Commerce Exam > Commerce Notes > Mathematics (Maths) Class 11 > Sample Solution Paper 1 - Math, Class 11
Sample Solution Paper 1 - 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 – 1 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 1 Solution
SECTION – A
1. [sin(x+1)]’ = cos (x+1) .1 = cos (x+1)
2. Giving one counter example is enough to prove the falsehood of a statement. Here
counter example is: The real number 1 is neither prime nor composite. So the statement
is false.
3.
? ? ? ?
2 2 2
1 3i 1 2i 1 6 3i 2i 5 5i 5 5i 5 5i
1i
1 2i 1 2i 1 4 5
1 4i
1 2i
? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ?
?
?
OR
i
n
+ i
n + 1
+ i
n + 2
+ i
n + 3
= i
n
(1 + i + i
2
+ i
3
) = i
n
(1 + i – 1 – i) = 0 ? i
2
= -1 and i
3
= -i
4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for atleast one head are {HH, HT, TH}.
Required probability =
3
4
SECTION – B
5. n(A - B) = 14 + x, n(B - A) = 3x and n(A ? B) = x
n(A) = n(B)
n(A) = n(A - B) + n(A ? B);
n(B) = n(B - A) + n(A ? B)
? n(A - B) + n(A ? B) = n(B - A) + n(A ? B)
? 14 + x + x = 3x + x
?14 = 2 x ?x = 7
Page 2
CBSE XI | Mathematics
Sample Paper – 1 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 1 Solution
SECTION – A
1. [sin(x+1)]’ = cos (x+1) .1 = cos (x+1)
2. Giving one counter example is enough to prove the falsehood of a statement. Here
counter example is: The real number 1 is neither prime nor composite. So the statement
is false.
3.
? ? ? ?
2 2 2
1 3i 1 2i 1 6 3i 2i 5 5i 5 5i 5 5i
1i
1 2i 1 2i 1 4 5
1 4i
1 2i
? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ?
?
?
OR
i
n
+ i
n + 1
+ i
n + 2
+ i
n + 3
= i
n
(1 + i + i
2
+ i
3
) = i
n
(1 + i – 1 – i) = 0 ? i
2
= -1 and i
3
= -i
4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for atleast one head are {HH, HT, TH}.
Required probability =
3
4
SECTION – B
5. n(A - B) = 14 + x, n(B - A) = 3x and n(A ? B) = x
n(A) = n(B)
n(A) = n(A - B) + n(A ? B);
n(B) = n(B - A) + n(A ? B)
? n(A - B) + n(A ? B) = n(B - A) + n(A ? B)
? 14 + x + x = 3x + x
?14 = 2 x ?x = 7
CBSE XI | Mathematics
Sample Paper – 1 Solution
6. Let a be any element which belongs to set A, i.e a ? A
P(A) is the set of all subsets of the set A. Therefore {a} belongs to P(A)
i.e {a} ? P(A)
But P(A) = P(B) [ Given ]
? {a} ? P(B)
? a ? B
So a ? A ? a ? B, Hence A ? B
Similarly, we can prove that A ? B
? A = B
OR
aN = {ax : x ?N}
3N = {3x : x ? N} = {3, 6, 9,….}
And 7N = {7x : x ? N} = {7, 14, 21, 28,…}
3N n 7N = {21, 42, …} = {21x : x ? N} = 21N
7. Let A be the set of teachers who teach math and B be the set of teachers who tech
physics. Then,
n(A ? B) = 11, n(A) = 7 & n(A ? B) = 3
n(A ? B) = n(A) + n(B) - n(A ? B)
11 = 7 - n(B) + 3
n(B) = 7
OR
n(A) = 20, n(A ? B) = 42 and n(A n B) = 4
n(A ? B) = n(A) + n(B) – n(An B)
42 = 20 + n(B) – 4
n(B) = 42 – 20 + 4 = 26
n(A – B) = n(A) – n(An B)
= 20 – 4 = 16
8. A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
Since A × B contains 4 elements therefore it has 2
4
= 16 subsets
Subsets are
{ },
{(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)},
{(1,3), (1,4)}, {(1,3), (2,3)}, {(1,3), (2,4)}, {(1,4), (2,3)}, {(1,4), (2,4)},
{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 4), (2, 3), (2, 4)},
{(1, 3), (1, 4), (2, 3), (2, 4)}.
Page 3
CBSE XI | Mathematics
Sample Paper – 1 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 1 Solution
SECTION – A
1. [sin(x+1)]’ = cos (x+1) .1 = cos (x+1)
2. Giving one counter example is enough to prove the falsehood of a statement. Here
counter example is: The real number 1 is neither prime nor composite. So the statement
is false.
3.
? ? ? ?
2 2 2
1 3i 1 2i 1 6 3i 2i 5 5i 5 5i 5 5i
1i
1 2i 1 2i 1 4 5
1 4i
1 2i
? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ?
?
?
OR
i
n
+ i
n + 1
+ i
n + 2
+ i
n + 3
= i
n
(1 + i + i
2
+ i
3
) = i
n
(1 + i – 1 – i) = 0 ? i
2
= -1 and i
3
= -i
4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for atleast one head are {HH, HT, TH}.
Required probability =
3
4
SECTION – B
5. n(A - B) = 14 + x, n(B - A) = 3x and n(A ? B) = x
n(A) = n(B)
n(A) = n(A - B) + n(A ? B);
n(B) = n(B - A) + n(A ? B)
? n(A - B) + n(A ? B) = n(B - A) + n(A ? B)
? 14 + x + x = 3x + x
?14 = 2 x ?x = 7
CBSE XI | Mathematics
Sample Paper – 1 Solution
6. Let a be any element which belongs to set A, i.e a ? A
P(A) is the set of all subsets of the set A. Therefore {a} belongs to P(A)
i.e {a} ? P(A)
But P(A) = P(B) [ Given ]
? {a} ? P(B)
? a ? B
So a ? A ? a ? B, Hence A ? B
Similarly, we can prove that A ? B
? A = B
OR
aN = {ax : x ?N}
3N = {3x : x ? N} = {3, 6, 9,….}
And 7N = {7x : x ? N} = {7, 14, 21, 28,…}
3N n 7N = {21, 42, …} = {21x : x ? N} = 21N
7. Let A be the set of teachers who teach math and B be the set of teachers who tech
physics. Then,
n(A ? B) = 11, n(A) = 7 & n(A ? B) = 3
n(A ? B) = n(A) + n(B) - n(A ? B)
11 = 7 - n(B) + 3
n(B) = 7
OR
n(A) = 20, n(A ? B) = 42 and n(A n B) = 4
n(A ? B) = n(A) + n(B) – n(An B)
42 = 20 + n(B) – 4
n(B) = 42 – 20 + 4 = 26
n(A – B) = n(A) – n(An B)
= 20 – 4 = 16
8. A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
Since A × B contains 4 elements therefore it has 2
4
= 16 subsets
Subsets are
{ },
{(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)},
{(1,3), (1,4)}, {(1,3), (2,3)}, {(1,3), (2,4)}, {(1,4), (2,3)}, {(1,4), (2,4)},
{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 4), (2, 3), (2, 4)},
{(1, 3), (1, 4), (2, 3), (2, 4)}.
CBSE XI | Mathematics
Sample Paper – 1 Solution
9.
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?
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 cos A sin A
cos A sin A
cosA sin A cosA sin A
cos A sin A
OR
cot
4
? + cot
2
? = (cot
2
?)
2
+ cot
2
?
= (cosec
2
? - 1)
2
+ cosec
2
? - 1
= cosec
4
? - 2cosec
2
? + 1 + cosec
2
? - 1
= cosec
4
? - cosec
2
?
10.
1. If a number is not divisible by 3, it is not divisible by 9.
2. If you are not citizen of India, then you were not born in India.
11. ? ? ? ? ?
3 3 3 3
The given series : 10 11 12 .. 20
? ? ? ?
??
?? ??
?
??
? ? ? ? ?
?
??
?
? ? ? ?
??
? ? ? ? ? ?
?
?
??
20 10
33
r1
3 3 3 3 3 3
r1
3
2
3
2
rr
20(
Can be written as
1 2 3 .. 20 1 2 3
20 1) 9(9
.. 9
1)
22
42075
12. ? ? ? ? ? ? ? Vertices of a parallelogram ABCD are A 4, 11 , B 5, 3 & C 2, 15
? ?
? ? ? ??
?
??
??
?? ??
?
??
??
let D x,y
4 2 11 15
then mid point of diagonal AC ,
22
5 x 3 y
also mid point of diagonal BD ,
22
Page 4
CBSE XI | Mathematics
Sample Paper – 1 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 1 Solution
SECTION – A
1. [sin(x+1)]’ = cos (x+1) .1 = cos (x+1)
2. Giving one counter example is enough to prove the falsehood of a statement. Here
counter example is: The real number 1 is neither prime nor composite. So the statement
is false.
3.
? ? ? ?
2 2 2
1 3i 1 2i 1 6 3i 2i 5 5i 5 5i 5 5i
1i
1 2i 1 2i 1 4 5
1 4i
1 2i
? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ?
?
?
OR
i
n
+ i
n + 1
+ i
n + 2
+ i
n + 3
= i
n
(1 + i + i
2
+ i
3
) = i
n
(1 + i – 1 – i) = 0 ? i
2
= -1 and i
3
= -i
4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for atleast one head are {HH, HT, TH}.
Required probability =
3
4
SECTION – B
5. n(A - B) = 14 + x, n(B - A) = 3x and n(A ? B) = x
n(A) = n(B)
n(A) = n(A - B) + n(A ? B);
n(B) = n(B - A) + n(A ? B)
? n(A - B) + n(A ? B) = n(B - A) + n(A ? B)
? 14 + x + x = 3x + x
?14 = 2 x ?x = 7
CBSE XI | Mathematics
Sample Paper – 1 Solution
6. Let a be any element which belongs to set A, i.e a ? A
P(A) is the set of all subsets of the set A. Therefore {a} belongs to P(A)
i.e {a} ? P(A)
But P(A) = P(B) [ Given ]
? {a} ? P(B)
? a ? B
So a ? A ? a ? B, Hence A ? B
Similarly, we can prove that A ? B
? A = B
OR
aN = {ax : x ?N}
3N = {3x : x ? N} = {3, 6, 9,….}
And 7N = {7x : x ? N} = {7, 14, 21, 28,…}
3N n 7N = {21, 42, …} = {21x : x ? N} = 21N
7. Let A be the set of teachers who teach math and B be the set of teachers who tech
physics. Then,
n(A ? B) = 11, n(A) = 7 & n(A ? B) = 3
n(A ? B) = n(A) + n(B) - n(A ? B)
11 = 7 - n(B) + 3
n(B) = 7
OR
n(A) = 20, n(A ? B) = 42 and n(A n B) = 4
n(A ? B) = n(A) + n(B) – n(An B)
42 = 20 + n(B) – 4
n(B) = 42 – 20 + 4 = 26
n(A – B) = n(A) – n(An B)
= 20 – 4 = 16
8. A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
Since A × B contains 4 elements therefore it has 2
4
= 16 subsets
Subsets are
{ },
{(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)},
{(1,3), (1,4)}, {(1,3), (2,3)}, {(1,3), (2,4)}, {(1,4), (2,3)}, {(1,4), (2,4)},
{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 4), (2, 3), (2, 4)},
{(1, 3), (1, 4), (2, 3), (2, 4)}.
CBSE XI | Mathematics
Sample Paper – 1 Solution
9.
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?
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 cos A sin A
cos A sin A
cosA sin A cosA sin A
cos A sin A
OR
cot
4
? + cot
2
? = (cot
2
?)
2
+ cot
2
?
= (cosec
2
? - 1)
2
+ cosec
2
? - 1
= cosec
4
? - 2cosec
2
? + 1 + cosec
2
? - 1
= cosec
4
? - cosec
2
?
10.
1. If a number is not divisible by 3, it is not divisible by 9.
2. If you are not citizen of India, then you were not born in India.
11. ? ? ? ? ?
3 3 3 3
The given series : 10 11 12 .. 20
? ? ? ?
??
?? ??
?
??
? ? ? ? ?
?
??
?
? ? ? ?
??
? ? ? ? ? ?
?
?
??
20 10
33
r1
3 3 3 3 3 3
r1
3
2
3
2
rr
20(
Can be written as
1 2 3 .. 20 1 2 3
20 1) 9(9
.. 9
1)
22
42075
12. ? ? ? ? ? ? ? Vertices of a parallelogram ABCD are A 4, 11 , B 5, 3 & C 2, 15
? ?
? ? ? ??
?
??
??
?? ??
?
??
??
let D x,y
4 2 11 15
then mid point of diagonal AC ,
22
5 x 3 y
also mid point of diagonal BD ,
22
CBSE XI | Mathematics
Sample Paper – 1 Solution
? ? ? ? ? ? ? ? ?
?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ?
but midpoint of AC BD coincides
so,
4 2 11 15 5 x 3 y
,,
2 2 2 2
5 x 4 2........... x 1
3 y 11 15......... y 1
SECTION – C
13. f(x) 2x 1,g(x) 2x 3;x R ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
22
f g (x) f(x) g(x) 2x 1 2x 3 4x 2;x R
f g (x) f(x) g(x) 2x 1 2x 3 4
(fg)(x) f(x)g(x) 2x 1 2x 3 4x 2x 6x 3 4x 4x 3
f f(x) 2x 1 3
(x) ;x R
g g(x) 2x 3 2
? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ?
?? ? ??
? ? ? ? ?
??
??
?
?? ??
14.
{(a, b), a = b
4
, a, b ? N}
(i) (a, a) ? R ? a = a
4
,
which is true for a = 1 only, not for other values of a ? N
? Relation is not reflexive
(ii) {(a, b), a = b
4
, a, b ? N} and {(b, a),
b = a
4
, a, b ? N}
a = b
4
and b =
a
4
cannot be true simultaneously
? Relation is not symmetric.
(iii) {(a, b), a =b
4
, a, b ? N}; {(b, c), b =c
4
, b, c ?N}
? a = b
4
= c
16
So a ? c
4
? (a, c) ? R
? Relation is not transitive.
(iv) Since the relation is not reflexive, not symmetric, and also not transitive,
? Relation is not an equivalence relation
15. We know,
2 2 2
2 2 2
b c a 25 49 9 65 13
cos A =
2bc 2(5)(7) 70 14
a c b 9 49 25 33
cos B =
2ac 2(3)(7) 42
? ? ? ?
? ? ?
? ? ? ?
??
Page 5
CBSE XI | Mathematics
Sample Paper – 1 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 1 Solution
SECTION – A
1. [sin(x+1)]’ = cos (x+1) .1 = cos (x+1)
2. Giving one counter example is enough to prove the falsehood of a statement. Here
counter example is: The real number 1 is neither prime nor composite. So the statement
is false.
3.
? ? ? ?
2 2 2
1 3i 1 2i 1 6 3i 2i 5 5i 5 5i 5 5i
1i
1 2i 1 2i 1 4 5
1 4i
1 2i
? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ?
?
?
OR
i
n
+ i
n + 1
+ i
n + 2
+ i
n + 3
= i
n
(1 + i + i
2
+ i
3
) = i
n
(1 + i – 1 – i) = 0 ? i
2
= -1 and i
3
= -i
4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for atleast one head are {HH, HT, TH}.
Required probability =
3
4
SECTION – B
5. n(A - B) = 14 + x, n(B - A) = 3x and n(A ? B) = x
n(A) = n(B)
n(A) = n(A - B) + n(A ? B);
n(B) = n(B - A) + n(A ? B)
? n(A - B) + n(A ? B) = n(B - A) + n(A ? B)
? 14 + x + x = 3x + x
?14 = 2 x ?x = 7
CBSE XI | Mathematics
Sample Paper – 1 Solution
6. Let a be any element which belongs to set A, i.e a ? A
P(A) is the set of all subsets of the set A. Therefore {a} belongs to P(A)
i.e {a} ? P(A)
But P(A) = P(B) [ Given ]
? {a} ? P(B)
? a ? B
So a ? A ? a ? B, Hence A ? B
Similarly, we can prove that A ? B
? A = B
OR
aN = {ax : x ?N}
3N = {3x : x ? N} = {3, 6, 9,….}
And 7N = {7x : x ? N} = {7, 14, 21, 28,…}
3N n 7N = {21, 42, …} = {21x : x ? N} = 21N
7. Let A be the set of teachers who teach math and B be the set of teachers who tech
physics. Then,
n(A ? B) = 11, n(A) = 7 & n(A ? B) = 3
n(A ? B) = n(A) + n(B) - n(A ? B)
11 = 7 - n(B) + 3
n(B) = 7
OR
n(A) = 20, n(A ? B) = 42 and n(A n B) = 4
n(A ? B) = n(A) + n(B) – n(An B)
42 = 20 + n(B) – 4
n(B) = 42 – 20 + 4 = 26
n(A – B) = n(A) – n(An B)
= 20 – 4 = 16
8. A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
Since A × B contains 4 elements therefore it has 2
4
= 16 subsets
Subsets are
{ },
{(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)},
{(1,3), (1,4)}, {(1,3), (2,3)}, {(1,3), (2,4)}, {(1,4), (2,3)}, {(1,4), (2,4)},
{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 4), (2, 3), (2, 4)},
{(1, 3), (1, 4), (2, 3), (2, 4)}.
CBSE XI | Mathematics
Sample Paper – 1 Solution
9.
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?
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 cos A sin A
cos A sin A
cosA sin A cosA sin A
cos A sin A
OR
cot
4
? + cot
2
? = (cot
2
?)
2
+ cot
2
?
= (cosec
2
? - 1)
2
+ cosec
2
? - 1
= cosec
4
? - 2cosec
2
? + 1 + cosec
2
? - 1
= cosec
4
? - cosec
2
?
10.
1. If a number is not divisible by 3, it is not divisible by 9.
2. If you are not citizen of India, then you were not born in India.
11. ? ? ? ? ?
3 3 3 3
The given series : 10 11 12 .. 20
? ? ? ?
??
?? ??
?
??
? ? ? ? ?
?
??
?
? ? ? ?
??
? ? ? ? ? ?
?
?
??
20 10
33
r1
3 3 3 3 3 3
r1
3
2
3
2
rr
20(
Can be written as
1 2 3 .. 20 1 2 3
20 1) 9(9
.. 9
1)
22
42075
12. ? ? ? ? ? ? ? Vertices of a parallelogram ABCD are A 4, 11 , B 5, 3 & C 2, 15
? ?
? ? ? ??
?
??
??
?? ??
?
??
??
let D x,y
4 2 11 15
then mid point of diagonal AC ,
22
5 x 3 y
also mid point of diagonal BD ,
22
CBSE XI | Mathematics
Sample Paper – 1 Solution
? ? ? ? ? ? ? ? ?
?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ?
but midpoint of AC BD coincides
so,
4 2 11 15 5 x 3 y
,,
2 2 2 2
5 x 4 2........... x 1
3 y 11 15......... y 1
SECTION – C
13. f(x) 2x 1,g(x) 2x 3;x R ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
22
f g (x) f(x) g(x) 2x 1 2x 3 4x 2;x R
f g (x) f(x) g(x) 2x 1 2x 3 4
(fg)(x) f(x)g(x) 2x 1 2x 3 4x 2x 6x 3 4x 4x 3
f f(x) 2x 1 3
(x) ;x R
g g(x) 2x 3 2
? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ?
?? ? ??
? ? ? ? ?
??
??
?
?? ??
14.
{(a, b), a = b
4
, a, b ? N}
(i) (a, a) ? R ? a = a
4
,
which is true for a = 1 only, not for other values of a ? N
? Relation is not reflexive
(ii) {(a, b), a = b
4
, a, b ? N} and {(b, a),
b = a
4
, a, b ? N}
a = b
4
and b =
a
4
cannot be true simultaneously
? Relation is not symmetric.
(iii) {(a, b), a =b
4
, a, b ? N}; {(b, c), b =c
4
, b, c ?N}
? a = b
4
= c
16
So a ? c
4
? (a, c) ? R
? Relation is not transitive.
(iv) Since the relation is not reflexive, not symmetric, and also not transitive,
? Relation is not an equivalence relation
15. We know,
2 2 2
2 2 2
b c a 25 49 9 65 13
cos A =
2bc 2(5)(7) 70 14
a c b 9 49 25 33
cos B =
2ac 2(3)(7) 42
? ? ? ?
? ? ?
? ? ? ?
??
CBSE XI | Mathematics
Sample Paper – 1 Solution
16. Let 5 12i x yi ? ? ?
? ?
22
5 12i x y 2xy i ? ? ? ? ?
Equating real and imaginary parts, we get
x
2
– y
2
= 5…….(i) and
2xy = -12..…….(ii)
Now (x
2
+ y
2
)
2
= (x
2
- y
2
)
2
+ (2xy)
2
= 5
2
+ 12
2
= 169
? x
2
+ y
2
= 13 ………………..(iii)
From (i) and (iii), we get
2x
2
= 18 ? x = ? 3
and y = ? 2
From equation (ii) we can say that xy is negative.
As xy is negative ? when x = 3, y = -2
and when x = -3, y = 2
The required square roots are (3 - 2i) and (-3 + 2i)
or ? (3 - 2i)
17. Total number of possible sets of 7 cards =
52
C7
Number of sets of 7 with all 4 aces =
4
C4 ?
48
C3
(4 aces from among 4 aces and other 3 cards must be chosen from the rest 48 cards)
Hence the probability that the 7 cards drawn contain 4 aces =
4 48
43
52
7
CC
C
?
=
1
7735
18. We have
2 3 4 5 6 3 5 2 4 6
2
22
1 1 1 1 1 1 1 1 1 1 1 1
33
5 3 5 3 5 3 3 5 5 5
1
1
1 9 1 25
35
11
3 8 25 24
11
35
3 1 10 5
8 24 24 12
? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ?
??
? ? ? ?
Read More
|
75 videos|238 docs|91 tests
|
FAQs on Sample Solution Paper 1 - Math, Class 11 - Mathematics (Maths) Class 11 - Commerce
| 1. What are the major topics covered in the Class 11 Math exam? |  |
Ans. The major topics covered in the Class 11 Math exam include sets, relations, and functions, algebra, trigonometry, coordinate geometry, calculus, and statistics.
| 2. What is the importance of studying math in Class 11? | |
Ans. Studying math in Class 11 is important as it forms the foundation for higher-level math courses. It helps in developing logical and analytical thinking, problem-solving skills, and provides a strong base for various competitive exams.
| 3. How can I prepare effectively for the Class 11 Math exam? | |
Ans. To prepare effectively for the Class 11 Math exam, you can follow these tips:
- Understand the concepts thoroughly by referring to textbooks and class notes.
- Practice solving a variety of problems from different sources, including previous year question papers.
- Create a study schedule and allocate specific time slots for each topic.
- Seek help from your teachers or classmates if you have any doubts or difficulty in understanding a concept.
- Regularly revise the topics and solve sample papers to improve your speed and accuracy.
| 4. Are there any online resources available for Class 11 Math exam preparation? | |
Ans. Yes, there are several online resources available for Class 11 Math exam preparation. You can find video tutorials, practice questions, and study materials on educational websites, YouTube channels, and online learning platforms. Some popular websites include Khan Academy, BYJU's, and Vedantu.
| 5. How can I improve my problem-solving skills in math? | |
Ans. To improve your problem-solving skills in math, you can follow these strategies:
- Understand the problem statement carefully and identify the key information given.
- Break down the problem into smaller parts and try to solve them individually.
- Use logical reasoning and apply appropriate mathematical concepts or formulas to solve the problem.
- Practice solving a variety of problems from different topics to enhance your problem-solving abilities.
- Analyze your mistakes and learn from them to avoid similar errors in the future.
About this Document
|
4.61/5 Rating |
|
Dec 26, 2024 Last updated |
Document Description: Sample Solution Paper 1 - Math, Class 11 for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation.
The notes and questions for Sample Solution Paper 1 - Math, Class 11 have been prepared according to the Commerce exam syllabus. Information about Sample Solution Paper 1 - Math, Class 11 covers topics
like and Sample Solution Paper 1 - Math, Class 11 Example, for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Sample Solution Paper 1 - Math, Class 11.
Introduction of Sample Solution Paper 1 - Math, Class 11 in English is available as part of our Mathematics (Maths) Class 11
for Commerce & Sample Solution Paper 1 - 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 1 - Math, Class 11 | Mathematics (Maths) Class 11 - Commerce
Description
Full syllabus notes, lecture & questions for Sample Solution Paper 1 - 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 1 - Math, Class 11
In this doc you can find the meaning of Sample Solution Paper 1 - Math, Class 11 defined & explained in the simplest way possible. Besides explaining types of
Sample Solution Paper 1 - Math, Class 11 theory, EduRev gives you an ample number of questions to practice Sample Solution Paper 1 - 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
Sample Solution Paper 1 - Math
,Sample Paper
,study material
,shortcuts and tricks
,practice quizzes
,mock tests for examination
,Sample Solution Paper 1 - Math
,Important questions
,video lectures
,Class 11 | Mathematics (Maths) Class 11 - Commerce
,MCQs
,Sample Solution Paper 1 - Math
,Semester Notes
,Viva Questions
,ppt
,Previous Year Questions with Solutions
,Summary
,Exam
,past year papers
,Extra Questions
,Objective type Questions
,Class 11 | Mathematics (Maths) Class 11 - Commerce
,Free
,Class 11 | Mathematics (Maths) Class 11 - Commerce
;
Additional Information about Sample Solution Paper 1 - Math, Class 11 for Commerce Preparation
Sample Solution Paper 1 - Math, Class 11 Free PDF Download
The Sample Solution Paper 1 - 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 1 - Math, Class 11 now and kickstart your journey towards success in the Commerce exam.
Importance of Sample Solution Paper 1 - Math, Class 11
The importance of Sample Solution Paper 1 - 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 1 - Math, Class 11 Notes
Sample Solution Paper 1 - 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 1 - 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 1 - Math, Class 11 Notes on EduRev are your ultimate resource for success.
Sample Solution Paper 1 - Math, Class 11 Commerce Questions
The "Sample Solution Paper 1 - 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 1 - Math, Class 11 on the App
Students of Commerce can study Sample Solution Paper 1 - 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 1 - 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 1 - Math, Class 11 is prepared as per the latest Commerce syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup