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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