Test: Relations & Functions- 3 - JEE MCQ

For f to be defined term under square root has to be greater than zero.
⇒ cosx − 1 ≥ 0
⇒ cosx ≥ 1
Only possible if cosx = 1
⇒ x = 2nπ ∀ n∈Z

Let y has some fractional part
x = 6.5
y = 0.9
[x] = 6, [y] = 0, [x]+[y] = 6
[x + y] = [7.4] = 7
[x + y] ≠ [x]+[y]
Let x, y be integers
x = 6.5
y = 1
[x] = 6, [y] = 1, [x]+[y] = 7
[x + y] = [7.5] = 7
[x + y] = [x]+[y]
So, [x + y] = [x]+[y] holds for y ϵ I

f(x) = x
g(x) = cosx 
D(f) = R
D(g) = R
D(f+g) = D(f) ⋂ D(g)
So , D(f+g) = R
Therefore, given function is defined for real numbers.

 lim(x → π) sinx/x-π
lim(h → 0) sin(π-h)/π-h-π
lim(h → 0) -sinh/h
⇒ -1

Clearly/is both one - one and onto Because if n is odd, values are set of all non-negative integers and if n is an even, values are set of all negative integers.

f(x) = 3.sin(√π²/16 − x²)
since the quantity within the square root can not be negative.
therefore
π²/16 − x² ≥ 0
x² ≤ π²/16
−π/4 ≤ x ≤ π/4
the minimum value of the function is 3.sin0=0
and the maximum value of the function is 3sinπ/4 = 3/√2
therefore range is [0, 3/√2]

 y = (x² - 3x + 2)/(x² + x -6)
= [(x-2)(x-1)]/[(x-2)(x+3)]
= (x-1)/(x+3)
= 1 - 4/(x+3)
x is not equal to 2 and -3
For x = 2, y = 1/5
for x = -3, y → 1
Range (y): R - {1, 1/5}

f(x) = log[(1+x)/(1-x)]    ……….(1)
Substitute 2x/1+x² in place of x in equation(1)
f(2x1+x²)=log⁡(1+2x1+x²)(1−2x1+x²)
⟹f(2x/(1+x²))=log[1+(2x/(1+x²))/(1-(2x/1+x²))]
⟹f(2x/(1+x²))=log⁡((1+x²+2x)/(1+x²−2x))
⟹f(2x/(1+x²))=log((1+x)² / (1−x))²
⟹f(2x/ (1+x²))=log⁡(1+x / 1−x)²
We know that log⁡mⁿ=nlog⁡m
⟹f(2x/(1+x²))=2log⁡(1+x / 1−x)
From equation (1):
f(2x/(1+x²))=2f(x)

f(x) = (x-3)(x-1)/[(x2 -4)^1/2]
= x2 - 4 = 0
(x-2)(x+2) = 0
x = 2 and -2
(-∞, -2 ) U (2,∞)

For a function to have its inverse in a given domain, it should be continuous in that domain and should be a one-one function in that domain.
If the function is one-one in the domain, then it has to be strictly monotonic.
For example y=sin(x) has its domain in xϵ[−π/2,π/2] since it is strictly monotonic and continuous in that domain.

Given that n(A)=3 and n(B)=4, the number of injections or one-one mapping is given by.

yx​² + xy + y = x² - x + 1
(y-1)x² + (y+1)x + y-1 = 0
if y = -1
-2x² -1 -1 = 0
-2x² -2 = 0
Therefore, y cannot be equal to -1 since x will be a complex value
Now, y not equal to -1. 
(y-1)x² + (y+1)x + y-1 = 0   has real roots
hence, (y+1)² - 4 (y-1)(y-1) ≥ 0
y2 + 2y + 1 - 4y² + 8y - 4 ≥ 0
-3y² + 10y - 3 ≥ 0
3y² - 10y + 3 ≤ 0
3y² - 9y - y + 3 ≤ 0
3y (y-3) - 1 (y-3) ≤0
(3y-1)(y-3) ≤ 0
y ∈ [1/3,3] 
Hence, range of function is [1/3,3]

f(x) = {x/2 x is even,     0 x is odd}
f(1) = f(3) = f(5) = f(odd) = 0
Hence f is not one - one. 
column = Z
Range = {0, even}
= {0,2,4,6,8,10.......}
f(x) = x/2 implies even
f(odd) = 0
Range is not equal to codomain 
Hence, onto

 f(x) = 3x - 4
Let f(x) = y
y = 3x - 4
y + 4 = 3x
x = (y+4)/3
x → f^-1(x) and y → x
f-1(x) = (x+4)/3

The number of bijections from A to A, will be n! for n elements is A.

Now there are 106, elements. 

Hence the number of bijections will be n!=(106)!

aRb iff both a and b leave same remainder when divided by 5. 
a= 5x+n
a-1= 5x
b=5y+n
b-1 =5y
Equivalence class = {1,6,11,16,21}