Test: Modulus Function - Grade 12 MCQ

Concept:

Given: ax ≡ b (mod m) --(1)

Step 1:

Find d,

d = the GCD of (a,m) 

We have exactly 'd', incongruent solutions

Step 2: 

Divide (1) with 'd', on both sides

find x₁ as one solution of this congruency,
Step 3:
Now, find the incongruent solutions,

Where, 0 ≤ p ≤ (d-1)
Putting the values of 'p' in (2), we will get the required solutions. 

Analysis:
6 x ≡ 3 (mod 9) ---(3)
d = GCD(6, 9) = 3
We have exactly 3 incongruent solutions,
Now, dividing (3), with 3, we get

2x ≡ 1 (mod 3)
We can manually see that x = 2, 
4 - 1 ≡ 3 and (3 mod 3) = 0
One solution is x1 = 2,
Now, to find all the incongruent solutions,

Where, 0 ≤ p ≤ (3-1)
put, p = 0,
x = 2 + 0 = 2
put, p = 1,
x = 2 + 3 × 1 = 5
put, p = 2,
x = 2 + 3 × 2 = 8
The three incongruent solutions modulo 9 of this congruence are:  2, 5, 8

Concept:
Modulus of X (|x|): It is a function that gives the absolute value of a variable (x). It is defined as such that
|x| = x,  if x > 0 
|x| = x, if x < 0 
|x| = 0, if x = 0

Calculation:
Given equation is x² - 3 |x| + 2 = 0.
We can write this as
|x|² - 3 |x| + 2 = 0
⇒   |x|² -  |x|  - 2|x| + 2 = 0
⇒   |x|(|x| - 1) - 2( |x| - 1) = 0
⇒ ( |x| - 1)(|x| - 2) = 0
This is possible if, atleast one of the two factors is zero, i.e.
|x| - 1  = 0  or  |x| - 2  = 0
⇒ |x| = 1  or |x| = 2
⇒ x = ± 1 or x = ± 2
Clearly, we can see that there is four distinct value of x. 

Concept:
If f: A → B and g: B → C are functions then f 0 g (x) = f(g (x)) is a function from A to C.

Calculation:
Given: f(x) = |x - 1| andg(x) = tan x
As we know that, if f: A → B and g: B → C are functions then f o g (x) = f(g (x)) is a function from A to C.
⇒ f o g (7π/4) = f (g(3π/4)) 
⇒ g (3π/4) = tan (3π/4) = tan (π/2 + π/4))
As we know that, tan (π/2 + θ) = - tan θ
⇒ g (3π/4) = - tan (π/4) = - 1
⇒ f o g (7π/6) = f(- 1)
∵ f(x) = |x - 1| so, f( - 1) = |-1 - 1| = |- 2|
Hence, f o g (3π/4) = 2

Concept:

|x| ≥ 0 for every x ∈ R

Calculation:

Let f(x) = |x - 4| + 2

As we know that |x| ≥ 0 for every x ∈ R

∴ |x - 4| ≥ 0

The minimum value of function is attained when |x - 4| = 0

Hence, Minimum value of f(x) = 0 + 2 = 2 

Concept:
The modulus of a given function gives the magnitude of that function. Modulus Function is defined as the real valued function.
The real function f: R → R defined by f (x) = |x|= x, x > 0, and f (x) = |x|= -x, if x < 0. ∀ x ∈ R is called the modulus function.

Calculations:
Given |x² – 12x + 32| + |x² – 9x + 20| = 0.
Every modulus function is a non-negative function and if two non-negative functions add up to get zero then individual function itself equal to zero simultaneously.
x² – 12x + 32 for x = 4 or 8
x² – 9x + 20 for x = 4 or 5
Both the equations are zero at x = 4
So, x = 4 is the only solution for this equation.

Concept:
|x| ≥ 0 for every x ∈ R

Calculation:
Let f(x) = |x - 4| + 2
As we know that |x| ≥ 0 for every x ∈ R
∴ |x - 4| ≥ 0
The minimum value of function is attained when |x - 4| = 0
Hence, Minimum value of f(x) = 0 + 2 = 2 

Calculation:
Given: f(x) = 3^{1+ x}
Then,
⇒ f(y) = 3^{1+ y}
⇒ f(z) = 3^{1+ z}
We have to find out value of f(x) f(y) f(z),
⇒ f(x) f(y) f(z) = 3^{1+ x} × 3^{1+ y} × 3^{1+ z}
⇒ f(x) f(y) f(z) = 3^{3+ x + y + z} = 3^{1 + 2 + x + y + z} = f(x + y + z + 2)

Concept:
The minimum value of modulus functions is zero.

Calculations:
Given:
f(x) = |x - 1|
Since the minimum value of modulus functions is zero. 
Hence, the minimum value of |x - 1|, where x ∈ R is 0

Concept:
If f: A → B and g: B → C are functions then f 0 g (x) = f(g (x)) is a function from A to C.

Calculation:
Given: f(x) = |x - 1| andg(x) = tan x
As we know that, if f: A → B and g: B → C are functions then f o g (x) = f(g (x)) is a function from A to C.
⇒ f o g (7π/4) = f (g(3π/4)) 
⇒ g (3π/4) = tan (3π/4) = tan (π/2 + π/4))
As we know that, tan (π/2 + θ) = - tan θ
⇒ g (3π/4) = - tan (π/4) = - 1
⇒ f o g (7π/6) = f(- 1)
∵ f(x) = |x - 1| so, f( - 1) = |-1 - 1| = |- 2|
Hence, f o g (3π/4) = 2

Concept:
The modulus of a given function gives the magnitude of that function. Modulus Function is defined as the real valued function.
The real function f: R → R defined by f (x) = |x|= x, x > 0, and f (x) = |x|= -x, if x < 0. ∀ x ∈ R is called the modulus function.

Calculations:
Given |x² – 12x + 32| + |x² – 9x + 20| = 0.
Every modulus function is a non-negative function and if two non-negative functions add up to get zero then individual function itself equal to zero simultaneously.
x² – 12x + 32 for x = 4 or 8
x² – 9x + 20 for x = 4 or 5
Both the equations are zero at x = 4
So, x = 4 is the only solution for this equation.