Solution
PT-1 MOCK (DUPLICATE)
Class 11 - Mathematics
1. (b) A
Explanation: (A ∩ B ) = A ′
′
⇒ A ∩ (A ∩ B ) = A ∩ A = A
2. (a) None of these
Explanation: Given that f (x) = 1
√x−|x|
x − x = 0 if x ≥ 0
Where x − |x| = {
2x if x < 0
Thus 1
is not defined for any x ∈ R.
√x−|x|
Hence f is not defined for any x ∈ R, i.e. Domain of f is none of the given options.
3. (c) null set
2x−1 3x
Explanation: 3
−
5
+ 1 < 0
2x−1
⇒ 15 ⋅
3
− 15 ⋅
3x
5
+ 15 < 0 [Multiply the inequality throughout by the L.C.M]
⇒ 5(2x - 1) -3(3x) + 15 < 0
⇒ 10x - 5 - 9x + 15 < 0
⇒ x + 10 < 0
⇒ x < -10, but given x ∈ W
Hence the solution set will be null set.
4. (b) Both A and R are true but R is not the correct explanation of A.
Explanation: We have, |3x - 5| > 9
⇒ 3x - 5 < -9 or 3x - 5 > 9
⇒ 3x < -4 or 3x > 14
−4
⇒ x < or x >
3
14
−4 14
∴ x ∈ (−∞, ) ∪ ( , ∞)
3 3
Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
′
5. To Prove: (A ∩ B ) ∪ (B ∩ C ) = A ∪ B
′ ′
′
LHS = (A ∩ B ) ∪ (B ∩ C ) ′
∪ (B ) ) ∪ (B ∩ C ) [According to DeMorgan's Law]
′ ′ ′
= (A
′
= (A ∪ B) ∪ (B ∩ C )
′ ′
= ((A ∪ B) ∪ B) ∩ ((A ∪ B) ∪ C )
′ ′
= (A ∪ (B ∪ B)) ∩ (A ∪ B ∪ C )
′ ′
= (A ∪ B) ∩ (A ∪ B ∪ C )
′
= (A ∪ B) = RHS
Hence Proved.
6. i. We have
A = (A ∩ U)
⇒ A = A ∩ (B ∪ B') [∵ B ∪ B' = U]
⇒ A = (A ∩ B) ∪ (A ∩ B') [∵ ∩ is distributive over union], therefore we get
⇒ A = (A ∩ B) ∪ ϕ [∵ A ∩ B' = ϕ ]
⇒ A = A ∩ B
∴ A ⊂ B
ii. From (i), we have
A ∩ B' = ϕ
⇔ (A ∩ B')' = ϕ '
⇔ A' ∪ (B')' = U [∵ ϕ ' = U]
⇔ A' ∪ B = U [∵ (B')' = B]
1/6
Therefore ,A ∩ B' = ϕ ⇔ A' ∪ B = U and, A ∩ B' = ϕ ⇒ A ⊂ B
∴ A' ∪ B = U ⇒ A ⊂ B
7. Here we have, f = {(x, 1
2
) : x ∈ R and x ≠ ±1}
1−x
Clearly, f(x) is defined for all real values of x for which (1 - x2) ≠ 0.
Now, (1 - x2) = 0 ⇒ (1 + x )(1 - x) = 0 ⇒ x = - 1 or x = ± 1
Thus, f(x) is defined for all values of x e R except ± 1.
∴ dom (f) = R - {-1, 1}
Let y = f(x). Then, we have
y= 1
2
⇒ y - x2y = 1 ⇒ x2y = y -1
1−x
−−−
y−1 y−1
⇒ x
2
=
y
⇒ x = ±√
y
............(i)
y−1
It is clear from equation (i) that x will take real values only when y
≥ 0
y−1
Now, y
≥ 0 ⇔ (y - 1 ≤ 0 and y < 0) or (y - 1 ≥ 0 and y > 0)
⇔ (y ≤ 1 and y < 0) or (y ≥ 1 and y > 0)
⇔ (y < 0) or (y ≥ 1)
⇔ y ∈ (- ∞ , 0) or [1, ∞ )
∴ range (f) = (- ∞ , 0) ∪ [1, ∞ )
Hence, dom (f) = R - {-1, 1} and range (f) = (- ∞ , 0) ∪ [1, ∞ )
2
x −2x+5
8. Solve 2
>
1
2
3x −2x−5
2
x −2x+5 1
>
2 2
3x −2x−5
2
x −2x+5 1
⇒ − > 0
2 2
3x −2x−5
2 2
2(x −2x+5)−(3x −2x−5)
⇒ > 0
2
2(3x −2x−5)
2
− x −2x+15
⇒ > 0
2
2(3x −2x−5)
2
−( x +2x−15)
⇒ > 0
2
2(3x −2x−5)
Multiplying inequality by -ve sign
2
x +2x−15
⇒ < 0
2
2(3x −2x−5)
2
x +2x−15
⇒ < 0
2
3x −2x−5
(x+5)(x−3)
⇒ < 0
(x+1)(3x−5)
On equating all factors to zero, we get x = - 5, -1 ,5 / 3, 3. Plotting these points on number line
5
x ∈ (−5, −1) ∪ ( , 3)
3
9. When,
|x - 2| ≤ 1
Then,
x – 2 ≤ -1 and x -2 ≥ 1
Now when,
x–2≤-1
Adding 2 to both the sides in above equation
==> x – 2 + 2 ≤ -1 + 2
==> x ≤ 1
Now when,
x–2≥1
Adding 2 to both the sides in above equation
==> x – 2 + 2 ≥ 1 + 2
==> x ≥ 3
For |x – 2| ≥ 1 <==> x ≤ 1 or x ≥ 3
When,
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|x - 2| ≤ 3
Then,
x – 2 ≥ - 3 and x – 2 ≤ 3
Now when,
x – 2 ≥ -3
Adding 2 to both the sides in above equation
==> x – 2 + 2 ≥ -3 + 2
==> x ≥ -1
Now when,
x–2≤3
Adding 2 to both the sides in above equation
==> x – 2 + 2 ≤ 3 + 2
==> x ≤ 5
For |x – 2| ≤ 3: x ≥ -1 or x ≤ 5
Combining the intervals:
x ≤ 1 or x ≥ 3 and x ≥ -1 or x ≤ 5
Merging the overlapping intervals:
-1 ≤ x ≤ 1 and 3 ≤ x ≤ 5
Therefore,
x ∈ [-1 ,1] ∪ [3, 5]
10. Let P, C and M be the sets of students who have taken physics,
chemistry and mathematics respectively.
Let a, b, c, d, e, f and g denote the number of students in the respective
regions, as shown in the adjoining Venn diagram.
As per data given, we have
,
,
,
,
,
,
.
abcd
bcef
cdfg
bc
cd
cf
c
12
11
15
4
9
5
3
3/6
Z
[
\
]]]]]]]]]]]]]]]]]]]]]]]]
From these equations, we get
c 3, f 2, d 6, b 1.
Now, c d f g 15 & 3 6 2 g 15 & g 4;
b c e f 11 & 1 3 e 2 11 & e 5;
c
ab
d
PC
M
e
f
g
Sets 43
a b c d 12 & a 1 3 6 12 & a 2;
a 2, b 1, c 3, d 6, e 5, f 2 and g 4.
So, we have:
(i) Number of students who offered physics only a 2.
(ii) Number of students who offered chemistry only e 5.
(iii) Number of students who offered mathematics only g 4.
(iv) Number of students who offered physics and chemistry but
not mathematics b 1.
(v) Number of students who offered physics and mathematics but
not chemistry d 6.
(vi) Number of students who offered only one of the given subjects
(a e g) (2 5 4) 11.
(vii) Number of students who offered at least one of the given
subjects (a b c d e f g) (2 1 3 6 5 2 4) 23.
(viii) Number of students who offered none of the three given
subjects (25 23) 2.
EXERCISE 1G
11. i. To determine A × (B ∪ C)
B ∪ C = {b, c, e} ∪ {b, c, f} = {b, c, e, f}
∴ A× (B ∪ C) = {a, d} × {b, c, e, f}
= {(a, b), (a, c), (a, e), (a, f), (d, b), (d, c), (d, e), (d, f)} ...(i)
To determine (A × B) ∪ (A × C)
A × B = {a, d} × {b, c, e}
= {(a, b), (a, c), (a, e), (d, b), (d, c), (d, e)}
A× C = {a, d} × {b, c, f}
= {(a, b), (a, c), (a, f), (d, b), (d, c), (d, f)}
∴ (A× B) ∪ (A× C)
= {(a, b), (a, c), (a, e), (a, f), (d, b), (d,c), (d,e),(d,f)} ...(ii)
From Eqs. (i) and (ii), we get
A× (B∪C) = (A× B)∪(A× C)
Hence verified.
ii. To determine A × (B ∩ C)
(B ∩ C) = {b, c, e} ∩ {b, c, f} = {b, c}
∴ A × (B ∩ C) = {a, d} × {b, c}
= {(a, b), (a, c), (d, b), (d, c)} ...(iii)
To determine (A× B)∩(A× C)
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A × B = {(a, b), (a, c), (a, e), (d, b), (d, c), (d, e)}
A × C = {(a, b), (a, c), (a, f), (d, b), (d, c), (d, f)}
∴ (A× B)∩ (A× C) = {(a, b), (a, c), (d, b), (d, c)} ...(iv)
From Eqs. (iii) and (iv), we get
A× (B∩C) = (A× B)∩(A× C)
Hence verified.
|x+3|+x
12. We have, x+2
>1
|x+3|+x
⇒
x+2
-1>0
|x+3|+x−x−2
⇒
x+2
>0
|x+3|−2
⇒
x+2
>0
Let x + 3 = 0
⇒ x=-3
∴ x = - 3 is a critical point.
So, here we have two intervals (−∞, −3) and [−3, ∞)
Case I: When - 3 ≤ x < ∞ , then |x + 3| = (x + 3)
|x+3|−2
∴
x+2
>0
x+3−2
⇒
x+2
>0
x+1
⇒
x+2
>0
2
> 0 × (x + 2)2
(x+1)(x+2)
⇒
(x+2)
⇒ (x + 1) (x + 2) > 0
Product of (x + 1) and (x + 2) will be positive, if both are of same sign.
∴ (x + 1) > 0 and (x + 2) > 0
or (x + 1) < 0 and (x + 2) < 0
⇒ x > - 1 and x > - 2
or x < - 1 and x < - 2
On number line, these inequalities can be represented as,
Thus, - 1 < x < ∞ or - ∞ < x < - 2
But, here - 3 ≤ x < ∞
∴ - 1 < x < ∞ or - 3 ≤ x < - 2
Then, solution set in this case is
x ∈ [- 3, - 2) ∪ (- 1, ∞ )
Case II: When x < - 3, then |x + 3| = - (x + 3)
|x+3|−2
∴
x+2
>0
−x−3−2
⇒
x+2
>0
−(x+5)
⇒
x+2
>0
x+5
⇒
x+2
<0
2
< 0 × (x + 2)2
(x+5)(x+2)
⇒
x+2
⇒ (x + 5) (x + 2) < 0
Product of (x + 5) and (x + 2) will be negative, if both are of opposite sign.
∴ (x + 5) > 0 and (x + 2) < 0
or (x + 5) < 0 and (x + 2) > 0
⇒ x > - 5 and x < - 2
or x < - 5 and x > - 2
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On number line, these inequalities can be represented as,
Thus, - 5 < x < - 2 i.e., solution set in the case is x ∈ (- 5, - 2).
On combining cases I and II, we get the required solution set of given inequality, which is
x ∈ (- 5, - 2) ∪ ( - 1, ∞ )
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