Test: Range of a Function - SAT MCQ

Concept:

The range of f(x) is all the y-values where there is a number x with y=f(x).

Calculations: 

To find the range of the function  first split the function.

We know that, the range of f(x) is all the y-values where there is a number x with y=f(x).

Now to find the range, take the limit of the function.

The range of the function 

Given:

f(x) = x²

Calculation:

f(x) = x²

⇒ Range of f = [0, ∞] = R⁺

⇒ Positive real numbers 

∴ The range of f will be positive real numbers

The number e, it is called the Euler's number, is an important mathematical constant approximately equal to 2.71828.

The approx. value of e is 2.71828

∴ 2 < e < 3

Concept:

• For a quadratic equation ax² + bx +c = 0 to have real roots: D = b² – 4ac ≥ 0

Calculation:

On solving,

y + yx² = x²

⇒ (y - 1) x² + y = 0

⇒(y - 1) x² + y = 0

Also for the quadratic equation to exist y – 1 = 0 is not possible

Which is a quadratic equation in x, on comparing ax² + bx +c = 0 we get a = (y - 1), b = 0, c = y.

For x ∈ R considering D ≥ 0 so b² – 4ac ≥ 0

0 – 4 y ( y – 1 ) ≥ 0

⇒ y ( y - 1 ) ≤ 0

Which gives the solution y ∈ [0, 1)

Concept:

Let, y = f(x), then interval of y is the minimum and maximum value range of y.

Calculation:

For x = 0, y = 0, for x = -1, y = + ½ etc.

So here y ≥ 0

For x = 1, y = ½

For x = 1/2, y = 4/17 < ½

For x = 2, y = 4/17 etc.

So here, y ≤ ½

Concept:

The cartesian coordinate obtained by measuring parallel to the y-axis. 

Calculation:

Given: y = 2 - sin x  

The function sin x has all real numbers in its domain, but Range is -1 ≤ sin x ≤ 1 or  -1 ≤ -sin x ≤ 1. 

 -1 ≤ -sin x ≤ 1

Adding 2 both sides, we get

⇒ 2 - 1 ≤ 2 - sin x ≤ 2 + 1

⇒ 1 ≤ y ≤ 3

∴ y ∈ [ 1 , 3 ] 

Concept:

Below are steps to find out range of function:

1. Write down y = f(x) and then solve the equation for x, giving something of the form x=g(y).

Find the domain of g(y), and this will be the range of f(x).

If you can't seem to solve for x, then try graphing the function to find the range.

Calculation:

Given that,

Cross multiply

⇒ y(1 + x² ) = x²

⇒ y + y . x² = x²

⇒ x² (1 -y) = y

⇒ x² = y/(1 -y)

Now we find out domain of g(y) which will be range of f(x)

So as we know inside root always positive values comes so,

By showing on number line we will get range [0, 1)

Concept:

The domain of a relation R = {(x, y)} is the set of all values of x, and the range of R is the set of all values of y.

Calculation:

Since R is a relation on ℕ, the elements x and y must be positive integers.

We have x + 2y = 8.

For y to be positive and an integer (y ∈ N), we conclude that x must be divisible by 2 and 

⇒ x < 8.

The only numbers less than 8 which are divisible by 2 are 2, 4 and 6.

∴ x ∈ {2, 4, 6} which is the required domain.

Concept:

 Period of a function:

• If a function repeats over at a constant period we say that is a periodic function.
• It is represented like f(x) = f(x + T), T is the real number and this is the period of the function.
• The period of sin x and cos x is 2π

Calculation:

To Find: Period of 4cos3 x - 3cos x

As we know 4cos3 x - 3cos x = cos 3x

Period of cos x is 2π

Therefore, the Period of cos 3x is 2π/3

Given function,

Domain = (-∞, ∞) \{-3}
 

where y is in the range of h(x).

⇒ (x + 3)y = x - 2

⇒ xy - x + 3y + 2 = 0

⇒ x(y - 1) = -(3y + 2)

y can take any real value but y ≠ 1

So, the range of h(x) is (-∞, 1) ∪ (1, ∞)