RS AGGARWAL CLASS 11 MATHS SOLUTION FUNCTIONS-1

RS AGGARWAL CLASS 11 MATHS SOLUTION FUNCTIONS-1

Function Exercise 3A, 3B, 3C

Exercise 3A

Q. 1. Define a function as a set of ordered pairs.

Answer: Function as a set of ordered pairs:
A function can be defined as a set of ordered pairs (x, y), where for each x-value (input), there is exactly one y-value (output).  i.e. that no two ordered pairs have the same first component and a different second component.

Q. 2. Define a function as a correspondence between two sets.

Answer: Function as a correspondence between two sets:
A function is a special type of correspondence between two sets, called the domain and the range, where all elements in the domain is assigned to exactly one element in the range. 
This means every input has a unique output. 

Q. 3. What is the fundamental difference between a relation and a function? Is every relation a function?

Answer: Fundamental difference between Relation and Function:
A relation is a set of ordered pairs that relates elements from one set (the domain) to elements in another set (the codomain or range). 
A function is a special type of relation where every element from one set (the domain) is associated with exactly one element in another set (the codomain or range). 

A relation can map an input to multiple outputs, while a function maps each input to exactly one output.
Every function is a relation, but not every relation is a function. 
Example:
The relation {(1, 2), (1, 3)} is not a function because the input 1 is paired with two different outputs, 2 and 3.
The relation {(1, 2), (2, 3)} is a function because each input is paired with exactly one output. 

Q. 4. Let X = {1, 2, 3, 4}, Y = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}.
Are the following true?
(i) f is a relation from X to Y
(ii) f is a function from X to Y.
Justify your answer in following true?

Answer: X = {1, 2, 3, 4} and Y = {1, 5, 9, 11, 15, 16}
and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
Therefore,
A x B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A x B .
Here f is a subset of A x B.
Thus, f is a relation from A to B.
(ii) Here, (2, 9), (2, 11) ∈ f
∴ 2 is coming twice and has two different images i.e., 9 and 11.
Hence, it does not have a unique (one) image.
So, it is not a function.

Q. 5. Let X = {-1, 0, 3, 7, 9} and f : X → R : f(x) = x³ + 1.
Express the function f as set of ordered pairs.

Solution: Given: f: X → R, f(x) = x³ + 1
Here, X = {-1, 0, 3, 7, 9}
If x = -1 then f(-1) = (-1)³ + 1 = 0
If x = 0 then f(0) = (0)³ + 1 = 1
x = 3 then f(-1) = (3)³ + 1 = 28
If x = 7 then f(-1) = (7)³ + 1 = 344
If x = 9 then f(-1) = (9)³ + 1 = 730
∴ the ordered pairs are (-1, 0), (0, 1), (3, 28), (7, 344), (9, 730)

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Q. 6. Let A = {–1, 0, 1, 2} and B = {2, 3, 4, 5}.
Find which of the following are functions from A to B. Give reason.
(i) f = {(–1, 2), (-1, 3), (0, 4), 1,5)}
(ii) g = {(0, 2), (1, 3), (2, 4)}
(iii) h = {(-1, 2), (0, 3), (1, 4), (2, 5)}

Solution: (i) Given: A = {-1, 0, 1, 2} and B = {2, 3, 4, 5} )
f = {(–1, 2), (-1, 3), (0, 4), 1,5)}
all elements of the first set are not associated with the elements of the second set(B).
∴ f is not a function.
(ii) Given: A = {-1, 0, 1, 2} and B = {2, 3, 4, 5}g = {(0, 2), (1, 3), (2, 4)}
Here, -1 is not associated with any element of set B
∴ g is not a function.

(iii) Given: A = {-1, 0, 1, 2} and B = {2, 3, 4, 5} h = {(-1, 2), (0, 3), (1, 4), (2, 5)}
All elements of first set is associated with the elements of second set and every element of set A has a unique element in set B.
∴ h is a function.

Q. 7. Let A = {1, 2} and B = {2, 4, 6}. Let f = {(x, y) : x ∈ A, y ∈ B and y > 2x + 1}.
Write f as a set of ordered pairs.
Show that f is a relation but not a function from A to B.

Solution: Given: A = {1, 2} and B = {2, 4, 6}
f = {(x, y): x ∈ A, y ∈ B and y > 2x + 1}
If x = 1 then y = 2.1 + 1 = 3 > 1
∴ 1 ∈ A then 4 ∈ B
∴ 1 ∈ A then 6 ∈ B
If x = 2 then y = 2.2 + 1 = 5 > 2
∴ 2 ∈ A then 6 ∈ B

∴ f as a set of ordered pairs:
f = {(1, 4), (1, 6), (2, 6)}
Since f is a set of ordered pairs where the first element comes from set A and the second element comes from set B.
So f is a subset of A×B
Therefore, f is a relation from A to B.
f = {(1, 4), (1, 6), (2, 6)}
Here, 1 is coming twice.
Hence, it does not have a unique (one) image.
So, f is not a function.

Q. 8. Let A = {0, 1, 2} and B = {3, 5, 7, 9}. Let f = {(x, y) : x ∈ A, y ∈ B and y = 2x + 3}.
Write f as a set of ordered pairs.
Show that f is function from A to B.
Find dom (f) and range (f).

Solution: Given: A = {0, 1, 2} and B = {3, 5, 7, 9}
f = {(x, y): x ∈ A, y ∈ B and y = 2x + 3}
∵ y = 2x + 3
If x = 0 then y = 2×0 + 3 = 3 ∈ B
x = 1 then y = 2×1 + 3 = 5 ∈ B
If x = 2 then y = 2×2 + 3 = 7 ∈ B
∴ f as a set of ordered pairs:
f = {(0, 3), (1, 5), (2, 7)}
All elements of first set A is associated with the elements of second set B and every element of set A has a unique element in set B.
∴ f is a function from A to B.
Dom (f) = {0, 1, 2}
Range (f) = {3, 5, 7}

Q. 9. Let A = {2, 3, 5, 7} and B = {3, 5, 9, 13, 15}. Let f = {(x, y) : x ∈ A, y ∈ B and y = 2x – 1}.
Write f in roster form.
Show that f is a function from A to B. Find the domain and range of f.

Solution: Given: A = {2, 3, 5, 7} and B = {3, 5, 9, 13, 15}
f = {(x, y): x ∈ A, y ∈ B and y = 2x – 1}
∵ y = 2x – 1
If x = 2 then y = 2×2 – 1 = 3 ∈ B
If x = 3 then y = 2×3 – 1 = 5 ∈ B
x = 5 then y = 2×5 – 1 = 9 ∈ B
If x = 7 then y = 2×7 – 1 = 13 ∈ B

∴ f in roster form:
f = {(2, 3), (3, 5), (5, 9), (7, 13)}
All elements of first set A is associated with the elements of second set B and every element of set A has a unique element in set B.
∴ f is a function from A to B.
Dom (f) = {2, 3, 5, 7}
Range (f) = {3, 5, 9, 13}

Q. 10. Let g = {(1, 2), (2, 5), (3, 8), (4, 10), (5, 12), (6, 12)}.
Is g a function?
If yes, find its domain and range.
If no, give reason.

Solution: Given: g = {(1, 2), (2, 5), (3, 8), (4, 10), (5, 12), (6, 12)}
All the first elements of the ordered pairs associated with a unique second elements of the ordered pairs of the given set ‘g’  that means all the first elements has a unique image.
So, g is a function in the given set.
Dom (g) = {1, 2, 3, 4, 5, 6}
Range (g) = {2, 5, 8, 10, 12}

Q. 11. Let f = {(0, -5), (1, -2), (2, 1), (3, 4), (4, 7)} be a linear function from Z into Z. Write an expression for f.

Solution: Given: f = {(0, -5), (1, -2), (3, 4), (4, 7)} be a linear function from Z to Z.
Let the linear function is y = mx + b ……(i)
At (0, -5) from (i) we get,
-5 = m×0 + b
⇒ b = -5
At (1, -2) from (i) we get,
-2 = m×1 + b
⇒ -2 = m×1 + (-5) ……. [∵ b = -5]
⇒ m = 3
The linear function is y = 3x – 5
Now verify the function at (3, 4):
4 = 3×3 – 5 = 4
∴ f(x) = 3x – 5

Solution: Given: f(x) = x²
∴ f(5) = (5)²
⇒ f(5) = 25
Similarly, f(1) = (1)²
⇒ f(1) = 1

Solution:
Given: f(x) = x²
∴ f(1.1) = (1.1)²
⇒ f(1.1) = 1.21
Similarly, f(1) = (1)²
⇒ f(1) = 1

Q. 14. Let A = {12, 13, 14, 15, 16, 17} and f : A → Z : f(x) = highest prime factor of x. Find range (f)

Solution: Given: A = {12, 13, 14, 15, 16, 17}
f : A → Z : f(x) = highest prime factor of x and x ∈ A,
f(12) = the highest prime factor of 12 = 3
f(13) = the highest prime factor of 13 = 13
∴ f(14) = the highest prime factor of 14 = 7
∴ f(15) = the highest prime factor of 15 = 5
f(16) = the highest prime factor of 16 = 2
f(17) = the highest prime factor of 17 = 17
Hence, range of f = {2, 3, 5, 7, 13, 17}

Q. 15. Let R⁺ be the set of all positive real numbers. 
Let f : R⁺ → R : f(x) = log_ex.
Find
(i) Range (f)
(ii) {x : x ∈ R⁺ and f(x) = -2}.
(iii) Find out whether f(xy) = f(x). f(y) for all x, y ∈ R⁺.

Solution: Given that f: R⁺→ R : f(x) = log_ex
Here domain of f is R⁺ and codomain of f is R
∴ 0 < x < ∞
⇒ -∞ < log_ex < +∞
⇒ -∞ < f(x) < +∞(i)
(i) ∴ Range of f(x) = (-∞, +∞)
(ii) {x : x ∈ R⁺ and f(x) = -2}
Here f(x) = log_ex
f(x) = -2
∴ log_ex = -2
⇒ x = e^-2
⇒ x = ¹/_e-2
∴ {x : x ∈ R⁺ and f(x) = -2} = {¹/_e-2}

(iii) f(x) = log_ex
put x = xy
∴ f(xy) = log_e(xy)
⇒  f(xy) = log_ex + log_ey
⇒ f(xy) = f(x) + f(y)

Q. 16. Let f : R → R : f(x) = 2^x. Find
(i) Range (f)
(ii) {x : f(x) = 1}.
(iii) Find out whether f(x + y) = f(x). f(y) for all x, y ∈ R.

Solution: Given: f: R → R such that f(x) = 2^x
(i) Clearly domain of f(x) is (-∞, +∞)
∴ -∞ < x < +∞
⇒ 2^-∞ < 2^x < 2^+∞
⇒ ¹/_2∞ < 2^x < ∞
∴ 0 < 2^x < ∞
∴ Range f(x) is (0, ∞)
(ii) Given:  {x : f(x) = 1}.
f(x) = 1
∴ 2^x = 1
⇒ log₂2^x = log₂1
⇒ x.log₂2 = 0
∴ x = 0
∴ {x : f(x) = 1} = {0}
(iii) f(x) = 2^x and f(y) = 2^y
Putting x = x + y
∴ f(x + y) = 2^x+y
⇒ f(x + y) = 2^x × 2^y
⇒ f(x + y) = f(x).^.f(y)
∴ f(x + y) = f(x). f(y) for all x, y ∈ R

Q. 17. Let f : R → R : f(x) = x² and g : C → C: g(x) = x², where C is the set of all complex numbers. Show that f ≠ g.

Solution: It is given that f : R → R and g : C → C
Thus, Domain (f) = R and Domain (g) = C
We know that, Real numbers ≠ Complex Number
∵ Domain (f) ≠ Domain (g)
∴ f(x) and g(x) are not equal functions
∴ f ≠ g (Proved)

Q. 18. f, g and h are three functions defined from R to R as follows:
(i) f(x) = x²
(ii) g(x) = x² + 1
(iii) h(x) = sin x
Then, find the range of each function.

Solution: (i) f: R → R such that f(x) = x²
Since x² ≥ 0
∴ Range of f(x) is [0, ∞)
(ii) g: R → R such that g(x) = x² + 1
Since x² ≥ 0
⇒ x² + 1 ≥ 0 +1 ≥ 1
∴ Range of g(x) is [1, ∞)
(iii) h: R → R such that h(x) = sinx
We know, sin (x) always lies between -1 to 1
∴ Range of h(x) is (-1, 1)

Q. 19. Let f : R → R : f(x) = x² + 1.
Find (i) f^-1{10}
(ii) f^-1{–3}.

Solution: Given: f(x) = x² + 1
Let y = f(x) = x² + 1
⇒ x² + 1 = y
⇒ x² = y – 1

Q. 20. The function f(x) = ^9x/₅ + 32 is the formula to convert x°C to Fahrenheit units. Find
(i) F(0),
(ii) F(–10),
(iii) The value of x when f(x) = 212.
Interpret the result in each case.

Solution: Given: f(x) = ^9x/₅ + 32 …….(i)
Substituting the value of x = 0 in equation (i)
f(0) = ^9.0/₅ + 32 = 32
∴ 0° C = 32° F
Substituting the value of x = -10 in equation (i)
f(-10) = ⁹/₅ ×(-10) + 32 = -18 + 32 = 14
∴ -10° C = 14° F

∵ f(x) = 212
⇒ ^9x/₅ + 32 = 212
⇒ ^9x/₅ + 32 = 212 – 32
∴ ^9x/₅ = 180
⇒ 9x = 180×5
⇒ x = 100
∴ Value of x = 100°C

Exercise 3B

Q. 1. If f(x) = x² – 3x + 4 and f(x) = f(2x + 1), find the values of x.

Solution: Given: f(x) = x² – 3x + 4……….. (i)
and f(x) = f(2x + 1)
Substitute x by (2x + 1) in equation (i) we get,
∴ f(2x + 1) = (2x + 1)² – 3(2x + 1) + 4
⇒ f(2x + 1) = 4x² + 4x + 1 – 6x – 3 + 4
⇒ f(2x + 1) = 4x² – 2x + 2
According to the given problem,
f(x) = f(2x + 1)
∴ x² – 3x + 4 = 4x² – 2x + 2
⇒ x² – 3x + 4 – 4x² + 2x – 2 = 0
⇒ -3x² – x + 2 = 0 ⇒ 3x² + x – 2 = 0
∴ 3x² + 3x – 2x – 2 = 0
⇒ 3x(x + 1) – 2(x + 1) = 0
⇒ (3x – 2)(x + 1) = 0
So either (3x – 2) = 0 or (x + 1) = 0
∴  x = ²/₃ or x = -1
Therefore, the value of x is either ²/₃   or -1

(i) Now replacing x by ¹/_x
We get,

(ii) Now replacing x by ^-1/_x
We get,

Solution: Given: f(x) = x³ – ¹/_x3
Replacing x by ¹/_x

RS AGGARWAL CLASS 11 MATHS SOLUTION FUNCTIONS

Replacing x by f(x),
∴ f{f(x)}

Replacing x by f(x),
∴ f{f(x)}

Replacing x by f(x),
∴ f{f(x)}

Again replacing x by f{f(x)},
∴ f[f{f(x)}]

Replacing x by tan θ,
∴ f(tan θ)

Replacing x by y,
∴ f(y)

Exercise 3C

Solution: The function is undefined when the denominator is zero.
 ∴ x² – 9 = 0
⇒ x²  = 9
⇒ x = ± 3
So the domain of the function is the set of all the real numbers except -3 and 3.
Therefore domain of the function,
 D_f(x) = (- ∞, -3) ∪ (-3, 3) ∪ (3, ∞).

Solution: The function is undefined when the denominator is zero.
 ∴ x² + x – 2 = 0
⇒ x²  + 2x – x – 2 = 0
∴ x(x  + 2) – 1(x + 2) = 0
⇒ (x  + 2)(x – 1) = 0
Either (x  + 2) = 0      or (x – 1) = 0
                     ∴ x = – 2                ∴ x = 1
So the domain of the function is the set of all the real numbers except – 2 and 1.
Therefore domain of the function,
 D_f(x) = (- ∞, -2) ∪ (-2, 1) ∪ (1, ∞).

Solution: The function is undefined when the denominator is zero.
 ∴ x² – 8x + 12 = 0
⇒ x^{2  =} 6x – 2x + 12 = 0
∴ x(x  ^– 6) – 2(x – 6) = 0
⇒ (x  – 6)(x – 2) = 0
Either (x  – 6) = 0      or (x – 2) = 0
               ∴ x = 6               ∴ x = 2
So the domain of the function is the set of all the real numbers except 2 and 6.
Therefore domain of the function,
 D_f(x) = (- ∞, 2) ∪ (2, 6) ∪ (6, ∞).

RS AGGARWAL CLASS 11 MATHS SOLUTION

Solution: The function is undefined when the denominator is zero.
 ∴ x² – 1 = 0
⇒ x²  = 1
⇒ x = ± 1
So the domain of the function is the set of all the real numbers except -1 and 1.
Therefore domain of the function,
 D_f(x) = (- ∞, -1) ∪ (-1, 1) ∪ (1, ∞).

2. Find the domain and the range of the following real function: f(x) = ¹/_x

Solution: The function is undefined when the denominator is zero.
 ∴ x = 0
So the domain of the function is the set of all the real numbers except 0.
Therefore domain of the function,
 D_f(x) = (- ∞, 0) ∪ (0, ∞).

Let y = f(x) = 1/x
∴ x = 1/y
If y = 0, then x is undefined.
So the range of the function is the set of all the real numbers except 0.
Therefore range of the function,
 R_f(x) = (- ∞, 0) ∪ (0, ∞).

3. Find the domain and the range of the following real function: f(x) = ¹/_x-5

Solution: The function is undefined when the denominator is zero.
 ∴ x – 5 = 0
⇒ x  = 5
So the domain of the function is the set of all the real numbers except 5.
Therefore domain of the function,
   D_f(x) = (- ∞, 5) ∪ (5, ∞).
Let y = f(x) = ¹/_x-5
⇒ y = ¹/_x-5
⇒ x – 5 = ¹/_y
∴ x = ¹/_y + 5
If y = 0, then x is undefined.
So the range of the function is the set of all the real numbers except 0.
Therefore range of the function,
   R_f(x) = (- ∞, 0) ∪ (0, ∞).

Solution: The function is undefined when the denominator is zero.
 ∴ 2 – x = 0
⇒ x  = 2
So the domain of the function is the set of all the real numbers except 5.
Therefore domain of the function,
   D_f(x) = (- ∞, 2) ∪ (2, ∞).

Let y = ^{x – 3}/_{2 – x}
⇒ 2y – xy = x – 3
⇒ xy + x = 2y – 3
∴ x(y + 1) = 2y – 3
⇒ x = ^{2y – 3}/_{(y + 1)}
   x is undefined,
 If (y + 1) = 0
⇒ y = -1
So the range of the function is the set of all the real numbers except -1.
Therefore range of the function,
   R_f(x) = (- ∞, -1) ∪ (-1, ∞).

RS AGGARWAL CLASS 11 MATHS SOLUTION

Solution: The function is undefined when the denominator is zero.
 ∴ x + 2 = 0
⇒ x  = -2
So the domain of the function is the set of all the real numbers except -2.
Therefore domain of the function,
    D_f(x) = R – {-2}.

Let y = ^{3x – 2}/_{x + 2}
⇒ xy + 2y = 3x – 2
⇒ xy – 3x = 2y – 3
x(y – 3) = 2y – 3
⇒ x = ^{2y – 3}/_{(y – 3)}
   x is undefined,
 If (y – 3) = 0
⇒ y = 3
So the range of the function is the set of all the real numbers except 3.
Therefore range of the function,
    R_f(x) = R – {3}.

Solution: f(x) undefined when
    x – 4 = 0
⇒ x  = 4
Therefore domain of the function,
    D_f(x) = R – {4}

Since x ≠ 4
Therefore,
f(x) ≠ 4 + 4 ≠ 8
∴ Range of the function,
    R_f(x) = R – {8}.

 Solution:  f(x) is defined when,
2x – 3 > 0
⇒ 2x  > 3
⇒ x  > ³/₂ Therefore domain of the function,
    D_f(x) = (³/₂, ∞).
Again,
√2x – 3 > 0
⇒ 0 < ¹/_{√2x – 3} < ∞
∴ Range of the function,
    R_f(x) = (0, ∞).

8. Find the domain and the range of the following real function:

Solution: The function is undefined when the denominator is zero.
  ∴ cx – d = 0
 ⇒ x  = ^d/_c
 So the domain of the function is the set of all the real numbers except ^d/_c.
 Therefore domain of the function,
    D_f(x) = R – {^d/_c}.

 Let y = ^{ax – b}/_{cx – d}
 ⇒ cxy – dy = ax – b
 ⇒ cxy – ax = dy – b
  x(cy – a) = dy – b
 ⇒ x = ^{dy – b}/_{(cy – a)}
    x is undefined,
  If (cy – a) = 0
 ⇒ y = ^a/_c
 So the range of the function is the set of all the real numbers except ^a/_c.
 Therefore range of the function,
    R_f(x) = R – {^a/_c}.

Solution:  f(x) is defined when,
    3x – 5 ≥ 0
⇒ 3x  ≥ 5
⇒ x  ≥ ⁵/₃
Therefore domain of the function,
    D_f(x) = [⁵/₃, ∞).
  ∵ 3x – 5 ≥ 0
⇒ √3x – 5 ≥ 0
∴ f(x)  ≥ 0
∴ Range of the function,
    R_f(x) = (0, ∞).

Solution:  f(x) is defined as,
    3 – x ≠ 0
⇒ x ≠ 3 and

Therefore domain of the function,
    D_f(x) = (3, 5]

When 3 < x < 5 then,
 f(x) will be positive real number.
∴ Range of the function,
    R_f(x) = [0, ∞).

NCERT CLASS 11 MATHS SOLUTION

Solution: f(x) is defined when,
    x² – 1 > 0
⇒ (x + 1)(x – 1) > 0 ……. (i)
Now,
If x < -1, then x + 1 < 0 and x – 1 < 0
So (x + 1)(x – 1) > 0
If -1 < x < 1, then x + 1 > 0 and x – 1 < 0
So (x + 1)(x – 1) > 0
If x > 1, then x + 1 > 0 and x – 1 > 0
So (x + 1)(x – 1) > 0
∴ Equation (i) satisfied when x < -1 and x > 1
Therefore domain of the function,
    D_f(x) = (-∞, -1) ∪ (1, ∞)

Again

∴ Range of the function,
    R_f(x) = (0, ∞).

12. Find the domain and the range of each of the following real function: f(x) = 1- |x – 2|

Solution: Given: f(x) = 1 – |x – 2|
|x – 2| is defined for all real number.
So f(x) = 1 – |x – 2| also defined for all real number.
Therefore the domain of the function,
     D_f(x) = (-∞, ∞).

Now ,
f(x) = 1 – |x – 2|,
∵  |x – 2 | ≥ 0
∴ -|x – 2 | ≤ 0
⇒ 1 – |x – 2 | ≤ 1
Therefore, the range of the function, R_f(x) = (-∞, 1]

Solution: The function is undefined when the denominator is zero.
   ∴ x – 4 = 0
 ⇒ x  = 4
 So the domain of the function is the set of all the real numbers except 4.
 Therefore domain of the function,
    D_f(x) = R – {4}.

Therefore, the range of the function,
R_f(x) = { -1 , 1 }

Solution: f(x) is undefined when,
   x – 3 = 0
⇒ x  = 3
Therefore domain of the function,
    D_f(x) = R – {3}

Since x ≠ 3 therefore ,
f(x) ≠ 3 + 3 ≠ 6
∴ Range of the function,
    R_f(x) = R – {6}.

CBSE CLASS 11 MATHS SOLUTION

Solution:

    -1 ≤ sin3x ≤ 1 for all x ∈ R
⇒ -1 ≤ -sin3x ≤ 1 for all x ∈ R
2 -1 ≤ 2 – sin3x ≤ 2 + 1 for all x ∈ R
⇒ 1 ≤ 2 – sin3x ≤ 3
Therefore domain of the function,
    D_f(x) = R

∴ Range of the function,
    R_f(x) = [¹/₃, 1]