KENDRIYA VIDYALAYA SANGATHAN

REGIONAL OFFICE VARANASISESSION ENDING EXAMINATION 2014-15

CLASS: XISUB: MATHS

TIME: 3Hrs Max Marks- 100

General Instructions:-

I. All questions are compulsory.II. The question paper consists of 26 questions divided into three sections A, B

and C. Section A comprises 6 questions of 1 mark each, section B comprises 13 questions of four marks each and section C comprises 7 questions of six marks each.

III. All questions in section A are to be answered in one word, one sentence or as per the exact requirement of the question.

IV. There is no overall choice. However an internal choice has been provided in 4 questions of four marks each and 2 questions of six mark each. You have to attempt only one of the alternatives in all such questions.

V. Use of calculator is not permitted.

(Section A)

Q:1 If A and B are two sets such that A B ,then what is A∩ B ?

Q:2 Find the sum to infinity of the G.P.

1+ 12+122

+ 123

+…

Q:3 Evaluate limx→0

sin 2 x3x .

Q: 4 Write the negation of the statement

“Sum of 2 and 5 is equal to 8”

Q: 5 Identify the quantifier in the following statement

“There exists a capital for every state in India”

Q:6 Write the contrapositive of the following statement

“If x is an even number implies that x is divisible by 4.”

(Section B)

Q:7 If P(x)= p(y),show that x=y.

Q:8 If R={(a,b):a,b ∈ W,2a+b=8} then find domain and range of R.

Q:9 The function of f is defined by

f(x)¿{1−x ,x<01, x=0x+1, x>0

Draw the graph of f(x).

Q:10 Prove that cos7 x+cos5 xsin 7 x−sin 5x =cot x.

Q:11 Using mathematical induction to prove that 32n+2−8n−9 is divisible by 8.

OR

Using mathematical induction to prove that

1.3+2. 32+3.33+…n. 3n = (2n−1 ) 3n+1+34

Q:12 Convert the complex number 1+7 i(2−i)2 in polar term.

OR

Find the square root of -7-24i.

Q:13 Solve the following system of inequalities graphically

x+2y≤10x+y≥1x-y≤0

x≥0y≥0

Q:14 If 22pr+1 : 22pr+2 = 11:52. Find r

OR

A boy has 3 library tickets and 8 books of his interest in the library. Of these 8, he does not want to borrow mathematics part II, unless mathematics part I is also borrowed. In how many ways can he choose the three books to be borrowed?

Q15: How many words with or without meaning each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER.

Q16: If S be the sum, P the product and R the sum of reciprocals of n terms of a G.P, prove that p2Rn = Sn

Q17: If P and q are length of the perpendicular from the origin the lines

x cosθ – y sinθ = k cos 2θ and x secθ + y cosecθ = k respectively, prove that p2 + 4q2= k2

OR

Show that the product of perpendicular on the line xacos θ+ yb sin θ=1 from

the points (±√a2−b2 ,0) is b2.

Q:18 Find the equation of the set of points P ,the sum of whose distances from A(4,0,0)and B(-4,0,0)is equal to 10.

Q: 19 Find the derivative of f(x)=x sin x from first principle.

. (Section C)

Q:20 In a survey it is found that 105 people take tobacco ,130 take cannabis and 145 take opium. If 70 people take tobacco as well as cannabis ,75 take cannabis as well as opium , 60 take tobacco as well as opium and 40 take all three, find how many people are surveyed? How many take opium only?

All the three are dangerous for health, what should be done to get rid of such harmful products?

Q:21 The angle of elevation of the top point A of the vertical tower AB of height h from a point C is 450 and from a point D ,the angle elevation is 600 where D

is a point at a distance ‘d’ from the point C measured along the line CD which makes an angle 300 with CB.

Then Prove that d=h(√3 -1).

Q:22 Show that the middle term in the expansion of (1+x)2n is

1.3.5… (2n−1)n!

.2n xn

Q23 Find the sum to n terms of series: 5+11+19+29+41+…

OR

Let x be the arithmetic mean and y, z be two geometric means between any two positive numbers. Then prove that

y3+z3

xyz=2

Q:24 Find the co-ordinates of the foci the vertices , the lengths of major and minor axis and eccentricity of the ellipse 9x2+4y2=36

Q:25 If 4 digit numbers greater than 5000 are randomly formed from the digit 0,1,3,5 and 7. What is the probability of forming a number divisible by 5 when

I. The digit may be repeatedII. The repetitions of digits is not allowed.

Q:26 Calculate the standard deviation for the following distribution giving the age distribution of persons:

Age in years 20-30 30-40 40-50 50-60 60-70 70-80 80-90Number of persons

3 61 132 153 140 51 2

Or

The mean of 5 observations in 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.

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KENDRIYA VIDYALAYA SANGATHANREGIONAL OFFICE VARANASI

MARKING SCHEME – SESSION ENDING EXAMClass XI

Sub MathsTime 3hrs Max.Marks 100

(Section A)

Q1 A

Q2 2

Q3 2/3

Q4 Sum of 2 and 5 is not equal to 8.

Q5 There exists

Q6 If x is not divisible by 4 then x is not an even number.

(Section B )

Q 7 For all x∈ X⇒ {x }∁ x⇒ {x }ϵP (x)

⇒{x}∈P ( y )(∴P ( x )=P( y))

={x}Y

=x∈Y 2 Mark

=x Y

Similarly Y X 121Mark

Thus X=Y 1/2Mark

Q8 Domain R={0,1,2,3,4} 2 Mark

Range R={8,6,4,2,0} 2 Mark

Q9 Drawing x+1, 2Marks

Drawing x+1, 2Marks

Q10 LHS 2cos 7 x+5 x

2.cos 7 x−5 x

2

2cos 7 x+5 x2

. sin 7 x−5 x2

3

Mark

=cos xsin x = cotx 1 Mark

Q11 Let p(n) be the statement given by

Y’

X’x

yF(x)=1-x f(x)=x+1

21

-1-2-3

1 2 3 -3 -2 -1

P(n):32n+2-8n-9 is divisible by 8.

For p(1) to be true 1 Mark

Assuming P(k) 1 Mark

Proving p(k+1) is true 2 Mark

Or

Let p(n)be the statement given by ½ Mark

P(n): 1.3+1.23+1.33+………….n.3n= (2n−1 )3n+1+34 .1 1 Mark

For p(1) to be true.

For p(k)true ⇒ p(k+1)true 2 Mark

As p(1)hold and p(k)hold ⇒p(k+1)holds. Hence from PMI proved ½ Mark

Q12 we have

1+7 i(2−i)2

= 1+7 i3−4 i

=-1+i 1 Mark

Let z=-1+i

r=√2 1 Mark

Principal argument =3 π/4 1 Mark

Polar form =√2(cos3π4 +I sin3π4 ) 1Mark

Or

Let x+iy=√−7−24 i 1/2Mark

Then (x+iy)2= -7-24i X2-y2 =-7 121Mark

2xy=-24(x2+y2)2= (x2-y2)2+(2xy)2

x2+y2 =25 1Mark

x=±3 y=±4

Product xy is negative, we have 1/2Mark

X=3,y=-4 or x=-3,y=4 1/2marks

Square root are 3-4i or -3+4i 1/2marks

Q13 Drawing graph 2 marks

Shading solution 2 marks

Q14 Have 22pr+1 : 20pr+2 = 11 : 52

⇒ 22!(21−r) ! :

20 !(18−r )! = 11:52 1Mark

X+y=1

5 4 3 2 1

X+2y=10

Y’

x' x

y

1 2 3 4 5 6 7 8 9 10 X-y=0

⇒ 22×21

(21−r ) (20−r )(19−r) =1152 1Mark

⇒ (21−r ) (20−r )(19−r ) =12×13×14

⇒ (21−r ) (20−r )(19−r ) =(21−7 ) (20−7 )(19−7)

⇒ r=7

Or

Required number of ways

6c1+7c3

= 6+7×6×53×2×1

= 41

Q15 Required number of words

3c2×5c3×5!

Q16 set a be the first term and r be the common ratio of given GP

S=a+ar+ar2+………… n term =1−rn

1−r

P= a.ar………………arn-1=anrn(n−1)2

R= 1a+ 1ar +…………+1

arn−1

=1a .⦋1−( 1

r)n

1−1r

⦌

R= 1−r n

a . rn−1(1−r )

LHS p2 Rn

Substituting the value of P & R solve it, and get the required result

2Mark

2Mark

1 Mark

1Markwrite combination 2 marks, permutation 1 marks

1Mark

1Mark

1Mark

1Mark

=sn RHS

Q17 P= length of perpendicular form (0,0) to x cosθ-ysinθ-k cos2θ=0

P=0.cosθ−0 sinθ−kcos2θ

√cos2θ+√sin 2θ P=k cos2θ

Similarly q=k sinθ×cosθ

Now p2+4q2

= k2cos22θ+4.k2sin2θ.cos2θ

=k2

Or

Similarly marks will be awarded.

Q18 ∵ PA+PB=10

√(x−4)2+ y2+z2+√(x+4)2+ y2+z2=10

√(x−4)2+ y2+z2=10−√(x+4)2+ y2+ z2

Squaring both sides and solve it we get

9x2+25y2+25z2=225

Q19 ∵ f’(x) =limh→0

f (x+h )−f (x)h

=limh→ 0

( x+h ) sin ( x+h )−xsinxh

=limh→ 0

xsin x (cosh−1 )h +lim

h→0xcosx sinh

h +limh→ 0(sinx .cosh+sinh .cosx)

=x cosx + sinx 1Mark

1Mark

1Mark

1Mark

1Mark

1Mark

1Mark

1Mark

1Mark

1Mark

1Mark

1Mark

(Section C )

Q20

As show in figure let a= no of people taking all the three =40

Also a+b=70

b =30 ½ marks

a+d=75

112for correct

Venn diagram

d =35

½ marks

a+c=60

c=20∵e+c+a+b=105

a +b+d+f=130

g+c+a+d=145

total people surveyed =215

All narcotics and other products which are harmful to health should be

banned strictly.

Q21

Applying Sine formula in ∆ ACD we get

e =15

f =25

g =50

½ marks

½ marks

½ marks

1 marks

1 marks for value based answer

D√2 hh hE C B A15⁰150⁰ 6030⁰15⁰45⁰

2 marks for correct & levelled diagram

C=20

d

ACsin 150⁰= CD

sin 15⁰= ADsin 15⁰ 2

marks

√2hsin 150⁰= d

sin 15⁰

d = sin 150∗√2h

sin 150⁰

d=(√3 -1)h

Q22 Middle term of the expansion

(1+x)2n is (n+1)th term

Tn+1 =2ncn xn

(2n)!n! n!

xn

= [1.3.5…. (2n−1 ) ] n !2n xnn!n !

[1.3.5…. (2n−1 ) ] 2n× xnn!

Q23 Let sn =5+11+19+29+……………………an-1+an

Or sn=5+11+19+……+an-1+an

Subtracting

0=5+[6+8+10+………….(n-1)term] -an

an=5+ (n−1 ) [12+(n−2 )∗2]2

an =n2+3n+1

Hence sn¿∑k=1

n

ak=∑k=1

n

(k2+3k+1) 1 marks

1 marks

1 marks

1 marks

2 marks

1 marks

1 marks

1 marks

1 marks

1 marks

1½ marks

=n (n+1 )(2n+1)6 +

3n(n+1)2 +n

=n (n+2 )(n+4)3 ½ marks

Or

Let a & b be any two positive number.

Then x=a+b2

∵ a ,y ,z,b are in GP

c.r=( ba)13

y=ar=a(ba¿¿

13 =b

13 .a

23

z=ar2=a.(ba )2/3 =b2/3.a1/3

y3+z3=ab(a+b)

yz=ab

∴ y3+z3

yz=a+b

y3+z3

yz=2 x

y3+z3

xyz=2

Q24 Equation of ellipse in standard form as

x2

4+ y

2

4 =1

e =√53 1 ½

marks

1 marks

1 marks

1 marks

1 marks

1 marks

1 marks

1 marks

foci →(0 , ±√5) 1 marks

vertices→(0 , ±3)

length of major axis =6

length of minor axis=4

Q25 (i)when repetition of digits is allowed

Number of 4 digit numbers greater

than 5000=250-1=249 1 marks

number of 4 digit number greater than 5000 and divisible by 5=100-1=99

required probability=99/249

(ii) when repetation of digits is not allowed

Number of 4 digit number greater than 5000=2×3×4×2=48

number of 4 digit number greater than 5000 and divisible by 5=12+6=18

required probability =1848=38

Q26

Age Class mark xi

Frequence fi

di x1−5510di

2 fidi fidi2

20-30 25 3 -3 9 -9 2730-40 35 61 -2 4 -132 24440-50 45 132 -1 1 -132 13250-60 55 153 0 0 0 060-70 65 140 1 1 140 14070-80 75 51 2 4 102 20480-90 85 2 3 9 6 18

542 -25 765

Here ∑ fi di2=765,∑ fi di❑=−25∑ fi=542 3 marks

1 marks

1 marks

1 marks

2 marks

1 ½ marks

1 marks

1 marks

S.D=h√∑ fi di2

∑ fi−(∑ fi di❑

∑ fi)2

¿10×√1.4093

=11.87

Or

Let other observation be x and y

Mean=1+2+6+x+ y5

X+y=13----------------------------------(1)

Variance=1n∑i=15

(xi−x)2

8.24=15[(3.4)2+(2.4)2+(1.6)2+x2+y2-2×4.4×(x+y)+2×(4.4)2]

X2+y2=97----------------------------------(2)

Solving x=9,y=4,x-y=5

X=4,y=9 x-y=-5

∴Remaining observation are 4& 9.

1 marks

3 marks

1 marks

1 marks

1 marks

1 marks

1 marks