Sample Solution Paper 4 - Math, Class 11 | Mathematics (Maths) Class 11 - Commerce PDF Download

 Page 1


  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
CBSE Board 
Class XI Mathematics 
Sample Paper – 4 Solution 
 
SECTION – A 
 
1. Such a function is called the greatest integer function. 
         From the definition [x] = –1 for –1 = x < 0 
           [x] = 0 for 0 = x < 1 
          [x] = 1 for 1 = x < 2 
          [x] = 2 for 2 = x < 3 and so on. 
 
2. 4x + i(3x – y) = 3 + i(–6) 
Equating real and the imaginary parts of the given equation, we get, 
4x = 3, 3x – y = – 6,  
Solving simultaneously, we get x =
3
4
and y = 
33
4
 
 
3. Given A = {1, 2} and B = {3, 4} A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}. 
Since n(A × B) = 4, the number of subsets of A × B is 2
4
. 
Therefore, the number of relations from A to B will be 16. 
OR 
(x + 3, 5) = (6, 2x + y) 
x + 3 = 6  
x = 3                …….(i) 
Also, 
2x + y = 5 
6 + y = 5          from (i) 
y = -1 
 
4. The negation of the given statement : 
It is false that Australia is a continent. 
Or 
Australia is not a continent. 
 
 
 
 
 
 
 
 
 
Page 2


  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
CBSE Board 
Class XI Mathematics 
Sample Paper – 4 Solution 
 
SECTION – A 
 
1. Such a function is called the greatest integer function. 
         From the definition [x] = –1 for –1 = x < 0 
           [x] = 0 for 0 = x < 1 
          [x] = 1 for 1 = x < 2 
          [x] = 2 for 2 = x < 3 and so on. 
 
2. 4x + i(3x – y) = 3 + i(–6) 
Equating real and the imaginary parts of the given equation, we get, 
4x = 3, 3x – y = – 6,  
Solving simultaneously, we get x =
3
4
and y = 
33
4
 
 
3. Given A = {1, 2} and B = {3, 4} A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}. 
Since n(A × B) = 4, the number of subsets of A × B is 2
4
. 
Therefore, the number of relations from A to B will be 16. 
OR 
(x + 3, 5) = (6, 2x + y) 
x + 3 = 6  
x = 3                …….(i) 
Also, 
2x + y = 5 
6 + y = 5          from (i) 
y = -1 
 
4. The negation of the given statement : 
It is false that Australia is a continent. 
Or 
Australia is not a continent. 
 
 
 
 
 
 
 
 
 
  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
SECTION – B 
 
 
5. Given 
2
2
2
()
1
22
(2)
2 1 5
2
2 10
5
( (2))
5 29
2
1
5
x
fx
x
f
f f f
?
?
? ? ?
?
??
? ? ? ?
??
??
??
?
??
??
  
OR 
f(x) = 3x
4
 – 5x
2
 + 9  
f(x – 1) = 3(x – 1)
4
 – 5(x – 1)
2
 + 9  
f(x – 1) = 3(x
4
 – 4x
3
 + 6x
2
 – 4x + 1) – 5(x
2
 – 2x + 1) + 9 
f(x – 1) = 3x
4
 – 12x
3
 + 18x
2
 – 12x + 3 – 5x
2
 + 10x – 5 + 9 
f(x – 1) = 3x
4
 – 12x
3
 + 13x
2
 – 2x + 7 
 
  
6.   
? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
?
? ? ? ? ? ? ?
?
? ? ? ? ? ? ? ? ? ? ?
3
33
3
2
2
2
given that cos3 8cos 0
4cos 3cos 8cos 0
12cos 3cos 0
3cos (4cos 1) 0
cos 0 or (4cos 1) 0
cos 0 ............. (2n 1) ,n I
2
1
(4cos 1) 0..............cos ....... n ,n I
23
 
 
7.  Discriminant of given equation is 0…….real and equal 
? ? ? ? ? ?
? ? ? ?
? ? ?
?
2 2 2 2 2
4 2 2 2
22
0 ( 2b(a c)) 4(a b )(b c )
0 4( b a c 2b ac)
0 4(b ac)
b ac
hence they are in GP
 
 
 
 
 
 
 
Page 3


  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
CBSE Board 
Class XI Mathematics 
Sample Paper – 4 Solution 
 
SECTION – A 
 
1. Such a function is called the greatest integer function. 
         From the definition [x] = –1 for –1 = x < 0 
           [x] = 0 for 0 = x < 1 
          [x] = 1 for 1 = x < 2 
          [x] = 2 for 2 = x < 3 and so on. 
 
2. 4x + i(3x – y) = 3 + i(–6) 
Equating real and the imaginary parts of the given equation, we get, 
4x = 3, 3x – y = – 6,  
Solving simultaneously, we get x =
3
4
and y = 
33
4
 
 
3. Given A = {1, 2} and B = {3, 4} A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}. 
Since n(A × B) = 4, the number of subsets of A × B is 2
4
. 
Therefore, the number of relations from A to B will be 16. 
OR 
(x + 3, 5) = (6, 2x + y) 
x + 3 = 6  
x = 3                …….(i) 
Also, 
2x + y = 5 
6 + y = 5          from (i) 
y = -1 
 
4. The negation of the given statement : 
It is false that Australia is a continent. 
Or 
Australia is not a continent. 
 
 
 
 
 
 
 
 
 
  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
SECTION – B 
 
 
5. Given 
2
2
2
()
1
22
(2)
2 1 5
2
2 10
5
( (2))
5 29
2
1
5
x
fx
x
f
f f f
?
?
? ? ?
?
??
? ? ? ?
??
??
??
?
??
??
  
OR 
f(x) = 3x
4
 – 5x
2
 + 9  
f(x – 1) = 3(x – 1)
4
 – 5(x – 1)
2
 + 9  
f(x – 1) = 3(x
4
 – 4x
3
 + 6x
2
 – 4x + 1) – 5(x
2
 – 2x + 1) + 9 
f(x – 1) = 3x
4
 – 12x
3
 + 18x
2
 – 12x + 3 – 5x
2
 + 10x – 5 + 9 
f(x – 1) = 3x
4
 – 12x
3
 + 13x
2
 – 2x + 7 
 
  
6.   
? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
?
? ? ? ? ? ? ?
?
? ? ? ? ? ? ? ? ? ? ?
3
33
3
2
2
2
given that cos3 8cos 0
4cos 3cos 8cos 0
12cos 3cos 0
3cos (4cos 1) 0
cos 0 or (4cos 1) 0
cos 0 ............. (2n 1) ,n I
2
1
(4cos 1) 0..............cos ....... n ,n I
23
 
 
7.  Discriminant of given equation is 0…….real and equal 
? ? ? ? ? ?
? ? ? ?
? ? ?
?
2 2 2 2 2
4 2 2 2
22
0 ( 2b(a c)) 4(a b )(b c )
0 4( b a c 2b ac)
0 4(b ac)
b ac
hence they are in GP
 
 
 
 
 
 
 
  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
8. Graph of function |x + 2| - 1  
 
x 0 -2 -1 1 2 
f(x) 1 -1 0 2 3 
 
 
 
 
9. Since the denominator of x
2
 is greater than the denominator of y
2
, the major axis is     
along the x-axis. Comparing the given equation with the standard equation of the 
ellipse 
 
22
22
xy
1
ab
??
 
a = 5, b =3 
? ? ? ? ?
22
ae a b 25 9 4 
So the foci are (4, 0) and (-4, 0) 
OR 
Let the equation of ellipse be 
22
22
xy
1
ab
?? and e be the eccentricity. 
Given that 
Latus rectum = (minor axis)/2 
2
22
2b 1
2b
a2
2b a
4b a
??
?
?
 
? ?
2 2 2
2
4a 1 e a
4 4e 1
??
??
 
2
33
ee
42
? ? ? 
Page 4


  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
CBSE Board 
Class XI Mathematics 
Sample Paper – 4 Solution 
 
SECTION – A 
 
1. Such a function is called the greatest integer function. 
         From the definition [x] = –1 for –1 = x < 0 
           [x] = 0 for 0 = x < 1 
          [x] = 1 for 1 = x < 2 
          [x] = 2 for 2 = x < 3 and so on. 
 
2. 4x + i(3x – y) = 3 + i(–6) 
Equating real and the imaginary parts of the given equation, we get, 
4x = 3, 3x – y = – 6,  
Solving simultaneously, we get x =
3
4
and y = 
33
4
 
 
3. Given A = {1, 2} and B = {3, 4} A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}. 
Since n(A × B) = 4, the number of subsets of A × B is 2
4
. 
Therefore, the number of relations from A to B will be 16. 
OR 
(x + 3, 5) = (6, 2x + y) 
x + 3 = 6  
x = 3                …….(i) 
Also, 
2x + y = 5 
6 + y = 5          from (i) 
y = -1 
 
4. The negation of the given statement : 
It is false that Australia is a continent. 
Or 
Australia is not a continent. 
 
 
 
 
 
 
 
 
 
  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
SECTION – B 
 
 
5. Given 
2
2
2
()
1
22
(2)
2 1 5
2
2 10
5
( (2))
5 29
2
1
5
x
fx
x
f
f f f
?
?
? ? ?
?
??
? ? ? ?
??
??
??
?
??
??
  
OR 
f(x) = 3x
4
 – 5x
2
 + 9  
f(x – 1) = 3(x – 1)
4
 – 5(x – 1)
2
 + 9  
f(x – 1) = 3(x
4
 – 4x
3
 + 6x
2
 – 4x + 1) – 5(x
2
 – 2x + 1) + 9 
f(x – 1) = 3x
4
 – 12x
3
 + 18x
2
 – 12x + 3 – 5x
2
 + 10x – 5 + 9 
f(x – 1) = 3x
4
 – 12x
3
 + 13x
2
 – 2x + 7 
 
  
6.   
? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
?
? ? ? ? ? ? ?
?
? ? ? ? ? ? ? ? ? ? ?
3
33
3
2
2
2
given that cos3 8cos 0
4cos 3cos 8cos 0
12cos 3cos 0
3cos (4cos 1) 0
cos 0 or (4cos 1) 0
cos 0 ............. (2n 1) ,n I
2
1
(4cos 1) 0..............cos ....... n ,n I
23
 
 
7.  Discriminant of given equation is 0…….real and equal 
? ? ? ? ? ?
? ? ? ?
? ? ?
?
2 2 2 2 2
4 2 2 2
22
0 ( 2b(a c)) 4(a b )(b c )
0 4( b a c 2b ac)
0 4(b ac)
b ac
hence they are in GP
 
 
 
 
 
 
 
  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
8. Graph of function |x + 2| - 1  
 
x 0 -2 -1 1 2 
f(x) 1 -1 0 2 3 
 
 
 
 
9. Since the denominator of x
2
 is greater than the denominator of y
2
, the major axis is     
along the x-axis. Comparing the given equation with the standard equation of the 
ellipse 
 
22
22
xy
1
ab
??
 
a = 5, b =3 
? ? ? ? ?
22
ae a b 25 9 4 
So the foci are (4, 0) and (-4, 0) 
OR 
Let the equation of ellipse be 
22
22
xy
1
ab
?? and e be the eccentricity. 
Given that 
Latus rectum = (minor axis)/2 
2
22
2b 1
2b
a2
2b a
4b a
??
?
?
 
? ?
2 2 2
2
4a 1 e a
4 4e 1
??
??
 
2
33
ee
42
? ? ? 
  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
 
10. Let T represent the students who drink tea and C represents students who drink 
coffee and x represent the number of students who drink tea or coffee.  
So, x = n(T ? C) 
 
n(T) = 150; n(C) = 225  
n (T ?C) = n(T)+ n(C) - n(T ?C)  
Substituting the values,  
x = 150 + 225 – 100 = 275 
Number of students who drink neither tea nor coffee   
= 600 – 275 = 325.  
OR 
Let H denote the set of people speaking Hindi and E denote the set of people 
speaking English. 
n(H) = 550, n(E) = 450 and n(H ? E) = 800 
n(H n E) = n(H) + n(E) – n(H ? E) 
                  = 550 + 450 – 800  
                  = 200 
Hence, 200 persons can speak both Hindi and English. 
 
11.  R: { (a, b): a, b ? A, a divides b}, Also 
A = {l, 2, 3, 4, 6} 
(i) R = {(1, 1), (1, 2), (1, 3), (1, 4),(1, 6), (2, 2),(2, 4), (2,6), (3, 3), (3, 6),(4, 4), (6, 6)}  
(ii) Domain of R = {1, 2, 3, 4, 6} 
(iii) Range of R = {1, 2, 3, 4, 6} 
 
 
 
 
 
 
 
 
Page 5


  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
CBSE Board 
Class XI Mathematics 
Sample Paper – 4 Solution 
 
SECTION – A 
 
1. Such a function is called the greatest integer function. 
         From the definition [x] = –1 for –1 = x < 0 
           [x] = 0 for 0 = x < 1 
          [x] = 1 for 1 = x < 2 
          [x] = 2 for 2 = x < 3 and so on. 
 
2. 4x + i(3x – y) = 3 + i(–6) 
Equating real and the imaginary parts of the given equation, we get, 
4x = 3, 3x – y = – 6,  
Solving simultaneously, we get x =
3
4
and y = 
33
4
 
 
3. Given A = {1, 2} and B = {3, 4} A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}. 
Since n(A × B) = 4, the number of subsets of A × B is 2
4
. 
Therefore, the number of relations from A to B will be 16. 
OR 
(x + 3, 5) = (6, 2x + y) 
x + 3 = 6  
x = 3                …….(i) 
Also, 
2x + y = 5 
6 + y = 5          from (i) 
y = -1 
 
4. The negation of the given statement : 
It is false that Australia is a continent. 
Or 
Australia is not a continent. 
 
 
 
 
 
 
 
 
 
  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
SECTION – B 
 
 
5. Given 
2
2
2
()
1
22
(2)
2 1 5
2
2 10
5
( (2))
5 29
2
1
5
x
fx
x
f
f f f
?
?
? ? ?
?
??
? ? ? ?
??
??
??
?
??
??
  
OR 
f(x) = 3x
4
 – 5x
2
 + 9  
f(x – 1) = 3(x – 1)
4
 – 5(x – 1)
2
 + 9  
f(x – 1) = 3(x
4
 – 4x
3
 + 6x
2
 – 4x + 1) – 5(x
2
 – 2x + 1) + 9 
f(x – 1) = 3x
4
 – 12x
3
 + 18x
2
 – 12x + 3 – 5x
2
 + 10x – 5 + 9 
f(x – 1) = 3x
4
 – 12x
3
 + 13x
2
 – 2x + 7 
 
  
6.   
? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
?
? ? ? ? ? ? ?
?
? ? ? ? ? ? ? ? ? ? ?
3
33
3
2
2
2
given that cos3 8cos 0
4cos 3cos 8cos 0
12cos 3cos 0
3cos (4cos 1) 0
cos 0 or (4cos 1) 0
cos 0 ............. (2n 1) ,n I
2
1
(4cos 1) 0..............cos ....... n ,n I
23
 
 
7.  Discriminant of given equation is 0…….real and equal 
? ? ? ? ? ?
? ? ? ?
? ? ?
?
2 2 2 2 2
4 2 2 2
22
0 ( 2b(a c)) 4(a b )(b c )
0 4( b a c 2b ac)
0 4(b ac)
b ac
hence they are in GP
 
 
 
 
 
 
 
  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
8. Graph of function |x + 2| - 1  
 
x 0 -2 -1 1 2 
f(x) 1 -1 0 2 3 
 
 
 
 
9. Since the denominator of x
2
 is greater than the denominator of y
2
, the major axis is     
along the x-axis. Comparing the given equation with the standard equation of the 
ellipse 
 
22
22
xy
1
ab
??
 
a = 5, b =3 
? ? ? ? ?
22
ae a b 25 9 4 
So the foci are (4, 0) and (-4, 0) 
OR 
Let the equation of ellipse be 
22
22
xy
1
ab
?? and e be the eccentricity. 
Given that 
Latus rectum = (minor axis)/2 
2
22
2b 1
2b
a2
2b a
4b a
??
?
?
 
? ?
2 2 2
2
4a 1 e a
4 4e 1
??
??
 
2
33
ee
42
? ? ? 
  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
 
10. Let T represent the students who drink tea and C represents students who drink 
coffee and x represent the number of students who drink tea or coffee.  
So, x = n(T ? C) 
 
n(T) = 150; n(C) = 225  
n (T ?C) = n(T)+ n(C) - n(T ?C)  
Substituting the values,  
x = 150 + 225 – 100 = 275 
Number of students who drink neither tea nor coffee   
= 600 – 275 = 325.  
OR 
Let H denote the set of people speaking Hindi and E denote the set of people 
speaking English. 
n(H) = 550, n(E) = 450 and n(H ? E) = 800 
n(H n E) = n(H) + n(E) – n(H ? E) 
                  = 550 + 450 – 800  
                  = 200 
Hence, 200 persons can speak both Hindi and English. 
 
11.  R: { (a, b): a, b ? A, a divides b}, Also 
A = {l, 2, 3, 4, 6} 
(i) R = {(1, 1), (1, 2), (1, 3), (1, 4),(1, 6), (2, 2),(2, 4), (2,6), (3, 3), (3, 6),(4, 4), (6, 6)}  
(ii) Domain of R = {1, 2, 3, 4, 6} 
(iii) Range of R = {1, 2, 3, 4, 6} 
 
 
 
 
 
 
 
 
  
 
CBSE XI | Mathematics 
Sample Paper – 4 Solution 
 
     
12.  Given Signum function: 
 
The two branches 
 Bold bullet at (0, 0)  
Circle at (0, 1) and (0, -1) 
 
SECTION – C 
 
13.  Let 1 = rcos?, v3 = rsin? 
By squaring and adding, we get, 
r²(cos²? + sin²?) = 4, ? r = 2 
, So 
3
13
cos , sin
22
?
?? ? ? ? ?
 
Therefore, required polar form is 
z 2 cos isin
33
?? ??
??
??
??
 
OR 
        
? ? ? ?
2
18
25
22
4 4 2 2
4 6 1
2
2
2
2
2
1
i
i
11
ii
i
i
1
1
i
11
1 2 1
ii
1 2 1 2
11
i 1 i
i
2 2i
2i 0 2i
ii
??
??
??
?
??
??
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ?
??
??
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ?
?
?
? ? ? ? ?