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Page 1
CBSE XI | Mathematics
Sample Paper – 9 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 9 Solution
SECTION – A
1.
x0
x0
sinax
lim
bx
sinax a
lim
ax b
a
b
?
?
??
?
2. The sentence is false because a month can’t have more than 31 days. Hence it is a
statement.
3.
? ?
2
1
i1
1
1i
1 1 i
1 i 1 i
1i
1i
1i
11
1i
2
1i
22
?
?
??
??
??
? ? ? ?
??
?
?
??
?
??
??
?
?
??
a = -1/2 and b = -1/2
OR
z = 2i
Comparing with a + bi we get a = 0 and b = 2
2
z 0 2 2 ? ? ?
Page 2
CBSE XI | Mathematics
Sample Paper – 9 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 9 Solution
SECTION – A
1.
x0
x0
sinax
lim
bx
sinax a
lim
ax b
a
b
?
?
??
?
2. The sentence is false because a month can’t have more than 31 days. Hence it is a
statement.
3.
? ?
2
1
i1
1
1i
1 1 i
1 i 1 i
1i
1i
1i
11
1i
2
1i
22
?
?
??
??
??
? ? ? ?
??
?
?
??
?
??
??
?
?
??
a = -1/2 and b = -1/2
OR
z = 2i
Comparing with a + bi we get a = 0 and b = 2
2
z 0 2 2 ? ? ?
CBSE XI | Mathematics
Sample Paper – 9 Solution
4. According to the question,
Variance of 20 observations is 5.
New variance = 2
2
× 5 = 20
SECTION – B
5. A × C = {(1, 5),(1, 6), (2, 5), (2, 6)}….(i)
B × D = {(1, 5),(1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), ( 3, 6), (3, 7), (3, 8),
(4, 5), (4, 6), (4, 7), (4, 8)}…….(ii)
A × C is a subset of B × D all the elements of A × C are contained in B × D.
6. According to the question,
f(x) = g(x)
f(-4) = g(-4)
-4 – 4 = ?
? = -8
OR
? ?
2
x9
fx
x3
?
?
?
f(x) is not defined for x – 3 = 0 i. e. x = 3. Therefore, domain (f) = r – {3}
Let f(x) = y then
2
x9
y
x3
?
?
?
x + 3 = y
It follows from the above relation that y takes all real values except 6 when x takes
values in the set R – {3}. Therefore, Range (f) = R – {6}
7. Let h be the required height in metres. Here h can be considered as the arc of a circle of
radius 12 m, which subtends an angle of 5’ at its centre.
? = 5’ =
c
51
60 12 180
?
? ? ? ? ?
??
? ? ? ?
? ? ? ?
and r = 12 m
arc h
r 12 180 12
h 1.7 m
180
?
? ? ? ?
?
?
??
OR
Page 3
CBSE XI | Mathematics
Sample Paper – 9 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 9 Solution
SECTION – A
1.
x0
x0
sinax
lim
bx
sinax a
lim
ax b
a
b
?
?
??
?
2. The sentence is false because a month can’t have more than 31 days. Hence it is a
statement.
3.
? ?
2
1
i1
1
1i
1 1 i
1 i 1 i
1i
1i
1i
11
1i
2
1i
22
?
?
??
??
??
? ? ? ?
??
?
?
??
?
??
??
?
?
??
a = -1/2 and b = -1/2
OR
z = 2i
Comparing with a + bi we get a = 0 and b = 2
2
z 0 2 2 ? ? ?
CBSE XI | Mathematics
Sample Paper – 9 Solution
4. According to the question,
Variance of 20 observations is 5.
New variance = 2
2
× 5 = 20
SECTION – B
5. A × C = {(1, 5),(1, 6), (2, 5), (2, 6)}….(i)
B × D = {(1, 5),(1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), ( 3, 6), (3, 7), (3, 8),
(4, 5), (4, 6), (4, 7), (4, 8)}…….(ii)
A × C is a subset of B × D all the elements of A × C are contained in B × D.
6. According to the question,
f(x) = g(x)
f(-4) = g(-4)
-4 – 4 = ?
? = -8
OR
? ?
2
x9
fx
x3
?
?
?
f(x) is not defined for x – 3 = 0 i. e. x = 3. Therefore, domain (f) = r – {3}
Let f(x) = y then
2
x9
y
x3
?
?
?
x + 3 = y
It follows from the above relation that y takes all real values except 6 when x takes
values in the set R – {3}. Therefore, Range (f) = R – {6}
7. Let h be the required height in metres. Here h can be considered as the arc of a circle of
radius 12 m, which subtends an angle of 5’ at its centre.
? = 5’ =
c
51
60 12 180
?
? ? ? ? ?
??
? ? ? ?
? ? ? ?
and r = 12 m
arc h
r 12 180 12
h 1.7 m
180
?
? ? ? ?
?
?
??
OR
CBSE XI | Mathematics
Sample Paper – 9 Solution
Let r1 and r2 be the radii of the given circles and let their arcs of same length s subtend
angles of 60° and 75° at their centres.
60 60
180 3
5
75 75
180 12
??
? ? ? ? ?
??
? ? ? ? ?
1 1 2 2
12
1
2
1
2
s r r
5
rr
3 12
r 5
3
r 12
r 5
r4
? ? ? ?
??
?
??
?
8. n(U) = 600, n(A) = 460, n(B) = 390 and n(A n B) = 325
n(A ? B) = n(A) + n(B) – n(A n B)
= 460 + 390 – 325
= 525
n(A ? B)’ = n(U) – n(A ? B)
= 600 – 525
= 75
9. sin(? + 30°) = sin ? cos 30° + cos 30° sin ?
= (sin ? cos 30° – cos ? sin 30°) + 2cos ? sin 30°
= sin (? – 30°) + 2cos ? × ½
= cos ? + sin (? – 30°)
? sin(? + 30°) = cos ? + sin (? – 30°)
OR
sin7A sin5A
cos5A cos7A
7A 5A 7A 5A
2cos sin
22
7A 5A 7A 5A
2cos cos
22
cos6Asin A
cos6Acos A
sin A
cos A
tan A
?
?
??
?
??
?
?
?
Page 4
CBSE XI | Mathematics
Sample Paper – 9 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 9 Solution
SECTION – A
1.
x0
x0
sinax
lim
bx
sinax a
lim
ax b
a
b
?
?
??
?
2. The sentence is false because a month can’t have more than 31 days. Hence it is a
statement.
3.
? ?
2
1
i1
1
1i
1 1 i
1 i 1 i
1i
1i
1i
11
1i
2
1i
22
?
?
??
??
??
? ? ? ?
??
?
?
??
?
??
??
?
?
??
a = -1/2 and b = -1/2
OR
z = 2i
Comparing with a + bi we get a = 0 and b = 2
2
z 0 2 2 ? ? ?
CBSE XI | Mathematics
Sample Paper – 9 Solution
4. According to the question,
Variance of 20 observations is 5.
New variance = 2
2
× 5 = 20
SECTION – B
5. A × C = {(1, 5),(1, 6), (2, 5), (2, 6)}….(i)
B × D = {(1, 5),(1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), ( 3, 6), (3, 7), (3, 8),
(4, 5), (4, 6), (4, 7), (4, 8)}…….(ii)
A × C is a subset of B × D all the elements of A × C are contained in B × D.
6. According to the question,
f(x) = g(x)
f(-4) = g(-4)
-4 – 4 = ?
? = -8
OR
? ?
2
x9
fx
x3
?
?
?
f(x) is not defined for x – 3 = 0 i. e. x = 3. Therefore, domain (f) = r – {3}
Let f(x) = y then
2
x9
y
x3
?
?
?
x + 3 = y
It follows from the above relation that y takes all real values except 6 when x takes
values in the set R – {3}. Therefore, Range (f) = R – {6}
7. Let h be the required height in metres. Here h can be considered as the arc of a circle of
radius 12 m, which subtends an angle of 5’ at its centre.
? = 5’ =
c
51
60 12 180
?
? ? ? ? ?
??
? ? ? ?
? ? ? ?
and r = 12 m
arc h
r 12 180 12
h 1.7 m
180
?
? ? ? ?
?
?
??
OR
CBSE XI | Mathematics
Sample Paper – 9 Solution
Let r1 and r2 be the radii of the given circles and let their arcs of same length s subtend
angles of 60° and 75° at their centres.
60 60
180 3
5
75 75
180 12
??
? ? ? ? ?
??
? ? ? ? ?
1 1 2 2
12
1
2
1
2
s r r
5
rr
3 12
r 5
3
r 12
r 5
r4
? ? ? ?
??
?
??
?
8. n(U) = 600, n(A) = 460, n(B) = 390 and n(A n B) = 325
n(A ? B) = n(A) + n(B) – n(A n B)
= 460 + 390 – 325
= 525
n(A ? B)’ = n(U) – n(A ? B)
= 600 – 525
= 75
9. sin(? + 30°) = sin ? cos 30° + cos 30° sin ?
= (sin ? cos 30° – cos ? sin 30°) + 2cos ? sin 30°
= sin (? – 30°) + 2cos ? × ½
= cos ? + sin (? – 30°)
? sin(? + 30°) = cos ? + sin (? – 30°)
OR
sin7A sin5A
cos5A cos7A
7A 5A 7A 5A
2cos sin
22
7A 5A 7A 5A
2cos cos
22
cos6Asin A
cos6Acos A
sin A
cos A
tan A
?
?
??
?
??
?
?
?
CBSE XI | Mathematics
Sample Paper – 9 Solution
10. The compound statements are
p : It is raining
q : It is cold.
11. f(x) = x
2
? ? ? ?
22
f 1.1 f 1
1.1 1
1.1 1
1.1 1
1.21 1
0.1
2.1
?
?
?
?
?
?
?
?
s
12. (x1, y1) = (-1, 3) and (x2, y2) = (4, -2)
The equation of line is y – 3 =
? ? 32
14
??
??
(x + 1)
y – 3 = -x – 1
x + y – 2 = 0
SECTION – C
13.
? ?
22
22
22
22
cos 33 cos 57
21 69
sin sin
22
1 sin 33 1 sin 57
21 69
sin sin
22
? ? ?
??
?
? ? ? ? ?
?
??
?
? ? ? ?
22
22
sin 57 sin 33
21 69
sin sin
22
sin 57 33 sin 57 33
21 69 21 69
sin sin
2 2 2 2
? ? ?
?
??
?
? ? ? ? ? ?
?
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?
? ?
sin90 sin24
sin45 sin 24
sin24
2
1
sin24
2
??
?
? ? ?
?
? ? ?
?
?
Page 5
CBSE XI | Mathematics
Sample Paper – 9 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 9 Solution
SECTION – A
1.
x0
x0
sinax
lim
bx
sinax a
lim
ax b
a
b
?
?
??
?
2. The sentence is false because a month can’t have more than 31 days. Hence it is a
statement.
3.
? ?
2
1
i1
1
1i
1 1 i
1 i 1 i
1i
1i
1i
11
1i
2
1i
22
?
?
??
??
??
? ? ? ?
??
?
?
??
?
??
??
?
?
??
a = -1/2 and b = -1/2
OR
z = 2i
Comparing with a + bi we get a = 0 and b = 2
2
z 0 2 2 ? ? ?
CBSE XI | Mathematics
Sample Paper – 9 Solution
4. According to the question,
Variance of 20 observations is 5.
New variance = 2
2
× 5 = 20
SECTION – B
5. A × C = {(1, 5),(1, 6), (2, 5), (2, 6)}….(i)
B × D = {(1, 5),(1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), ( 3, 6), (3, 7), (3, 8),
(4, 5), (4, 6), (4, 7), (4, 8)}…….(ii)
A × C is a subset of B × D all the elements of A × C are contained in B × D.
6. According to the question,
f(x) = g(x)
f(-4) = g(-4)
-4 – 4 = ?
? = -8
OR
? ?
2
x9
fx
x3
?
?
?
f(x) is not defined for x – 3 = 0 i. e. x = 3. Therefore, domain (f) = r – {3}
Let f(x) = y then
2
x9
y
x3
?
?
?
x + 3 = y
It follows from the above relation that y takes all real values except 6 when x takes
values in the set R – {3}. Therefore, Range (f) = R – {6}
7. Let h be the required height in metres. Here h can be considered as the arc of a circle of
radius 12 m, which subtends an angle of 5’ at its centre.
? = 5’ =
c
51
60 12 180
?
? ? ? ? ?
??
? ? ? ?
? ? ? ?
and r = 12 m
arc h
r 12 180 12
h 1.7 m
180
?
? ? ? ?
?
?
??
OR
CBSE XI | Mathematics
Sample Paper – 9 Solution
Let r1 and r2 be the radii of the given circles and let their arcs of same length s subtend
angles of 60° and 75° at their centres.
60 60
180 3
5
75 75
180 12
??
? ? ? ? ?
??
? ? ? ? ?
1 1 2 2
12
1
2
1
2
s r r
5
rr
3 12
r 5
3
r 12
r 5
r4
? ? ? ?
??
?
??
?
8. n(U) = 600, n(A) = 460, n(B) = 390 and n(A n B) = 325
n(A ? B) = n(A) + n(B) – n(A n B)
= 460 + 390 – 325
= 525
n(A ? B)’ = n(U) – n(A ? B)
= 600 – 525
= 75
9. sin(? + 30°) = sin ? cos 30° + cos 30° sin ?
= (sin ? cos 30° – cos ? sin 30°) + 2cos ? sin 30°
= sin (? – 30°) + 2cos ? × ½
= cos ? + sin (? – 30°)
? sin(? + 30°) = cos ? + sin (? – 30°)
OR
sin7A sin5A
cos5A cos7A
7A 5A 7A 5A
2cos sin
22
7A 5A 7A 5A
2cos cos
22
cos6Asin A
cos6Acos A
sin A
cos A
tan A
?
?
??
?
??
?
?
?
CBSE XI | Mathematics
Sample Paper – 9 Solution
10. The compound statements are
p : It is raining
q : It is cold.
11. f(x) = x
2
? ? ? ?
22
f 1.1 f 1
1.1 1
1.1 1
1.1 1
1.21 1
0.1
2.1
?
?
?
?
?
?
?
?
s
12. (x1, y1) = (-1, 3) and (x2, y2) = (4, -2)
The equation of line is y – 3 =
? ? 32
14
??
??
(x + 1)
y – 3 = -x – 1
x + y – 2 = 0
SECTION – C
13.
? ?
22
22
22
22
cos 33 cos 57
21 69
sin sin
22
1 sin 33 1 sin 57
21 69
sin sin
22
? ? ?
??
?
? ? ? ? ?
?
??
?
? ? ? ?
22
22
sin 57 sin 33
21 69
sin sin
22
sin 57 33 sin 57 33
21 69 21 69
sin sin
2 2 2 2
? ? ?
?
??
?
? ? ? ? ? ?
?
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?
? ?
sin90 sin24
sin45 sin 24
sin24
2
1
sin24
2
??
?
? ? ?
?
? ? ?
?
?
CBSE XI | Mathematics
Sample Paper – 9 Solution
14.
? ?
x1
fx
x1
?
?
?
? ? fx
x1
1 x 1
?
?
?
? ?
? ?
f x 1
x 1 x 1
f x 1 x 1 x 1
?
? ? ?
?
? ? ? ?
? ?
? ?
f x 1
x
f x 1
?
??
?
? ?
? ?
f x 1
x
1 f x
?
?
?
? ?
2x 1
f 2x
2x 1
?
?
?
? ?
? ?
? ?
? ?
? ?
f x 1
21
1 f x
f 2x
f x 1
21
1 f x
?
??
?
?
?
??
?
? ?
? ? ? ?
? ? ? ?
2 f x 2 1 f x
f 2x
2 f x 2 1 f x
? ? ? ?
?
? ? ? ?
? ?
? ?
? ?
3f x 1
f 2x
f x 3
?
?
?
15. i. f(x) = x
2
+ 3
f(x) = 28
x
2
+ 3 = 28
x
2
= 25
x = ? 5
{x : f(x) = 28} = {5, -5}
ii. Let x be the pre-image of 39. Then,
f(x) = 39
x
2
+ 3 = 39
x
2
= 36
x = ? 6
So, pre-image of 39 are -6 and 6.
Let x be the pre-image of 2. Then,
f(x) = 2
x
2
+ 3 = 2
x
2
= -1
We find that no real value of x satisfies the equation. Therefore, 2 does not have any
pre-image under f.
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FAQs on Sample Solution Paper 9 - Math, Class 11 - Mathematics (Maths) Class 11 - Commerce
| 1. What is the quadratic formula and how is it used to solve quadratic equations? |  |
Ans. The quadratic formula is a formula used to find the solutions (roots) of a quadratic equation of the form ax^2 + bx + c = 0. It is given by the formula: x = (-b ± √(b^2 - 4ac)) / (2a). To use the quadratic formula, substitute the values of a, b, and c from the quadratic equation into the formula and then simplify to find the values of x.
| 2. How do you determine if a quadratic equation has real or complex solutions? | |
Ans. To determine if a quadratic equation has real or complex solutions, you can evaluate the discriminant of the quadratic equation. The discriminant is the value under the square root in the quadratic formula, b^2 - 4ac.
- If the discriminant is positive, the equation will have two distinct real solutions.
- If the discriminant is zero, the equation will have one real solution (a repeated root).
- If the discriminant is negative, the equation will have two complex conjugate solutions.
| 3. Can a quadratic equation have no solutions? | |
Ans. Yes, a quadratic equation can have no solutions. This occurs when the discriminant (b^2 - 4ac) is negative, indicating that the quadratic equation does not intersect the x-axis and does not have any real solutions. In other words, the graph of the quadratic equation will not intersect or touch the x-axis.
| 4. Is it possible for a quadratic equation to have only one solution? | |
Ans. Yes, it is possible for a quadratic equation to have only one solution. This occurs when the discriminant (b^2 - 4ac) is equal to zero, indicating that the quadratic equation has one real solution that is repeated (a repeated root). In other words, the graph of the quadratic equation will touch the x-axis at one point.
| 5. Can the quadratic formula be used to solve any type of quadratic equation? | |
Ans. Yes, the quadratic formula can be used to solve any type of quadratic equation, regardless of the values of a, b, and c. It provides a straightforward method for finding the solutions (roots) of quadratic equations, even if they cannot be easily factored or solved by other methods. However, it is important to note that the quadratic formula may yield complex solutions if the discriminant is negative.
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Nov 21, 2024 Last updated |
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Sample Solution Paper 9 - Math, Class 11 Commerce Questions
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The content of Sample Solution Paper 9 - Math, Class 11 is prepared as per the latest Commerce syllabus.
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