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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
? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ?
??
? ? ? ?
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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.
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