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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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Class 11 | Mathematics (Maths) Class 11 - Commerce

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Sample Paper

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Viva Questions

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Semester Notes

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mock tests for examination

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study material

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practice quizzes

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