1321619488 microsoftword-classxi math topper sample paper 4 r0

8/3/2019 1321619488 MicrosoftWord-ClassXI Math Topper Sample Paper 4 R0

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TOPPER SAMPLE PAPER - 4CLASS 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 onemark each, section B comprises of 12 questions of four marks eachand section C comprises 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 beenprovided in some questions

5. Use of calculators is not permitted. You may ask for logarithmictables, if required.

SECTION A

Q.1 Identify the function that the given graph represents.

Sol.Such a function is called the greatest integer function.

From the definition [x] = 1 for 1 x < 0[x] = 0 for 0 x < 1[x] = 1 for 1 x < 2[x] = 2 for 2 x < 3 and so on (1 Mark)

Q.2 Let f(x) = x2 and g(x) = 2x + 1 be two real valued functions. Find(fg) (x).

Sol.f(x) =x2 , g(x) = (2x+1)(fg) (x) = f(x)g(x) = x2 (2x + 1) = 2x3 + x2 (1 Mark)

Q.3 If 4x + i(3x y) = 3 + i ( 6), where x and y are real numbers,then find x and y.

Sol.


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4x + i (3x y) = 3 + i (6)Equating real and the imaginary parts of the given equation, weget,

4x = 3, 3x y = 6,Solving simultaneously, we get x =

3

4and y =

33

4(1Mark)

Q.4 Find the distance between the points P(1, 3, 4) and Q ( 4, 1, 2).

Sol.The distance PQ between the points P (1,3, 4) and Q ( 4, 1, 2) is

2 2 2PQ ( 4 1) (1 3) (2 4)

25 16 4

45 3 5 units

= + + +

= + +

= =

(1 Mark)

Q.5 Let A = {1, 2} and B = {3, 4}. Find the number of relations from Ato B.

Sol.Given A = {1,2} B = {3,4}A B = {(1, 3), (1, 4), (2, 3), (2, 4)}.Since n (A B ) = 4, the number of subsets of A B is 24Therefore, the number of relations from A to B will be 16 (1 Mark)

Q.6 Find what the equation x2 + y2 + xy = 0 becomes when the origin isshifted to the point (1, 1)?

Sol.Let the coordinate of point P(x,y) when origin is shifted are P(X,Y),Then x = X+1 and y = Y+1Thus equation becomes:(X+1)2 + (Y+1)2 + (X+1)(Y+1) = 0X2 + Y2 + XY + 3X +3Y +3 = 0 (1 Mark)

Q.7 Write the component statements of the following compound

statement and check whether the compound statement is true orfalse.P:0 is less than every positive integer and every negative integer

Sol.The component statements arep: 0 is less than every positive integer.q: 0 is less than every negative integer.The second statement is false.Therefore, the compound statement is false. (1 Mark)

Q.8 Find the coordinates of the foci of the ellipse2 2

x y1

25 9+ =

Sol.


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Since denominator of x2 is larger than the denominator of y2 the majoraxis is along the x-axis. Comparing the given equation with the

standard equation of the ellipse

2 2

2 2

x y

1a b+ = a = 5 b =9

c = 2 2a b 25 16 3 = = So the foci is (3,0) and (-3,0) (1 Mark)

Q.9 Write the contrapositive of the statement: If a number is divisibleby 9, then it is divisible by 3.

Sol.The contrapositive of the statements is:

If a number is not divisible by 3, it is not divisible by 9. (1 Mark)

Q.10 Given below is a pair of statements:Combine these two statements using if and only if .p: If a rectangle is a square, then all its four sides are equal.q: If all the four sides of a rectangle are equal, then the rectangle isa square.

Sol.A rectangle is a square if and only ifall its four sides are equal.

(1 Mark)

SECTION B

Q.11 In a survey of 600 students in a school, 150 students were found tobe taking tea and 225 taking coffee, 100 were taking both tea andcoffee. Find how many students were taking neither tea nor coffee?

Sol.Let T represents the students taking tea and C represents studentstaking coffeeLet x represents the number of students taking tea or coffee

So, x = n(T C)

n(T) = 150 n(C) = 225

n (TC) = n(T)+n(C)- n(TC) (1 Mark)

Substituting the values

x = 150 + 225 100 = 275 (2 Marks)


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Number of students taking neither of two drinks tea or coffee= 600 275 = 325.

(1 Mark)

Q.12 Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by

{(a, b): a, b A , a divides b}

(i) Write in the roster form(ii) Find the domain of R(iii) Find the range of R

Sol.R: { (a, b): a, b A, a divides b}, Also

A = {l, 2, 3, 4, 6}(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5),(1, 6), (2, 2),(2, 4),

(2,6), (3, 3), (3, 6),(4, 4), (6, 6)} (2 Marks)(ii) Domain of R= {1, 2, 3, 4, 6}. (1 Mark)(iii) Range of R = {1, 2, 3, 4, 5, 6} (1 Mark)

Q.13 In any triangle ABC, prove that:2 2 2 2 2 2

2 2 2

b c c a a bsin2A sin2B sin2C 0

a b c

+ + =

Sol.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

b c c a a bLHS sin2A sin2B sin2C

a b c

b c c a a b(2sinAcosA) (2sinBcosB) (2sinCcosC) [1]a b c

b c b c a c a a c b a b2(ka) 2(kb) 2(kc)

a 2bc b 2ac c

= + +

= + +

+ + = + +

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

2 2 2

2 2 2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2

4

b a c[2]

2ba

Using sine and cosine laws

b c b c a b c c a c a b c ak

abc a b a b c a b

kb

abc

+

+ + + = + +

=4 2 2 2 2 4 4 2 2 2 2 4 4 2 2 2 2c a b a c c a b c b a a b c a c b

0 [1]

RHS

+ + + + +

=

=

Q.14 Represent the complex number z = 1 + i 3 in the polar form.OR

Find the square root of the complex number 3 + 4 i.

Sol.Let 1 = r cos , 3 = r sin (1 Mark)By squaring and adding, we get,r(cos + sin) = 4, r = 2 (1 Mark)

1 3cos , sin2 2 = = , so = /3 (1 Mark)


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Therefore, required polar form is z 2 cos isin3 3

= +

(1 Mark)

OR

( ) ( )

2 2

2 2

2 22 2 2 2 2 2

Let 3 4i x iy 3 4i x y i(2xy)

By comparing, x y = 3 ... (1)

and 2xy = 4 ... (2) [1]

Now x y x y 4x y

+ = + + = +

+ = +

2 2

2 2

2

9 16 25

x y 5 ... (3) [1]

Adding (1) and (3), we get2x 8 x 4 x 2

from (2), y 5 4 1 y 1

= + =

+ =

= = =

= = = [1]

But 2xy = 4 >0, therefore

x=2, y=1 or x =-1 or y=-1

Hence 3 4i 2 i or -2 -i [1]+ = +

Q.15 Find n given that n-1P3 :nP4 = 1: 9

Sol.

( ) ( )

( )

( )

n 1 n3 4P : P 1: 9

(n 1)! n! 1: (1 Mark)

n 4 ! n 4 ! 9

n 4 !(n 1)! 1(1 Mark)

n 4 ! n! 9

1 1

n 9

=

=

=

= (1 Mark)

n 9 (1 Mark) =

Q.16 Solve the given equation 2 cos2x + 3 sin x = 0OR

Prove thatcos 4x cos3x cos2x

cot3xsin4x sin3x sin2x

+ +=

+ +

Sol.


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2

2 2

2

2cos x+ 3 sin x = 0

Using cos x = 1-sin x

2sin x - 3 sinx - 2 =0 (1 Mark)(sin x - 2)(2sinx +1) =0 (1 Mark)

sin x = 2 (not possibl

n

1e) and sin x = - (1 Mark)

2

sinx sin sin( )6 6

7sinx sin (1 Mark)

67

x n ( 1)6

= = +

=

= +

ORcos4x cos3x cos2x

LHS =sin4x sin3x sin2x

(cos4x cos2x) cos3x= (1 Mark)

(sin4x sin2x) sin3x

2cos3xcosx cos3x

2sin3xcosx sin3x

+ +

+ +

+ +

+ +

+=

+(1 Mark)

cos3x(2cosx 1)(1 Mark)

sin3x(2cosx 1)

cot3x (1 Mark)

+=

+

=

Q.17 Draw the graph of function |x+2|-1.

Sol.


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Graph of function |x+2|-1

x 0 -2 -1 1 2

f(x) 1 -1 0 2 3

Q.18 The income of a person is Rs. 3, 00, 000, in the first year and hereceives an increment of Rs.10, 000 to his income per year for thenext 19 years. Find the total amount, he received in 20 years.

Sol.The sequence is an A.P. with a = 3,00,000, d = 10,000, andn = 20

(1 Mark)

Using the sum formula, we get,S20 =

20

2[600000 +19 10000] (1 Mark)

= 10 (790000) = 79,00,000. (1 Mark)Hence, the person received Rs. 79,00,000 as the total amount atthe end of 20 years. (1 Mark)


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Q.19 Insert three numbers between 1 and 256 so that the resultingsequence is a GP

OR

Prove that11 1

82 46 .6 .6 .... 6= Sol.

Let G1, G2, G3 be three numbers between 1 and 256 such that(1 Mark)

1, G1, G2,G3 ,256 is a G.P.Now 1 is the I term and 256 is the v term of the GPLet r be the common difference of the GP256 = r4 giving r = 4 (Taking real roots only) (1 Mark)For r = 4, we have G1 = ar = 4, G2 = ar

2 = 16, G3 = ar3 = 64

(1 Mark)Similarly, for r = 4, numbers are 4,16 and 64.Hence, we can insert, 4, 16, 64 or 4, 16, 64, between 1 and 256so that the resulting sequences are in G.P. (1 Mark)

OR

1 1 1 11 1...

8 2 4 82 4LHS = 6 .6 .6 .... 6 (1 Mark)

1 1 1Consider, ...

2 4 8

1 1Which is GP, with a = and r=2 2

+ + +

=

+ + +

1

(1 Mark)

1a 2S 1 (2 Marks)

11 r 12

Hence, LHS = 6 6 RHS

= = =

= =

Q.20 Find the equation of the circle which passes through the points

(2,-2), and (3, 4) and whose centre lies on the line x + y = 2.

Sol.Let the equation of the circle be (x h)2 + (y k)2 = r2Since the circle passes through (2, 2) and (3,4), we have,(2 h)2 + (2 k)2= r2... (1)and (3 h)2 + (4 k)2 = r2... (2)Also since the centre lies on the line x + y = 2, we have,h + k = 2 ... (3). (1 Mark)Solving the equations (1), (2) and (3), we get,h = .7, k = 1.3 and r2 = 12.58 (2 Marks)

Hence, the equation of the required circle is(x .7)2 + (y 1.3)2 = 12.58. (1 Mark)


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Q.21 Solve 5 x + x + 5 = 0.

Sol.Given equation is , 5 x + x + 5 = 0.a = 5 b = 1 c = 5

discriminant of the equation is

b2 - 4ac= 1 - 4 x 5 x 5 = 1 20 = -19 (2 Marks)

Therefore, the roots are1 19 1 19i 1 19i

,2 5 2 5 2 5

+ = (2 Marks)

Q.22 Find the rth term from the end in the expansion of (x + a)n

ORFind the coefficient of x6y3 in the expansion of (x + 2y)9

Sol.

There are (n + 1) terms in the expansion of (x + a)n (1 Mark)Observing the terms we can say that the first term from the end isthe last term,i.e., (n + 1)th term of the expansion and n + 1 = (n + 1) (1 1).The second term from the end is the nth term of the expansion, andn = (n + 1) (2 1).The third term from the end is the (n 1)th term of the expansionandn 1 = (n + 1) (3 1) and so on.

Thus rth

term from the end will be term number (n + 1) (r 1) =(n r + 2)th of the expansion. (2 Marks)the (n r + 2)th term is nC n r + 1 x

r-1 a n r + 1 (1 Mark)

OR

Tr + 1 =9Cr x

9 r(2y)r = 9Cr2r . x9 r . yr (1 Mark)

Comparing the indices of x as well as y in x6y3 and in T r + 1, we getr = 3.Thus, the coefficient of x6y3is (1 Mark)9 3 3 3

3

9! 9 8 7C 2 .2 .2 6723!6! 3.2

= = = 2 Marks)

SECTION C

Q.23 Prove that9 3 5

2cos cos cos cos 013 13 13 13

+ + =

Sol.


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Consider L.H.S:9 3 5

2cos cos cos cos13 13 13 13

+ +

9 9 3 5

cos cos cos cos13 13 13 3 13 13

= + + + +

(1 Mark)10 8 3 5

cos cos cos cos13 13 13 13

= + + + (2 Marks)

3 5 3 5cos cos cos cos

13 13 13 13

= + + +

(2 Marks)

5 5 3 5cos cos cos cos 0

13 13 13 13

= + + = = RHS (1 Mark)

Thus LHS=RHS

Q.24 Find the solution region for the following system of inequations:

x + 2y 10, x + y 1, x - y 0, x 0, y 0

Sol.Given inequations:

x + 2y 10, x + y 1, x - y 0, x 0, y 0,

Consider the corresponding equations x + 2y = 10, x + y = 1 andx y = O. (1 Mark)On plotting these equations on the graph, we get the graph asshown.Also we find the shaded portion by substituting (0, 0) in the inequations. (2 Mark)

(3 Marks)

Q.25 Find the mean deviation about the mean for the followingcontinuous frequency distribution, using short cut method forfinding mean

Marks Obtained Number of Students

0 - 10 1210 - 20 1820 - 30 27


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30 - 40 2040 - 50 1750 - 60 6

ORThe scores of 48 children in an intelligence test are shown in the following

frequency table .Calculate the variance 2 and find out the percentage of

children whose scores lie between x and x +

Score Frequency71 476 379 483 586 689 592 497 4101 3103 3107 3110 2114 2

Sol.

Marks Obtained Number of Students

0-10 1210-20 1820-30 2730-40 2040-50 1750-60 6

Let assumed mean =a = 25xi fi

di =

ix a

h

fi di

5 12 -2 -2415 18 -1 -1825 27 0 035 20 1 2045 17 2 3455 6 3 18

Tota100

30

n

i i

i 1

f d

x a hN== +


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30 1025 28

100

= + =

(2 Marks)

xi fi id

fii

d

5 12 23 27615 18 13 23425 27 3 8135 20 7 14045 17 17 28955 6 27 162

(2 Marks)n n n

i i i i ii 1 i 1 i 1

n n

i i ii 1 i 1

n n

i ii 1 i 1

f = 100 x xf d f 1182

x x f d f 1182

M.D (x) 11.82100

f f

= = =

= =

= =

= =

= = = =

(2 Marks)

OR

.

xi fi di= xi-a di2 fi di fi di

2

71 4 -19 361 -76 144476 3 -14 196 -42 58879 4 -11 121 -44 48483 5 -7 49 -35 24586 6 -4 16 -24 9689 5 -1 1 -5 592 4 2 4 8 1697 4 7 49 28 196101 3 11 121 33 363103 3 13 169 39 507107 3 17 289 51 867

110 2 20 400 40 800114 2 24 576 48 1152Total 48 21 6763

(2 Marks)

Mean:

n

i ii 1

n

ii 1

fd

x a

f

=

=

= +


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21x 90 90 0.44 90.44

48= + = + =

(1 Mark)

Variance:

2n n2

i i i i2 i 1 i 1

n n

i ii 1 i 1

f d f d=

f f

= =

= =

( )

22 6763 21

48 48

140.896 .191 140.705 nearly (1Mark)

Hence, 140.705 11.86

x 90.44 11.86 78.58

andx 90.44 11.86 102.30 (1

=

= =

= =

= =

+ = + = Mark)

The number of observations which lie between x and x + = 4 + 5 + 6 + 5 + 4 + 4 + 3 = 31

Percentage of these observations =31

100 64.58(nearly)48

=

(1 Mark)

Q.26 Find the equations of the lines through the point (3, 2) which are at

an angle of 45 with the line x -2y = 3.Sol.Let line through (3,2) be y - 2 =m(x- 3)... (i)

Slope of line x - 2y =3 is1

2.

Now,

( )

1m 2m 12tan 45 1

m 2 m12

= =+

+

(2 Marks)

Case I: 2m 1 1 2m 1 2 m2 m

= = ++

, so m = 3 (2 Marks)


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Equation of line is y - 2 = 3 (x - 3).Therefore 3x - y - 7 = 0 is the required equation

Case II: 2m 1 1 2m 1 2 m2 m

= = +

, 3m = - 1

m =1

3

Now the equation is1

y 2 (x 3)3

=

3y-6 =-x + 3x + 3y - 9 = 0 (2 Marks)

Q.27 Find the derivative using the first principle of f(x), where f (x) is

given by 1f(x) xx

= +

Sol.The function is not defined at x = 0.At all other points we have,

h 0 h 0

1 1x h x

f(x h) f(x) x h xf '(x) lim lim

h h

+ + +

+ + = = (2 Marks)

h 0

1 1 1lim h

h x h x

= + +

(1 Mark)

h 0 h 0

2h 0

1 x x h 1 1lim h lim h 1 (2 Mark)

h x(x h) h x(x h)1 1

lim 1 1 (1 Mark)x(x h) x

= + =

+ +

= = +

Q.28 For all n 1, prove using Principle of Mathematical Induction

1 1 1 1 n

1.2 2.3 3.4 n(n 1) n 1+ + + + =

+ +

Sol. Let1 1 1 1 n

P(n) :1.2 2.3 3.4 n(n 1) n 1

+ + + + =+ +

P(1):1 1

1.2 1 1=

+, which is true. Thus P(n) is true for n = 1.

(1 Mark)Assume that P(k) is true for some natural number k,

P(k):1 1 1 1 k

1.2 2.3 3.4 k(k 1) k 1+ + + + =

+ +(1 Mark)

We need to prove that P(k + 1) is true whenever P(k) is true.We have, P(k+1) =


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1 1 1 1 1 (k 1)

1.2 2.3 3.4 k(k 1) (k 1)(k 2) (k 2)

++ + + + + =

+ + + +(1 Mark)

LHS of P(k+1) =

1 1 1 1 1

1.2 2.3 3.4 k(k 1) (k 1)(k 2)

+ + + + +

+ + + (1 Mark)

k 1

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

+ + +=

2k 2k 1 k 1

(k 1)(k 2) k 2

+ + +=

+ + += RHS (1 Mark)

So P(k+1) is true whenever P(k) is trueSo the result holds for all natural numbers. (1 Mark)

Q.29 One card is drawn from a well shuffled deck of 52 cards. If each

outcome is equally likely, calculate the probability that the card willbe(i) A diamond(ii) An ace(iii) A black card(iv) not a diamond(v) not a black card.(vi) not an ace

ORA committee of two persons is selected from two men and two women.What is the probability that the committee will have (a) no man? (b) one

man?(c) two men?Sol.

When a card is drawn from a well shuffled deck of 52 cards, thenumber of possible outcomes is 52.(i) Let A be the event 'the card drawn is a diamond'Clearly the number of elements in set A is 13.Therefore, P(A) = 13/52= 1/4i.e. Probability of a diamond card = 1/4 (1 Mark)(ii) We assume that the event Card drawn is an ace is BClearly the number of elements in set B is 4.So P(B)= 4/52 = 1/13

(1 Mark)

(iii) Let C denote the event card drawn is black cardTherefore, number of elements in the set C = 26i.e. P(C) = 26/52= Thus, Probability of a black card = (1 mark)(iv) We assumed in (i) above that A is the event card drawn is adiamond,so the event card drawn is not a diamond may be denoted as A' or

not ANow, P(not A) = 1 P(A) = 1 -1/4 = (1 Mark)

(v) The event card drawn is not a black card may be denoted as C

or not C.We know that P (not C) = 1 P(C) = 1 1/2 =1/2.


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Therefore, Probability of not a black card = 1/2 (1 Mark)(vi)Card drawn is not an ace should be B. from (ii)We know that P(B) = 1 P(B) = 1-1/13 = 12/13

Therefore the probability that a Card drawn is not an ace = 12/13(1 Mark)OR

The total number of persons = 2 + 2 = 4. Out of these four person,two can be selected in 4C2 ways.(a) No men in the committee of two means there will be twowomen in the committee.Out of two women, two can be selected in 2 C2 =1 way

2

24

2

C 2 1P(no man)

4 3 6C

= = =

(2 Marks)

(b) One man in the committee means that there is one woman. Oneman out of 2 can be selected in 2C1 ways and one woman out of 2can be selected in2C1 ways.Together they can be selected in 2C1

2C1 ways.2 2

1 14

2

C C 2 2 2P(one man)

2 3 3C

= = =

(2 Marks)

(c) Two men can be selected in 2 C2 ways2

24 4

2 2

C1 1P(Two men) 6C C

= = = (2 Marks)

1321619488 microsoftword-classxi math topper sample paper 4 r0

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