Topper Sample PaperCLASS 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

2x 43. Find the domain of the function f(x) =

x 8x 124. 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 statementsymbolically.‘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 theconjugate of (-1 - 3i)

9. Find the focus of a parabola whose axis lies along the X- axis andthe parabola passes through the point (5,10)

10. Write the equation, coordinates of foci, vertices and the eccentricityof the conic section given below.

Section B11. If the general term of a sequence is a linear polynomial, show that

it is an AP.

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 Σn = 210, then find Σn2 .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

22 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)Imz z z

2 1022. Solve the equation 3 x 4x 07

23. Plot the given linear inequations and shade the region that iscommon to the solution of all inequations x 0, y 0, 5x + 3y500; 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 B32 1928 3147 4863 5371 6739 9010 1060 6296 4014 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

31 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 cos4y

sinx28. 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 of5 cm at the centre and the deflected beam is in the shape of aparabola. How far from the centre is the deflection 2 cm?

29. Prove by using the principle of mathematical induction thatx 2n – y 2n is divisible by (x + y).

CLASS 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 1.21f(1.1) f(1) 1.21 1 0.21 2.1 [ 1Mark]

(1.1 1) 0.1 0.1

3. f(x) =2

2x 4

x 8x 12For f(x) to be defined x2-8x+12 must be non zero i.e x2-8x+12 0

( x-2)(x-6) 0ie x 2 and y 6

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 8solving the two equations , we get a = 5, b=-2a = 5, b=-2 also satisfy the other two ordered pairsf( 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 exists”Statement 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 ).

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 14AB BC ACTherefore, the points A, B , C are collinear points. [1Mark]

2

2

8. (x iy)(3 5i) 1 3i 1 3i1 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 17 176 7x ;y [ 1Mark]17 17

9 The parabola is y² = 4axThe point (5, 10) lies on it

(10²) = 4a (5)

100a 5 [ 1Mark]20

2 2

2 2

10. Given conic sec tion represents a hyperbolax y 1a b

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

SECTION B11. Since, the general term of the AP is a linear polynomial

So 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

an = n (n+1) = n2 + n [ 1 Mark]n n n

2 2n

k 1 k 1 k 1S k + k k k

n(n 1)(2n 1) n(n 1) [1mark]6 2

n(n 1) (2n 1) 12 3

n(n 1)(n 2) [1mark]3

12. Let H denote the set of people who can speak Hindi, and E denotethe set of people who can speak English.Given everyone can speak atleast one language

Therefore, n(H U E) = 400 and n(H) = 250, n(E) = 200[1Mark]

n(H U E) = n(H) + n(E) - n(H E) [1Mark]

n(H E) = n(H) + n(E) - n(H U E) [1Mark]

n(H E) = 250 + 200 - 400 = 50 [1Marks]

13. Σn=210n(n 1)

2= 210 [1Mark]

n(n+1) = 42021 x 20 = 420 so n= 20 [1Mark]

Σn2 =n(n 1)(2n 1)6

= 420 416

= 2870 [2Marks]

14.

Let the two vertices of the triangle be Q and R

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

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

2

2

k 1 k ktan30 RTRT RT3 3

kR(k, ) [1Mark]3

Now R lies on the parabola : y = 4 ax

k 4 a(k)3

k 4a3k 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 x3x+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

2sinx cosxsin 2x 2sinx cosxsin2x=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 cos2 2 2 2

2sin4xcos3x ......(i) [1Mark]3x 5x 3x 5xsin3x + sin 5x 2sin cos 2

2 28x 2xsin cos2 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 2cos2 2 2

2xcos2

4sin4xcos2xcosx [2Marks]

16.Length of pendulum 36 cm long

Angle of oscillation = 10 degrees180 degrees = radians

so, 10 degrees= radians18

θ= radians (1 Mark)18

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

θ= radians18

(1 Mark)

we get,

=36 x π18

=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]

2

2

2

2 2

22 2 2 2 2

2

2

2

x18. f(x)x 1

x Let y = f(x)x 1

x 0 x 1 1 Denominator Numerator y 1xNow,y y x 1 x yx y x x (y 1) y [1Mark]

x 1yx

1 y1 y 0y 1

yNow,x 0 [1Mark]1 y

Case1: y 0;1 y 0 y 0;1 y or y 1,but y 1i.e y [0,1) [1Mark]Case2 : y 0;1 y 0 y 0;1 y or y 1Not possible [1 Mark]

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

524

134

52! 52.51.50.49= C4!48! 1.2.3.4

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

13! 13.12.11.10= C4!9!

7151.2.3.4

Now, there are 4 suits to choose from, soThe 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

tThe number of ways of choosing one card from each suit= C C C C 13 13 13 13 13 [1Mark](iii) are face cardsThere are 12 face cardsThe 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 13 252!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. 99To 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+110010000500 5001000019509900499 [2 Marks]

OR

9

r 9 r9r 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 63 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 92!7!9!a 7a 7aCoefficient of x C a 3 a. C

[1Mark]

22

22

222

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 2 222 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 yc d

a b a b x y [1Mark]c dc dOR

We have,z1 = 2 – i and z2 = -2 + i

1 2

1

z zi To find Rez

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 i6 3i 8i 4 i

4 16 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)Imz z

2

1 2

1 1 12 i 2 i 4 2i 2i iz z

1 1 .. [1Mark]4 1 5

1 2

1 1Im Im 0 .. [1Mark]5z z

2

2

2

2

1022. 3 x 4x 07

21x 28x 10 01D 28 4 21 10 784 840 56 0 …[ Mark]2

The 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 14i 14 14i, …[2Marks]21 21

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 5x3

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 1 x 62 5 31 13x 20 x 6

10 3Multiplying both sides by LCM of 10 and 3, i.e 30, we get30 303x 20 x 610 3

3(3x 20) 10 x 69x 60 10x 6060 60 10x 9x120 xx 120x ( ,120]

Thus, all real numbers less than or equal to 120 are the sthe 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 Bx di= x -

50di2 y fi = x -

50fi2

32 -18 324 19 -31 96128 -22 484 31 -19 36147 -3 9 48 -2 463 13 169 53 3 971 21 441 67 17 28939 -11 121 90 40 160010 -40 1600 10 -40 160060 10 100 62 12 14496 46 2116 40 -10 10014 -36 1296 80 30 900Total -40 6660 Tota

l0 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 =

(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 of0,1,3,5,7 can come(i)When repetition of digits is allowedThe number of choices for place 4 =

Th T UH

2The number of choices for place 3 = 5The number of choices for place 2 = 5The number of choices for place 1 = 5Total 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 inthe units placeThe number of choices for place 4 = 2The number of choices for place 3 = 5The number of choices for place 2 = 5The number of choices for place 1 = 2Total number of choices = 2 x 5 x 5 x 2 = 100 [1Mark]

100P(number divisible by 5 is formed when digits are repeated ) =250

2 [1Mark]5

(ii) When repetition of digits is not allowedThe number of choices for place 4 = 2The number of choices for next place = 4The number of choices for next place = 3The number of choices for next place = 2Total number of choices = 2 x 4 x 3 x 2 = 48 [1Mark]For the number to be divisible by 5 there should be either 0 or 5 in the unitsplace , giving rise to 2 cases.Case I : there is 0 in the units placeThe number of choices for place 4 = 2Total number of choices for remaining places = 3 x 2 x 1=6Total number of choices = 2 x3 x 2 x 1= 12Case II : there is 5 in the units placeThen there is 7 in place 4The number of choices for place 4 = 1Total number of choices for remaining places = 3 x 2 x 1=6Total number of choices=6From 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

3

2

2

2 2

3

26. x Q III quadrant and cos x =-3

cos2 2cos 1xcosx 2cos 12

1 x 2 x- 1 2cos cos3 2 3 2 2x 1cos [1Mark]2 3xsec 32

Now,x Q32n x 2n 232n2n x 2

2 2 2x 3n n

2 2 4Case I:W

2

4

hen n is even 2k(say)x 32k 2k2 2 4

x Q [1Mark]2Case I:When n is odd 2k 1(say)

x 32k 1 2k 12 2 4x 32k 2k

2 2 43 x 72k 2k2 2 4

x Q [1Mark]2

x xsin 1 cos2 2

22

2

4

1 1 21 1 [1Mark]3 3 3

x 2InQ sin [1Mark]2 3x 2InQ sin [1Mark]2 3

x 2So sin2 3

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 162sin sin [1Mark]

2 2

2x+2sin

x 0 x 0

x -x8 sin

2 2

2x+ x - 2x+ x -x x8 82sin sin sin sinf(x+ x)-f(x) 2 2 2 2 [1Mark]xx x

2xsinx 2lim sin x+ - lim x2 16

2

sin x - [1Mark]16

22

2 2

2

2

2

x2

x cosx4(ii)y cos

sinx 4 sinx

d dsinx (x ) x (sinx)dy dx dxcos [1Mark]dx 4 sin x

dy 2x.sinx x .cosxcosdx 4 sin x

dy 2x.sinx xcosdx 4

2

2x

2

2

2x2

2

2x

2

.cosx [1Mark]sin x

2 .sin .cosdy 2 2 2 2cosdx 4 sin

2

2 .1 .02 4dy 1cos

dx 4 1 2 [1Mark]

2

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)aregivenThe point of intersection of (1) and ( 2) can be obtained by solvingy = 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 + cm x + c - m x - c 0m - m x c - c

c - cx [1Mark]m - m

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

c - c m c - m cThe 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 itis 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 cm - 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

Therefore it must satisfy the equation of parabola i.e 49i.e a = 49/20 (1 mark)So the equation of parabola becomesx2 = 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 =9.8 3 = 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 + ywhich 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]