JEE Advanced Level Test: Relations and Functions- 3 - JEE MCQ

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]

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,∞). 

Given function is defined if ¹⁰C_x-1 >3  ¹⁰C_x

Moreover, 10 > x – 1 and 10 > x
⇒ x > 9 but x < 10 ⇒ x = 9, 10. 
[∴ x ∈N]

Let y = cot^-1 (2x - x²)
⇒ y = cot^-1 {1- (x - 1)²}

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

 


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).

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

x² - [x]² > 0 ⇒ x² > [x]²

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

f(7) + f(-7) = -10
⇒ f(7) = -17 
⇒ f(7) + 17 cos x = -17 + 17 cos x which has the range [-34, 0] 

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]

For f(x) to get defined {sin x} + {- sin x} ≠ 0
⇒ sin x ≠ integer
⇒ sin x ≠ + 1,0

Hence the domain is   
 

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

Case I
0 < |x| - 1 < 1 ⇒ 1 < |x| < 2,then
x² +4x+4 < 1
⇒ x²+4x+3 < 0
⇒ -3 < x < -1          ......(1)
So x∈ (-2,-1)
(I) Case II
Ix| - 1 > 1 ⇒ |x| > 2, then x² +4x+4 > 1
⇒ x²+4x+3 > 0
= x > -1 or x < -3
So,x∈ (-∞,-3) ∪(2,∞)           ......(2)
From(1) and (2),x∈ (-∞,-3) ∪(-2,-1) ∪ (2,∞)

For f(x) to be defined, cosx > 0

We must have

Thus, from domain point of view,

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

sin−1[2x² - 1]
We know that the range of sin⁻¹ is [−π/2,π/2] and domain is [−1,1]
−1≤2x²−3≤2
2≤2x²≤5
2x²−2≥0; 2x²−5≤0
x² ≥1 ; x²≤5/2
∴ xϵ[−(5/2)^1/2,−1]⋃[1,(5/2)^1/2]

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.

Hence, the range is (0, 1]. 

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, 

R is reflexive and transitive but not symmetric

R is symmetric and transitive but not reflexive

Total number of relations = 2n²

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. 

(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. 

Let W = {CAT, TOY, YOU, ……..)
Clearly, R is reflexive and symmetric but not transitive.
Since, CAT^RTOY, TOY^RYOU ⇒ CAT^RYOU

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 

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.