Page 1
CBSE XI | Mathematics
Sample Paper – 3 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 3 Solution
SECTION – A
1. ? ? ? Since m C 90 ,therefore
a 18 3
sin A
c 30 5
? ? ?
2. ?? 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 = 8
Solving the two equations, we get a = 5, b = -2
a = 5, b = -2 also satisfy the other two ordered pairs
f(-2) = 5(-2) – 2 = -12 ? (-2, -12)
f(-1) = 5(-1) – 2 = -7 ? (-1, -7)
Therefore the values are a = 5 and b = -2.
3. f(x) =
2
2
x4
x 8x 12
?
??
For f(x) to be defined, x
2
- 8x + 12 must be non-zero i.e. x
2
- 8x + 12 ? 0
(x - 2)(x - 6) ? 0
i.e. x ? 2 and x ? 6
Therefore domain will be ? ? 2,6 R ?
So domain of f = R - {2, 6}
4. ( ? p ? ? q)
OR
There exists a living person who is 150 years.
SECTION – B
5. (x iy)(3 5i) 1 3i 1 3i ? ? ? ? ? ? ? ?
? ?
2
2
1 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 17
67
x ;y
17 17
? ? ?
? ? ? ? ? ?
? ? ? ? ?
? ? ?
?
? ? ? ? ? ?
? ? ? ? ?
?
? ? ? ?
Page 2
CBSE XI | Mathematics
Sample Paper – 3 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 3 Solution
SECTION – A
1. ? ? ? Since m C 90 ,therefore
a 18 3
sin A
c 30 5
? ? ?
2. ?? 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 = 8
Solving the two equations, we get a = 5, b = -2
a = 5, b = -2 also satisfy the other two ordered pairs
f(-2) = 5(-2) – 2 = -12 ? (-2, -12)
f(-1) = 5(-1) – 2 = -7 ? (-1, -7)
Therefore the values are a = 5 and b = -2.
3. f(x) =
2
2
x4
x 8x 12
?
??
For f(x) to be defined, x
2
- 8x + 12 must be non-zero i.e. x
2
- 8x + 12 ? 0
(x - 2)(x - 6) ? 0
i.e. x ? 2 and x ? 6
Therefore domain will be ? ? 2,6 R ?
So domain of f = R - {2, 6}
4. ( ? p ? ? q)
OR
There exists a living person who is 150 years.
SECTION – B
5. (x iy)(3 5i) 1 3i 1 3i ? ? ? ? ? ? ? ?
? ?
2
2
1 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 17
67
x ;y
17 17
? ? ?
? ? ? ? ? ?
? ? ? ? ?
? ? ?
?
? ? ? ? ? ?
? ? ? ? ?
?
? ? ? ?
CBSE XI | Mathematics
Sample Paper – 3 Solution
OR
(3 + 4i)(4 + i) = 12 + 3i + 16i + 4i
2
= 12 + 19i – 4 = 8 + 19i
22
z 8 19 64 361 425 5 17 ? ? ? ? ? ?
6. Length of pendulum is 36 cm long
Angle of oscillation = 10 degrees
180 degrees = radians
so, 10 degrees= radians
18
?= radians
18
?
?
?
?
So using this formula l = r? and substituting the values of r = 36, ? = radians
18
?
we get,
?=36 x
p
18
=2 x (3.14) = 6.28 cm
OR
Area of sector =
2
1
r
2
?
2
2
2
2
1
r 5.024
2
1 36
r 5.024
2 180
180 2
r 5.024
36 3.14
r 16
r 4 cm
??
? ? ?
?
??
?
?
?
7. Let a be the first term and d be the common difference
Tn =2n+1
a=3………. (T1)
T2 = 5
d = 2 ……… (T2 - a)
? ?
? ?
? ? ?
? ? ? ? ? ?
n
n
n
S 2a (n 1)d
2
19
S 2 3 (19 1) 2 399
2
Page 3
CBSE XI | Mathematics
Sample Paper – 3 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 3 Solution
SECTION – A
1. ? ? ? Since m C 90 ,therefore
a 18 3
sin A
c 30 5
? ? ?
2. ?? 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 = 8
Solving the two equations, we get a = 5, b = -2
a = 5, b = -2 also satisfy the other two ordered pairs
f(-2) = 5(-2) – 2 = -12 ? (-2, -12)
f(-1) = 5(-1) – 2 = -7 ? (-1, -7)
Therefore the values are a = 5 and b = -2.
3. f(x) =
2
2
x4
x 8x 12
?
??
For f(x) to be defined, x
2
- 8x + 12 must be non-zero i.e. x
2
- 8x + 12 ? 0
(x - 2)(x - 6) ? 0
i.e. x ? 2 and x ? 6
Therefore domain will be ? ? 2,6 R ?
So domain of f = R - {2, 6}
4. ( ? p ? ? q)
OR
There exists a living person who is 150 years.
SECTION – B
5. (x iy)(3 5i) 1 3i 1 3i ? ? ? ? ? ? ? ?
? ?
2
2
1 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 17
67
x ;y
17 17
? ? ?
? ? ? ? ? ?
? ? ? ? ?
? ? ?
?
? ? ? ? ? ?
? ? ? ? ?
?
? ? ? ?
CBSE XI | Mathematics
Sample Paper – 3 Solution
OR
(3 + 4i)(4 + i) = 12 + 3i + 16i + 4i
2
= 12 + 19i – 4 = 8 + 19i
22
z 8 19 64 361 425 5 17 ? ? ? ? ? ?
6. Length of pendulum is 36 cm long
Angle of oscillation = 10 degrees
180 degrees = radians
so, 10 degrees= radians
18
?= radians
18
?
?
?
?
So using this formula l = r? and substituting the values of r = 36, ? = radians
18
?
we get,
?=36 x
p
18
=2 x (3.14) = 6.28 cm
OR
Area of sector =
2
1
r
2
?
2
2
2
2
1
r 5.024
2
1 36
r 5.024
2 180
180 2
r 5.024
36 3.14
r 16
r 4 cm
??
? ? ?
?
??
?
?
?
7. Let a be the first term and d be the common difference
Tn =2n+1
a=3………. (T1)
T2 = 5
d = 2 ……… (T2 - a)
? ?
? ?
? ? ?
? ? ? ? ? ?
n
n
n
S 2a (n 1)d
2
19
S 2 3 (19 1) 2 399
2
CBSE XI | Mathematics
Sample Paper – 3 Solution
8. We have 5! = 5 × 4! And 6! = 6 × 5 × 4!
LCM of 4!, 5! and 6! = LCM of {4!, 5 × 4!, 6 × 5 × 4!} = 4! 6 × 5 = 6! = 720
OR
? ?
2 2
11
4 4i i
2i
?
??
?
? ?
2
11
4 4i 1
2i
?
??
?
? ?
2
11
3 4i
2i
?
?
?
? ?
2
1 1 3 4i
3 4i 3 4i
2i
?
??
??
?
? ?
2 2
1 3 4i
9 16i
2i
?
?
?
?
? ?
2
1 3 4i
9 16
2i
?
?
?
?
? ?
2
1 3 4i
25
2i
?
?
?
? ?
2
1 3 4i
25 25
2i
??
?
9.
To make a rectangle we need to select 2 vertical lines from given 6 lines
and 2 horizontal lines from given 5 line
so the number of rectangles so formed =
5
C2 ×
6
C2 =150
Page 4
CBSE XI | Mathematics
Sample Paper – 3 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 3 Solution
SECTION – A
1. ? ? ? Since m C 90 ,therefore
a 18 3
sin A
c 30 5
? ? ?
2. ?? 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 = 8
Solving the two equations, we get a = 5, b = -2
a = 5, b = -2 also satisfy the other two ordered pairs
f(-2) = 5(-2) – 2 = -12 ? (-2, -12)
f(-1) = 5(-1) – 2 = -7 ? (-1, -7)
Therefore the values are a = 5 and b = -2.
3. f(x) =
2
2
x4
x 8x 12
?
??
For f(x) to be defined, x
2
- 8x + 12 must be non-zero i.e. x
2
- 8x + 12 ? 0
(x - 2)(x - 6) ? 0
i.e. x ? 2 and x ? 6
Therefore domain will be ? ? 2,6 R ?
So domain of f = R - {2, 6}
4. ( ? p ? ? q)
OR
There exists a living person who is 150 years.
SECTION – B
5. (x iy)(3 5i) 1 3i 1 3i ? ? ? ? ? ? ? ?
? ?
2
2
1 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 17
67
x ;y
17 17
? ? ?
? ? ? ? ? ?
? ? ? ? ?
? ? ?
?
? ? ? ? ? ?
? ? ? ? ?
?
? ? ? ?
CBSE XI | Mathematics
Sample Paper – 3 Solution
OR
(3 + 4i)(4 + i) = 12 + 3i + 16i + 4i
2
= 12 + 19i – 4 = 8 + 19i
22
z 8 19 64 361 425 5 17 ? ? ? ? ? ?
6. Length of pendulum is 36 cm long
Angle of oscillation = 10 degrees
180 degrees = radians
so, 10 degrees= radians
18
?= radians
18
?
?
?
?
So using this formula l = r? and substituting the values of r = 36, ? = radians
18
?
we get,
?=36 x
p
18
=2 x (3.14) = 6.28 cm
OR
Area of sector =
2
1
r
2
?
2
2
2
2
1
r 5.024
2
1 36
r 5.024
2 180
180 2
r 5.024
36 3.14
r 16
r 4 cm
??
? ? ?
?
??
?
?
?
7. Let a be the first term and d be the common difference
Tn =2n+1
a=3………. (T1)
T2 = 5
d = 2 ……… (T2 - a)
? ?
? ?
? ? ?
? ? ? ? ? ?
n
n
n
S 2a (n 1)d
2
19
S 2 3 (19 1) 2 399
2
CBSE XI | Mathematics
Sample Paper – 3 Solution
8. We have 5! = 5 × 4! And 6! = 6 × 5 × 4!
LCM of 4!, 5! and 6! = LCM of {4!, 5 × 4!, 6 × 5 × 4!} = 4! 6 × 5 = 6! = 720
OR
? ?
2 2
11
4 4i i
2i
?
??
?
? ?
2
11
4 4i 1
2i
?
??
?
? ?
2
11
3 4i
2i
?
?
?
? ?
2
1 1 3 4i
3 4i 3 4i
2i
?
??
??
?
? ?
2 2
1 3 4i
9 16i
2i
?
?
?
?
? ?
2
1 3 4i
9 16
2i
?
?
?
?
? ?
2
1 3 4i
25
2i
?
?
?
? ?
2
1 3 4i
25 25
2i
??
?
9.
To make a rectangle we need to select 2 vertical lines from given 6 lines
and 2 horizontal lines from given 5 line
so the number of rectangles so formed =
5
C2 ×
6
C2 =150
CBSE XI | Mathematics
Sample Paper – 3 Solution
10. 1.2 + 2. 3 + 3. 4 +…
an = n (n + 1) = n
2
+ n
? ?
n n n
22
n
k 1 k 1 k 1
S k + k k k
n(n 1)(2n 1) n(n 1)
62
n(n 1) (2n 1)
1
23
n(n 1)(n 2)
3
? ? ?
? ? ? ? ? ?
? ? ?
??
?? ??
??
??
??
??
?
11. Let H denote the set of people who can speak Hindi, and E denote the 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
n(H U E) = n(H) + n(E) - n(H ? E)
n(H ? E) = n(H) + n(E) - n(H U E)
n(H ? E) = 250 + 200 - 400 = 50
50 persons can speak both Hindi and English.
12. Sn = 210
n(n 1)
210
2
?
?
n(n + 1) = 420
20 x 21 = 420 so n = 20
Sn
2
=
n(n 1)(2n 1)
6
??
=
420 41
= 2870
6
?
SECTION – C
13. Let the two vertices of the triangle be Q and R
Points Q and R will have the same x-coordinate = k(say)
Page 5
CBSE XI | Mathematics
Sample Paper – 3 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 3 Solution
SECTION – A
1. ? ? ? Since m C 90 ,therefore
a 18 3
sin A
c 30 5
? ? ?
2. ?? 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 = 8
Solving the two equations, we get a = 5, b = -2
a = 5, b = -2 also satisfy the other two ordered pairs
f(-2) = 5(-2) – 2 = -12 ? (-2, -12)
f(-1) = 5(-1) – 2 = -7 ? (-1, -7)
Therefore the values are a = 5 and b = -2.
3. f(x) =
2
2
x4
x 8x 12
?
??
For f(x) to be defined, x
2
- 8x + 12 must be non-zero i.e. x
2
- 8x + 12 ? 0
(x - 2)(x - 6) ? 0
i.e. x ? 2 and x ? 6
Therefore domain will be ? ? 2,6 R ?
So domain of f = R - {2, 6}
4. ( ? p ? ? q)
OR
There exists a living person who is 150 years.
SECTION – B
5. (x iy)(3 5i) 1 3i 1 3i ? ? ? ? ? ? ? ?
? ?
2
2
1 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 17
67
x ;y
17 17
? ? ?
? ? ? ? ? ?
? ? ? ? ?
? ? ?
?
? ? ? ? ? ?
? ? ? ? ?
?
? ? ? ?
CBSE XI | Mathematics
Sample Paper – 3 Solution
OR
(3 + 4i)(4 + i) = 12 + 3i + 16i + 4i
2
= 12 + 19i – 4 = 8 + 19i
22
z 8 19 64 361 425 5 17 ? ? ? ? ? ?
6. Length of pendulum is 36 cm long
Angle of oscillation = 10 degrees
180 degrees = radians
so, 10 degrees= radians
18
?= radians
18
?
?
?
?
So using this formula l = r? and substituting the values of r = 36, ? = radians
18
?
we get,
?=36 x
p
18
=2 x (3.14) = 6.28 cm
OR
Area of sector =
2
1
r
2
?
2
2
2
2
1
r 5.024
2
1 36
r 5.024
2 180
180 2
r 5.024
36 3.14
r 16
r 4 cm
??
? ? ?
?
??
?
?
?
7. Let a be the first term and d be the common difference
Tn =2n+1
a=3………. (T1)
T2 = 5
d = 2 ……… (T2 - a)
? ?
? ?
? ? ?
? ? ? ? ? ?
n
n
n
S 2a (n 1)d
2
19
S 2 3 (19 1) 2 399
2
CBSE XI | Mathematics
Sample Paper – 3 Solution
8. We have 5! = 5 × 4! And 6! = 6 × 5 × 4!
LCM of 4!, 5! and 6! = LCM of {4!, 5 × 4!, 6 × 5 × 4!} = 4! 6 × 5 = 6! = 720
OR
? ?
2 2
11
4 4i i
2i
?
??
?
? ?
2
11
4 4i 1
2i
?
??
?
? ?
2
11
3 4i
2i
?
?
?
? ?
2
1 1 3 4i
3 4i 3 4i
2i
?
??
??
?
? ?
2 2
1 3 4i
9 16i
2i
?
?
?
?
? ?
2
1 3 4i
9 16
2i
?
?
?
?
? ?
2
1 3 4i
25
2i
?
?
?
? ?
2
1 3 4i
25 25
2i
??
?
9.
To make a rectangle we need to select 2 vertical lines from given 6 lines
and 2 horizontal lines from given 5 line
so the number of rectangles so formed =
5
C2 ×
6
C2 =150
CBSE XI | Mathematics
Sample Paper – 3 Solution
10. 1.2 + 2. 3 + 3. 4 +…
an = n (n + 1) = n
2
+ n
? ?
n n n
22
n
k 1 k 1 k 1
S k + k k k
n(n 1)(2n 1) n(n 1)
62
n(n 1) (2n 1)
1
23
n(n 1)(n 2)
3
? ? ?
? ? ? ? ? ?
? ? ?
??
?? ??
??
??
??
??
?
11. Let H denote the set of people who can speak Hindi, and E denote the 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
n(H U E) = n(H) + n(E) - n(H ? E)
n(H ? E) = n(H) + n(E) - n(H U E)
n(H ? E) = 250 + 200 - 400 = 50
50 persons can speak both Hindi and English.
12. Sn = 210
n(n 1)
210
2
?
?
n(n + 1) = 420
20 x 21 = 420 so n = 20
Sn
2
=
n(n 1)(2n 1)
6
??
=
420 41
= 2870
6
?
SECTION – C
13. Let the two vertices of the triangle be Q and R
Points Q and R will have the same x-coordinate = k(say)
CBSE XI | Mathematics
Sample Paper – 3 Solution
Now in the right ?PRT, right angled at T.
? ?
2
2
k k k
tan60 3 RT
RT RT
3
k
R k,
3
Now R lies on the parabola : y = 4 ax
k
4 a(k)
3
k
4a
3
k 12a
12a
k
Length of side of the triangle = 2(RT)=2. 2. 8 3a
33
?
? ? ? ? ?
??
?
??
??
??
??
??
??
??
??
??
14. Consider L.H.S. = (cos3x – cosx) cosx + (sin3x + sinx) sinx
? ? ? ?
3x+x 3x-x 3x+x 3x-x
= 2sin sin cosx 2sin cos sinx
2 2 2 2
2sin2xcosx sinx 2sin2xsinx cosx
2sinxcosxsin 2x 2sinxcosxsin2x=0
? ? ? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ?
? ? ?
??
OR
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