An Ideal Institute Of Mathematics
Test Paper Date : 05-07-15, 11th Class Solution
1.
(c)
1
f (x ) 
x  2 2x  4
1
f (11) 

1
3 2

2
1
11  2 18
1
3 2

15. (a)
3 2 3 2 6

 .
7
7
7
(c)
3.
(b)Therefore log| x 2  9 | does not exist at x   3, 3.
(d)
 7 n  n C1 7 n 1  n C 2 7 n  2  .....  n Cn 1 7  n Cn  7 n  1
(d)
n
1  x  0  x  1 ; 1  x  0  x  1, x  0
 8 n  7n  1 is a multiple of 49 for n  2
For n  1 , 8 n  7n  1  8  7  1  0 ;
For n  2, 8 n  7n  1  64  14  1  49
f (x )  x  x 2  4  x  4  x
 Domain of f  (, 4]  [4, )  [0,1]  [0,1] .
n(C) = 20, n(B) = 50, n(C  B) = 10
Now n(C  B) = n(C) + n(B) – n(C  B)
= 20 + 50 – 10 = 60.
Hence, required number of persons = 60%.
 8 n  7n  1 is a multiple of 49 for all n  N.
 X contains elements which are multiples of 49
and clearly Y contains all multiplies of 49. 
X Y .
16. (b)
N 3  N 4  {3, 6, 9,12,15 ......}  {4, 8,12,16, 20,.....}
= {12, 24, 36......} = N12 .
Trick : N 3  N 4  N 12
[  3, 4 are relatively prime numbers].
17. (b)
n(A  B) = n(A) + n(B) – n(A  B) =
3  6  n( A  B)
Since, maximum number of elements in
7. (a) It is distributive law.
8. (b)
It is De' Morgan law.
9.
(d) It is obvious.
10.
(b) n( A  B)  n( A)  n (B)  n( A  B) .
11.
n
 49 [n C 2  n C 3 (7)  ......  n Cn 7 n  2 ]
4x 0  x  4
x (1  x )  0  x  0 and x  1
(c)
n
Hence domain of function is R   3, 3.
Clearly f (x ) is defined, if 4  x  0  x  4
6.
,
( C 0  C n , C1  C n 1 etc.)
n
Hence domain is [1, 1]  {0} .
5.
n
n
Since 8  7 n  1  (7  1)  7n  1
 n C 2 7 2  n C 3 7 3  ..  n Cn 7 n
For x  3, 3, | x 2  9 |  0
2.
4.
Number of relations on the set A  Number of
subsets of A  A  2 n , [ n( A  A)  n 2 ] .
14. (a)
It is obvious.
x  2 2x  4

11  2 18
13. (c)
1

AB  3
(c)
Let A denote the set of Americans who like
cheese and let B denote the set of Americans who
like apples.
Let Population of American be 100.
 Minimum
AB 9 3 6 .
18. (d)
number
of
elements
in
A = Set of all values (x, y) : x 2  y 2  25  5 2
x
Then n ( A)  63, n (B)  76
2
(12)
Now, n ( A  B)  n(A)  n(B)  n(A  B)
2

y
2
(4 )
2
1
 63  76  n( A  B)
 n ( A  B)  n( A  B)  139
x2 + y2 = 52
 n ( A  B)  139  n( A  B)
But n ( A  B)  100
B=
 n ( A  B)  100
 139  n (A  B)  139  100  39
 n( A  B)  39 i.e., 39  n( A  B)
.....(i)
19.
Again, A  B  A, A  B  B
 n ( A  B)  n ( A)  63 and n ( A  B)  n (B)  76
 n( A  B)  63
12.
20. (d)
(a) Let n (P) = Number of teachers in Physics.
n (M ) = Number of teachers in Maths
n (P  M)  n(P)  n (M)  n (P  M)
20  n (P)  12  4  n (P)  12 .
P. S. Kushwah
CONTACT-
Clearly, A  B consists of four points.
(c) Here A and B sets having 2 elements in common,
so A  B and B  A have 2 2 i.e., 4 elements in
common.
Hence, n [(A  B)  (B  A)]  4 .
…..(ii)
Then, 39  n ( A  B)  63  39  x  63 .
x2
y2
x2
y2

 1 i.e.,
+
1.
2
144 16
(4 )2
(12 )
n (M)  55, n (P)  67, n (M  P)  100
Now, n (M  P)  n (M)  n (P)  n (M  P)
100  55  67  n (M  P)
 n (M  P)  122  100  22
M.sc. (Maths)B.ed
Study Center- CL-69 D. D. Nagar Gwalior, MOB- 9893668836 Email – mathsbyp.skushwah@gmil.com
1
Now
21.
n
(P
only)
=
n (P)  n(M  P)
For n  2, 8 n  7n  1  64  14  1  49
 67  22  45 .
 8 n  7n  1 is a multiple of 49 for all n  N.
(c) In general, A  B  B  A
 X contains elements which are multiples
A  B  B  A is true, if A = B.
22.
(b) From De’ morgan’s law, ( A  B)  A   B  .
23.
(d)
of 49 and clearly Y contains all multiplies of
49.  X  Y .
n( A)  4 , n(B)  3
34. (b)
N 3  N 4  {3, 6, 9,12,15 ......}  {4, 8,12,16, 20,.....}
n( A)  n(B)  n(C )  n( A  B  C )
4  3  n(C )  24  n(C ) 
24.
= {12, 24, 36......} = N12 .
24
2.
12
Trick : N 3  N 4  N12
[  3, 4 are relatively prime numbers].
(c) Given set is {(a, b) : 2a 2  3b 2  35 , a, b  Z}
35. (b)
We can see that, 2(2)2  3(3)2  35
3  6  n( A  B)
and 2(4 )  3(1)  35
2
n(A  B) = n(A) + n(B) – n(A  B) =
2
Since, maximum number of elements in
 (2, 3), (2, –3), (–2, –3), (–2, 3), (4, 1), (4, –1),
AB  3
(–4, –1), (–4, 1) are 8 elements of the set. 

Minimum
number
of
elements
n 8.
AB 9 3 6 .
25. (a)
It is distributive law.
36. (b)
26. (b)
It is De' Morgan law.
37. (a)
27. (c)
(A – B)  (B – A) = (A  B) – (A  B).
28. (c)
n( A  B)  pq .
29. (c)
B  C = {c, d}  (d, e} = {c, d, e}
38. (b) Solution: We have, 4 - x < 2 – 3
⇒
3x – 9 – 12x < 6x – 6 – 4x + 8
⇒
- 11x < 11
⇒
-x<1
⇒
x>-1
…… (i)
and
2 – x > 2x – 8
10
⇒
- 3x > - 10
⇒
x<
3
……. (ii)
From Eqs. (i) and (ii), the solution of the
given system of inequalities is given by x ∈
10
(−1, 3 ).
Hence, (b) is the correct answer.
(x−1)(x−2)
39. (b) Solution:
≤0
(x−3)
x−3
 A × (B  C) = {a, b} × {c, d, e}
= {(a, c), (a, d), (a, e), (b, c), (b, d), (b,
e)}.
30. (a,b) R  (P c  Q c )c  R  [(P c )c  (Q c )c ]
= R  ( P  Q )  ( R  P )  (R  Q ) = (R  Q )  ( R  P ) .
31. (d)
It is fundamental concept.
32. (b) It is fundamental concept.
33. (a)
x−1
in
x−2
Since 8 n  7n  1  (7  1)n  7n  1
 7 n  n C1 7 n 1  n C 2 7 n  2  .....  n Cn 1 7  n Cn  7 n  1
,
 n C 2 7 2  n C 3 7 3  ..  n Cn 7 n
(n C 0  n C n , n C1  n C n 1 etc.)
 49 [n C 2  n C 3 (7)  ......  n Cn 7 n  2 ]
 8 n  7n  1 is a multiple of 49 for n  2
x ≤ 1 or 2 ≤ x < 3 ⇒ (- ∞, 1] ∪ [2, 3]
Here, interval is closed at 1 and 2 as numerator
becomes zero so inequality hold true but open
at 3 as (x – 3) is in the denominator, so x = 3
makes left hand side undefined.
Hence, (b) is the correct answer.
x−1
40. (d) Solution:
-2≥0
x
For n  1 , 8  7n  1  8  7  1  0 ;
n
P. S. Kushwah
CONTACT-
⇒
⇒
x−1−2x
≥
x
1+x
≤0
x
0⇒
− 1− x
≥
x
0
M.sc. (Maths)B.ed
Study Center- CL-69 D. D. Nagar Gwalior, MOB- 9893668836 Email – mathsbyp.skushwah@gmil.com
2
⇒ x ∈ [-1, 0)
Here, interval is closed at -1 as
numerator becomes zero so the
inequality holds true but open at 0 as
(x) is in the denominator, so x = 0
makes left hand side undefined.
Hence, (d) is the correct answer.
41. (a) Solution: Squaring both sides, we get
|x + 3|2 > |2x − 1|2
or {(x + 3) − (2x − 1)} {(x + 3) + (2x − 1)} > 0
{(−x + 4)(3x + 2)} > 0
⇒
2
⇒
x ∈ (− , 4)
3
Hence, (a) is the correct answer.
42. (c) Solution:Case – I When x 2 + 3x ≥ 0
⇒
x(x + 3) ≥ 0,
ie,
x ∈ (- ∞, - 3] ∪ [0, ∞) ….. (i)
In this case inequality becomes
x 2 + 3x + x 2 - 2 ≥ 0
or 2x 2 + 3x – 2 ≥ 0 or (x + 2) (2x – 1) ≥ 0
or
x ∈(- ∞, - 2] ∪ [ ½, ∞) ….. (ii)
From eqs. (i) and (ii) the solution in this
case is given by
1
x ∈ (-∞ , - 3) ∪ [2, ∞)
Case- II When x 2 + 3x < 0
ie,
x ∈ (-3, 0)
….. (i)
In this case, inequality becomes
- x 2 - 3x + x 2 - 2 ≥ 0
⇒
- 3x – 2 ≥ 0
2
⇒
x ≤ - 3 ….. (ii)
From eqs. (i) and (ii) the solution in this
case is given by
x ∈ (-3, – 2/3].
Combining the solutions of both cases, we have
x ∈ (-∞ , – 2/3] ∪ [1/ 2, ∞)
Hence, (c) is the correct answer.
43. (d)
44. (a)
45. (b)
46. (d)
47. (c)
48. (b)
49. (b)
50. (b)
P. S. Kushwah
CONTACT-
M.sc. (Maths)B.ed
Study Center- CL-69 D. D. Nagar Gwalior, MOB- 9893668836 Email – mathsbyp.skushwah@gmil.com
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