Solution

11TH MATHEMATICS REVISION

Class 11 - Mathematics
Section A
1.
(c) A
Explanation: The set is {a, e, i, o, u}
2.
(b) A ⊆ B
Explanation: A ⊆ B
⇒ since set A is totally contained in Set B.

3.
(b) A
Explanation: (A ∩ B ) = A′

′
⇒ A ∩ (A ∩ B ) = A ∩ A = A

4.
(c) 7, 4
Explanation: Now to find value of m and n
The number of subsets of a set containing x elements is given by 2x
According to question: 2m – 2n = 112
⇒ 2n (2m-n – 1) = 16 × 7
⇒ 2n (2m-n – 1) = 24 × 7
On comparing on both sides 2n = 24 and 2m-n – 1 = 7
⇒ n = 4 and 2m-n = 8
⇒ 2m-n = 23
⇒ m – n = 3

⇒ m – 4 = 3

⇒ m = 7

Therefore,the value of m and n is 7 and 4 respectively

5. (a) 63
Explanation: 63
The no. of proper subsets = 2n - 1
Here n(A) = 6
In case of the proper subset, the set itself is excluded that's why the no. of the subset is 63. But if it is asked no. of improper or
just no. of subset then you may write 64
So no. of proper subsets = 63
6.
(d) four points
Explanation: From A, x2 + y2 = 25 and from B, x2 + 9y2 = 144
∴ From B, (x2 + y2) + 8y2 = 144
⇒ 25 + 8y2 = 144
⇒ 8y2 = 119
−−
−
⇒ y = ±√ 119

∴ x2 + y2 = 25 ⇒ x2 = 25 - y2 = 25 - 119

8
= 81

8
−−
⇒ x = ±√ 81

Since we solved equations simultaneously, therefore A ∩ B has four points A has 2 elements & B has 2 elements.

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 7.
(d) B ⊂ A
c c

Explanation: Let A ⊂ B
To prove Bc ⊂ Ac, it is enough to show that x ∈ Bc ⇒ x ∈ Ac
Let x ∈ Bc
⇒ x ∉ B

⇒ x ∉ A since A ⊂ B

⇒ x ∈ Ac
Hence Bc ⊂ Ac
8.
(b) {3, 6, 9, 12, 18, 21, 24, 27}
Explanation: Since set B represent multiple of 5 so from Set A common multiple of 3 and 5 are excluded.
9.
(d) C - D = E
Explanation: C - D = {a, b, c} - {c, d} = {a, b}
But E = {d}
Hence C - D ≠ E
10.
(b) None of these
Explanation: 4 ∉ A
{4} ⊄ A
B ⊄ A
Therefore, we can say that none of these options satisfy the given relation.
11. (a) C = ϕ
Explanation: ϕ is denoted as null set.
12.
(c) A
Explanation: We have to find (A')' = ?
Now, A = U\A
⇒ (A')' = (U\A)' = U'\A'

⇒ (A')' = U'\(U\A)
⇒ (A')' = U'\(U\A)

⇒ (A')' = A

13. (a) an infinite set

Explanation: Set A = {2, 3, 5, 7,...} so it is infinite.
14.
(d) {1, 2, 3, 4}
Explanation: Given A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
B ∩ C = {4}
A ∪ (B ∩ C) = {1, 2, 3, 4}
15.
(d) S ∪ T ∪ C = S
Explanation: Given,
S = set of points inside the square,
T = the set of points inside the triangle and
C = the set of points inside the circle
Since, the triangle and circle intersect each other and are contained in a square.

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 Clearly,hence S ∪ T ∪ C = S
Section B
16. (a) (6, 8) ∈ R
Explanation: (6, 8) ∈ R
as b - 2 = 8 - 2 = 6 and b>6 .
17.
(c) R – {-2, 3}
2
x +2x+1
Explanation: We have, f(x) = 2
x −x−6

f(x) is not defined, if x-6=0 x2 -

⇒ (x - 3)(x + 2) = 0

∴ x = -2, 3

∴ Domain of f = R - {-2, 3}

18.
(b) (−∞, 1] ∪ [2, ∞)
Explanation: ∵ f : R → R defined by
−−−−−−−−−
2
f (x) = √x − 3x + 2

Here, x2 - 3x + 2 ≥ 0
(x - 1) (x - 2) ≥ 0
x ≤ 1 or x ≥ 2
∴ Domain of f = (−∞, 1] ∪ [2, ∞)

19.
(d) Domain = R, Range = (−∞, 2]
Explanation: We have, f(x) = 2 − |x − 5|
Clearly, f(x) is defined for all x∈R
∴ Domain of f = R

Now, |x − 5| ≥ 0, ∀x ∈ R
⇒ −|x − 5| ≤ 0

⇒ 2 − |x − 5| ≤ 2

∴ f (x) ≤ 2

∴ Range of f = (−∞, 2]
20.
(d) 0
Explanation: Since f (x) = x 3
−
1

3
x
1 1 1 1 3
f ( )= − = − x
x 3 1 3
x x
3
x
1 1 1
Hence, f (x) + f ( x
)= x
3
−
3
+
3
3
− x = 0
x x

21.
(d) (−∞, −1]
−−−−−
Explanation: f(x) = log (1 - x) + √x − 1 2

Solving inequality,
log (1 - x) ≥ 0
⇒ 1 - x ≥ e0 (log taken to the opposite side of the equation becomes e)
⇒ x ≤ 1 .....(i)

x2 - 1 should be positive

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 ∴ x2 - 1 ≥ 0
and −1 ≤ x ≥ 1 ......(ii)
On taking intersection of equation (i) and (ii), we get
Domain of the f(x) is (-∞ , -1]
22.
(d) (−∞, 1] ∪ (4, ∞)
−−−
x−1
Explanation: f(x) = √ x+4
−−−
x−1
domain of the function can be defined for √ x+4
≥ 0

−−−
x−1
⇒ √ ≥ 0
x+4

x−1
⇒ ≥ 0
x+4

⇒ x-1≥0
⇒ x ≥ 1 and x ≠ 4

The intersection of above two equations gives (−∞, 1) ∪ (4, ∞)

Therefore, domain of f(x) is (−∞, 1) ∪ (4, ∞)
23.
(c) R - {3}, {1, -1}
|x−3|
Explanation: The given function is f(x) = x−3

This function is well defined for all real numbers other than 3.
∴ Its domain is R - {3}
|x−3|
Now, f(x) = x−3
x−3
⎧ : x > 3
x−3 1 : x > 3
= ⎨ = {
−(x−3)
⎩ −1 : x < 3
: x < 3
x−3

⇒ Range of function f is {1, -1}

24.
(d) {(8,11), (10,13)}
Explanation: Since, y = x – 3;
Therefore, for x = 11, y = 8.
For x = 12, y = 9. [ But the value y = 9 does not exist in the given set.]
For x = 13, y =10.
So, we have R = {(11, 8), (13, 10)}
Now, R−1 = {(8, 11), (10, 13)}.
25.
(d) [1, ∞)
Explanation: f(x) = 1

2
(1− x )

Range of f(x) can be found out by putting f(x) = y

y= 1

2
(1− x )

⇒ y - yx2 = 1
⇒ yx2 = y - 1
x2 =
y−1
⇒
y
−−−
y−1
⇒ x=√ y
−−−
y−1
⇒ √ ≥ 0
y

⇒ y≥1
Hence, Range of the f(x) = [1, ∞ )
26.
(c) Domain = [1, ∞) , Range = [0, ∞)
−−−−−
Explanation: We have, f(x) = √x − 1
Clearly, f(x) is defined if x − 1 ≥ 0

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 ⇒ x ≥ 1

∴ Domain of f = [1, ∞)
Now for x ≥ 1, x − 1 ≥ 0
−−−−−
⇒ √x − 1 ≥ 0

⇒ Range of f = [0, ∞)
27. (a) ϕ
Explanation: For function to be defined,
x−2
≥ 0, x ≠ −2
x+2

x ∈ (−∞, −2) ∪ [2, ∞) … (1)

1−x
and 1+x
≥ 0, x ≠ −1

x−1
≤ 0
x+1

x ∈ (−1, 1] … (2)

Taking common of both the solutions we get x ∈ ϕ

28.
(c) 2-1
1

Explanation: fof(x) = f(f(x)) = {f (25 − x 4

) 4 }

1 1

= {25 − (25 − x 4
)} 4
= (x 4
) 4
=x
∴ f (f ( )) =
1 1

2 2

= 2-1
29.
(d) {1, 5, 9}
Explanation: The set of elements related to 1 is {1, 5, 9}.
Since, |1 - 1| = 0 is multiple of 4
|5 - 1| = 4 is a multiple of 4
and |9 - 1| = 8 is a multiple of 4
30. (a) R - { - 1 / 2 ,1 }
2

Explanation: Let y = x −x

2
x +2x

y(x2+2x)= x2-x
yx(x+2)= x(x-1)
y(x+2)= x-1
x(y − 1) = −(1 + 2y)

(1+2y)
x = −
y−1

Value of x can’t be zero or it cannot be not defined.Therefore,

y = 1, -1/2 are not possible.
So, range = R - { -1/2, 1}
31.
(d) {(2,4), (3, 4)}
Explanation: x2 - 5x + 6 = 0
⇒ x2 - 2x - 3x + 6 = 0
⇒ x(x - 2) - 3(x - 2) = 0

⇒ (x - 3)(x - 2) = 0
∴ A = {2, 3}; Also, B = {2, 4}, c = {4, 5}
Now, B ∩ C = {4}
∴ A × B ∩ C = {2, 3} × {4}

= {(2, 4), (3, 4)}

32. (a) {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 2), (2, 3)}
Explanation: R = {(x, y) : |x2 - y2| < 7}
R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 2), (2, 3)}

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 2

33. (a) 2 n

Explanation: The number of elements in A × A is n × n = n2. Hence, the number of relations on A = number of subsets of A
A = 2n× n = 2 .
2
n
×

34. (a) {(1, 4)}

Explanation: (A – B) = {1, 2, 4} – {2, 4, 5} = {1}
and (B - C) = {2, 4, 5} – {2, 5}
= {4}
Now, (A−B) × (B − C) = {1} × {4}
= {(1, 4)}
Since, it is a set so it is written in curly braces.
Therefore, option B is correct.
35. (a) (-3, -2) ∪ (-2, -1) ∪ (1, ∞ )
Explanation: Given function is f(x) = log3+x (x2 - 1)

It is obvious that f(x) is defined when x2 - 1 > 0, 3 + x > 0 and 3 + x ≠ 1.

Now, x2 - 1 > 0 ⇒ x2 > 1
⇒ x < -1 or x > 1

3 + x > 0 ⇒ x > -3
3 + x ≠ 1 ⇒ x ≠ -2
Therefore, domain of the function f(x) = (-3, -2) ∪ (-2, -1) ∪ (1, ∞ )
Section C
36.
(c) ± 1

√3

Explanation: 7 sin 2
θ + 3 cos
2
θ= 4

⇒ 7 tan
2
θ + 3 = 4 sec
2
θ [dividing by cos2θ]
θ) + 3 = 0
2 2
⇒ 7 tan θ − 4 (1 + tan

2 2 2
⇒ 7 tan θ − 4 tan θ −1 = 0 ⇒ 3 tan θ= 1

2 1 1
⇒ tan θ= ⇒ tan θ = ±
3 √3

37.
3
(d) 2

Explanation: In quadrant II, cos θ < 0.

−−
−4
Now, cos 2
θ = (1 − sin
2
θ) = (1 −
25
9
) =
16

25
⇒ cos θ = − √
16

25
=
5

−5 −4 −4
∴ sec θ =
4
and cot θ = cos θ

sin θ
=
5
×
5

3
=
3
(−5) (−4) −5 3
∴ (2 sec θ − 3 cot θ) = {2 × − 3 × }= ( + 4) =
4 3 2 2

38. (a) tan x

(7x +5x ) (7x −5x )
2 cos sin
sin 7x−sin 5x
Explanation: cos 7x+cos 5x
=
2

(7x +5x )
2

(7x −5x )
=
sin x

cos x
= tan x
2 cos cos
2 2

39.
(d) tan x
(1−cos 2x)+sin 2x 2
2 sin x+2 sin x cos x
Explanation: Given exp. = =
2
(1+cos 2x)+sin 2x 2 cos x+2 sin x cos x

2 sin x(sin x+cos x)

= =
sin x

cos x
= tan x.
2 cos x(cos x+sin x)

40.
(b) π

4
tan A+tan B
Explanation: tan (A + B) = 1−tan A tan B
a 1
+
a+1 2a+1
= a
1−
(a+1)(2a+1)

2
2a +a+a+1
=
2
2a +3a+1−a

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 2
2a +2a+1
=
2
2a +2a+1

=1
Therefore, A + B = = tan −1
(1) =
π

41.
(b) π

Explanation: Here, angles of triangle are in A.P.

So, Let angles of triangle are a, a + d , a + 2d.
We know that , sum of angles of triangle is π.
∴ a + a + d + a + 2d = π

∴ 3a + 3d = π

∴ 3(a + d) = π
π
∴ a+ d =
3

Also , by our assumption, a + d is one angle of triangle.

So, required measure of one of the angles is . π

42.
(b) 6.6 cm
Explanation: Angle traced by the minute hand in 60 min = (2π) c

c r

Angle traced by the minute hand in 45 min = ( 2π

60
× 45) = (
3π

2
)

c
3π
∴ r = 1.4 cm and θ = ( 2
)

⇒ l = rθ = (1.4 ×
3π

2
) cm = (1.4 ×
3

2
×
22

7
) cm = 6.6 cm

43.
(b) - cot θ
Explanation: cos(π + θ) = − cos θ , cos(−θ) = cos θ , cos(π − θ) = − cos θ , cos( π

2
+ θ) = − sin θ

(− cos θ)(cos θ)
∴ given exp. = = − cot θ
(− cos θ)(− sin θ)

44.
(c) r2
Explanation: (x2 + y2 + z2) = r 2
cos
2
α cos
2
β + r
2
cos
2
α sin
2
β + r
2
sin
2
α

2 2 2 2 2 2
= r cos α (cos β+ sin β) + r sin α

2 2 2 2 2 2 2 2
= r cos α + r sin α = r (cos α + sin α) = r

45.
(c) sin 4β
Explanation: Given tan α = 1

7
and tan β = 1

Now to find the value of cos 2α

2
1− tan x
[By using cos 2x = ]
1+ tan2 x
2
1− tan α
cos 2α =
2
1+ tan α

[as tan α = 1

7
is given]
2
1
1−( )
7
cos 2α =
2
1
1+( )
7

49−1
=
49+1
48 24
= =
50 25

Hence cos 2α = 24

25

The same value is obtained for sin 4β. [By sin 2x = 2 sin x cos x]
sin 4α = 2 sin 2α cos 2α
2
2 tan x 1− tan x
[using sin 2x = 2
and cos 2x = 2
]
1+ tan x 1+ tan x

We have
2
2 tan β 1− tan β
sin 4β = 2 ( )( )
1+ tan2 β 1+ tan2 β

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 As tan β = 1

3
2
1 1
2( ) 1−( )
3 3
sin 4β = 2 ( )( )
2 2
1 1
1+( ) 1+( )
3 3

6 9−1
= 2( )( )
9+1 9+1

48 48 24
= 2( ) = =
100 50 25

46. (a) tan 3A tan 2A tan A

Explanation: tan 3A = tan(2A + A)
tan 2A+tan A
tan 3A = 1−tan 2A⋅tan A

⇒ tan 3A(1 - tan 2A.tan A) = tan 2A + tan A

⇒ tan 3A - tan 3A.tan 2A.tan A = tan 2A + tan A
⇒ tan 3A - tan 2A - tan A = tan 3A.tan 2A.tan A

47.
−1
(d)
√2

Explanation: cos 135° = cos (90° + 45°) = - sin 45° = −1

√2

48.
(c) 1

Explanation: 2 cos 5π

12
cos
π

12
= cos(
5π

12
+
π

12
) + cos(
5π

12
−
12
π
) [ using 2cosAcosB= cos (A + B) + cos (A - B) ]
π π 1 1
= (cos + cos ) = (0 + )=
2 3 2 2

49.
(c) tan( π

4
+
x

2
)

− −−− −− −−− 1
 π π x 1
2 2
−−−−−  1−cos( +x) 2 sin ( + )
1+sin x
Explanation: √
2 4 2 2 π x 2 π x
=  = { } = {tan ( + )} = tan( + )
1−sin x π π x 4 2 4 2
⎷ 1+cos( +x) 2 cos
2
( + )
2 4 2

50. (a) 1

Explanation: In quadrant II, cos θ < 0.

4 2 2 16 9
sin θ = ⇒ cos θ = (1 − sin θ) = (1 − ) =
5 25 25
−−
9 −3
⇒ cos θ = − √ =
25 5

sin θ 4 5 −4 −3
∴ tan θ = = × = ⇒ cot θ =
cos θ 5 (−3) 3 4

−5
cosec θ =
1

sin θ
=
5

4
and sec θ = 1

cos θ
=
3
(−3) 5 (−4)
{5× }+(4× )+{3× }
(5 cos θ+4 cosec θ+3 tan θ) 5 4 3

∴ =
(4 cot θ+3 sec θ+5 sin θ) (−3) (−5) 4
{4× }+{3× }+{ 5× }
4 3 5

(−3+5−4) −2 1
= = =
(−3−5+4) −4 2

51.
(c) cot x
sin 2x 2 sin x cos x cos x
Explanation: 1−cos 2x
=
2
=
sin x
= cot x
2 sin x

52.
√3
(b) 2

31π π π √3
Explanation: sin( 3
)= sin(10π +
3
) = sin
3
=
2

53.
√3
(c) 2

Explanation: In quadrant III, we have: sin x < 0 and cos x < 0

cos x = (1 - sin2 x) = (1 −
−1 3
sin x = 2
⇒
1

4
) =
4

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 −
−
3 √3
⇒ cos x = − √ = −
4 2

(−1) (− √3) √3
∴ sin 2x = 2sin x cos x = 2 × 2
×
2
=
2

54.
(b) 1
Explanation: sin θ + sin 2
θ = 1 ⇒ sin θ = (1 − sin
2
θ) = cos
2
θ ⇒ sin
2
θ = cos
4
θ

∴ 1 − cos
2
θ = cos
4
θ ⇒ cos
2
θ + cos
4
θ= 1 .
(2+ √3)
55. (a) 2

Explanation: 2 sin 5π

12
cos
π

12

= sin(
5π

12
+
π

12
) + sin(
5π

12
−
π

12
) [ using 2sinASin B= sin (A + B) + sin (A - B) ]
π π √3 (2+ √3)
= (sin + sin ) = (1 + ) =
2 3 2 2

56.
(b) 2
π
Explanation: α + β =
4

taking tan both sides,

π
⇒ tan(α + β) = tan
4
tan α+tan β
⇒ = 1
1−tan α⋅tan β

⇒ tan α + tan β = 1- tan α . tan β

⇒ tan α + tan β + tan α .tan β = 1
On adding 1 both sides, we get
⇒ 1+ tan α + tan β + tan α . tan β = 1 + 1

⇒ 1(1 + tan α ) + tan β (1 + tan α ) = 2

⇒ (1+ tan α ) (1 + tan β ) = 2

√3
57. (a) 2

Explanation: Using (3 sin θ − 4 sin 3

θ) = sin 3θ , we get
(3 sin 40° - 4 sin340°) = sin(3 × 40 )
∘
= sin 120°
√3
= sin (180° - 60°) = sin 60° = 2

58.
( √5+1)
(b) 4
( √5+1)
Explanation: sin 54° = sin (90° - 36°) = cos 36° = 4

59.
(d) 8
Explanation: We have:
2 ∘ ∘

x sin 45° cos 60° = tan 60

∘
cosec 30

2 ∘
sec 45 cot 30
2
2 ( √3) ×(2)
1 1
⇒ x × ( ) × ( ) =
√2 2 2
( √2)×( √3)

x 6
⇒ =
4√2 3√2
– 6
⇒ x = (4√2) [ ]
(3√2)

⇒ 4(2)
⇒ x = 8

60.
(d) tan 55o
∘ ∘
cos 10 +sin 10
Explanation: ∘
cos 10 −sin 10
∘

[Dividing the numerator and denominator by cos 10o]

1+tan 10
= ∘
1−tan 10
∘ ∘
tan 45 +tan 10
= ∘ ∘
1−tan 45 ×tan 10

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 = tan (45o + 10o) [ Using
tan A+tan B
tan(A + B) = ]
1−tan A tan B

= tan 55o
Section D
61.
(b) (6, 8),(8, 10), (10, 12)
Explanation: Let the consecutive even positive integers be x and x + 2.
By data, x > 5 and x + (x + 2) < 23
Now x + (x + 2) < 23
⇒ 2x + 2 < 23

⇒ 2x < 21

⇒ x < = 10
21 1

2 2

So we have the least possible value of x is 6 and the maximum value of x is 10.
Therefore the possible pairs of consecutive even positive integers are (6, 8), (8, 10), (10, 12).
62.
(d) {0, 1, 2, 3}
Explanation: Given 2(x - 1) < 3x - 1
⇒ 2x - 2 < 3x - 1

⇒ 2x - 2 + 2 < 3x - 1 + 2

⇒ 2x < 3x + 1

⇒ 2x - 3x < 3x + 1 - 3x

⇒ -x < + 1
⇒ x > -1 but x ∈ Z
Hence A = {0, 1, 2, 3, 4, ....}
Now 4x - 3 ≤ 8 + x
⇒ 4x - 3 + 3 ≤ 8 + x + 3

⇒ 4x ≤ 11 + x
⇒ 4x - x ≤ 11 + x - x
⇒ 3x ≤ 11
3x 11
⇒ ≤
3 3

⇒ x≤ 11

⇒ x ≤ 3 , but x ∈ Z
2

Therefore B = {....., -2, -1, 0, 1, 2, 3}

Hence A ∩ B = {0, 1, 2, 3}
63.
(d) x ∈ (2, ∞)
|x−2|
Explanation: Since x−2
≥ 0 , for |x - 2| ≥ 0, and x - 2 ≠ 0 solution set (2,∞ )

64.
(c) null set
2x−1
Explanation: 3
−
3x

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.

65.
(c) x ∈ [ - 11 , 7 ]
Explanation: |x + 2| ≤ 9
⇒ −9 ≤ x + 2 ≤ 9 [∵ |x| ≤ a ⇔ −a ≤ x ≤ a]

⇒ −9 − 2 ≤ x + 2 − 2 ≤ 9 − 2

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 ⇒ −11 ≤ x ≤ 7

xϵ [−11, 7]

66. (a) x > 5

Explanation: x - 5 > 0
⇒ x > 5

⇒ x ∈ (5, ∞)
2x−4
Now x+2
<2
2x−4

x+2
-2<0
2x−4−2(x+2)
⇒
x+2
<0
2x−4−2x−4
⇒
x+2
<0
−8
⇒
x+2
<0
⇒ x + 2 > 0 [∵ a

b
< 0, a < 0 ⇒ b > 0]
⇒ x > -2
⇒ x ∈ (−2, ∞)

Hence the solution set is (5, ∞) ∩ (−2, ∞) = (5, ∞)

Which means x > 5.
67.
(c) no solution
x+7
Explanation: x−8
> 2

x+7
⇒ − 2 > 0
x−8

x+7−2(x−8)
⇒
x−8
>0
x+7−2x+16
⇒
x−8
>0
(23−x) a
⇒
x−8
> 0 [∵ b
> 0 ⇒ (a > 0 and b > 0) or (a < 0 and b < 0)]
⇒ (23 - x > 0 and x - 8 > 0) or (23 - x < 0 and x - 8 < 0)
⇒ (x < 23 and x > 8) or (x > 23 and x < 8)

⇒ 8 < x < 23 [Since x > 23 and x < 8 is not possible]

⇒ xϵ (8, 23)
2x+1
Now 7x−1
>5
2x+1
⇒
7x−1
-5>0
2x+1−5(7x−1)
⇒
7x−1
>0
2x+1−35x+5
⇒
7x−1
>0
(6−33x)
⇒
7x−1
> 0 [∵ a

b
> 0 ⇒ (a > 0 and b > 0) or (a < 0 and b < 0)]
⇒ (6 - 33x > 0 and 7x - 1 > 0) or (6 - 33x < 0 and 7x - 1 < 0)
⇒ (x < and x > ) or (x > and x < )
6 1 2 1

33 7 11 7
1 2
⇒ < x <
7 11

⇒ x ∈ (
1

7
,
2

11
) [Since x > 2

11
and x < 1

7
is not possible]
x+7 2x+1
Hence, the solution of the system x−8
> 2,
7x−1
> 5 will be (8, 23) ∩ ( 1

7
,
2

11
) = ϕ

68.
(b) -23 < x ≤ 2
3(x−2)
Explanation: −15 < 5
≤ 0

3(x−2)
5 5 5
⇒ −15 ⋅ < ⋅ ≤ 0 ⋅
3 5 3 3

⇒ -25 < (x - 2) ≤ 0 + 2
⇒ -25 + 2 < x - 2 + 2 ≤ 2

⇒ -23 < x ≤ 2

69.
(d) x ∈ (– ∞ , – 4] ∪ [3, ∞ )
Explanation: Common solution of the inequalities is from – ∞ to – 4 and 3 to ∞
{(– ∞ , – 4] ∪ [3, ∞ )} ∩ {(– ∞ , – 3] ∪ [1, ∞ )} = (– ∞ , – 4] ∪ [3, ∞ )

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70.
(c) ( 3

2
,
9

2
)

2(3−x)
Explanation: ∣∣ 5
∣
∣
<
3

2(3−x)
3 3
⇒ − < <
5 5 5

3 5 2(3−x) 5 3 5
⇒ − ⋅ < ⋅ < ⋅
5 2 5 2 5 2
3 3
⇒ − < 3 − x <
2 2
3 3
⇒ − − 3 < 3 − x − 3 < − 3 [∵ |x| < a ⇔ −a < x < a]
2 2
9 −3
⇒ − < −x <
2 2
9 3
⇒ > x >
2 2

3 9
⇒ x ∈ ( , )
2 2

71.
(d)

Explanation:

72.
(d) 8 ≤ x ≤ 22
Explanation: Let the length of the shortest piece be x cm. Then we have the length of the second and third pieces are x + 3 and
2x centimeters respectively.
According to the question,
x + (x + 3) + 2x ≤ 91
⇒ 4x + 3 ≤ 91

⇒ 4x ≤ 88
⇒ x ≤ 22

Also 2x ≥ (x + 3) + 5
⇒ 2x ≥ x + 8

⇒ x ≥ 8

⇒ 8 ≤ x ≤ 22

Hence the shortest piece may be atleast 8 cm long but it cannot be more than 22cm in length.
73. (a) no solution
Explanation: We have given: 4x + 3 ≥ 2x + 17
⇒ 4x − 2x ≥ 17 − 3 ⇒ 2x ≥ 1

⇒ x ≥
14

2
[Dividing by 2 on both sides]
⇒ x ≥ 7 ..... (i)

Also we have 3x - 5 < -2

⇒ 3x < −2 + 5 ⇒ 3x < 3

⇒ x < 1

On combining (i) and (ii), we see that solution is not possible because nothing is common between these two solutions.(i.e., x <
1, x ≥ 7)
74.
(b) ac < bc
Explanation: The sign of the inequality is to be reversed (< to > or > to <) if both sides of an inequality are multiplied by the
same negative real number.

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75.
(b) (−∞, −2) ∪(−1, 1) ∪(2, ∞)
|x|−1
Explanation: Given ≥ 0, x ≠ ±2
|x|−2

|x|−1
≥ 0
|x|−2

⇒ |x| - 1 ≥ 0 and |x| - 2 ≥ 0 or |x| - 1 ≤ 0 and |x| - 2 ≤ 0 [∵ a

b
≥ 0 ⇒ (a ≥ 0 and b ≥ 0) or (a ≤ 0 and b ≤ 0)]
⇒ |x| ≥ 1 and |x| ≥ 2 or |x| ≤ 1 and |x| ≤ 2
⇒ |x| ≥ 2 or |x| ≤ 1 [∵ |x| ≥ a ⇒ x ≥ a or x ≤ -a and |x| ≤ a ⇒ −a ≤ x ≤ a ]
⇒ x ≥ 2 or x ≤ -2 or −1 ≤ x ≤ 1
⇒ x ∈ (2, ∞) or x ∈ (−∞, −2) or xϵ(−1, 1)

⇒ x ∈ (2, ∞) ∪ (−∞, −2) ∪ (−1, 1)

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