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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