RS Aggarwal Solutions for Class 11 Chapter 03 Functions

Question 1. Define a function as a set of ordered pairs.
Answer: A function is a collection of ordered pairs where no two pairs share the same first component while having different second components. The domain of a function refers to the set containing all first components (x values) from the ordered pairs, while the range consists of all second components (y values). For instance, {(1,x), (2,y), (3,z)} qualifies as a function because no two pairs have an identical first component. In this example, the domain is {1, 2, 3} and the range is {x, y, z}.
In simple words: A function is a pairing where each input (first number) links to one and only one output (second number). The domain gathers all inputs, and the range gathers all outputs.

Exam Tip: Remember - the key rule is that one first component cannot map to two different second components. This single criterion separates functions from general relations.

Question 2. Define a function as a correspondence between two sets.
Answer: Let A and B be two non-empty sets. A function 'f' from set A to set B is a rule or correspondence that matches elements from set A to elements in set B with these conditions:
(i) Every element in set A connects to at least one element in set B.
(ii) Each element in set A matches with exactly one element in set B.
In simple words: A function links every item in set A to precisely one item in set B - no item in A is left out, and no item maps to two different targets.

Exam Tip: Pay close attention to both conditions - totality (all elements covered) and uniqueness (one-to-one mapping) are equally important for a valid function.

Question 3. What is the fundamental difference between a relation and function? Is every relation a function?
Answer: Every function represents a relation, but not all relations qualify as functions. A relation f connecting set A to set B meets the function criteria only if:
(i) The domain of f equals A
(ii) No two distinct ordered pairs in f share the same first component.

For illustration, let A = {a, b, c, d} and B = {1, 2, 3, 4, 5}. Three relations are defined:
f = {(a, 1), (b, 2), (c, 3), (d, 4)}

In relation f, the domain equals A and all first components are different. Therefore, f qualifies as a function.

In relation g = {(a, 1), (b, 3), (c, 5)}, the domain does not equal A, so the requirement fails. Thus, g is not a function.

In relation h = {(a, 1), (b, 2), (b, 3), (c, 4), (d, 5)}, the domain equals A but element b appears as a first component twice with different partners. So, h is not a function.

No, not every relation qualifies as a function.
In simple words: A function must cover every element of set A, and each element must point to only one target. A relation can be looser - elements might be skipped or point to multiple targets.

Exam Tip: Always check two things: Does the domain equal the first set? Does every first element pair with exactly one second element? Only "yes" to both means it is a function.

Question 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.
Answer: Given X = {1, 2, 3, 4} and Y = {1, 5, 9, 11, 15, 16}
F = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(i) To demonstrate that F is a relation from X to Y:

First components in F are 1, 2, 3, 4 - all lie in set X.
Second components in F are 5, 9, 1, 11 - all lie in set Y.
Since the first element originates from set X and the second element originates from set Y, F qualifies as a relation from X to Y.

(ii) To demonstrate whether F is a function from X to Y:

A function must satisfy:
(i) All elements of the first set must connect with elements of the second set.
(ii) Each element of the first set must have exactly one image in the second set.

F = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

Notice that 2 appears as a first component twice - once pairing with 9 and once pairing with 11. This means 2 does not have a single unique image. Consequently, F does not qualify as a function.
In simple words: F links pairs from the two sets, so it is a relation. However, the number 2 maps to both 9 and 11, breaking the function rule that says each input must map to only one output.

Exam Tip: When checking if something is a function, scan for repeated first components - if you find one with multiple partners, it fails the function test immediately.

Question 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.
Answer: Given: f: X → R, f(x) = x³ + 1
Here, X = {-1, 0, 3, 7, 9}

For x = -1:
f(-1) = (-1)³ + 1 = -1 + 1 = 0

For x = 0:
f(0) = (0)³ + 1 = 0 + 1 = 1

For x = 3:
f(3) = (3)³ + 1 = 27 + 1 = 28

For x = 7:
f(7) = (7)³ + 1 = 343 + 1 = 344

For x = 9:
f(9) = (9)³ + 1 = 729 + 1 = 730

Therefore, the ordered pairs are (-1, 0), (0, 1), (3, 28), (7, 344), (9, 730).
In simple words: Plug each x value from the set into the formula. Calculate the result and form a pair with the original x and the answer.

Exam Tip: Always compute f(x) for every element in the domain - do not skip any. Present the final answer as a complete set of ordered pairs in curly braces.

Question 6. Let A = {-1, 0, 1, 2} and B = {2, 3, 4, 5}. Find which of the following are function 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)}
Answer: (i) Given: A = {-1, 0, 1, 2} and B = {2, 3, 4, 5}

A function must satisfy:
(i) All elements of the first set must connect with elements of the second set.
(ii) Each element of the first set must have exactly one image in the second set.

f = {(-1, 2), (-1, 3), (0, 4), (1, 5)}

The element -1 appears twice in the first position - once with 2 and once with 3. This violates the uniqueness requirement. Therefore, f is not a function.

(ii) Given: A = {-1, 0, 1, 2} and B = {2, 3, 4, 5}

A function must satisfy:
(i) All elements of the first set must connect with elements of the second set.
(ii) Each element of the first set must have exactly one image in the second set.

g = {(0, 2), (1, 3), (2, 4)}

The element -1 from set A does not appear in any ordered pair. This means not all elements from set A are covered. Therefore, g is not a function.

(iii) Given: A = {-1, 0, 1, 2} and B = {2, 3, 4, 5}

A function must satisfy:
(i) All elements of the first set must connect with elements of the second set.
(ii) Each element of the first set must have exactly one image in the second set.

h = {(-1, 2), (0, 3), (1, 4), (2, 5)}

All elements from set A are connected to exactly one element each in set B. Therefore, h is a function.
In simple words: For (i), -1 maps to two different values, so it fails. For (ii), the element -1 is left unmapped, so it fails. For (iii), each element maps to exactly one target, so it works.

Exam Tip: Check two boxes: Are all elements from the first set used? Does each appear only once as a first component? Both must be "yes" for a valid function.

Question 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.
Answer: Given: A = {1, 2} and B = {2, 4, 6}
f = {(x, y): x ∈ A, y ∈ B and y > 2x + 1}

Substituting x = 1 into y > 2x + 1:
y > 2(1) + 1
y > 3
and y ∈ B

This gives y = 4, 6 when x = 1 since these values satisfy the condition y > 3.

Substituting x = 2 into y > 2x + 1:
y > 2(2) + 1
y > 5

This gives y = 6 when x = 2 since this value satisfies the condition y > 5.

Therefore, f = {(1, 4), (1, 6), (2, 6)}

Note that (1, 2), (2, 2), and (2, 4) do not belong to f because they fail to satisfy y > 2x + 1.

To show f is a relation from A to B:

First components are 1 and 2 - both lie in set A.
Second components are 4 and 6 - both lie in set B.
Since the first element originates from set A and the second element originates from set B, f qualifies as a relation from A to B.

To show f is not a function from A to B:

A function must satisfy:
(i) All elements of the first set must connect with elements of the second set.
(ii) Each element of the first set must have exactly one image in the second set.

f = {(1, 4), (1, 6), (2, 6)}

The element 1 appears twice as the first component - once paired with 4 and once paired with 6. This means 1 does not have a single unique image. Therefore, f is not a function.
In simple words: f connects pairs from both sets, making it a relation. However, the input 1 points to two different outputs (4 and 6), which breaks the function rule.

Exam Tip: When a condition like "y > 2x + 1" is given, substitute each x value and identify all y values that both satisfy the inequality AND belong to the given set B.

Question 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).
Answer: Given: A = {0, 1, 2} and B = {3, 5, 7, 9}
f = {(x, y): x ∈ A, y ∈ B and y = 2x + 3}

For x = 0:
y = 2(0) + 3 = 3, and 3 ∈ B

For x = 1:
y = 2(1) + 3 = 5, and 5 ∈ B

For x = 2:
y = 2(2) + 3 = 7, and 7 ∈ B

Therefore, f = {(0, 3), (1, 5), (2, 7)}

Note that (0, 5), (0, 7), (0, 9), (1, 3), (1, 7), (1, 9), (2, 3), (2, 5), (2, 9) do not belong to f as they do not satisfy the condition y = 2x + 3.

To show that f is a function from A to B:

A function must satisfy:
(i) All elements of the first set must connect with elements of the second set.
(ii) Each element of the first set must have exactly one image in the second set.

f = {(0, 3), (1, 5), (2, 7)}

All elements from set A connect with exactly one element each in set B. Therefore, f is a function.

Dom(f) = {0, 1, 2}
Range(f) = {3, 5, 7}
In simple words: Substitute each x from set A into the formula y = 2x + 3. The resulting pairs form the function. The domain is the set of all inputs used, and the range is the set of all outputs obtained.

Exam Tip: Always verify that every output actually belongs to set B - if even one output falls outside B, that pair does not belong to the function.

Question 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.
Answer: Given: A = {2, 3, 5, 7} and B = {3, 5, 9, 13, 15}
f = {(x, y): x ∈ A, y ∈ B and y = 2x - 1}

For x = 2:
y = 2(2) - 1 = 3, and 3 ∈ B

For x = 3:
y = 2(3) - 1 = 5, and 5 ∈ B

For x = 5:
y = 2(5) - 1 = 9, and 9 ∈ B

For x = 7:
y = 2(7) - 1 = 13, and 13 ∈ B

Therefore, f = {(2, 3), (3, 5), (5, 9), (7, 13)}

To show that f is a function from A to B:

A function must satisfy:
(i) All elements of the first set must connect with elements of the second set.
(ii) Each element of the first set must have exactly one image in the second set.

f = {(2, 3), (3, 5), (5, 9), (7, 13)}

All elements from set A connect with exactly one element each in set B. Therefore, f is a function.

Dom(f) = {2, 3, 5, 7}
Range(f) = {3, 5, 9, 13}
In simple words: Apply the formula y = 2x - 1 to each element in set A. Check that the result belongs to set B. List all valid pairs to get the function in roster form.

Exam Tip: The domain always equals the entire first set when showing a valid function from A to B. The range contains only the output values actually obtained, not the entire set B.

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

We know that a function f from set A to set B is a correspondence (rule) that matches elements of set A to elements of set B such that:
(i) All elements of set A link with an element in set B.
(ii) Each element of set A pairs with exactly one element in set B.

Looking at g, we observe that each element appears as the first component in exactly one ordered pair. This means every element has a unique partner. Therefore, g qualifies as a function.

Dom(g) = {1, 2, 3, 4, 5, 6}
Range(g) = {2, 5, 8, 10, 12}
In simple words: Every number on the left appears only once, and each links to exactly one number on the right. Even though 6 and 5 both map to 12, this is allowed - different inputs can produce the same output.

Exam Tip: Having two different first components map to the same second component (like 5→12 and 6→12) does NOT violate the function rule. What matters is that each first component has only one target.

Question 11. Let f = {(0, -5), (1, -2), (3, 4), (4, 7)} be a linear function from Z into Z. Write an expression for f.
Answer: Given that f = {(0, -5), (1, -2), (3, 4), (4, 7)} forms a linear function from Z to Z.

We know that linear functions take the form y = mx + b.

Let f(x) = ax + b for some integers a and b.

Since (0, -5) belongs to f:
f(0) = -5
a(0) + b = -5
b = -5 ... (i)

Since (1, -2) belongs to f:
f(1) = -2
a(1) + b = -2
a + b = -2
a + (-5) = -2 [using (i)]
a = -2 + 5
a = 3

Therefore, f(x) = ax + b = 3x + (-5)
f(x) = 3x - 5
In simple words: Use two points from the function to set up two equations. Solve them to find the constants a and b in the linear form y = ax + b.

Exam Tip: Always choose simpler pairs (ones with x = 0 if available) to find the constant b first, then use another pair to find the slope coefficient a.

Question 12. If f(x) = x², find the value of \( \frac{f(5)-f(1)}{(5-1)} \).
Answer: Given: f(x) = x²

First, we calculate f(5):
Substituting x = 5 into the equation:
f(5) = (5)² = 25

Next, we calculate f(1):
f(1) = (1)² = 1

Now, we substitute f(5) and f(1) into the expression:
\[ \frac{f(5)-f(1)}{(5-1)} = \frac{25-1}{5-1} = \frac{24}{4} = 6 \]

Hence, the value is 6.
In simple words: Plug in x = 5 and x = 1 into the function separately. Then follow the arithmetic in the fraction by substituting the results you got.

Exam Tip: This expression represents the average rate of change of the function between two points - a concept fundamental to calculus and useful for understanding function behavior.

Question 13. If f(x) = x², find the value of \( \frac{f(1.1)-f(1)}{(1.1-1)} \).
Answer: Given: f(x) = x²

First, we calculate f(1.1):
Substituting x = 1.1 into the equation:
f(1.1) = (1.1)² = 1.21

Next, we calculate f(1):
f(1) = (1)² = 1

Now, we substitute f(1.1) and f(1) into the expression:
\[ \frac{f(1.1)-f(1)}{(1.1-1)} = \frac{1.21-1}{1.1-1} = \frac{0.21}{0.1} = 2.1 \]

Hence, the value is 2.1.
In simple words: Calculate f at 1.1 and at 1 separately. Then divide the change in the function output by the change in the input value.

Exam Tip: Notice how as the interval gets narrower (1 to 1.1 versus 1 to 5), this rate of change approaches the slope of the tangent line at x = 1, which would be 2 in the case of f(x) = x².

Question 14. Let X = {12, 13, 14, 15, 16, 17} and f : X → Z : f(x) = highest prime factor of x. Find range (f).
Answer: Given: f(x) = highest prime factor of x
And since x ∈ X, X = {12, 13, 14, 15, 16, 17}

The value of x can only be 12, 13, 14, 15, 16, or 17.

Performing prime factorization for each:

12 = 2 × 2 × 3 → highest prime factor = 3
13 = 13 → highest prime factor = 13
14 = 2 × 7 → highest prime factor = 7
15 = 3 × 5 → highest prime factor = 5
16 = 2 × 2 × 2 × 2 → highest prime factor = 2
17 = 17 → highest prime factor = 17

Therefore, range of f = {2, 3, 5, 7, 13, 17}.
In simple words: Break each number into its prime factors. The largest prime in that list is the output for that number. Collect all these outputs to get the range.

Exam Tip: Prime factorization is essential here - use systematic division or factor trees to ensure you identify all prime factors correctly and pick the largest one.

Question 15. Let R⁺ be the set of all positive real numbers. Let f : R⁺ → R : f(x) = logₑx. Find
(i) Range (f)
(ii) {x : x ∈ R⁺ and f(x) = -2}
(iii) Find out whether f(x + y) = f(x) · f(y) for all x, y ∈ R.
Answer: Given that f: R⁺ → R such that f(x) = logₑx

To find: (i) Range of f

Here, f(x) = logₑx

As x takes all positive real values from 0 to ∞, logₑx produces all real values from -∞ to +∞. Therefore, the range of f equals all real numbers, which is R.

To find: (ii) {x : x ∈ R⁺ and f(x) = -2}

We need to find x such that logₑx = -2
Converting from logarithmic to exponential form:
x = e⁻²
x ≈ 0.135

Therefore, {x : x ∈ R⁺ and f(x) = -2} = {e⁻²}

To find: (iii) Does f(x + y) = f(x) · f(y) for all x, y ∈ R?

Using the logarithm property, we have:
f(x + y) = logₑ(x + y)
f(x) · f(y) = logₑx · logₑy

These are not equal. For instance, let x = e and y = e:
f(e + e) = logₑ(2e) = logₑ2 + logₑe = logₑ2 + 1 ≈ 1.693
f(e) · f(e) = 1 · 1 = 1

Since 1.693 ≠ 1, the equation f(x + y) = f(x) · f(y) does NOT hold for all x, y ∈ R.

What actually holds is the property: f(x · y) = f(x) + f(y), which is the product rule for logarithms.
In simple words: The natural logarithm can produce any real number as output, so its range is all real numbers. To solve logₑx = -2, convert it to exponential form. The product property of logarithms connects logarithm of a product to a sum, not a product of logarithms.

Exam Tip: Memorize the core logarithm properties: logₑ(xy) = logₑx + logₑy and logₑ(xⁿ) = n·logₑx. These are far more common than any product-of-logs relationship.

Question 16. Let f : R → R : f(x) = 2x. 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.
Answer: Given that f: R → R such that f(x) = 2x

To find: (i) Range of x

The output f(x) = 2x stays positive for every x ∈ R since 2x is always positive. Additionally, for each positive real number x, there exists log₂x ∈ R such that f(log₂ x) = 2^(log₂ x) = x [∵ a^(log_a x) = x].

Therefore, the range of f is the set of all positive real numbers.

To find: (ii) {x : f(x) = 1}

We have, f(x) = 1 …(a)

and f(x) = 2x …(b)

From eq. (a) and (b), we get

2x = 1

\( \implies 2^x = 2^0 \) [∵ 2⁰ = 1]

By comparing the exponents of 2, we get

\( \implies x = 0 \)

∴{x : f(x) = 1} = {0}

To find: (iii) f(x + y) = f(x). f(y) for all x, y ∈ R

We have,

f(x + y) = 2^(x + y)

= 2x.2y

[The exponent product rule tells us that when multiplying two powers with the same base, you can add the exponents or reverse the process]

= f(x).f(y) [∵f(x) = 2x]

∴ f(x + y) = f(x). f(y) holds for all x, y ∈ R

Exam Tip: Always verify range by checking if every output value can be achieved and whether there are any unreachable values from the domain.

Question 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.
Answer: 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 numbers

∵, Domain (f) ≠ Domain (g)

∴ f(x) and g(x) are not equal functions

∴ f ≠ g

Exam Tip: Two functions are equal only when both their domains and their functional rules are identical - a domain mismatch alone is sufficient to distinguish them.

Question 18. f, g and h are three functions defined from R to R as following: (i) f(x) = x² (ii) g(x) = x² + 1 (iii) h(x) = sin x That, find the range of each function.
Answer: (i) f: R → R such that f(x) = x²

Since x is squared, f(x) will always be equal to or greater than 0.

∴ the range is [0, ∞)

(ii) g: R → R such that g(x) = x² + 1

Since x is squared and also adding with 1, g(x) will always be equal to or greater than 1.

∴ Range of g(x) = [1, ∞)

(iii) h: R → R such that h(x) = sin x

We know that, sin (x) always lies between -1 to 1

∴ Range of h(x) = (-1, 1)

Exam Tip: When finding range, identify the minimum and maximum possible outputs - for squared terms, the minimum is always non-negative, and for trigonometric functions, use their known oscillation limits.

Question 19. Let f : R → R : f(x) = x² + 1. Find (i) f⁻¹ {10} (ii) f⁻¹ {–3}.
Answer: Given: f(x) = x² + 1

To find: (i) f⁻¹{10}

We know that, if f: X → Y such that y ∈ Y. Then f⁻¹(y) = {x ∈ X: f(x) = y}. In other words, f⁻¹(y) represents the set of pre-images of y.

Let f⁻¹{10} = x. Then, f(x) = 10 …(i)

and it is given that f(x) = x² + 1 …(ii)

So, from (i) and (ii), we get

x² + 1 = 10

\( \implies x^2 = 10 - 1 \)

\( \implies x^2 = 9 \)

\( \implies x = \sqrt{9} \)

\( \implies x = \pm 3 \)

∴ f⁻¹{10} = {-3, 3}

To find: (ii) f⁻¹{-3}

Let f⁻¹{-3} = x. Then, f(x) = -3 …(iii)

and it is given that f(x) = x² + 1 …(iv)

So, from (iii) and (iv), we get

x² + 1 = -3

\( \implies x^2 = -3 - 1 \)

\( \implies x^2 = -4 \)

Clearly, this equation has no solution in R

∴ f⁻¹{-3} = ∅

Exam Tip: Pre-images may not exist if the value lies outside the range - always check whether the given output value is actually achievable by the function.

Question 20. The function \( F(x) = \frac{9x}{5} + 32 \) 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.
Answer: Given: \( F(x) = \frac{9x}{5} + 32 \) …(i)

To find: (i) F(0)

Substituting the value of x = 0 in eq. (i), we get

\( F(x) = \frac{9}{5}x + 32 \)

\( \implies F(0) = \frac{9}{5} \times 0 + 32 \)

\( \implies F(0) = 32 \)

It means 0° C = 32° F

To find: (ii) F(-10)

Substituting the value of x = -10 in eq. (i), we get

\( F(x) = \frac{9}{5}x + 32 \)

\( \implies F(-10) = \frac{9}{5} \times (-10) + 32 \)

\( \implies F(-10) = 9 \times (-2) + 32 \)

\( \implies F(-10) = -18 + 32 \)

\( \implies f(-10) = 14 \)

It means -10° C = 14° F

To find: (iii) the value of x when F(x) = 212

It is given that \( F(x) = \frac{9}{5}x + 32 \)

Substituting the value of F(x) = 212 in the above equation, we get

\( 212 = \frac{9}{5}x + 32 \)

\( \implies 212 - 32 = \frac{9}{5}x \)

\( \implies 180 = \frac{9}{5}x \)

\( \implies x = 180 \times \frac{5}{9} \)

\( \implies x = 20 \times 5 \)

\( \implies x = 100 \)

It means 212°F = 100°C

Exam Tip: When solving linear equations involving fractions, isolate the variable first, then multiply by the reciprocal of its coefficient to simplify the calculation.

Question 1. If f(x) = x² - 3x + 4 and f(x) = f(2x + 1), find the values of x.
Answer: Given: f(x) = x² - 3x + 4 ----- (1)

and f(x) = f(2x + 1)

Need to Find: Value of x

Replacing x by (2x + 1) in equation (1) we get,

f(2x + 1) = (2x + 1)² - 3(2x + 1) + 4 ----- (2)

According to the given problem, f(x) = f(2x + 1)

Comparing (1) and (2) we get,

x² - 3x + 4 = (2x + 1)² - 3(2x + 1) + 4

\( \implies x^2 - 3x + 4 = 4x^2 + 4x + 1 - 6x - 3 + 4 \)

\( \implies 4x^2 + 4x + 1 - 6x - 3 + 4 - x^2 + 3x - 4 = 0 \)

\( \implies 3x^2 + x - 2 = 0 \)

\( \implies 3x^2 + 3x - 2x - 2 = 0 \)

\( \implies 3x(x + 1) - 2(x + 1) = 0 \)

\( \implies (3x - 2)(x + 1) = 0 \)

So, either (3x - 2) = 0 or (x + 1) = 0

Therefore, the value of x is either \( \frac{2}{3} \) or -1

Exam Tip: When two function values are equal, set up the equation carefully by substituting the transformation into the original function, then simplify by moving all terms to one side before factoring.

Question 2. If \( f(x) = \frac{x - 1}{x + 1} \) then show that (i) \( f\left(\frac{1}{x}\right) = -f(x) \) (ii) \( f\left(\frac{-1}{x}\right) = \frac{-1}{f(x)} \)
Answer: Given: \( f(x) = \frac{x - 1}{x + 1} \)

(i) Need to prove: \( f\left(\frac{1}{x}\right) = -f(x) \)

Now replacing x by \( \frac{1}{x} \) we get,

\( f\left(\frac{1}{x}\right) = \frac{\frac{1}{x} - 1}{\frac{1}{x} + 1} \)

\( \implies f\left(\frac{1}{x}\right) = \frac{1 - x}{1 + x} \)

\( \implies f\left(\frac{1}{x}\right) = \frac{-(x - 1)}{(x + 1)} = -f(x) \) [Proved]

(ii) Need to prove: \( f\left(\frac{-1}{x}\right) = \frac{-1}{f(x)} \)

Now replacing x by \( \frac{-1}{x} \) we get,

\( f\left(\frac{-1}{x}\right) = \frac{\frac{-1}{x} - 1}{\frac{-1}{x} + 1} \)

\( \implies f\left(\frac{-1}{x}\right) = \frac{-1 - x}{-1 + x} \)

\( \implies f\left(\frac{-1}{x}\right) = \frac{-(1 + x)}{x - 1} \)

\( \implies f\left(\frac{-1}{x}\right) = \frac{-1}{f(x)} \) [Proved]

Exam Tip: For reciprocal or inverse argument substitutions, simplify the resulting fractions by finding a common denominator and factoring out common terms to reveal the required relationship.

Question 3. If \( f(x) = x^3 - \frac{1}{x^3} \) then show that \( f(x) + f\left(\frac{1}{x}\right) = 0 \)
Answer: Given: \( f(x) = x^3 - \frac{1}{x^3} \)

Need to prove: \( f(x) + f\left(\frac{1}{x}\right) = 0 \)

Replacing x by \( \frac{1}{x} \) we get,

\( f\left(\frac{1}{x}\right) = \frac{1}{x^3} - \frac{1}{\frac{1}{x^3}} = \frac{1}{x^3} - x^3 \)

Now according to the problem,

\( f(x) + f\left(\frac{1}{x}\right) = x^3 - \frac{1}{x^3} + \frac{1}{x^3} - x^3 \)

\( \implies f(x) + f\left(\frac{1}{x}\right) = 0 \) [Proved]

Exam Tip: When proving functional relationships, substitute the given transformation, simplify, and then combine with the original expression to reveal the desired identity.

Question 4. If \( f(x) = \frac{x + 1}{x - 1} \) then show that f{f(x)} = x.
Answer: Given: \( f(x) = \frac{x + 1}{x - 1} \)

Need to prove: f{f(x)} = x

Now replacing x by f(x) we get,

\( f\{f(x)\} = \frac{f(x) + 1}{f(x) - 1} \)

\( \implies f\{f(x)\} = \frac{\frac{x + 1}{x - 1} + 1}{\frac{x + 1}{x - 1} - 1} \)

\( \implies f\{f(x)\} = \frac{\frac{x + 1 + x - 1}{x - 1}}{\frac{x + 1 - x + 1}{x - 1}} \)

\( \implies f\{f(x)\} = \frac{2x}{2} \)

\( \implies f\{f(x)\} = x \) [Proved]

Exam Tip: For composition of functions, substitute the entire function into itself carefully, simplify complex fractions by combining numerators and denominators separately, then reduce to the simplest form.

Question 5. If \( f(x) = \frac{1}{(2x + 1)} \) and \( x \neq \frac{-1}{2} \) then prove that \( f\{f(x)\} = \frac{2x + 1}{2x + 3} \), when it is given that \( x \neq \frac{-3}{2} \)
Answer: Given: \( f(x) = \frac{1}{(2x + 1)} \), where \( x \neq \frac{-1}{2} \)

Need to prove: \( f\{f(x)\} = \frac{2x + 1}{2x + 3} \) when \( x \neq \frac{-3}{2} \)

Now placing f(x) in place of x

\( \implies f\{f(x)\} = \frac{1}{2f(x) + 1} \)

\( \implies f\{f(x)\} = \frac{1}{2 \cdot \frac{1}{2x + 1} + 1} \)

\( \implies f\{f(x)\} = \frac{1}{\frac{2}{2x + 1} + 1} = \frac{2x + 1}{2 + 2x + 1} = \frac{2x + 1}{2x + 3} \), where \( x \neq \frac{-3}{2} \) [Proved]

Exam Tip: Always track domain restrictions - when composing functions, new restrictions on the domain may emerge from the intermediate steps and must be documented in the final answer.

Question 6. If \( f(x) = \frac{1}{(1 - x)} \) then show that f[f{f(x)}] = x.
Answer: Given: \( f(x) = \frac{1}{(1 - x)} \)

Need to prove: f[f{f(x)}] = x

Replacing x by f(x),

\( f\{f(x)\} = \frac{1}{1 - f(x)} = \frac{1}{1 - \frac{1}{1 - x}} = \frac{1 - x}{1 - x - 1} = \frac{1 - x}{-x} \)

Now again replacing x by f(x) we get,

\( f[f\{f(x)\}] = \frac{1}{1 - f\{f(x)\}} = \frac{1}{1 - \frac{1 - x}{-x}} \)

\( \implies f[f\{f(x)\}] = \frac{1}{\frac{-x - 1 + x}{-x}} = \frac{-x}{-1} = x \) [Proved]

Exam Tip: For triple compositions, work through each layer systematically, simplifying at each stage to avoid algebraic errors - keep track of complex fractions by using common denominators.

Question 7. If \( f(x) = \frac{2x}{(1 + x^2)} \) then show that f(tan θ) = sin 2θ.
Answer: Given: \( f(x) = \frac{2x}{(1 + x^2)} \)

Need to prove: f(tan θ) = sin 2θ

\( f(\tan \theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} \)

\( \implies f(\tan \theta) = \frac{2\tan\theta}{\sec^2 \theta} \) [as 1 + tan² θ = sec² θ]

\( \implies f(\tan \theta) = 2 \frac{\sin \theta}{\cos \theta} \cos^2 \theta \) [as \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \sec \theta = \frac{1}{\cos \theta} \)]

\( \implies f(\tan \theta) = 2 \sin \theta \cos \theta = \sin 2\theta \) [Proved]

Exam Tip: When working with trigonometric function compositions, apply standard identities immediately - recognize sec² θ and convert to simpler forms early to streamline the proof.

Question 8. If \( y = f(x) = \frac{3x + 1}{5x - 3} \), prove that x = f(y).
Answer: Given: \( y = f(x) = \frac{3x + 1}{5x - 3} \)

Need to prove: x = f(y)

Replacing x by y in the function,

\( f(y) = \frac{3y + 1}{5y - 3} \)

Now, given in the problem that \( y = f(x) = \frac{3x + 1}{5x - 3} \)

\( \implies f(y) = \frac{3 \cdot \frac{3x + 1}{5x - 3} + 1}{5 \cdot \frac{3x + 1}{5x - 3} - 3} \)

\( \implies f(y) = \frac{\frac{9x + 3 + 5x - 3}{5x - 3}}{\frac{15x + 5 - 15x + 9}{5x - 3}} \)

\( \implies f(y) = \frac{14x}{14} = x \)

\( \implies x = f(y) \) [Proved]

Exam Tip: For inverse relationship proofs, substitute the expression for y back into the function definition - the result should simplify to x, confirming the inverse relationship.

Question 1. Find the domain of each of the following real function. (i) \( f(x) = \frac{3x + 5}{x^2 - 9} \) (ii) \( f(x) = \frac{2x - 3}{x^2 + x - 2} \) (iii) \( f(x) = \frac{x^2 - 2x + 1}{x^2 - 8x + 12} \) (iv) \( f(x) = \frac{x^3 - 8}{x^2 - 1} \)
Answer: (i) Given: \( f(x) = \frac{3x + 5}{x^2 - 9} \)

Need to find: Where the functions are defined.

To find the domain of the function f(x) we need to equate the denominator to 0.

Therefore,

\( x^2 - 9 = 0 \)

\( \implies x^2 = 9 \)

\( \implies x = \pm 3 \)

It means that the denominator is zero when x = 3 and x = -3

So, the domain of the function is the set of all the real numbers except +3 and -3.

The domain of the function, D_f(x) = (- ∞, -3) ∪ (-3, 3) ∪ (3, ∞).

(ii) Given: \( f(x) = \frac{2x - 3}{x^2 + x - 2} \)

Need to find: Where the functions are defined.

To find the domain of the function f(x) we need to equate the denominator to 0.

Therefore,

\( x^2 + x - 2 = 0 \)

\( \implies x^2 + 2x - x - 2 = 0 \)

\( \implies x(x + 2) - 1(x + 2) = 0 \)

\( \implies (x + 2)(x - 1) = 0 \)

\( \implies x = -2 \text{ and } x = 1 \)

It means that the denominator is zero when x = 1 and x = -2

So, the domain of the function is the set of all the real numbers except 1 and -2.

The domain of the function, D_f(x) = (- ∞, -2) ∪ (-2, 1) ∪ (1, ∞).

(iii) Given: \( f(x) = \frac{x^2 - 2x + 1}{x^2 - 8x + 12} \)

Need to find: Where the functions are defined.

To find the domain of the function f(x) we need to equate the denominator to 0.

Therefore,

\( x^2 - 8x + 12 = 0 \)

\( \implies x^2 - 2x - 6x + 12 = 0 \)

\( \implies x(x - 2) - 6(x - 2) = 0 \)

\( \implies (x - 2)(x - 6) = 0 \)

\( \implies x = 2 \text{ and } x = 6 \)

It means that the denominator is zero when x = 2 and x = 6

So, the domain of the function is the set of all the real numbers except 2 and 6.

The domain of the function, D_f(x) = (- ∞, 2) ∪ (2, 6) ∪ (6, ∞).

(iv) Given: \( f(x) = \frac{x^3 - 8}{x^2 - 1} \)

Need to find: Where the functions are defined.

To find the domain of the function f(x) we need to equate the denominator to 0.

Therefore,

\( x^2 - 1 = 0 \)

\( \implies x^2 = 1 \)

\( \implies x = \pm 1 \)

It means that the denominator is zero when x = -1 and x = 1

So, the domain of the function is the set of all the real numbers except -1 and +1.

The domain of the function, D_f(x) = (- ∞, -1) ∪ (-1, 1) ∪ (1, ∞).

Exam Tip: To find domain, set the denominator equal to zero and solve - the solutions are the values excluded from the domain; express the domain as the union of intervals excluding these points.

Question 2. Find the domain and the range of each of the following real function: \( f(x) = \frac{1}{x} \)
Answer: Given: \( f(x) = \frac{1}{x} \)

Need to find: Where the functions are defined.

Let, \( f(x) = \frac{1}{x} = y \) ---- (1)

To find the domain of the function f(x) we need to equate the denominator of the function to 0.

Therefore,

\( x = 0 \)

So, the domain of the function is the set of all the real numbers except 0. The domain of the function, D_f(x) = (- ∞, 0) ∪ (0, ∞).

To find the range of the function, from equation (1), we get \( x = \frac{1}{y} \).

For x to be defined, y must not be zero. So, the range of the function is the set of all the real numbers except 0. The range of the function, R_f(x) = (- ∞, 0) ∪ (0, ∞).

Exam Tip: To find range, express x in terms of y from the function equation, then determine which y values make x defined - the excluded values for y form the complement of the range.

Question 1. Find the domain and the range of the real function \( f(x) = \frac{1}{x} \).
Answer: Given the function \( f(x) = \frac{1}{x} \), we need to determine where it is defined. Setting the denominator equal to zero: \( x = 0 \). The denominator becomes zero at \( x = 0 \), so this value must be excluded. Therefore, the domain includes all real numbers except 0: \( D_{f(x)} = (-\infty, 0) \cup (0, \infty) \). To find the range, we swap \( x \) and \( y \) in the equation \( y = \frac{1}{x} \), which gives \( x = \frac{1}{y} \). Setting the denominator equal to zero: \( y = 0 \). The range also excludes 0: \( R_{f(x)} = (-\infty, 0) \cup (0, \infty) \).

Exam Tip: For rational functions with a variable in the denominator, always set the denominator equal to zero to find excluded values from both domain and range. This is a key concept examiners test frequently.

Question 2. Find the domain and the range of the real function \( f(x) = \frac{1}{x - 5} \).
Answer: Given: \( f(x) = \frac{1}{x - 5} = y \). To determine the domain, equate the denominator to zero: \( x - 5 = 0 \) gives \( x = 5 \). The denominator is zero when \( x = 5 \), so the domain includes all real numbers except 5: \( D_{f(x)} = (-\infty, 5) \cup (5, \infty) \). For the range, swap \( x \) and \( y \): \( \frac{1}{y - 5} = x \). Rearranging: \( 1 = x(y - 5) \), so \( y = \frac{1}{x} + 5 \). Setting the denominator to zero: \( x = 0 \). Thus, the range excludes 0: \( R_{f(x)} = (-\infty, 0) \cup (0, \infty) \).

Exam Tip: When finding the range, interchanging variables and solving for the new dependent variable reveals which values cannot be attained by the function.

Question 3. Find the domain and the range of the real function \( f(x) = \frac{x - 3}{2 - x} \).
Answer: Given: \( f(x) = \frac{x - 3}{2 - x} = y \). For the domain, set the denominator equal to zero: \( 2 - x = 0 \) gives \( x = 2 \). The domain excludes 2: \( D_{f(x)} = (-\infty, 2) \cup (2, \infty) \). To find the range, interchange \( x \) and \( y \): \( \frac{y - 3}{2 - y} = x \). Cross-multiply: \( y - 3 = x(2 - y) = 2x - xy \). Rearranging: \( y + xy = 2x + 3 \), so \( y(1 + x) = 2x + 3 \), giving \( y = \frac{2x + 3}{1 + x} \). Setting the denominator to zero: \( 1 + x = 0 \) gives \( x = -1 \). The range excludes -1: \( R_{f(x)} = (-\infty, -1) \cup (-1, \infty) \).

Exam Tip: When rearranging after interchanging variables, collect all terms containing the new dependent variable on one side to identify excluded values.

Question 4. Find the domain and the range of the real function \( f(x) = \frac{3x - 2}{x + 2} \).
Answer: Given: \( f(x) = \frac{3x - 2}{x + 2} = y \). For the domain, equate the denominator to zero: \( x + 2 = 0 \) gives \( x = -2 \). The domain excludes -2: \( D_{f(x)} = (-\infty, -2) \cup (-2, \infty) \). To find the range, swap \( x \) and \( y \): \( \frac{3y - 2}{y + 2} = x \). Cross-multiply: \( 3y - 2 = x(y + 2) = xy + 2x \). Rearranging: \( 3y - xy = 2x + 2 \), so \( y(3 - x) = 2x + 2 \), giving \( y = \frac{2x + 2}{3 - x} \). Setting the denominator to zero: \( 3 - x = 0 \) gives \( x = 3 \). The range excludes 3: \( R_{f(x)} = (-\infty, 3) \cup (3, \infty) \).

Exam Tip: For rational functions where the numerator and denominator both contain the variable, use the interchange method to identify the excluded range value efficiently.

Question 5. Find the domain and the range of the real function \( f(x) = \frac{x^2 - 16}{x - 4} \).
Answer: Given: \( f(x) = \frac{x^2 - 16}{x - 4} = y \). For the domain, set the denominator equal to zero: \( x - 4 = 0 \) gives \( x = 4 \). The domain excludes 4: \( D_{f(x)} = (-\infty, 4) \cup (4, \infty) \). Observe that \( x^2 - 16 = (x - 4)(x + 4) \), so the function simplifies to \( f(x) = x + 4 \) for \( x \neq 4 \). When we input any value from the domain set, the output will be either negative or positive, but never 8 (which would require \( x = 4 \), excluded from the domain). Therefore, the range excludes 8: \( R_{f(x)} = (-\infty, 8) \cup (8, \infty) \).

Exam Tip: Factor the numerator when possible - this reveals cancelled terms and clarifies which output values the function cannot produce.

Question 6. Find the domain and the range of the real function \( f(x) = \frac{1}{\sqrt{2x - 3}} \).
Answer: Given: \( f(x) = \frac{1}{\sqrt{2x - 3}} = y \). For the function to be defined, the expression under the square root must be positive (strictly greater than zero, since it also appears in the denominator): \( 2x - 3 > 0 \) gives \( x > \frac{3}{2} \). The domain consists of all real numbers greater than \( \frac{3}{2} \): \( D_{f(x)} = \left(\frac{3}{2}, \infty\right) \). When any value of \( x \) from the domain is substituted, the resulting function value is always a fraction whose denominator is nonzero. The range of the function is \( R_{f(x)} = (0, 1) \).

Exam Tip: For square root functions in the denominator, the radicand must be strictly positive - this is a stricter condition than just non-negative.

Question 7. Find the domain and the range of the real function \( f(x) = \frac{ax - b}{cx - d} \).
Answer: Given: \( f(x) = \frac{ax - b}{cx - d} = y \). For the domain, set the denominator equal to zero: \( cx - d = 0 \) gives \( x = \frac{d}{c} \). The domain excludes \( \frac{d}{c} \): \( D_{f(x)} = \left(-\infty, \frac{d}{c}\right) \cup \left(\frac{d}{c}, \infty\right) \). To find the range, swap \( x \) and \( y \): \( \frac{ay - b}{cy - d} = x \). Cross-multiply: \( ay - b = x(cy - d) = cxy - dx \). Rearranging: \( ay - cxy = b - dx \), so \( y(a - cx) = b - dx \), giving \( y = \frac{b - dx}{a - cx} \). Setting the denominator to zero: \( a - cx = 0 \) gives \( x = \frac{a}{c} \). The range excludes \( \frac{a}{c} \): \( R_{f(x)} = \left(-\infty, \frac{a}{c}\right) \cup \left(\frac{a}{c}, \infty\right) \).

Exam Tip: For general rational functions of the form \( \frac{ax + b}{cx + d} \), the excluded range value is \( \frac{a}{c} \) - memorizing this pattern saves calculation time.

Question 8. Find the domain and the range of the real function \( f(x) = \sqrt{3x - 5} \).
Answer: Given: \( f(x) = \sqrt{3x - 5} \). For the function to be defined, the expression under the square root must be non-negative: \( 3x - 5 \geq 0 \) gives \( x \geq \frac{5}{3} \). The domain consists of all real numbers greater than or equal to \( \frac{5}{3} \): \( D_{f(x)} = \left[\frac{5}{3}, \infty\right) \). When \( x = \frac{5}{3} \) is substituted into the function, we get \( f(x) = 0 \). The range of the function includes all values greater than or equal to 0: \( R_{f(x)} = [0, \infty) \).

Exam Tip: For square root functions without denominators, the radicand need only be non-negative, allowing the endpoint to be included in both domain and range.

Question 9. Find the domain and the range of the real function \( f(x) = \sqrt{\frac{x - 5}{3 - x}} \).
Answer: Given: \( f(x) = \sqrt{\frac{x - 5}{3 - x}} \). For the function to be defined, two conditions must hold: the numerator must be non-negative (\( x - 5 \geq 0 \)) and the denominator must be positive (\( 3 - x > 0 \)). This gives \( x \geq 5 \) and \( x < 3 \). These conditions cannot be satisfied simultaneously, as there is no value of \( x \) that is both at least 5 and less than 3. Therefore, the domain is empty, and there is no range for this function.

Exam Tip: When multiple conditions are required for a function to be defined, check whether they can all be satisfied at the same time - some combinations are mathematically impossible.

Question 10. Find the domain and the range of the real function \( f(x) = \sqrt{\log(x - 1)} \).
Answer: Given: \( f(x) = \sqrt{\log(x - 1)} \). For the function to be defined, two conditions must be satisfied. First, the argument of the logarithm must be positive: \( x - 1 > 0 \) gives \( x > 1 \). Second, the expression under the square root must be non-negative: \( \log(x - 1) \geq 0 \). Since \( \log(1) = 0 \), this inequality holds when \( x - 1 \geq 1 \), meaning \( x \geq 2 \). Combining these: the domain is \( x > 1 \) (for the logarithm) and \( x \geq 2 \) (for the square root). The more restrictive condition is \( x > 1 \) overall; however, focusing on both together gives \( D_{f(x)} = (1, \infty) \). When \( x = 2 \), we have \( f(2) = \sqrt{\log(1)} = \sqrt{0} = 0 \). As \( x \) increases beyond 2, \( \log(x - 1) \) increases, so \( f(x) \) increases without bound. The range is \( R_{f(x)} = [0, \infty) \).

Exam Tip: For composite functions involving logarithms and roots, apply each function's constraints separately, then combine them to find the most restrictive overall condition.

Question 11. Find the domain and the range of the real function \( f(x) = \frac{1}{\sqrt{x^2 - 1}} \).
Answer: Given: \( f(x) = \frac{1}{\sqrt{x^2 - 1}} \). For the function to be defined, the expression under the square root must be strictly positive (greater than zero, since it appears in the denominator): \( x^2 - 1 > 0 \) gives \( x^2 > 1 \), so \( x > 1 \) or \( x < -1 \). The domain excludes the interval \( [-1, 1] \): \( D_{f(x)} = (-\infty, -1) \cup (1, \infty) \). When any value of \( x \) from the domain is substituted, the resulting output is always a positive fraction whose denominator is nonzero. The range of the function is \( R_{f(x)} = (0, 1) \).

Exam Tip: For functions combining roots and fractions, the radicand must be strictly positive to avoid division by zero - this is more restrictive than a simple inequality.

Question 12. Find the domain and the range of the real function \( f(x) = 1 - |x - 2| \).
Answer: Given: \( f(x) = 1 - |x - 2| \). Since the absolute value function produces real output for all real values of \( x \), the domain includes all real numbers: \( D_{f(x)} = (-\infty, \infty) \). The term \( |x - 2| \) is always non-negative, so its maximum value is unbounded, meaning the minimum value of \( 1 - |x - 2| \) approaches \( -\infty \). The maximum value occurs when \( |x - 2| = 0 \), giving \( f(x) = 1 \). Therefore, the range of the function is \( R_{f(x)} = (-\infty, 1] \).

Exam Tip: For absolute value functions, identify when the expression inside equals zero to find the maximum or minimum value of the overall function.

Question 13. Find the domain and the range of the real function \( f(x) = \frac{|x - 4|}{x - 4} \).
Answer: Given: \( f(x) = \frac{|x - 4|}{x - 4} \). For the domain, set the denominator equal to zero: \( x - 4 = 0 \) gives \( x = 4 \). The domain excludes 4: \( D_{f(x)} = (-\infty, 4) \cup (4, \infty) \). Now, when \( x - 4 > 0 \) (i.e., \( x > 4 \)), we have \( |x - 4| = x - 4 \), so \( f(x) = \frac{x - 4}{x - 4} = 1 \). When \( x - 4 < 0 \) (i.e., \( x < 4 \)), we have \( |x - 4| = -(x - 4) = 4 - x \), so \( f(x) = \frac{4 - x}{x - 4} = -1 \). For any value of \( x \) from the domain set, the output is either \( +1 \) or \( -1 \). Therefore, the range of the function is a set containing -1 and +1: \( R_{f(x)} = \{-1, 1\} \).

Exam Tip: When an absolute value appears in a fraction with the same expression in the denominator, the result is always \( +1 \) or \( -1 \) depending on the sign of the expression inside.

Question 14. Find the domain and the range of the real function \( f(x) = \frac{x^2 - 9}{x - 3} \).
Answer: Given: \( f(x) = \frac{x^2 - 9}{x - 3} \). For the domain, set the denominator equal to zero: \( x - 3 = 0 \) gives \( x = 3 \). The domain excludes 3: \( D_{f(x)} = (-\infty, 3) \cup (3, \infty) \). Notice that \( x^2 - 9 = (x - 3)(x + 3) \), so the function simplifies to \( f(x) = x + 3 \) for \( x \neq 3 \). When we substitute any value from the domain, the output will be either negative or positive, but never 6 (which would occur if \( x = 3 \), a value excluded from the domain). Therefore, the range excludes 6: \( R_{f(x)} = (-\infty, 6) \cup (6, \infty) \).

Exam Tip: Factor rational functions to identify hidden cancellations - these reveal which output values the function cannot achieve, even though they appear on the simplified curve.

Question 15. Find the domain and the range of the real function \( f(x) = \frac{1}{2 - \sin 3x} \).
Answer: Given: \( f(x) = \frac{1}{2 - \sin 3x} \). The angle \( 3x \) can range from 0 to \( 2\pi \), giving a maximum value of \( x = \frac{2\pi}{3} \). The minimum value of \( x \) is 0. Therefore, the domain is \( D_{f(x)} = \left(0, \frac{2\pi}{3}\right) \). The sine function ranges from a minimum of 0 to a maximum of 1. When \( \sin 3x = 0 \), the denominator equals 2, and when \( \sin 3x = 1 \), the denominator equals 1. Thus, the denominator ranges from 1 to 2. The function value is the reciprocal of this range: the range is \( R_{f(x)} = \left(\frac{1}{2}, 1\right) \).

Exam Tip: For trigonometric functions, recall the standard ranges of sine and cosine (0 to 1 and -1 to 1 respectively) to determine denominator bounds, then invert to find the function range.

Exercise 3D

Question 1. Consider the real function f: R \( \rightarrow \) R: f(x) = x + 5 for all x \( \in \) R. Find its domain and range. Draw the graph of this function.
Answer: Given: \( f(x) = x + 5 \) for all \( x \in \mathbb{R} \). The domain of the given function is all real numbers, since there is no real number that makes the expression undefined. As \( f(x) \) is a polynomial function, any value of \( x \) is permissible. Therefore, \( \text{Domain}(f) = (-\infty, \infty) \) or \( \{x | x \in \mathbb{R}\} \). Now, let \( y = f(x) \), so \( y = x + 5 \). Solving for \( x \): \( x = y - 5 \). Since \( x \) can be any real number, \( y \) can also be any real number. The range is the set of all valid values of \( y \): \( \text{Range}(f) = (-\infty, \infty) \) or \( \{y | y \in \mathbb{R}\} \).

Graph: The graph is a straight line with slope 1 and y-intercept at (0, 5). The line passes through points such as (-5, 0), (0, 5), and (1, 6).

Exam Tip: For polynomial functions like linear equations, the domain is always all real numbers unless otherwise restricted. The graph should pass through the y-intercept and extend infinitely in both directions.

Question 2. Consider the function f: R \( \rightarrow \) R, defined by \( f(x) = \begin{cases} 1 - x, & \text{when } x < 0 \\ x, & \text{when } x = 0 \\ x + 1, & \text{when } x > 0 \end{cases} \) Write its domain and range. Also, draw the graph of f(x).
Answer: Given the piecewise function as stated. For \( f(x) = 1 - x \) when \( x < 0 \): there is no value of \( x \) in the domain \( (-\infty, 0) \) that makes the expression undefined. Therefore, \( \text{Domain}(f) = (-\infty, 0) \). For \( f(x) = x \) when \( x = 0 \): the domain is exactly this point. Therefore, \( \text{Domain}(f) = \{0\} \). For \( f(x) = x + 1 \) when \( x > 0 \): there is no value of \( x \) in the domain \( (0, \infty) \) that makes the expression undefined. Therefore, \( \text{Domain}(f) = (0, \infty) \). Combining all three parts, the overall domain is \( \text{Domain}(f) = (-\infty, \infty) \). Now for the range: when \( f(x) = 1 - x \), solving for \( x \) gives \( x = 1 - f(x) \). As \( x \) ranges from \( -\infty \) to 0, \( f(x) \) ranges from 1 to \( \infty \). Therefore, \( \text{Range}(f) = (1, \infty) \). When \( f(x) = x \) at \( x = 0 \), we get \( f(0) = 0 \). Therefore, \( \text{Range}(f) = \{0\} \). When \( f(x) = x + 1 \), as \( x \) ranges from 0 to \( \infty \), \( f(x) \) ranges from 1 to \( \infty \). Therefore, \( \text{Range}(f) = (1, \infty) \). Combining all three parts, the overall range is \( \text{Range}(f) = (1, \infty) \cup \{0\} \cup (1, \infty) = \{0\} \cup (1, \infty) \).

Graph: The graph consists of three pieces. For \( x < 0 \), the line \( y = 1 - x \) slopes downward (negative slope), passing through points like (-1, 2) and approaching (0, 1) from the left. At \( x = 0 \), a single point is plotted at (0, 0). For \( x > 0 \), the line \( y = x + 1 \) slopes upward (positive slope), starting just above (0, 1) and passing through points like (1, 2) and (2, 3).

Exam Tip: For piecewise functions, analyze each piece separately to find its domain contribution and range, then combine all pieces to determine the overall domain and range. Pay attention to whether endpoints are included or excluded at transition points.

Question 1. Let f : R → R : f(x) = x + 1 and g : R → R : g(x) = 2x – 3. Find (i) (f + g) (x) (ii) (f – g) (x) (iii) (fg) (x) (iv) (f/g) (x)
Answer: Given: f(x) = x + 1 and g(x) = 2x - 3

(i) To find: (f + g) (x)

(f + g) (x) = f(x) + g(x) = (x + 1) + (2x - 3) = x + 1 + 2x - 3 = 3x - 2

Therefore, (f + g) (x) = 3x - 2

(ii) To find: (f - g) (x)

(f - g) (x) = f(x) - g(x) = (x + 1) - (2x - 3) = x + 1 - 2x + 3 = 4 - x

Therefore, (f - g) (x) = 4 - x

(iii) To find: (fg) (x)

(fg) (x) = f(x) · g(x) = (x + 1)(2x - 3) = x(2x) - 3(x) + 1(2x) - 1(3) = 2x² - 3x + 2x - 3 = 2x² - x - 3

Therefore, (fg) (x) = 2x² - x - 3

(iv) To find: (f/g) (x)

\( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x + 1}{2x - 3} \)

Therefore, \( \left(\frac{f}{g}\right)(x) = \frac{x + 1}{2x - 3} \)
In simple words: When adding two functions, combine their outputs. When subtracting, take the second function's output away from the first. For multiplication, expand the expressions using the distributive property. For division, place one function on top of the other as a fraction.

Exam Tip: Always simplify algebraic expressions completely after performing operations - combine like terms and reduce to lowest terms where possible.

Question 2. Let f : R → R : f(x) = 2x + 5 and g : R → R : g(x) = x² + x. Find (i) (f + g) (x) (ii) (f – g) (x) (iii) (fg) (x) (iv) (f/g) (x)
Answer: Given: f(x) = 2x + 5 and g(x) = x² + x

(i) To find: (f + g) (x)

(f + g) (x) = f(x) + g(x) = (2x + 5) + (x² + x) = x² + 3x + 5

Therefore, (f + g) (x) = x² + 3x + 5

(ii) To find: (f - g) (x)

(f - g) (x) = f(x) - g(x) = (2x + 5) - (x² + x) = 2x + 5 - x² - x = -x² + x + 5

Therefore, (f - g) (x) = -x² + x + 5

(iii) To find: (fg) (x)

(fg) (x) = f(x) · g(x) = (2x + 5)(x² + x) = 2x(x²) + 2x(x) + 5(x²) + 5(x) = 2x³ + 2x² + 5x² + 5x = 2x³ + 7x² + 5x

Therefore, (fg) (x) = 2x³ + 7x² + 5x

(iv) To find: (f/g) (x)

\( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{2x + 5}{x² + x} \)
In simple words: Add the outputs to get the sum. Subtract the second from the first to get the difference. Multiply by distributing each term in one expression across all terms in the other. Divide by placing the numerator function over the denominator function.

Exam Tip: When combining polynomials, arrange terms in decreasing order of degree to avoid missing or duplicating terms.

Question 3. Let f: R → R: f(x) = x³ + 1 and g: R → R: g(x) = (x + 1). Find: (i) (f + g) (x) (ii) (f – g) (x) (iii) (1/f) (x) (iv) (f/g) (x)
Answer: Given: f(x) = x³ + 1 and g(x) = x + 1

(i) To find: (f + g) (x)

(f + g) (x) = f(x) + g(x) = (x³ + 1) + (x + 1) = x³ + x + 2

Therefore, (f + g) (x) = x³ + x + 2

(ii) To find: (f - g) (x)

(f - g) (x) = f(x) - g(x) = (x³ + 1) - (x + 1) = x³ - x

Therefore, (f - g) (x) = x³ - x

(iii) To find: (1/f) (x)

\( \left(\frac{1}{f}\right)(x) = \frac{1}{f(x)} = \frac{1}{x³ + 1} \)

Therefore, \( \left(\frac{1}{f}\right)(x) = \frac{1}{x³ + 1} \)

(iv) To find: (f/g) (x)

\( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x³ + 1}{x + 1} \)

Using the factorization \( a³ + b³ = (a + b)(a² - ab + b²) \):

\( = \frac{(x + 1)(x² - x + 1)}{x + 1} = x² - x + 1 \)

Therefore, \( \left(\frac{f}{g}\right)(x) = x² - x + 1 \)
In simple words: Add or subtract the function outputs directly. To divide by a function, write it as a fraction with that function in the denominator. When dividing polynomials, look for common factors that can be cancelled out to simplify the result.

Exam Tip: Always factor polynomials before dividing - common factors in numerator and denominator will cancel, greatly simplifying the answer.

Question 4. Let f: R → R; f(x) = x/c, where c is a constant. Find (i) (cf) (x) (ii) (c²f) (x) (iii) ((1/c)f) (x)
Answer: Given: f(x) = x/c

(i) To find: (cf) (x)

(cf)(x) = c · f(x) = c · \( \frac{x}{c} \) = x

Therefore, (cf) (x) = x

(ii) To find: (c²f) (x)

(c²f)(x) = c² · f(x) = c² · \( \frac{x}{c} \) = cx

Therefore, (c²f) (x) = cx

(iii) To find: ((1/c)f) (x)

\( \left(\frac{1}{c}f\right)(x) = \frac{1}{c} \cdot f(x) = \frac{1}{c} \cdot \frac{x}{c} = \frac{x}{c²} \)

Therefore, \( \left(\frac{1}{c}f\right)(x) = \frac{x}{c²} \)
In simple words: To scale a function by a constant, multiply each output of the function by that constant. When scaling by c², the result is multiplied twice by c. When scaling by 1/c, the output is divided by c.

Exam Tip: Keep track of how constants combine when multiplying scaled functions - multiplying by c² is the same as applying the constant c twice.

Question 5. Let f:(2, ∞) → R: f(x) = √(x - 2), and g: (2, ∞) → R: g(x) = √(x + 2) Find: (i) (f + g) (x) (ii) (f - g) (x) (iii) (fg) (x)
Answer: Given: f(x) = √(x - 2) : x > 2 and g(x) = √(x + 2) : x > 2

(i) To find: (f + g) (x)

Domain(f) = (2, ∞), Range(f) = (0, ∞)
Domain(g) = (2, ∞), Range(g) = (2, ∞)

(f + g) (x) = f(x) + g(x) = √(x - 2) + √(x + 2)

Therefore, (f + g) (x) = √(x - 2) + √(x + 2)

(ii) To find: (f - g) (x)

Range(g) ⊆ Domain(f), so (f - g) (x) exists.

(f - g) (x) = f(x) - g(x) = √(x - 2) - √(x + 2)

Therefore, (f - g) (x) = √(x - 2) - √(x + 2)

(iii) To find: (fg) (x)

(fg) (x) = f(x) · g(x) = √(x - 2) · √(x + 2) = √[(x - 2)(x + 2)] = √(x² - 4)

Therefore, (fg) (x) = √(x² - 4)
In simple words: Add the two square root expressions directly by keeping them separate. Subtract one from the other in the same way. When multiplying square roots, combine what is under the radicals using algebra - the product √a · √b equals √(ab).

Exam Tip: Check that the domain restriction x > 2 is satisfied by both functions before performing operations - this ensures all outputs are real numbers.

Question 6. Find the set of values for which the function f(x) = 1 – 3x and g(x) = 2x² – 1 are equal.
Answer: Given: f(x) = 1 - 3x, g(x) = 2x² - 1

To find: Set of values of x for which f(x) = g(x)

Setting f(x) = g(x):

1 - 3x = 2x² - 1

2x² + 3x - 2 = 0

2x² + 4x - x - 2 = 0

2x(x + 2) - (x + 2) = 0

(x + 2)(2x - 1) = 0

x = -2 or x = 1/2

The set of values for which f(x) and g(x) have the same value is {-2, 1/2}.
In simple words: Set the two functions equal to each other and rearrange to get zero on one side. Factor the resulting quadratic expression. Each factor that equals zero gives you an x-value where the functions match.

Exam Tip: Always verify your solutions by substituting back into both original functions to confirm they produce the same output.

Question 7. Find the set of values for which the function f(x) = x + 3 and g(x) = 3x² – 1 are equal.
Answer: Given: f(x) = x + 3, g(x) = 3x² - 1

To find: Set of values of x for which f(x) = g(x)

Setting f(x) = g(x):

x + 3 = 3x² - 1

3x² - x - 4 = 0

3x² - 4x + 3x - 4 = 0

x(3x - 4) + (3x - 4) = 0

(3x - 4)(x + 1) = 0

x = 4/3 or x = -1

The set of values for which f(x) and g(x) have the same value is {4/3, -1}.
In simple words: Equate the two functions and move all terms to one side. Factor the resulting quadratic by grouping. Solve each factor to find the x-values where both functions produce identical outputs.

Exam Tip: Factor carefully by grouping - look for the greatest common factor in pairs of terms to identify the binomial factors.

Question 8. Let X = {–1, 0, 2, 5} and f : X → R Z: f(x) = x³ + 1. Then, write f as a set of ordered pairs.
Answer: Given: X = {-1, 0, 2, 5}, f : X → R Z: f(x) = x³ + 1

Computing f(x) for each value of x:

f(-1) = (-1)³ + 1 = -1 + 1 = 0
f(0) = (0)³ + 1 = 0 + 1 = 1
f(2) = (2)³ + 1 = 8 + 1 = 9
f(5) = (5)³ + 1 = 125 + 1 = 126

Therefore, f represented as ordered pairs is: f = {(-1, 0), (0, 1), (2, 9), (5, 126)}
In simple words: Plug each element of the set X into the function one at a time. Calculate the output for each input. Write the input-output pairs as ordered pairs (x, f(x)), and collect all pairs into a set.

Exam Tip: Arrange ordered pairs in order of increasing x-values for clarity - this makes it easier to spot patterns and verify all inputs have been included.

Question 9. Let A = {–2, –1, 0, 2} and f : A → Z: f(x) = x² – 2x – 3. Find f(A).
Answer: Given: A = {-2, -1, 0, 2}, f : A → Z: f(x) = x² - 2x - 3

Computing f(x) for each value of x:

f(-2) = (-2)² - 2(-2) - 3 = 4 + 4 - 3 = 5
f(-1) = (-1)² - 2(-1) - 3 = 1 + 2 - 3 = 0
f(0) = (0)² - 2(0) - 3 = 0 - 0 - 3 = -3
f(2) = (2)² - 2(2) - 3 = 4 - 4 - 3 = -3

Therefore, f(A) = {5, 0, -3} or as ordered pairs: f = {(-2, 5), (-1, 0), (0, -3), (2, -3)}
In simple words: Substitute each element from set A into the function f. Calculate the corresponding output. The range f(A) is the collection of all these output values. Note that different inputs may give the same output.

Exam Tip: When listing the range, include each distinct output value only once - even if two different inputs produce the same output.

Question 10. Let f : R → R : f(x) = x². Determine (i) range (f) (ii) {x : f(x) = 4}
Answer: Given: f(x) = x²

(i) Range(f):

To find the range, let y = f(x), so y = x².
Then x = √y.
Since x² is always non-negative for all real x, the value of y ≥ 0.

Therefore, Range(f) = [0, ∞)

(ii) Find {x : f(x) = 4}:

Let f(x) = 4
Then x² = 4
x = ±2

Therefore, {x : f(x) = 4} = {2, -2}
In simple words: The range includes all possible outputs - since squaring any real number gives zero or a positive result, the range starts at zero and goes up forever. To find which x-values produce a specific output, solve the equation x² equals that output value.

Exam Tip: Remember that x² = 4 has two solutions: x = 2 and x = -2 - never forget the negative solution when taking square roots.

Question 11. Let f : R → R : f(x) = x² + 1. Find f⁻¹ {10}.
Answer: Given: f : R → R : f(x) = x² + 1

To find: f⁻¹{10}, which means the set of all x-values where f(x) = 10

Setting f(x) = 10:

x² + 1 = 10
x² = 9
x = ±3

Therefore, f⁻¹{10} = {3, -3}
In simple words: The notation f⁻¹{10} asks: "What input values produce an output of 10?" Set the function equal to 10 and solve for x. Both positive and negative values are valid solutions.

Exam Tip: The symbol f⁻¹{10} does NOT mean the inverse function - it means the pre-image or the set of inputs that map to 10. Always solve the equation f(x) = 10 rather than trying to find an inverse function.

Question 7. Let f : R⁺ → R : f(x) = logₑ x. Find {x : f(x) = –2}.
Answer: Given, f : R⁺ → R : f(x) = logₑ x. We need to find all x values where f(x) equals -2. Setting the equation, logₑ x = -2. Taking the exponential (antilog) on both sides, x = e⁻². Therefore, the value of x for which f(x) = -2 is e⁻².
In simple words: When you take the natural log of a number and get -2, that number must be e to the power of -2, which is roughly 0.135.

Exam Tip: Remember that logₑ x = y means x = eʸ; this inverse relationship is key to solving logarithmic equations.

Question 8. Let A = {6, 10, 11, 15, 12} and let f : A → N : f(n) is the highest prime factor of n. Find range (f).
Answer: Given, A = {6, 10, 11, 15, 12} and f : A → N where f(n) gives the highest prime factor of n. Examining each element:
(1) When n = 6, the prime factorization is 2 × 3, so the greatest prime factor is 3. Thus, f(6) = 3.
(2) When n = 10, we have 2 × 5, so the greatest prime factor is 5. Thus, f(10) = 5.
(3) When n = 11, since 11 itself is prime, f(11) = 11.
(4) When n = 15, we have 3 × 5, so the greatest prime factor is 5. Thus, f(15) = 5.
(5) When n = 12, we have 2² × 3, so the greatest prime factor is 3. Thus, f(12) = 3.
Collecting all output values, the range of f is {3, 5, 11}.
In simple words: The range is the set of all highest prime factors you get from the set A, which are 3, 5, and 11.

Exam Tip: Always factorize completely and identify the largest prime divisor, not just any prime factor.

Question 9. Find the range of the function f(x) = sin x.
Answer: The sine function is periodic, repeating its pattern forever. The output values of sin(x) are bounded - they oscillate between -1 and +1 for all real inputs. The maximum value of 1 happens at π/2 + 2nπ (where n is odd), and the minimum value of -1 occurs at 3π/2 + 2nπ (where n is even). Since the function never exceeds 1 or falls below -1, the range of f(x) = sin x is the closed interval [-1, 1].
In simple words: No matter what x you plug in, sin(x) always gives you a result between -1 and +1, including those endpoints.

Exam Tip: Periodic functions like sine are bounded; always check the graph or use calculus to confirm maximum and minimum values.

Question 10. Find the range of the function f(x) = |x|.
Answer: The absolute value function is defined as: |x| = x when x ≥ 0, and |x| = -x when x < 0. Since squaring any real number (positive or negative) yields a non-negative result, the output of |x| can never be negative. The smallest value the absolute value can take is 0 (when x = 0), and it increases without bound as |x| grows larger. Therefore, the range of |x| is [0, ∞), meaning all non-negative real numbers.
In simple words: Absolute value removes the minus sign, so all outputs are zero or positive.

Exam Tip: The absolute value function always produces non-negative outputs; 0 is included in the range.

Question 11. Write the domain and the range of the function, f(x) = \(\sqrt{x - [x]}\).
Answer: Given f(x) = \(\sqrt{x - [x]}\), where [x] is the greatest integer function (floor function).
Note that x - [x] equals {x}, the fractional part of x.
So f(x) = \(\sqrt{\{x\}}\).

(i) Domain: The fractional part {x} is defined for all real numbers. Its value is always either zero or positive (ranging from 0 up to but not including 1). Since we are taking the square root of {x}, and the square root of a non-negative number is real, the domain of f(x) is all real numbers, R.

(ii) Range: The fractional part has range [0, 1). When we take the square root of values in [0, 1), the resulting outputs also stay in [0, 1). Thus, the range of f(x) is [0, 1).
In simple words: The fractional part of any number is always between 0 (inclusive) and 1 (exclusive), so its square root also stays in that same interval.

Exam Tip: Recall that {x} = x - [x] always lies in [0, 1); this fact directly constrains the range of the composite function.

Question 12. If f(x) = \(\frac{x - 5}{5 - x}\), then find dom (f) and range (f).
Answer: Given f(x) = \(\frac{x - 5}{5 - x}\).

(i) Domain: A rational function is defined everywhere except where the denominator is zero. Setting 5 - x = 0, we get x = 5. Therefore, the domain is all real numbers except 5, written as (-∞, ∞) - {5}.

(ii) Range: Let y = \(\frac{x - 5}{5 - x}\). Notice that x - 5 = -(5 - x), so y = \(\frac{-(5 - x)}{5 - x}\) = -1 for all x ≠ 5. This means the function is constant and always equals -1 on its domain. Thus, the range is the single-element set {-1}.
In simple words: This function simplifies to a constant: no matter what x-value you pick (except 5), you always get -1.

Exam Tip: Simplify rational functions algebraically before analyzing range; constants disguised in fraction form yield singleton ranges.

Question 13. Let f = {(1, 6), (2, 5), (4, 3), (5, 2), (8, –1), (10, –3)} and g = {(2, 0), (3, 2), (5, 6), (7, 10), (8, 12), (10, 16)}. Find (i) dom (f + g) (ii) dom \(\left(\frac{f}{g}\right)\).
Answer: Given the two relations f and g as sets of ordered pairs.

Domain of f = {1, 2, 4, 5, 8, 10}
Domain of g = {2, 3, 5, 7, 8, 10}

(i) For the sum (f + g), both functions must be defined at a given input. The domain of (f + g) consists of all x where x belongs to both domain of f and domain of g. This is the intersection: dom(f + g) = {2, 5, 8, 10}.

(ii) For the quotient (f/g), we need g(x) to be defined and non-zero. The domain of (f/g) is the intersection of the two domains, excluding any x where g(x) = 0. From the given g, g(2) = 0. Removing 2 from {2, 5, 8, 10}, we get dom(f/g) = {5, 8, 10}. (Note: We must also verify that 5, 8, and 10 have non-zero g-values: g(5) = 6 ≠ 0, g(8) = 12 ≠ 0, g(10) = 16 ≠ 0, so all three remain.)
In simple words: For f + g, find where both are defined. For f/g, find where both are defined AND g is not zero.

Exam Tip: Always check the zero-condition for quotient functions; removing zeros is as important as finding the intersection.

Question 14. If f(x) = \(\frac{x - 1}{x}\), find the value of \(f\left(\frac{1}{x}\right)\).
Answer: Given f(x) = \(\frac{x - 1}{x}\).
To find \(f\left(\frac{1}{x}\right)\), substitute \(\frac{1}{x}\) in place of x in the formula.

\(f\left(\frac{1}{x}\right) = \frac{\frac{1}{x} - 1}{\frac{1}{x}}\)

Simplify the numerator: \(\frac{1}{x} - 1 = \frac{1 - x}{x}\)

So \(f\left(\frac{1}{x}\right) = \frac{\frac{1 - x}{x}}{\frac{1}{x}} = \frac{1 - x}{x} \times \frac{x}{1} = 1 - x\)

Therefore, \(f\left(\frac{1}{x}\right) = 1 - x\).
In simple words: Plugging 1/x into the formula and simplifying the complex fraction gives you 1 - x.

Exam Tip: When substituting reciprocals, simplify complex fractions by multiplying by the reciprocal of the denominator.

Question 15. If f(x) = \(\frac{kx}{x + 1}\), where x ≠ –1 and f{f(x)} = x for x ≠ –1 then find the value of k.
Answer: Given f(x) = \(\frac{kx}{x + 1}\) and the condition that f(f(x)) = x.

First, compute f(f(x)):
\(f(f(x)) = f\left(\frac{kx}{x + 1}\right) = \frac{k \cdot \frac{kx}{x + 1}}{\frac{kx}{x + 1} + 1}\)

Simplify the denominator: \(\frac{kx}{x + 1} + 1 = \frac{kx + (x + 1)}{x + 1} = \frac{kx + x + 1}{x + 1}\)

So \(f(f(x)) = \frac{\frac{k^2 x}{x + 1}}{\frac{kx + x + 1}{x + 1}} = \frac{k^2 x}{kx + x + 1} = \frac{k^2 x}{(k + 1)x + 1}\)

Given that f(f(x)) = x:
\(\frac{k^2 x}{(k + 1)x + 1} = x\)

Cross-multiply: \(k^2 x = x[(k + 1)x + 1]\)
\(k^2 x = (k + 1)x^2 + x\)

Rearrange: \((k + 1)x^2 + x - k^2 x = 0\)
\((k + 1)x^2 + (1 - k^2)x = 0\)
\((k + 1)x^2 - (k^2 - 1)x = 0\)
\((k + 1)x^2 - (k - 1)(k + 1)x = 0\)
\((k + 1)x[x - (k - 1)] = 0\)

For this to hold for all x ≠ -1 and x ≠ 0, the coefficient (k + 1) must be zero, or we need k - 1 = 0. Testing k = -1: if k = -1, then f(x) = \(\frac{-x}{x + 1}\), and f(f(x)) = \(\frac{-\frac{-x}{x+1}}{\frac{-x}{x+1}+1}\) = \(\frac{\frac{x}{x+1}}{\frac{-x+x+1}{x+1}}\) = \(\frac{x}{1}\) = x. ✓

Therefore, k = -1.
In simple words: You need to find the value of k such that applying f twice gives you back your original x. Testing k = -1 works.

Exam Tip: When f(f(x)) = x, the function is an involution; verify your answer by substituting back.

Question 16. Find the range of the function, f(x) = \(\frac{x}{|x|}\).
Answer: The absolute value is defined as |x| = x when x ≥ 0, and |x| = -x when x < 0.

For x > 0: \(f(x) = \frac{x}{|x|} = \frac{x}{x} = 1\)

For x < 0: \(f(x) = \frac{x}{|x|} = \frac{x}{-x} = -1\)

Note that x = 0 is not in the domain (division by zero). Therefore, the function outputs only two values: 1 and -1. The range is {1, -1}.
In simple words: For any positive number, you get 1; for any negative number, you get -1.

Exam Tip: Functions defined using absolute value often have restricted ranges; analyze positive and negative cases separately.

Question 17. Find the domain of the function, f(x) = log |x|.
Answer: The logarithm function has domain R⁺ (all positive real numbers). When x is replaced by |x|, we must ensure |x| > 0.

Since |x| ≥ 0 for all real x, and |x| = 0 only when x = 0, we require |x| > 0, which means x ≠ 0.

The logarithm is not defined for zero or negative arguments. Thus, the domain of f(x) = log |x| is all real numbers except zero: R - {0}, or equivalently, (-∞, 0) ∪ (0, ∞).
In simple words: You can take the log of any non-zero number (positive or negative, since we use absolute value), but not of zero.

Exam Tip: Always remember that logarithms require positive arguments; replacing the argument with |x| still excludes zero.

Question 18. If \(f\left(x + \frac{1}{x}\right) = \left(x^2 + \frac{1}{x^2}\right)\) for all x ∈ R – {0} then write an expression for f(x).
Answer: Given \(f\left(x + \frac{1}{x}\right) = \left(x^2 + \frac{1}{x^2}\right)\).

Let \(y = x + \frac{1}{x}\).

Square both sides: \(y^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}\)

Rearrange: \(x^2 + \frac{1}{x^2} = y^2 - 2\)

From the original functional equation, \(f(y) = y^2 - 2\).

Replacing y with x: \(f(x) = x^2 - 2\).
In simple words: You have a function defined in terms of a complicated input; by letting y equal that input and using algebra, you isolate the output in terms of y, then rename y as x.

Exam Tip: For functional equations, substitute to create a new variable for the compound argument, then solve for the output algebraically.

Question 19. Write the domain and the range of the function, f(x) = \(\frac{ax + b}{bx - a}\).
Answer: Given f(x) = \(\frac{ax + b}{bx - a}\).

(i) Domain: A rational function is defined everywhere except where its denominator equals zero. Setting bx - a = 0, we get x = a/b. Therefore, the domain is all real numbers except a/b: R - {a/b}.

(ii) Range: Let \(y = \frac{ax + b}{bx - a}\).

Multiply both sides by the denominator: \(y(bx - a) = ax + b\)
\(byx - ay = ax + b\)
\(byx - ax = ay + b\)
\(x(by - a) = ay + b\)
\(x = \frac{ay + b}{by - a}\)

For x to be defined, the denominator by - a must be non-zero. So by - a ≠ 0, which gives y ≠ a/b.

Therefore, the range is R - {a/b}.
In simple words: Solve the equation for x in terms of y. If y makes the denominator zero, that y-value is excluded from the range.

Exam Tip: For rational functions of the form (ax + b)/(cx + d), both domain and range typically exclude the same value; use the rearrangement method to confirm.

Question 20. Write the domain and the range of the function, f(x) = \(\sqrt{x - 1}\).
Answer: Given f(x) = \(\sqrt{x - 1}\).

(i) Domain: For a square root to be real, the expression under the radical must be non-negative. So x - 1 ≥ 0, which gives x ≥ 1. The domain is [1, ∞).

(ii) Range: The square root function produces non-negative outputs. When x = 1, \(f(1) = \sqrt{0}\) = 0. As x increases beyond 1, \(f(x) = \sqrt{x - 1}\) also increases without bound, approaching infinity. Shifting the graph of \(\sqrt{x}\) to the right by 1 unit (the transformation x ↦ x - 1) does not alter the range—it remains [0, ∞).
In simple words: The radicand must be non-negative, so x starts at 1. The output starts at 0 and grows indefinitely.

Exam Tip: Horizontal shifts of functions do not change the range; focus on the vertical extent of the graph.

Question 21. Write the domain and the range of the function, f(x) = –|x|.
Answer: Given f(x) = -|x|.

(i) Domain: The absolute value function is defined for all real numbers. Since we are simply negating |x|, this does not restrict the domain. Thus, the domain is all real numbers: R.

(ii) Range: The absolute value |x| has range [0, ∞). When we negate every output, we get -|x|, which has range (-∞, 0]. In other words, the function outputs zero when x = 0, and outputs increasingly negative values as |x| grows larger.
In simple words: The absolute value is always non-negative, so its negative is always non-positive, ranging from 0 down to negative infinity.

Exam Tip: Negating a function flips its range across the x-axis; if the original range is [0, ∞), the negated range is (-∞, 0].