c2) = 0 is the requiredcondition for concurrence. (2 Marks)

(ii) Point of concurrency is 2 1 1 2 2 1

1 2 1 2

c - c m c - m c,

m - m m - m

( 1 Mark)

OR

Let the vertex be at the origin and the axis vertical along y axisTherefore general equation is of the form x2 = 4ay (1 mark)

Since deflection in centre is 5 cm, So (7,5) is a point on parabola


22/22

Therefore it must satisfy the equation of parabola i.e 49 = 20 ai.e a = 49/20 (1 mark)So the equation of parabola becomes

x2

= 49/5y = 9.8 y (1 mark)

Let the deflection of 2 cm be t cm away from the origin. Let AB be thedeflection of beam,So the coordinates of point B will be (t,3) (1 mark)Now, since parabola passes through ( t, 3), it must satisfy the equation ofparabola

Therefore t2 =9.83 = 29.4 (1 mark) t=5.422 cm (1 mark)

Therefore distance of the deflection from the is 5.422 cm

2n 2n

2 1 2 1

2 2

29. Let P(n): x - y is divisible by x + y

P(1) : x - y is divisible by x + y

P(1) : x - y =(x -y) (x+y) is divisible by x + y

which is true [1Mark]

Now we assume P(k

2k 2k

2 k+1 2 k+1

2 k+1 2 k+1 2k+2 2k+2 2 2k 2 2k 2 2k

) : x - y is divisible by x + y [1Mark]

To Prove :P(k 1) : x - y is divisible by x + y [1Mark]

x - y x - y x x - x y + x y -

2 2k

2 2k 2k 2k 2 2

2k 2k

2 2k 2k

2 2

2k 2 2

y y [1Mark]

x x - y y x - y

x - y is divisible by x + y from P(k)

x x - y is divisible by x + y from P(k)

x - y is divisible by x + y

y x - y is divisible by x

2 2k 2k 2k 2 2+ y

x x - y y x - y is divisible by x + y [2Marks]

1291109075 Classxi Math Toppersamplepaper 30

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