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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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FAQs on Sample Solution Paper 3 - Math, Class 11 - Mathematics (Maths) Class 11 - Commerce
| 1. What is the importance of studying math in Class 11? |  |
Ans. Studying math in Class 11 is important as it forms the foundation for higher-level math courses. It helps develop critical thinking, problem-solving, and logical reasoning skills, which are essential in various fields such as engineering, science, finance, and computer science. Additionally, math provides a solid base for understanding and analyzing real-world situations, making it relevant in everyday life.
| 2. How can I improve my math skills in Class 11? | |
Ans. To improve math skills in Class 11, it is important to practice regularly and consistently. Solve a variety of math problems, including textbook exercises, sample papers, and previous year's question papers. Seek help from teachers or classmates whenever you encounter difficulties. Additionally, utilize online resources, such as video tutorials or interactive math websites, to reinforce concepts and gain a deeper understanding of the subject.
| 3. What are the key topics covered in the Class 11 math exam? | |
Ans. The Class 11 math exam typically covers topics such as sets, relations, and functions, algebraic expressions, complex numbers, quadratic equations, linear inequalities, sequences and series, coordinate geometry, trigonometry, statistics, and probability. It is important to thoroughly understand these topics and practice solving related problems to perform well in the exam.
| 4. How can I manage time effectively during the Class 11 math exam? | |
Ans. To manage time effectively during the Class 11 math exam, it is advisable to create a study schedule and allocate specific time slots for each topic. Practice solving previous year's question papers or sample papers within the stipulated time frame to improve speed and accuracy. During the exam, read the questions carefully, prioritize them based on difficulty level, and attempt the easier ones first. If you get stuck on a particular question, move on to the next one and come back to it later if time permits.
| 5. Are there any useful online resources for Class 11 math preparation? | |
Ans. Yes, there are several useful online resources for Class 11 math preparation. Websites like Khan Academy, BYJU'S, and Mathway provide comprehensive video lessons, practice exercises, and interactive tools to enhance understanding and practice. Additionally, there are online forums and discussion boards where students can ask questions, clarify doubts, and engage in peer-to-peer learning. Many educational apps are also available for mobile devices, offering math-related quizzes and practice problems for convenient on-the-go studying.
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Dec 22, 2024 Last updated |
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,Class 11 | Mathematics (Maths) Class 11 - Commerce
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Additional Information about Sample Solution Paper 3 - Math, Class 11 for Commerce Preparation
Sample Solution Paper 3 - Math, Class 11 Free PDF Download
The Sample Solution Paper 3 - Math, Class 11 is an invaluable resource that delves deep into the core of the Commerce exam.
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Sample Solution Paper 3 - Math, Class 11 Notes
Sample Solution Paper 3 - Math, Class 11 Notes offer in-depth insights into the specific topic to help you master it with ease.
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Sample Solution Paper 3 - Math, Class 11 Commerce Questions
The "Sample Solution Paper 3 - Math, Class 11 Commerce Questions" guide is a valuable resource for all aspiring students preparing for the
Commerce exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
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Study Sample Solution Paper 3 - Math, Class 11 on the App
Students of Commerce can study Sample Solution Paper 3 - Math, Class 11 alongwith tests & analysis from the EduRev app,
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The content of Sample Solution Paper 3 - Math, Class 11 is prepared as per the latest Commerce syllabus.
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