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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
??
??
??
?
??
??
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ?
??
??
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ?
?
?
? ? ? ? ?
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FAQs on Sample Solution Paper 4 - Math, Class 11 - Mathematics (Maths) Class 11 - Commerce
| 1. What is the quadratic formula? |  |
Ans. The quadratic formula is a formula used to calculate the solutions of a quadratic equation. It is given by:
x = (-b ± √(b^2 - 4ac))/(2a)
where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.
| 2. How do I solve a quadratic equation using the quadratic formula? | |
Ans. To solve a quadratic equation using the quadratic formula, follow these steps:
1. Identify the coefficients a, b, and c of the quadratic equation ax^2 + bx + c = 0.
2. Substitute the values of a, b, and c into the quadratic formula: x = (-b ± √(b^2 - 4ac))/(2a).
3. Simplify and calculate the values of x using the formula.
4. The solutions will be the values of x obtained from the formula.
| 3. Can the quadratic formula be used for any quadratic equation? | |
Ans. Yes, the quadratic formula can be used to solve any quadratic equation of the form ax^2 + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. It provides the exact solutions for the equation, regardless of the values of the coefficients.
| 4. Are there any alternative methods to solve quadratic equations? | |
Ans. Yes, there are alternative methods to solve quadratic equations, such as factoring, completing the square, and graphing. These methods can be used when the quadratic equation is in a different form or when it is easier to use a specific method based on the given equation.
| 5. What are the discriminant and its significance in the quadratic formula? | |
Ans. The discriminant is a term in the quadratic formula given by b^2 - 4ac. It determines the nature and number of solutions of a quadratic equation.
1. If the discriminant is positive (b^2 - 4ac > 0), the quadratic equation has two distinct real solutions.
2. If the discriminant is zero (b^2 - 4ac = 0), the quadratic equation has one real solution (a repeated root).
3. If the discriminant is negative (b^2 - 4ac < 0), the quadratic equation has no real solutions (complex roots).
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Nov 24, 2024 Last updated |
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Sample Solution Paper 4 - Math, Class 11 Commerce Questions
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