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QUESTION: 1

The domain of the function

Solution:

Let g(x) = sin ^{-1} (3 - x)

= -1 __<__ 3 - x __<__ 1

The domain of g(x) is [2, 4] and let h(x) = log(|x| - 2)

⇒ | x | -2 > 0 or |x| > 2 and | x | -2 ≠ 1 ⇒ | x | ≠ 3 ⇒ x ≠ ±3

⇒ x < -2 or x > 2

⇒ (-∞, -2) ∪ (2, ∞)

Therefore, the domain of f(x)

= (2,4] - {3} = (2, 3) ∪ (3, 4]

QUESTION: 2

The domain of f (x) = log | logx | is

Solution:

f(x) = log|logx|, f(x) is defined if |log x| > 0 and x > 0,

i.e., if x > 0 and x ≠ 1

(∵ |log x| > 0 if x ≠ 1)

⇒ x ∈ (0,1)∪(1,∞).

QUESTION: 3

The domain of the function contains the points** **

Solution:

Given function is defined if ^{10}C_{x-1} >3 ^{10}C_{x}

Moreover, 10 __>__ x – 1 and 10 __>__ x

⇒ x __>__ 9 but x __<__ 10 ⇒ x = 9, 10.

[∴ x ∈N]

QUESTION: 4

The range of the function f (x) = cot^{-1} (2x- x^{2}) is

Solution:

Let y = cot^{-1} (2x - x^{2})

⇒ y = cot^{-1} {1- (x - 1)^{2}}

QUESTION: 5

The range of

Solution:

QUESTION: 6

The function where [x] denotes the greatest integer less than or equal to x, is defined for all x ∈

Solution:

The function sec-1 x is defined for all x ∈ R - (-1, 1) and the function is defined for all x ∈ R - Z

So the given function is defined for all

QUESTION: 7

The domain of

Solution:

The function defined if 3 - x > and x ≠ 2, i.e., if ≠ 2 and x > 3.

Thus, the domain of the given function is {x| - 6 __<__ x __<__ 6} ∩ {x| x ≠ 2, x < 3} = [-6,2)∪(2,3).

QUESTION: 8

The domain of the function

Solution:

Hence, the domain of f(x) = R - (nπ : n ∈ Z}.

QUESTION: 9

The domain of the function greatest integer less than or equal to x, is

Solution:

x^{2} - [x]^{2} __>__ 0 ⇒ x^{2} __>__ [x]^{2}

This is true for all positive values of x and all negative integer x.

QUESTION: 10

If f(x) = ax^{7} + bx^{3} + cx – 5; a, b, c are real constants and f (-7) = 7, then the range of f(7) + 17 cos x is

Solution:

f(7) + f(-7) = -10

⇒ f(7) = -17

⇒ f(7) + 17 cos x = -17 + 17 cos x which has the range [-34, 0]

QUESTION: 11

The range of the function

Solution:

Clearly,f(x) is identically zero if x __>__ 0

⇒ -1 < y < 1and y < 0 or y > 1

⇒ -1 < y < 0.

Combinning (1) and (2), we get -1 < y __<__ 0

⇒ Range = (-1, 0]

QUESTION: 12

The domain of the function where {.} denotes the fractional part, is

Solution:

For f(x) to get defined {sin x} + {- sin x} ≠ 0

⇒ sin x ≠ integer

⇒ sin x ≠ __+__ 1,0

Hence the domain is

QUESTION: 13

The range of the function f defined by respectively, denote the greatest integer and the fractional part] is

Solution:

∴ {x} ∈ [0, 1)

sin {x} ∈ (0, sin 1) as f(x) is defined if sin {x} ≠ 0

QUESTION: 14

Let f(x) (where {.} denotes the fractional part of x and X, Y are its domain and range, respectively), then

Solution:

QUESTION: 15

The domain of the function :

Solution:

**Case I**

0 < |x| - 1 < 1 ⇒ 1 < |x| < 2,then

x^{2} +4x+4 __<__ 1

⇒ x^{2}+4x+3 __<__ 0

⇒ -3 __<__ x __<__ -1 ......(1)

So x∈ (-2,-1)

**(I) Case II **

Ix| - 1 > 1 ⇒ |x| > 2, then x^{2} +4x+4 __>__ 1

⇒ x^{2}+4x+3 __>__ 0

= x __>__ -1 or x __<__ -3

So,x∈ (-∞,-3) ∪(2,∞) ......(2)

From(1) and (2),x∈ (-∞,-3) ∪(-2,-1) ∪ (2,∞)

QUESTION: 16

The domain of the function

Solution:

For f(x) to be defined, cosx __>__ 0

QUESTION: 17

The domain of f(x) where {.} denotes the fractional part in [-1,1], is

Solution:

We must have

QUESTION: 18

The range of where [.] denotes the greater integer function, is

Solution:

Thus, from domain point of view,

⇒ f(x) = sin^{-1}(1) + cos^{-1} (0) or sin^{-1} (0) + cos^{-1} (-1)

⇒ f(x) = {π}

QUESTION: 19

The domain of f(x) = sin ^{-1} [2x^{2} - 3], where [.] denotes the greatest integer function, is

Solution:

sin−1[2x^{2} - 1]

We know that the range of sin^{−1} is [−π/2,π/2] and domain is [−1,1]

−1≤2x^{2}−3≤2

2≤2x^{2}≤5

2x^{2}−2≥0; 2x^{2}−5≤0

x^{2} ≥1 ; x^{2}≤5/2

∴ xϵ[−(5/2)^{1/2},−1]⋃[1,(5/2)^{1/2}]

QUESTION: 20

The range of f(x) = [|sin x |+| cos x|] , where [.] denotes the greatest integer function, is

Solution:

Range of |sinx| is [0,1] and |cosx| is [0,1] for all x belongs to R.

So the range of f(x) is also [0,1] as sin0 =! cos0.

So [|sinx| + |cosx|] = 1 as per the definition of greatest integer function.

QUESTION: 21

If then the range of f(x) is

Solution:

Hence, the range is (0, 1].

QUESTION: 22

The range of where [.] denote the greatest integer function £ x , is

Solution:

Given f(x) = [sinx + [cosx + [tanx+ [secx]]]]

= [sinx + p], where p = [cosx + [tanx+ [sccx]]]

= [sinx] +p, (as p is an integer)

= [sinx] + [cosx + [tanx+ [secx]]]

= [sinx] + [cosx] + [tanx] + [secx]

Now, for x ∈ (0, π/4), sin x ∈

tan x ∈ (0, 1), sec x ∈ (1, √2)

⇒[sin x] = 0, [cos x] = 0, [tan x] = 0 and [sec x] = 1

⇒ The range of f(x) is 1,

QUESTION: 23

Let A = {1, 2, 3} and let R = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 2), (1, 2)}. Then R is

Solution:

R is reflexive and transitive but not symmetric

QUESTION: 24

Let A = {a, b, c} and let R = {(a, a)(a, b), (b, a)}. Then, R is

Solution:

R is symmetric and transitive but not reflexive

QUESTION: 25

Let A = {1, 2, 3} then total number of relations in

Solution:

Total number of relations = 2n^{2}

QUESTION: 26

Let S be the set of all straight lines in a plane. Let R be a relation on S defined by a R b ⇔ a ⊥ b. Then, R is

Solution:

a ⊥ a is not true. So, R is not reflexive

a ⊥ b and b ⊥ c does not imply a ⊥ c. So, R is not transitive

But, a ⊥ b ⇒ b ⊥ a is always true. So, R is symmetric.

QUESTION: 27

Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔ |a – b| __<__ 1. Then, R is

Solution:

(i) |a – a| = 0 __<__ 1 is always true

(ii) a R b ⇒ |a – b| __<__ 1 ⇒ |-(a – b)| __<__ 1 ⇒ |b – a| __<__ 1 ⇒ b R a. So, R is symmetric.

But, 2 is not related to 1/2. So, R is not transitive.

QUESTION: 28

Let W denote the words in the English dictionary. Define the relation R by R = {(x, y) ∈ W × W| the words x and y have at least one letter in common}. Then, R is

Solution:

Let W = {CAT, TOY, YOU, ……..)

Clearly, R is reflexive and symmetric but not transitive.

Since, CAT^{R}TOY, TOY^{R}YOU ⇒ CAT^{R}YOU

QUESTION: 29

Let R = {(3, 3), (6, 6), (9, 9), (3,6), (3, 9), (9, 12), (3,12), (6, 12), (12, 12)}, be a relation on the set A = {3, 6, 9, 12} Then the relation is

Solution:

R is reflexive

∴ (3, 3), (6,6), (9, 9), (12, 12) ∈ R

(6, 12) ∈ R but (12, 6) ∈ R Þ R is not symmetric

R is transitive

∴ (3, 6) ∈ R, (6, 12) ∈ R, (3, 12) ∈ R

QUESTION: 30

Let R be a relation on the set A of ordered pairs of positive integers defined by (x, y)R(u, v) if and only if xv = yu, then

Solution:

Clearly,(x, y) R(x,y),since xy = yx. This shows that R is reflexive.

Further, (x, y)R(u, v) ⇒ xv = yu ⇒ uy = vx and hence (u, v)R(x, y).

This shows that R is symmetric. Similarly, (x, y)R(u, v) and (u, v)R(a, b) ⇒ xv = yu and ub = va

ya and hence (x, y)R(a, b). Thus, R is transitive. Thus, R is an equivalent relation.

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