Test: Modulus Function - SAT MCQ

Test: Modulus Function - SAT MCQ

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10 Questions MCQ Test - Test: Modulus Function

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Test: Modulus Function - Question 1

Consider the linear congruence 6 x ≡ 3 (mod 9). Then the incongruent solutions modulo 9 of this congruence are:

Detailed Solution for Test: Modulus Function - Question 1

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,
x = x1 + m/dp     -----(2)
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
⇒ 6x/3 ≡ 33 (mod 9/3)
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,
x = x1 + m/d p
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

Test: Modulus Function - Question 2

If f(x) = |x - 1| and g(x) = tan x then f o g (3π/4)

Detailed Solution for Test: Modulus Function - Question 2

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

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Test: Modulus Function - Question 3

If |x2 – 12x + 32| + |x2 – 9x + 20| = 0. Then find the value of x.

Detailed Solution for Test: Modulus Function - Question 3

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.

Test: Modulus Function - Question 4

If f(x) = 31+x, then f(x) f(y) f(z) is equal to

Detailed Solution for Test: Modulus Function - Question 4

Calculation:
Given: f(x) = 31+ x
Then,
⇒ f(y) = 31+ y
⇒ f(z) = 31+ z
We have to find out value of f(x) f(y) f(z),
⇒ f(x) f(y) f(z) = 31+ x × 31+ y × 31+ z
⇒ f(x) f(y) f(z) = 33+ x + y + z = 31 + 2 + x + y + z = f(x + y + z + 2)

Test: Modulus Function - Question 5

The equation |x + 2| = -2 has:

Detailed Solution for Test: Modulus Function - Question 5

Concept:
The Modulus Function '| |' is defined as:

Calculation:
Since the modulus function '||' always returns a positive value or 0, its not possible to have -2 as the value of modulus of any expression.
Hence, |x + 2| = -2 has no solution.

Test: Modulus Function - Question 6

The number of real solutions of equation x2 - 3 |x| + 2 = 0 is:

Detailed Solution for Test: Modulus Function - Question 6

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.

Test: Modulus Function - Question 7

What is the minimum values of the function |x - 4| + 2?

Detailed Solution for Test: Modulus Function - Question 7

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

Test: Modulus Function - Question 8

What is the minimum value of |x - 1|, where x ∈ R ?

Detailed Solution for Test: Modulus Function - Question 8

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

Test: Modulus Function - Question 9

Which one of the following is correct in respect of the function f: R → R+ defined as f(x) = |x + 1|?

Detailed Solution for Test: Modulus Function - Question 9

Calculation:
Given that,
⇒ f(x) = |x + 1|
first option
⇒ f(x2) = |x2 + 1|
⇒ f(x)2 = |x + 1|2
so we can say that
⇒ f(x2) = |f(x)|2
⇒ |x + 1|2 ≠ |x2 + 1|
First option not correct.
Second option,
⇒ f(|x|) = |f(x)|
⇒ f(|x|) = ||x| + 1|
⇒ |f(x)| = ||x + 1|| = |x + 1|
f(|x|) ≠ |f(x)| because ||x| + 1| ≠ |x + 1| for real values of x.
Third option,
⇒ f(x + y) = f(X) + f(y)
⇒ f(x + y) = |(x + y) + 1|
⇒ f(y) = |y + 1|
so,
⇒ |(x + y) + 1| ≠ |x + 1| + |y + 1|
So f(x + y) ≠ f(x) + f(y)

Test: Modulus Function - Question 10

If f(x) = sin x and g(x) = |x| then g of (7π/6)

Detailed Solution for Test: Modulus Function - Question 10

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

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

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