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CBSE XI | Mathematics
Sample Paper – 6 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 6 Solution
SECTION – A
1. cos[sin v x]’ = -sin (sin v x) × cos v x × 1/2v x
2. He is not bald and he is not tall.
3.
2
2
6 6 i 6i
6i i 1
i i i i
? ? ? ? ? ?
? ? ?
OR
z = sin ? – i cos ?
Comparing with a + bi we get a = sin ? and b = - cos ?
2 2 2 2
z a b sin cos 1 ? ? ? ? ? ? ?
4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for two heads are {HH}.
Required probability =
1
4
SECTION – B
5. n(A) = 200, n(B) = 300 and n(A ? B) = 100
n(A’ ? B’) = n(A ? B)’
= 700 - n(A ? B)
= 700 – [n(A) + n(B) – n(A n B)]
= 700 – (200 + 300 – 100)
= 700 – 400
= 300
6. f(x) = vx
f(25) = 5, f(16) = 4 and f(1) = 1
Hence,
? ?
? ? ? ?
f 25
5
1
f 16 f 1 4 1
??
??
OR
Page 2
CBSE XI | Mathematics
Sample Paper – 6 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 6 Solution
SECTION – A
1. cos[sin v x]’ = -sin (sin v x) × cos v x × 1/2v x
2. He is not bald and he is not tall.
3.
2
2
6 6 i 6i
6i i 1
i i i i
? ? ? ? ? ?
? ? ?
OR
z = sin ? – i cos ?
Comparing with a + bi we get a = sin ? and b = - cos ?
2 2 2 2
z a b sin cos 1 ? ? ? ? ? ? ?
4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for two heads are {HH}.
Required probability =
1
4
SECTION – B
5. n(A) = 200, n(B) = 300 and n(A ? B) = 100
n(A’ ? B’) = n(A ? B)’
= 700 - n(A ? B)
= 700 – [n(A) + n(B) – n(A n B)]
= 700 – (200 + 300 – 100)
= 700 – 400
= 300
6. f(x) = vx
f(25) = 5, f(16) = 4 and f(1) = 1
Hence,
? ?
? ? ? ?
f 25
5
1
f 16 f 1 4 1
??
??
OR
CBSE XI | Mathematics
Sample Paper – 6 Solution
f(x) = x
2
– 1 and g(x) = vx
f ° g(x) = f[g(x)] = f(vx) = (vx)
2
– 1 = x – 1
7. f(x) = |x – 3|
f(x) is defined for all x ? R. Therefore, domain f = R
|x – 3| > 0 for all x ? R.
0 x 3 ? ? ? ? for all x ? R.
? ? ? ? f x 0, ?? for all x ? R.
Range of f is
? ? 0, ?
OR
n(A) = 50, n(A ? B) = 60 and n(A n B) = 10
n(A ? B) = n(A) + n(B) – n(An B)
60 = 50 + n(B) – 10
n(B) = 60 – 50 + 10 = 20
n(A – B) = n(A) – n(An B)
= 50 – 10 = 40
8. A × B = {(6, 3), (6, 5), (8, 3), (8, 5)}
A × A = {(6, 6), (6, 8), (8, 6), (8, 8)}
9. LHS =
cos A sin A cos A sin A
sin A cosA
1 tan A 1 cot A
11
cosA sin A
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?
22
cos A sin A
cosA sin A sin A cosA
? ? ? ?
??
? ? ? ?
? ? ? ?
??
? ? ? ?
? ?
22
cos A sin A
cosA sin A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ?
? ? ? ?
22
cos A sin A
cos A sin A
1
cos A sin A
?
?
?
?
?
= RHS
cos A sin A 1
1 tan A 1 cot A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ?
? ? ? ?
Page 3
CBSE XI | Mathematics
Sample Paper – 6 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 6 Solution
SECTION – A
1. cos[sin v x]’ = -sin (sin v x) × cos v x × 1/2v x
2. He is not bald and he is not tall.
3.
2
2
6 6 i 6i
6i i 1
i i i i
? ? ? ? ? ?
? ? ?
OR
z = sin ? – i cos ?
Comparing with a + bi we get a = sin ? and b = - cos ?
2 2 2 2
z a b sin cos 1 ? ? ? ? ? ? ?
4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for two heads are {HH}.
Required probability =
1
4
SECTION – B
5. n(A) = 200, n(B) = 300 and n(A ? B) = 100
n(A’ ? B’) = n(A ? B)’
= 700 - n(A ? B)
= 700 – [n(A) + n(B) – n(A n B)]
= 700 – (200 + 300 – 100)
= 700 – 400
= 300
6. f(x) = vx
f(25) = 5, f(16) = 4 and f(1) = 1
Hence,
? ?
? ? ? ?
f 25
5
1
f 16 f 1 4 1
??
??
OR
CBSE XI | Mathematics
Sample Paper – 6 Solution
f(x) = x
2
– 1 and g(x) = vx
f ° g(x) = f[g(x)] = f(vx) = (vx)
2
– 1 = x – 1
7. f(x) = |x – 3|
f(x) is defined for all x ? R. Therefore, domain f = R
|x – 3| > 0 for all x ? R.
0 x 3 ? ? ? ? for all x ? R.
? ? ? ? f x 0, ?? for all x ? R.
Range of f is
? ? 0, ?
OR
n(A) = 50, n(A ? B) = 60 and n(A n B) = 10
n(A ? B) = n(A) + n(B) – n(An B)
60 = 50 + n(B) – 10
n(B) = 60 – 50 + 10 = 20
n(A – B) = n(A) – n(An B)
= 50 – 10 = 40
8. A × B = {(6, 3), (6, 5), (8, 3), (8, 5)}
A × A = {(6, 6), (6, 8), (8, 6), (8, 8)}
9. LHS =
cos A sin A cos A sin A
sin A cosA
1 tan A 1 cot A
11
cosA sin A
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?
22
cos A sin A
cosA sin A sin A cosA
? ? ? ?
??
? ? ? ?
? ? ? ?
??
? ? ? ?
? ?
22
cos A sin A
cosA sin A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ?
? ? ? ?
22
cos A sin A
cos A sin A
1
cos A sin A
?
?
?
?
?
= RHS
cos A sin A 1
1 tan A 1 cot A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ?
? ? ? ?
CBSE XI | Mathematics
Sample Paper – 6 Solution
OR
tan
4
? + tan
2
? = (tan
2
?)
2
+ tan
2
?
= (sec
2
? - 1)
2
+ sec
2
? - 1
= sec
4
? - 2sec
2
? + 1 + sec
2
? - 1
= sec
4
? - sec
2
?
10. We observe that 9 is an odd number which is not prime. Similarly, 21, 25 etc are odd
integers which are not prime.
11. a, b, c are in AP
b
2
= ac
(b
2
)
n
= (ac)
n
b
2n
= a
n
c
n
log b
2n
= log a
n
c
n
log (b
n
)
2
= log a
n
+ log c
n
2log b
n
= log a
n
+ log c
n
log a
n
, log b
n
and log c
n
are in AP
12. y
2
= 12x comparing with y
2
= 4ax,we get a = 3
The focal distance of any point (x, y) on y
2
= 4ax is x + a.
The focal distance is x + 3.
?x + 3 = 4
?x = 1
Hence, the abscissa of the given point is 1.
SECTION – C
13. tan (pcos ?) = cot (psin ?)
? ?
? ?
? ?
? ?
sin cos cos sin
cos cos sin sin
? ? ? ?
?
? ? ? ?
sin(pcos ?) sin(psin ?) = cos (psin ?) cos (pcos ?)
cos (psin ?) cos (pcos ?) - sin(pcos ?) sin(psin ?) = 0
cos (pcos ? + psin ?) = 0
pcos ? + psin ? =
2
?
?
cos ? + sin ? =
1
2
?
1 1 1
cos sin
2 2 2 2
? ? ? ? ?
Page 4
CBSE XI | Mathematics
Sample Paper – 6 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 6 Solution
SECTION – A
1. cos[sin v x]’ = -sin (sin v x) × cos v x × 1/2v x
2. He is not bald and he is not tall.
3.
2
2
6 6 i 6i
6i i 1
i i i i
? ? ? ? ? ?
? ? ?
OR
z = sin ? – i cos ?
Comparing with a + bi we get a = sin ? and b = - cos ?
2 2 2 2
z a b sin cos 1 ? ? ? ? ? ? ?
4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for two heads are {HH}.
Required probability =
1
4
SECTION – B
5. n(A) = 200, n(B) = 300 and n(A ? B) = 100
n(A’ ? B’) = n(A ? B)’
= 700 - n(A ? B)
= 700 – [n(A) + n(B) – n(A n B)]
= 700 – (200 + 300 – 100)
= 700 – 400
= 300
6. f(x) = vx
f(25) = 5, f(16) = 4 and f(1) = 1
Hence,
? ?
? ? ? ?
f 25
5
1
f 16 f 1 4 1
??
??
OR
CBSE XI | Mathematics
Sample Paper – 6 Solution
f(x) = x
2
– 1 and g(x) = vx
f ° g(x) = f[g(x)] = f(vx) = (vx)
2
– 1 = x – 1
7. f(x) = |x – 3|
f(x) is defined for all x ? R. Therefore, domain f = R
|x – 3| > 0 for all x ? R.
0 x 3 ? ? ? ? for all x ? R.
? ? ? ? f x 0, ?? for all x ? R.
Range of f is
? ? 0, ?
OR
n(A) = 50, n(A ? B) = 60 and n(A n B) = 10
n(A ? B) = n(A) + n(B) – n(An B)
60 = 50 + n(B) – 10
n(B) = 60 – 50 + 10 = 20
n(A – B) = n(A) – n(An B)
= 50 – 10 = 40
8. A × B = {(6, 3), (6, 5), (8, 3), (8, 5)}
A × A = {(6, 6), (6, 8), (8, 6), (8, 8)}
9. LHS =
cos A sin A cos A sin A
sin A cosA
1 tan A 1 cot A
11
cosA sin A
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?
22
cos A sin A
cosA sin A sin A cosA
? ? ? ?
??
? ? ? ?
? ? ? ?
??
? ? ? ?
? ?
22
cos A sin A
cosA sin A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ?
? ? ? ?
22
cos A sin A
cos A sin A
1
cos A sin A
?
?
?
?
?
= RHS
cos A sin A 1
1 tan A 1 cot A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ?
? ? ? ?
CBSE XI | Mathematics
Sample Paper – 6 Solution
OR
tan
4
? + tan
2
? = (tan
2
?)
2
+ tan
2
?
= (sec
2
? - 1)
2
+ sec
2
? - 1
= sec
4
? - 2sec
2
? + 1 + sec
2
? - 1
= sec
4
? - sec
2
?
10. We observe that 9 is an odd number which is not prime. Similarly, 21, 25 etc are odd
integers which are not prime.
11. a, b, c are in AP
b
2
= ac
(b
2
)
n
= (ac)
n
b
2n
= a
n
c
n
log b
2n
= log a
n
c
n
log (b
n
)
2
= log a
n
+ log c
n
2log b
n
= log a
n
+ log c
n
log a
n
, log b
n
and log c
n
are in AP
12. y
2
= 12x comparing with y
2
= 4ax,we get a = 3
The focal distance of any point (x, y) on y
2
= 4ax is x + a.
The focal distance is x + 3.
?x + 3 = 4
?x = 1
Hence, the abscissa of the given point is 1.
SECTION – C
13. tan (pcos ?) = cot (psin ?)
? ?
? ?
? ?
? ?
sin cos cos sin
cos cos sin sin
? ? ? ?
?
? ? ? ?
sin(pcos ?) sin(psin ?) = cos (psin ?) cos (pcos ?)
cos (psin ?) cos (pcos ?) - sin(pcos ?) sin(psin ?) = 0
cos (pcos ? + psin ?) = 0
pcos ? + psin ? =
2
?
?
cos ? + sin ? =
1
2
?
1 1 1
cos sin
2 2 2 2
? ? ? ? ?
CBSE XI | Mathematics
Sample Paper – 6 Solution
1
cos cos sin sin
44
22
??
? ? ? ? ?
1
cos
4
22
? ??
? ? ? ?
??
??
14. For any a ?R we have 1 + a
2
> 0 for (a, a) ?R1
(a, a) ?R1 for all a ?R
Let (a, b) ?R1 then,
1 + ab > 0
? 1 + ba > 0
? (b, a) ?R1
(a, b) ?R1 ? (b, a) ?R1 for all a, b ?R
15. Let
9
?
??
Let C = cos cos2 cos3 cos4 ? ? ? ?
S = sin sin2 sin3 sin4 ? ? ? ?
C × S = ? ? ? ? ? ? ? ? sin cos sin2 cos2 sin3 cos3 sin4 cos4 ? ? ? ? ? ? ? ?
=
1 1 1 1
sin2 sin4 sin6 sin8
2 2 2 2
? ? ? ?
But
9
?
?? ? 9 ? ? ?
? ? ? sin8 sin sin ? ? ? ? ? ? ? and ? ? sin6 sin 3 sin3 ? ? ? ? ? ? ?
C × S =
44
11
sin2 sin4 sin3 sin S
22
? ? ? ? ? ? ?
C =
4
1
2
? S ? 0
16. Let
2
, ? ? ? ? ? ? then
2 3 2
, 1, 1 ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ?
22
2 3 2 4 2 3
2 2 2 3 4 3
2 2 2
22
33
xyz a b a b a b
a b a b a b
a b a ab ab b
a b a ab ab b 1,
a b a ab b
a b a ab b
xyz a b
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
??
??
Page 5
CBSE XI | Mathematics
Sample Paper – 6 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 6 Solution
SECTION – A
1. cos[sin v x]’ = -sin (sin v x) × cos v x × 1/2v x
2. He is not bald and he is not tall.
3.
2
2
6 6 i 6i
6i i 1
i i i i
? ? ? ? ? ?
? ? ?
OR
z = sin ? – i cos ?
Comparing with a + bi we get a = sin ? and b = - cos ?
2 2 2 2
z a b sin cos 1 ? ? ? ? ? ? ?
4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for two heads are {HH}.
Required probability =
1
4
SECTION – B
5. n(A) = 200, n(B) = 300 and n(A ? B) = 100
n(A’ ? B’) = n(A ? B)’
= 700 - n(A ? B)
= 700 – [n(A) + n(B) – n(A n B)]
= 700 – (200 + 300 – 100)
= 700 – 400
= 300
6. f(x) = vx
f(25) = 5, f(16) = 4 and f(1) = 1
Hence,
? ?
? ? ? ?
f 25
5
1
f 16 f 1 4 1
??
??
OR
CBSE XI | Mathematics
Sample Paper – 6 Solution
f(x) = x
2
– 1 and g(x) = vx
f ° g(x) = f[g(x)] = f(vx) = (vx)
2
– 1 = x – 1
7. f(x) = |x – 3|
f(x) is defined for all x ? R. Therefore, domain f = R
|x – 3| > 0 for all x ? R.
0 x 3 ? ? ? ? for all x ? R.
? ? ? ? f x 0, ?? for all x ? R.
Range of f is
? ? 0, ?
OR
n(A) = 50, n(A ? B) = 60 and n(A n B) = 10
n(A ? B) = n(A) + n(B) – n(An B)
60 = 50 + n(B) – 10
n(B) = 60 – 50 + 10 = 20
n(A – B) = n(A) – n(An B)
= 50 – 10 = 40
8. A × B = {(6, 3), (6, 5), (8, 3), (8, 5)}
A × A = {(6, 6), (6, 8), (8, 6), (8, 8)}
9. LHS =
cos A sin A cos A sin A
sin A cosA
1 tan A 1 cot A
11
cosA sin A
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?
22
cos A sin A
cosA sin A sin A cosA
? ? ? ?
??
? ? ? ?
? ? ? ?
??
? ? ? ?
? ?
22
cos A sin A
cosA sin A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ?
? ? ? ?
22
cos A sin A
cos A sin A
1
cos A sin A
?
?
?
?
?
= RHS
cos A sin A 1
1 tan A 1 cot A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ?
? ? ? ?
CBSE XI | Mathematics
Sample Paper – 6 Solution
OR
tan
4
? + tan
2
? = (tan
2
?)
2
+ tan
2
?
= (sec
2
? - 1)
2
+ sec
2
? - 1
= sec
4
? - 2sec
2
? + 1 + sec
2
? - 1
= sec
4
? - sec
2
?
10. We observe that 9 is an odd number which is not prime. Similarly, 21, 25 etc are odd
integers which are not prime.
11. a, b, c are in AP
b
2
= ac
(b
2
)
n
= (ac)
n
b
2n
= a
n
c
n
log b
2n
= log a
n
c
n
log (b
n
)
2
= log a
n
+ log c
n
2log b
n
= log a
n
+ log c
n
log a
n
, log b
n
and log c
n
are in AP
12. y
2
= 12x comparing with y
2
= 4ax,we get a = 3
The focal distance of any point (x, y) on y
2
= 4ax is x + a.
The focal distance is x + 3.
?x + 3 = 4
?x = 1
Hence, the abscissa of the given point is 1.
SECTION – C
13. tan (pcos ?) = cot (psin ?)
? ?
? ?
? ?
? ?
sin cos cos sin
cos cos sin sin
? ? ? ?
?
? ? ? ?
sin(pcos ?) sin(psin ?) = cos (psin ?) cos (pcos ?)
cos (psin ?) cos (pcos ?) - sin(pcos ?) sin(psin ?) = 0
cos (pcos ? + psin ?) = 0
pcos ? + psin ? =
2
?
?
cos ? + sin ? =
1
2
?
1 1 1
cos sin
2 2 2 2
? ? ? ? ?
CBSE XI | Mathematics
Sample Paper – 6 Solution
1
cos cos sin sin
44
22
??
? ? ? ? ?
1
cos
4
22
? ??
? ? ? ?
??
??
14. For any a ?R we have 1 + a
2
> 0 for (a, a) ?R1
(a, a) ?R1 for all a ?R
Let (a, b) ?R1 then,
1 + ab > 0
? 1 + ba > 0
? (b, a) ?R1
(a, b) ?R1 ? (b, a) ?R1 for all a, b ?R
15. Let
9
?
??
Let C = cos cos2 cos3 cos4 ? ? ? ?
S = sin sin2 sin3 sin4 ? ? ? ?
C × S = ? ? ? ? ? ? ? ? sin cos sin2 cos2 sin3 cos3 sin4 cos4 ? ? ? ? ? ? ? ?
=
1 1 1 1
sin2 sin4 sin6 sin8
2 2 2 2
? ? ? ?
But
9
?
?? ? 9 ? ? ?
? ? ? sin8 sin sin ? ? ? ? ? ? ? and ? ? sin6 sin 3 sin3 ? ? ? ? ? ? ?
C × S =
44
11
sin2 sin4 sin3 sin S
22
? ? ? ? ? ? ?
C =
4
1
2
? S ? 0
16. Let
2
, ? ? ? ? ? ? then
2 3 2
, 1, 1 ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ?
22
2 3 2 4 2 3
2 2 2 3 4 3
2 2 2
22
33
xyz a b a b a b
a b a b a b
a b a ab ab b
a b a ab ab b 1,
a b a ab b
a b a ab b
xyz a b
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
??
??
CBSE XI | Mathematics
Sample Paper – 6 Solution
17. Let A denote the event : both the balls are of the same colour.
B : both are white
C : both are red
A = B ? C ? D
P(A) = P (B ? C ? D)
= P(B) + P(C) + P(D) B, C and D are mutually exclusive.
P(B) =
? ?
? ?
7
2
16
2
nB
C 21
n S C 120
??
P(C) =
? ?
? ?
5
2
16
2
nC
C 10
n S C 120
??
P(D) =
? ?
? ?
4
2
16
2
nD
C 6
n S C 120
??
P(A) =
21 10 6 37
120 120 120 120
? ? ?
Required probability = 1 – P(A) =
83
120
18. Let A be the first term and D the common difference. It is given that Sp = a
? ?
p
2A p 1 D a
2
?? ? ? ?
??
? ? p1
a
AD
2p
?
?? ………..(i)
Similarly,
? ? q1
b
AD
2q
?
?? ……….(ii)
and
? ? r1
c
AD
2r
?
?? ……….(iii)
Multiplying by (i) by (q – r), (ii) by (r – p) and (iii) by (p – q) and adding
? ? ? ? ? ? ? ? ? ? ? ? ? ?
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FAQs on Sample Solution Paper 6 - Math, Class 11 - Mathematics (Maths) Class 11 - Commerce
| 1. What is the importance of studying mathematics in Class 11? |  |
Ans. Studying mathematics in Class 11 is important as it forms the foundation for higher-level mathematics courses. It helps in developing logical thinking and problem-solving skills. Additionally, mathematical concepts are widely used in various fields like science, engineering, economics, and computer science.
| 2. How can I improve my performance in Class 11 mathematics? | |
Ans. To improve performance in Class 11 mathematics, it is essential to practice regularly. Solve a variety of problems from textbooks and reference books. Seek help from teachers or classmates whenever necessary. Understanding the concepts thoroughly and revising regularly can also contribute to better performance.
| 3. What are the key topics covered in the Class 11 mathematics exam? | |
Ans. The Class 11 mathematics exam typically covers topics such as sets, relations, and functions, algebra, coordinate geometry, calculus, statistics, and probability. It is important to have a clear understanding of these topics and practice solving problems related to them.
| 4. How can I manage my time effectively during the Class 11 mathematics exam? | |
Ans. Time management during the Class 11 mathematics exam is crucial. Start by familiarizing yourself with the exam pattern and allocate time for each section accordingly. Practice solving previous year question papers within the given time limit. During the exam, read the questions carefully, prioritize the easier ones, and avoid spending too much time on a single question.
| 5. Are there any online resources available for studying Class 11 mathematics? | |
Ans. Yes, there are several online resources available for studying Class 11 mathematics. Websites like Khan Academy, BYJU'S, and NCERT's official website provide video lessons, practice exercises, and sample papers for Class 11 mathematics. Additionally, there are various educational apps and YouTube channels dedicated to teaching mathematics concepts for Class 11 students.
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Document Description: Sample Solution Paper 6 - Math, Class 11 for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation.
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