CODE: 7KKP2HP11MMG191
KENDRIYA VIDYALAYA SANGATHAN CHANDIGARH REGION
Class – XI
Subject – Mathematics
MARKING SCHEME
TIME:3hrs.
Instructions: 1.Evaluate the answer sheet as per the marking scheme.
2.For any alternative answer marks to be awarded.
MAX.MARKS:100
SECTION-A [1 MARK QUESTIONS]
1
2
3
4
1 mark for correct answer
Relation R = { ( 0,5), (1,6) ,( 2,7), (3,8),(4,9) ,(5,10) }
Range = {5, 6, 7, 8, 9, 10}
1
1
3
2
1
0+i
1
-
SECTION-B [2 MARK QUESTIONS].
5
6
1/2
2
½
2
2 (1-sin x) + 3 sin x = 0
1
sin x = 
sin x= 2
2
sin x = 2
not possible
7
x =
6
½
½
½
7
(i) = (ii) = (iii)=∅, (iv)= {5}.
8
The contrapositive statement of ‘p’ is “For any real numbers
x, y if 2x + a ≠ 2y + a, where a  Z, then x ≠ y.
Given 2x + a ≠ 2y + a
 2x ≠ 2y
 x≠y
Sn=3n2 + 5n ,
9
1/2 each
n=1, S1= 8,
n= 2, S2= 22, a1= S1, a1+a2=22
8+a2 =22,
a2= 14, d=6.
am= a+(m-1)d , m= 27.
1
1
1
1
10
a
, a, ar be three terms of GP
r
Let
a
13
3
4
+ ar+a =
, a = -1 , r =
or
r
12
4
3
4
3
3
4
, -1,
or , , -1,
3
4
4
3
G.P.
1 + 3𝑖 1 + 2𝑖
−5 + 5𝑖
×
=
1 − 2𝑖 1 + 2𝑖
5
11
= -1+i
=√2 (cos
12
1
3𝜋
4
+ 𝑖 cos
3𝜋
4
1
1
½
)
½
1 − 𝑐𝑜𝑠2𝑥
1 − 𝑐𝑜𝑠𝑥
lim
𝑥→0
= lim
2𝑠𝑖𝑛2 𝑥
½
𝑥
𝑥→0 2𝑠𝑖𝑛2 2
𝑠𝑖𝑛𝑥
=
lim ( 𝑥 ) 2
𝑥→0
1
𝑥
𝑠𝑖𝑛
( 𝑥2) 2
2
1×4
=
1
½
=4
SECTION-C [4 MARK QUESTIONS]
13
3+2𝑖𝑠𝑖𝑛𝜃
x
1+2𝑖𝑠𝑖𝑛𝜃
1
1−2𝑖𝑠𝑖𝑛𝜃
1+2𝑖𝑠𝑖𝑛𝜃
3+6𝑖𝑠𝑖𝑛𝜃+2𝑖𝑠𝑖𝑛𝜃−4𝑠𝑖𝑛2𝜃
=
=
1+4𝑠𝑖𝑛2 𝜃
8𝑖𝑠𝑖𝑛𝜃
( 3−4𝑠𝑖𝑛2𝜃)
1+4𝑠𝑖𝑛2 𝜃
+
1+4𝑠𝑖𝑛2 𝜃
1
If it is purely imaginary number than real part must be zero
∴
3−4𝑠𝑖𝑛2 𝜃
1+4𝑠𝑖𝑛2 𝜃
=0
3-4𝑠𝑖𝑛2 𝜃 = 0
4𝑠𝑖𝑛2 𝜃 = 3
√3
Sin𝜃 = 2
1
𝜋
𝜃 = 𝑛𝜋 + ( −1)𝑛 3 , 𝑛 ∈ I
14
Consider A=A∩(AUX)
= A∩(BUX) , Then proving A is subset of B
Again consider B=B∩(BUX)
= B∩(AUX) ,Then proving B is subset of A
Combining the result
OR
(i) Let A = {1, 2}, B = {2 , 3}, C = {2 , 4} or any other example
A ∩ B = {2}
A ∩ C = {2} but B  C [ for proof]
(ii) A  B  A  B  A  (B  B)
=A  A
15
1
2
1.5
½
2
2
f(x) is defined if (x-1)(x-2)  0
domain Df = (,1 2, 
To find range f(x) = y
1
 x 2  3x  (2  y2 )  0
1
x is real if y  0,
R f  0 ,  
1
1
16
1
1
1
17
n=1 , supposing n=k, proving for n=k
1
½+1+2
Hence by using the principle of mathematical induction statement is true for
alln ∈ N .
18
Correct Graph with regions
½
3
1
Showing feasible solution
19
Solving the equations we get,
x =-5/22
Line parallel to x- axis is of the form x=x1
Equation of line is x=-5/22 or 22x+5=0
OR
Let C divides the join of A(1 , 0) and B(2 , 3) in the ratio 1: n
3 
2n
 coordinate s of C are 
,

 1 n 1 n 
30
Slope of AB 
3
2 1
1
Slope of required line  
3
3
1
2n
Equation of required line is y 
  x 

1 n
3
1 n 
i.e. 1  n x  31  n y  n  11
20
To write the eq. in standard form
2
1
1
1
1
1
1
½
1
½
½
½
½
½
Or
2
½
2
Let eq. of circle x +y +2gx+2fy+c=0
8g+2f+c=-17, 12g+10f+c=-16 subtracting these we get
g+2f=-11
Also the Centre (-g, -f) lies on the line
Therefore
Eq. of circle
4g+f=-16
4x+y=11
, f= -4 , g= -3 ,c= 15
1
x2 + y2 + 2(-3) x+ 2 (-4) y +15=0
x2 + y2 -6 x-8 y +15=0
21
½
½
½
1
Let the YZ planedivide the line segment joining points (–2, 4, 7) and (3, –5, 8)
in the ratio k:1.Hence, by section formula, the coordinates of point of
1
intersection are given by
On the YZ plane, the x-coordinate of any point is zero.
1
1
Thus, the YZ plane divides the line segment formed by joining the given points
in the ratio 2:3.
1
22
𝐿𝑒𝑡 𝑦 = 𝑡𝑎𝑛𝑥
𝑑𝑦
𝑑𝑥
= lim
tan(𝑥+ℎ)−𝑡𝑎𝑛𝑥
1
ℎ
h→0
𝑆𝑖𝑛(𝑥+ℎ)𝑐𝑜𝑠𝑥−cos(𝑥+ℎ)𝑠𝑖𝑛𝑥
= lim
=lim
h→0
𝑠𝑖𝑛ℎ
h→0 ℎ
lim
ℎ𝑐𝑜𝑠(𝑥+ℎ)𝑐𝑜𝑠𝑥
1
h→0 𝑐𝑜𝑠(𝑥+ℎ)𝑐𝑜𝑠𝑥
=1/cos2x = sec2x
23
52
Total no. Of possible hands = C7
1
1
1
1
(i ) No. Of hands with 4 kings = 4C4 ×48C3
1
1
P(a hand will have 4 kings ) = 7735
9
(ii)
P(3kings) =1547
1
P(atleast three kings )= P (3kings or 4 kings )
9
1
46
1
=1547 +7735 = 7735
SECTION-D [6 MARK QUESTIONS]
24
For labeled venn diagram
2
i) 52
ii) 30
25
a cos A  b cos B  k sin A. cos A  k sin B. cos B
 sin 2A  sin 2B  sin 2A  sin 2B  0
 2 cosA  B.sin A  B  0

  is right angled .
2
sin A  B  0  A  B  0   is isosceles .
cosA  B  0  A  B 
26
(i)Words start with P and end with S =
2
2
2
1
1
1
1
10!
2!
2
= 18154400
8!
Vowels are together = × 5!
2!
= 2419200
There are always 4 letters between P and S
(ii)
(iii)
10!
=
2!
×25401600
2
2
OR
There will be as many ways of choosing 4 cards from 52 cards as there are
combinations of 52 different things, taken 4 at a time
∴ Required number
52!
of ways = 52c4 =4!48! =270725
1
(i)
Required number of ways =13c4+13c4 + =13c4+13c4.
13!
=4× 4!9! = 2860
1
(ii)
Required number of ways =13c1×13c1×13c1×13c1=134
1
(iii)
There are 12 face cards and 4 are to be selected out of these 12 cards.
12!
Required number of ways =4!8! =495
1
There are 26 red cards and 26 black cards. Required number of ways =
26
c2×26c2=105625
1
4 red cards can be selected out of 26 red cards in 26c4. 4 black cards can
be selected ou5t of 26 black cards in 26c4 ways. Required number of
ways = 26c4 + 26c4=29900
1
(iv)
(v)
27
.𝑇𝑟−1 = 𝑇𝑟−2+1 = 𝑛𝑐𝑟−2 𝑥 𝑛−𝑟+2 1𝑟−2
𝑇𝑟 = 𝑇𝑟−1+1 = 𝑛𝑐𝑟−1 𝑥 𝑛−𝑟+1 1𝑟−1
𝑇𝑟+1
= 𝑛𝑐𝑟 𝑥 𝑛−2 1𝑟
Coefficients are 𝑛𝑐𝑟−2 , 𝑛𝑐𝑟−1 ,𝑛𝑐𝑟 .
Coefficients are in the ratio is 1: 3: 5
𝑛𝑐𝑟−2
:
𝑛𝑐𝑟−1 : 𝑛𝑐𝑟 = 1 : 3 : 5
∴
𝑛𝑐𝑟−2
=
1
𝑛𝑐𝑟−1
𝑛𝑐𝑟
=
3
2
5
3 𝑛𝑐𝑟−2 = 𝑛𝑐𝑟−1 and 5 𝑛𝑐𝑟−1 = 3 𝑛𝑐𝑟
𝑛!
3 (𝑛−𝑟+2)!(𝑟−2)! =
3
( 𝑛−𝑟+2)(𝑛−𝑟+1)!(𝑟−2)!
And
1
𝑛!
(𝑛−𝑟+1)!(𝑟−1)!
=
1
( 𝑛−𝑟+1)!(𝑟−1)(𝑟−2)!
3r -3 = n – r+2
n -4r + 5= 0
5 𝑛𝑐𝑟−1 = 3
5
5
(𝑛−𝑟+1)(𝑛−𝑟)!(𝑟−1)!
𝑛𝑐𝑟
𝑛!
(𝑛−𝑟+1)!(𝑟−1)!
3
=
1
-(1)
=
3 𝑛!
(𝑛−𝑟)!𝑟!
(𝑛−𝑟)!𝑟(𝑟−1)!
5r = 3n-3r+3
3n-8r+3=0
On solving (1) &(2) we get n = 7 , r = 3
(2)
1
1
2
OR
(1 + 2𝑎)4= 4𝑐0 + 4𝑐1 (2a)2 + 4𝑐3 (2𝑎)3 + 4𝑐4 (2𝑎)4 .
= 1+ 4(2a) + 6 (4𝑎2 ) + 4 (8𝑎3 ) + 16𝑎4
= 1 + 8a + 24𝑎2 + 32𝑎3 + 16𝑎4
and (2 − 𝑎)5 = 5𝑐0 (2)5 - 5𝑐1 (2)4 (a) + 5 (2)3 (𝑎)2 - 5𝑐3 (2)2 (𝑎)3
+5𝑐4 (2) (𝑎)4- 5𝑐5 𝑎5
28
= 32 – 80 a + 80𝑎2 - 40𝑎3 + 10𝑎4 − 𝑎5
Thus product (1+2a)4 (2 − 𝑎)5 =
(1+8a+24𝑎2 + 32𝑎3 + 16𝑎4 ) (32-80a+80𝑎2 − 40𝑎3 + 10𝑎4 − 𝑎5 )
The coefficient of 𝑎4 is
1(10𝑎4 ) + 8a (-40𝑎3 ) + (24𝑎2 )(80𝑎2 ) + (32𝑎3 )(−80𝑎) + 16𝑎4 (32) =
−438𝑎4
Coefficient of 𝑎4 is - 438.
2
  k (k  1) 2 
 
 
2
13  23  33  ...  k 3
2   (k  1)


tk 


1  3  5  ...  2k  1)   k 2  2(k  1
4
2



3
Sn 

29
2
n
 tk 
k 1
n 
1 n 2 n
k

2
k


1

4 k 1
k 1
k 1 
2
1  n n  12n  1
n n  1
 n
2
 n 
2n 2  9n  13

4
6
2
24



Number of observations (n) = 100
Incorrect mean
Incorrect standard deviation (σ) = 3
∴ Incorrect sum of observations = 2000
⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940
1+1
6
∴ Correct Standard Deviation