Commerce Exam > Commerce Notes > Mathematics (Maths) Class 11 > Sample Solution Paper 9 - Math, Class 11
Sample Solution Paper 9 - Math, Class 11 | Mathematics (Maths) Class 11 - Commerce PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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.
Read More
|
75 videos|238 docs|91 tests
|
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.
About this Document
|
4.87/5 Rating |
|
Nov 24, 2024 Last updated |
Document Description: Sample Solution Paper 9 - Math, Class 11 for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation.
The notes and questions for Sample Solution Paper 9 - Math, Class 11 have been prepared according to the Commerce exam syllabus. Information about Sample Solution Paper 9 - Math, Class 11 covers topics
like and Sample Solution Paper 9 - Math, Class 11 Example, for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Sample Solution Paper 9 - Math, Class 11.
Introduction of Sample Solution Paper 9 - Math, Class 11 in English is available as part of our Mathematics (Maths) Class 11
for Commerce & Sample Solution Paper 9 - Math, Class 11 in Hindi for Mathematics (Maths) Class 11 course.
Download more important topics related with notes, lectures and mock test series for Commerce
Exam by signing up for free. Commerce: Sample Solution Paper 9 - Math, Class 11 | Mathematics (Maths) Class 11 - Commerce
Description
Full syllabus notes, lecture & questions for Sample Solution Paper 9 - Math, Class 11 | Mathematics (Maths) Class 11 - Commerce - Commerce | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 11 | Best notes, free PDF download
Information about Sample Solution Paper 9 - Math, Class 11
In this doc you can find the meaning of Sample Solution Paper 9 - Math, Class 11 defined & explained in the simplest way possible. Besides explaining types of
Sample Solution Paper 9 - Math, Class 11 theory, EduRev gives you an ample number of questions to practice Sample Solution Paper 9 - Math, Class 11 tests, examples and also practice Commerce
tests
|
Explore Courses for Commerce exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Free
,study material
,Sample Solution Paper 9 - Math
,Sample Solution Paper 9 - Math
,Important questions
,Sample Solution Paper 9 - Math
,Extra Questions
,Viva Questions
,past year papers
,video lectures
,Exam
,ppt
,Class 11 | Mathematics (Maths) Class 11 - Commerce
,MCQs
,shortcuts and tricks
,Class 11 | Mathematics (Maths) Class 11 - Commerce
,Previous Year Questions with Solutions
,Sample Paper
,Class 11 | Mathematics (Maths) Class 11 - Commerce
,Objective type Questions
,Summary
,Semester Notes
,mock tests for examination
,practice quizzes
;
Additional Information about Sample Solution Paper 9 - Math, Class 11 for Commerce Preparation
Sample Solution Paper 9 - Math, Class 11 Free PDF Download
The Sample Solution Paper 9 - Math, Class 11 is an invaluable resource that delves deep into the core of the Commerce exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Sample Solution Paper 9 - Math, Class 11 now and kickstart your journey towards success in the Commerce exam.
Importance of Sample Solution Paper 9 - Math, Class 11
The importance of Sample Solution Paper 9 - Math, Class 11 cannot be overstated, especially for Commerce aspirants.
This document holds the key to success in the Commerce exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Sample Solution Paper 9 - Math, Class 11 Notes
Sample Solution Paper 9 - Math, Class 11 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Sample Solution Paper 9 - Math, Class 11.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Sample Solution Paper 9 - Math, Class 11 Notes on EduRev are your ultimate resource for success.
Sample Solution Paper 9 - Math, Class 11 Commerce Questions
The "Sample Solution Paper 9 - Math, Class 11 Commerce Questions" guide is a valuable resource for all aspiring students preparing for the
Commerce exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Sample Solution Paper 9 - Math, Class 11 on the App
Students of Commerce can study Sample Solution Paper 9 - Math, Class 11 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Sample Solution Paper 9 - Math, Class 11,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Sample Solution Paper 9 - Math, Class 11 is prepared as per the latest Commerce syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup