Sample Solution Paper 3 - Math, Class 11

# Sample Solution Paper 3 - Math, Class 11 | Mathematics (Maths) Class 11 - Commerce PDF Download

``` 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
18
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
18
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
18
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
18
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
```

## Mathematics (Maths) Class 11

85 videos|243 docs|99 tests

### Up next

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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.

## Mathematics (Maths) Class 11

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