RS Aggarwal Solutions for Class 12 Chapter 02 Functions

Question 1. Define a function. What do you mean by the domain and range of a function? Give examples.
Answer: A relation R connecting set A to set B is referred to as a function when all elements from set A have a single, distinct image in B. The notation f:A→B means 'f maps A to B'. When f:A→B, the set A represents the domain of f, while set B represents the co-domain of f. The collection of all images produced by elements in A is called the range of f.

Domain of f = {a | a ∈ A, (a,f(a)) ∈ f}
Range of f = {f(a) | a ∈ A, f(a) ∈ B}

Example: For y = sin x, the domain covers all real numbers (R) since no restrictions apply to x values. The range stays between - 1 and 1, written as - 1 ≤ y ≤ 1.
In simple words: A function pairs each input to exactly one output. The domain is all possible inputs, and the range is all possible outputs that actually happen.

Question 2. Define each of the following:
(i) injective function
(ii) surjective function
(iii) bijective function
(iv) many - one function
(v) into function
Give an example of each type of functions.
Answer:
(i) Injective Function
Definition: A function f: A → B is one - one (or injective) when different elements from A produce different images in B. A function f is injective if whenever f(x) = f(y), then x = y holds true.
Example: The function f(x) = x + 9 from real numbers R to R is injective. When x = 3, then f(x) = 12; when f(y) = 8, the value of y can only be - 1, confirming x = y.

(ii) Surjective Function
Definition: The function f:A→B is surjective (or onto) when each element in the co-domain B is an image of at least one element in A. Thus, f: A→B is surjective if for all b∈B, there exists some a∈A such that f(a) = b.
Example: The function f(x) = 2x from natural numbers N to non-negative even numbers is surjective.

(iii) Bijective Function
Definition: A function f from set A to set B is bijective when, for each y in B, exactly one x in A exists such that f(x) = y. In other words, f is both one - to - one and onto simultaneously.
Example: The function f(x) = x^2 from positive real numbers to positive real numbers is both injective and surjective, making it bijective.

(iv) Many - One Function
Definition: A function f: A→B is many - one when two or more distinct elements from A share the same image in B. Trigonometric functions such as sin x are many - to - one since sin x = sin(2π + x) = sin(4π + x) and so forth.

(v) Into Function
Definition: If f:A→B has at least one element in the co-domain that is not the image of any element in the domain, then f is into. Let f(x) = y = x - 1000, so x = y + 1000 = g(y). Here g(y) is defined for all y∈I, but g(y) ∉ N when y ≤ - 1000, confirming f is into.
In simple words: Injective means no two inputs give the same output. Surjective means every possible output is actually used. Bijective combines both. Many - one means multiple inputs share one output. Into means some outputs are never used.

Question 3. Give an example of a function which is
(i) one - one but not onto
(ii) one - one and onto
(iii) neither one - one nor onto
(iv) onto but not one - one
Answer:
(i) One - one but not onto
Consider f(x) = 6x.
For one - one: Let f(x₁) = 6x₁ and f(x₂) = 6x₂. If f(x₁) = f(x₂), then 6x₁ = 6x₂, which gives x₁ = x₂. Therefore, f is one - one.
For onto: Let f(x) = y such that y∈N. Then 6x = y, so x = y/6. If y = 1, then x = 1/6 ≈ 0.166667, which is not in N. Therefore, f is not onto.

(ii) One - one and onto
Consider f(x) = x^5. Then y = x^5. The horizontal line test shows that no line cuts the curve at two equal y values, making f one - one. The range of f(x) = ( - ∞,∞) = R, matching the codomain. Therefore, f is onto. Thus, f is one - one and onto.

(iii) Neither one - one nor onto
Consider f(x) = x^2.
For one - one: Let f(x₁) = (x₁)² and f(x₂) = (x₂)². If f(x₁) = f(x₂), then (x₁)² = (x₂)², which means x₁ = x₂ or x₁ = - x₂. Since x₁ lacks a unique image, f is not one - one.
For onto: Let f(x) = y such that y∈R. Then x² = y, so x = √y. When y is negative, the square root is not real. Therefore, f is not onto. Thus, f is neither one - one nor onto.

(iv) Onto but not one - one
Consider f:Z→N where f(x) = |x|. Since Z maps to every single element in N twice (e.g., both - 2 and 2 map to 2), this function is onto but not one - one. (Z denotes integers; N denotes natural numbers.)
In simple words: One - one means different inputs always give different outputs. Onto means all possible outputs are actually produced. When neither holds, some inputs repeat the same output and some outputs are never reached.

Question 4. Let f : R → R be defined by
\( f(x) = \begin{cases} 2x + 3, & \text{when} \quad x < -2 \\ x^2 - 2, & \text{when} \quad -2 \leq x \leq 3 \\ 3x - 1, & \text{when} \quad x > 3 \end{cases} \)
Find (i) f(2) (ii) f(4) (iii) f( - 1) (iv) f( - 3).
Answer:
(i) f(2)
Since f(x) = x² - 2 when x = 2 falls in the range - 2 ≤ x ≤ 3, we have f(2) = (2)² - 2 = 4 - 2 = 2.

(ii) f(4)
Since f(x) = 3x - 1 when x = 4 (which satisfies x > 3), we get f(4) = (3×4) - 1 = 12 - 1 = 11.

(iii) f( - 1)
Since f(x) = x² - 2 when x = - 1 is in the range - 2 ≤ x ≤ 3, we have f( - 1) = ( - 1)² - 2 = 1 - 2 = - 1.

(iv) f( - 3)
Since f(x) = 2x + 3 when x = - 3 (satisfying x < - 2), we get f( - 3) = 2×( - 3) + 3 = - 6 + 3 = - 3.
In simple words: Check which rule applies based on the x value. Plug the number into the correct formula and compute the result.

Question 5. Show that the function f: R → R : f(x) = 1 + x² is many - one into.
Answer: To show: f: R → R : f(x) = 1 + x² is many - one into.
Proof: Let f(x) = 1 + x², so y = 1 + x². The horizontal line test shows that lines intersect the curve at two equal y values, confirming the function is many - one. The range of f(x) = [1,∞) does not equal R (the codomain), so f is into. Therefore, f: R → R : f(x) = 1 + x² is many - one into.
In simple words: The same y value appears for two different x values (like x = 2 and x = - 2 both give y = 5), making it many - one. Some y values below 1 are never reached, making it into.

Question 6. Show that the function f : R → R : f(x) = x⁴ is many - one and into.
Answer: To show: f: R → R : f(x) = x⁴ is many - one into.
Proof: Let f(x) = x⁴, so y = x⁴. Horizontal lines cut the curve at two equal y values, so the function is many - one. The range of f(x) = [0,∞) ≠ R (the codomain), making f into. Therefore, f: R → R : f(x) = x⁴ is many - one into.
In simple words: Both x = 2 and x = - 2 produce the same output 16, making it many - one. Negative y values are impossible since x⁴ is always non-negative, making it into.

Question 7. Show that the function f: R → R : f(x) = x⁵ is one - one and onto.
Answer: To show: f: R → R : f(x) = x⁵ is one - one and onto.
Proof: Let f(x) = x⁵, so y = x⁵. Horizontal lines do not cut the curve at two equal y values, making the function one - one. The range of f(x) = ( - ∞,∞) = R (matching the codomain), so f is onto. Therefore, f: R → R : f(x) = x⁵ is one - one and onto.
In simple words: Each x value produces a unique y value, and every y value is actually produced. The function covers the entire real number line.

Question 8. Let f : \( [0, \frac{\pi}{2}] \) → R : f(x) = sin x and g : \( [0, \frac{\pi}{2}] \) → R : g(x) = cos x. Show that each one of f and g is one - one but (f + g) is not one - one.
Answer: For f : \( [0, \frac{\pi}{2}] \) → R : f(x) = sin x: In this restricted range, horizontal lines do not intersect the curve at multiple equal y values, making f one - one.

For g : \( [0, \frac{\pi}{2}] \) → R : g(x) = cos x: In this restricted range, horizontal lines do not intersect the curve at multiple equal y values, making g one - one.

For (f + g) : \( [0, \frac{\pi}{2}] \) → R = sin x + cos x: In this range, horizontal lines intersect the curve at two equal y values, making the combined function not one - one. Therefore, each of f and g is one - one, but (f + g) is not one - one.
In simple words: Both sine and cosine are one - one on their restricted domain. However, when added together, the resulting curve allows different x values to produce identical sums, breaking the one - one property.

Question 9. Show that the function
(i) f : N → N : f(x) = x² is one - one into.
(ii) f : Z → Z : f(x) = x² is many - one into
Answer:
(i) f : N → N : f(x) = x² is one - one into.
Let f(x) = x², so y = x². Since f(x) is strictly increasing across the domain N → N, the function is one - one. The range of f(x) = (0,∞) ≠ N (the codomain), so f is into. Therefore, f : N → N : f(x) = x² is one - one into.

(ii) f : Z → Z : f(x) = x² is many - one into.
Let f(x) = x², so y = x². Horizontal lines cut the curve at two equal y values, making the function many - one. The range of f(x) = (0,∞) ≠ Z (the codomain), so f is into. Therefore, f : Z → Z : f(x) = x² is many - one into.
In simple words: When restricted to natural numbers, squaring is one - one because each natural number increases. Over all integers, negative and positive integers produce the same squares, making it many - one. In both cases, negative outputs are impossible, making both functions into.

Question 10. Show that the function
(i) f : N → N : f(x) = x³ is one - one into
(ii) f : Z → Z : f(x) = x³ is one - one into
Answer:
(i) f : N → N : f(x) = x³ is one - one into.
Let f(x) = x³. Since f(x) is strictly increasing from the domain N → N, the function is one - one. The range of f(x) = ( - ∞,∞) ≠ N (the codomain), so f is into. Therefore, f : N → N : f(x) = x³ is one - one into.

(ii) f : Z → Z : f(x) = x³ is one - one into.
Let f(x) = x³. Since f(x) is strictly increasing from the domain Z → Z, the function is one - one. The range of f(x) = ( - ∞,∞) ≠ Z (the codomain), so f is into. Therefore, f : Z → Z : f(x) = x³ is one - one into.
In simple words: Cubing preserves order - if one number is larger, its cube is also larger. This ensures each input has a unique output. Over N, many cubes fall outside the range of N, and similarly for Z, making both into functions.

Question 11. Show that the function f : R → R : f(x) = sin x is neither one - one nor onto.
Answer: Let f(x) = sin x, so y = sin x. Horizontal lines cut the curve at multiple equal y values throughout the real domain, making the function not one - one. The range of f(x) = [ - 1,1] ≠ R (the codomain), so f is not onto. Therefore, the function f : R → R : f(x) = sin x is neither one - one nor onto.
In simple words: Sine repeats the same values infinitely many times (sin 0 = sin π = sin 2π...), so it is not one - one. Sine can never reach values outside [ - 1,1], so it is not onto R.

Question 12. Prove that the function f : N → N : f(n) = (n² + n + 1) is one - one but not onto.
Answer: In the given range of N, f(x) is strictly increasing. Therefore, f(n) = n² + n + 1 is one - one. However, the range of f(n) = [0.75,∞) ≠ N (the codomain), so f(n) is not onto. Hence, the function f : N → N : f(n) = (n² + n + 1) is one - one but not onto.
In simple words: Since the formula always produces larger outputs for larger natural number inputs, no two different inputs yield the same result. However, many natural numbers (like 1, 2, 3) are never produced by the formula, making it not onto.

Question 13. Show that the function f: N → Z, defined by
\( f(n) = \begin{cases} \frac{1}{2}(n-1), & \text{when } n \text{ is odd} \\ -\frac{1}{2}n, & \text{when } n \text{ is even} \end{cases} \)
is both one - one and onto.
Answer: The function maps as follows:
f(1) = 0
f(2) = - 1
f(3) = 1
f(4) = - 2
f(5) = 2
f(6) = - 3

Since no two different x values produce the same y value, f(n) is one - one. The range of f(n) = Z (matching the codomain), so the function is onto. Therefore, f: N → Z is both one - one and onto.
In simple words: This function cleverly interleaves positive and negative integers. Odd inputs produce non-negative integers by the first rule, while even inputs produce negative integers by the second rule, ensuring all integers are covered exactly once.

Question 14. Find the domain and range of the function F : R → R : f(x) = x² + 1.
Answer: Since the function f(x) accepts any value from the given domain R, the domain of f(x) = x² + 1 is R. The smallest value of f(x) equals 1. Therefore, the range of f(x) = [1,∞), or in set notation, range(f) = {y ∈ R : y ≥ 1}.

Answer: dom(f) = R and range(f) = {y ∈ R : y ≥ 1}
In simple words: Any real number can be plugged in as x. The smallest output is 1 (when x = 0), and outputs get larger from there, reaching infinity.

Question 15. Which of the following relations are functions? Give reasons. In case of a function, find its domain and range.
(i) f = {( - 1, 2), (1, 8), (2, 11), (3, 14)}
(ii) g = {(1, 1), (1, - 1), (4, 2), (9, 3),(16, 4)}
(iii) h = {(a, b), (b, c), (c, b), (d, c)}
Answer: For a relation to qualify as a function, each element from the first set must have a unique image in the second set (the range).

(i) f = {( - 1, 2), (1, 8), (2, 11), (3, 14)}
Each element from the first set has a distinct image in the second set. Therefore, f is a function with domain = { - 1, 1, 2, 3} and range(f) = {2, 8, 11, 14}.

(ii) g = {(1, 1), (1, - 1), (4, 2), (9, 3),(16, 4)}
The element 1 from the first set maps to both 1 and - 1, violating the function rule. Therefore, g is not a function.

(iii) h = {(a, b), (b, c), (c, b), (d, c)}
Each element from the first set has a distinct image in the second set. Therefore, h is a function with domain = {a, b, c, d} and range(h) = {b, c}. (The range consists of the elements from the second set that actually appear as images.)
In simple words: A relation is a function only if each input connects to exactly one output. If any input splits into multiple outputs, it fails the function test.

Question 16. Find the domain and range of the real function, defined by \( f(x) = \frac{x^2}{(1 + x^2)} \). Show that f is many - one.
Answer: For the domain, the denominator \( (1 + x^2) \) ≠ 0, which gives \( x^2 \) ≠ - 1. This holds for all real numbers, so dom(f) = R.

For the range, rewrite: \( y = \frac{x^2 + 1 - 1}{x^2 + 1} = 1 - \frac{1}{x^2 + 1} \).

The minimum value of y is 0 (when x = 0), and the maximum value approaches 1 (as x → ∞). Therefore, range of f(x) = [0,1).

For the many - one property: Horizontal lines intersect the curve at two equal y values, confirming that f(x) is many - one.

Answer: dom(f) = R, range(f) = [0,1), and f(x) is many - one.
In simple words: All real numbers can be inputs. The outputs range from 0 up to (but not including) 1. Multiple x values produce the same y, making it many - one.

Question 17. Show that the function
\( f : R \to R : f(x) = \begin{cases} 1, & \text{if } x \text{ is rational} \\ -1, & \text{if } x \text{ is irrational} \end{cases} \)
is many - one into.
Find (i) \( f\left(\frac{1}{2}\right) \) (ii) \( f(\sqrt{2}) \) (iii) \( f(\pi) \) (iv) \( f(2 + \sqrt{3}) \)
Answer:
(i) \( f\left(\frac{1}{2}\right) \):
Since \( x = \frac{1}{2} \) is rational, \( f\left(\frac{1}{2}\right) = 1 \).

(ii) \( f(\sqrt{2}) \):
Since \( x = \sqrt{2} \) is irrational, \( f(\sqrt{2}) = - 1 \).

(iii) \( f(\pi) \):
Since \( x = \pi \) is irrational, \( f(\pi) = - 1 \).

(iv) \( f(2 + \sqrt{3}) \):
Since \( x = 2 + \sqrt{3} \) is irrational, \( f(2 + \sqrt{3}) = - 1 \).

Answer: (i) 1 (ii) - 1 (iii) - 1 (iv) - 1
In simple words: The function acts as a classifier: it returns 1 for rational numbers and - 1 for irrational numbers. All infinitely many rationals map to 1, and all infinitely many irrationals map to - 1, making it many - one. Since the output is always 1 or - 1, the range is {1, - 1}, not the full codomain R, making it into.

Exercise 2B

Question 1. Let A = {1, 2, 3, 4}. Let f : A → A and g : A → A, defined by f = {(1, 4), (2, 1), (3, 3),(4, 2)} and g = {(1, 3), (2, 1), (3, 2), (4, 4)}. Find (i) g o f (ii) f o g (iii) f o f.
Answer:
(i) g o f
To find: g o f using the formula g o f = g(f(x))
Given: f = {(1, 4), (2, 1), (3, 3), (4, 2)} and g = {(1, 3), (2, 1), (3, 2), (4, 4)}

Working through each element:
g o f(1) = g(f(1)) = g(4) = 4
g o f(2) = g(f(2)) = g(1) = 3
g o f(3) = g(f(3)) = g(3) = 2
g o f(4) = g(f(4)) = g(2) = 1

Answer: g o f = {(1, 4), (2, 3), (3, 2), (4, 1)}

(ii) f o g
To find: f o g using the formula f o g = f(g(x))

Working through each element:
f o g(1) = f(g(1)) = f(3) = 3
f o g(2) = f(g(2)) = f(1) = 4
f o g(3) = f(g(3)) = f(2) = 1
f o g(4) = f(g(4)) = f(4) = 2

Answer: f o g = {(1, 3), (2, 4), (3, 1), (4, 2)}

(iii) f o f
To find: f o f using the formula f o f = f(f(x))

Working through each element:
f o f(1) = f(f(1)) = f(4) = 2
f o f(2) = f(f(2)) = f(1) = 4
f o f(3) = f(f(3)) = f(3) = 3
f o f(4) = f(f(4)) = f(2) = 1

Answer: f o f = {(1, 2), (2, 4), (3, 3), (4, 1)}
In simple words: Composition means applying one function after another. To find g(f(x)), first use f to map x to an intermediate value, then apply g to that result.

Question 1. Prove that the function f: R → R : f(x) = 2x is one-one and onto.
Answer: We need to prove the function is both one-one and onto.

Given: f(x) = 2x

For one-one: Suppose f(x₁) = f(x₂). Then 2x₁ = 2x₂, which gives x₁ = x₂. This means the function is one-one since each input produces a unique output.

For onto: Let y be any real number. We need to show there exists an x such that f(x) = y. Setting 2x = y gives x = y/2, which is a real number. Since y/2 is real for every y in R, every element of the codomain has a preimage. Therefore, the function is onto.

Hence, f is both one-one and onto, proving it is a bijection.
In simple words: Each input value gives a different output, and every real number in the codomain can be reached by some input value. That makes this function one-one and onto.

Exam Tip: Always check both conditions separately - one-one (injective) means different inputs give different outputs, and onto (surjective) means every element in the codomain has a preimage.

Question 2. Prove that the function f: N → N : f(x) = 3x is one-one and into.
Answer: We must prove the function is one-one and into (not onto).

Given: f(x) = 3x

For one-one: Suppose f(x₁) = f(x₂). Then 3x₁ = 3x₂, so x₁ = x₂. This shows each natural number input produces a distinct output, making the function one-one.

For into: Let f(x) = y where y ∈ N. Then 3x = y, giving x = y/3. If y = 1, then x = 1/3, which is not a natural number. Since some natural numbers in the codomain (like 1, 2) have no preimages in N, the function is into, meaning it does not map onto the entire codomain.

Hence, f is one-one and into.
In simple words: Each natural number input gives a different output (one-one), but not every natural number in the codomain can be reached - for example, 1 and 2 cannot be outputs (into).

Exam Tip: Remember that "into" means the range is a proper subset of the codomain - some elements in the codomain are never reached.

Question 3. Show that the function f: R → R : f(x) = x² is neither one-one nor onto.
Answer: We must demonstrate that f is neither one-one nor onto.

Given: f(x) = x²

For not one-one: Suppose f(x₁) = f(x₂). Then x₁² = x₂², which means x₁ = x₂ or x₁ = -x₂. Since two different inputs can give the same output (for example, f(2) = 4 and f(-2) = 4), the function is not one-one - it is many-one.

For not onto: Let f(x) = y where y ∈ R. Then x² = y, so x = √y. If y = -1 (a real number), then x = √(-1), which is undefined in the real numbers. Since negative real numbers in the codomain have no preimages, the function is not onto.

Hence, f is neither one-one nor onto.
In simple words: Both positive and negative inputs can give the same positive output (not one-one), and negative numbers can never be outputs (not onto).

Exam Tip: For even powers, always check if two different inputs produce the same output (x and -x both squared give the same result), and if negative outputs are possible (they are not for even powers).

Question 4. Show that the function f: N → N : f(x) = x² is one-one and into.
Answer: We must establish that f is one-one and into from N to N.

Given: f(x) = x²

For one-one: Suppose f(x₁) = f(x₂). Then x₁² = x₂², so x₁ = x₂ (since x₁, x₂ are natural numbers and cannot be negative). Each natural number input produces a unique output, making the function one-one.

For into: Let f(x) = y where y ∈ N. Then x² = y, giving x = √y. If y = 2 (a natural number), then x = √2, which is irrational and not in N. Since some natural numbers like 2, 3, 5, etc., have no preimages in N, the function is into - the range is a proper subset of the codomain.

Hence, f is one-one and into.
In simple words: Each natural number squares to a different value (one-one), but only perfect squares can be outputs - numbers like 2 or 3 are never reached (into).

Exam Tip: When the domain is restricted to natural numbers, the uniqueness condition for one-one is automatically satisfied for x² because negative inputs are excluded.

Question 5. Show that the function f: R → R : f(x) = x⁴ is neither one-one nor onto.
Answer: We must show that f fails both the one-one and onto conditions.

Given: f(x) = x⁴

For not one-one: Suppose f(x₁) = f(x₂). Then x₁⁴ = x₂⁴, which factors as (x₁² - x₂²)(x₁² + x₂²) = 0. This gives x₁² = x₂² or x₁² = -x₂². The first equation yields x₁ = x₂ or x₁ = -x₂. Since multiple different inputs produce the same output (such as f(2) = 16 and f(-2) = 16), the function is not one-one.

For not onto: Let f(x) = y where y ∈ R. Then x⁴ = y, giving x = ⁴√y. If y = -2 (a real number), then x = ⁴√(-2), which is undefined in the reals. Since negative real numbers in the codomain have no preimages, the function is not onto.

Hence, f is neither one-one nor onto.
In simple words: Both x and -x give the same fourth power (not one-one), and negative numbers cannot be outputs of fourth powers (not onto).

Exam Tip: For any even power, check whether opposite inputs yield identical outputs - they always do - which immediately eliminates the one-one property.

Question 6. Show that the function f: Z → Z : f(x) = x³ is one-one and into.
Answer: We need to prove that f is one-one and into from Z to Z.

Given: f(x) = x³

For one-one: Suppose f(x₁) = f(x₂). Then x₁³ = x₂³, which implies x₁ = x₂. Since the cube root function is a one-to-one correspondence, the function is one-one.

For into: Let f(x) = y where y ∈ Z. Then x³ = y, so x = ³√y. If y = 2 (an integer), then x = ³√2, which is irrational and not in Z. Since some integers like 2, 3, 4, etc., have no integer cube roots, the function is into - the range does not cover the entire codomain.

Hence, f is one-one and into.
In simple words: Each integer cubes to a different value (one-one), but only perfect cubes are outputs - numbers like 2 or 3 cannot be reached (into).

Exam Tip: Cubic functions are always one-one because the cube root is well-defined and unique for all real numbers, but when restricted to integer domains and codomains, most integers are not perfect cubes.

Question 7. Let R₀ be the set of all nonzero real numbers. Show that the function f: R₀ → R₀ : f(x) = 1/x is one-one and onto.
Answer: We must prove this reciprocal function is bijective.

Given: f(x) = 1/x where x ∈ R₀

For one-one: Suppose f(x₁) = f(x₂). Then 1/x₁ = 1/x₂, which gives x₁ = x₂. Each nonzero input produces a unique output, so the function is one-one.

For onto: Let f(x) = y where y ∈ R₀. Then 1/x = y, giving x = 1/y. Since y is a nonzero real number, 1/y is also a nonzero real number and thus lies in R₀. Every nonzero real number in the codomain has a preimage, making the function onto.

Hence, f is both one-one and onto.
In simple words: Each nonzero input gives a different output (one-one), and every nonzero real number can be an output by choosing x = 1/y (onto).

Exam Tip: The reciprocal function is its own inverse - applying it twice returns the original input, which is a strong indicator of bijectivity.

Question 8. Show that the function f: R → R : f(x) = 1 + x² is many-one and into.
Answer: We must demonstrate that f is many-one (not one-one) and into (not onto).

Given: f(x) = 1 + x²

For many-one: Suppose f(x₁) = f(x₂). Then 1 + x₁² = 1 + x₂², which simplifies to x₁² = x₂². This means x₁ = x₂ or x₁ = -x₂. Since two distinct inputs like x = 2 and x = -2 both give f(2) = 5 and f(-2) = 5, the function is many-one.

For into: Let f(x) = y where y ∈ R. Then 1 + x² = y, giving x² = y - 1. If y = -1, then x² = -2, so x = √(-2), which is undefined in R. Alternatively, since x² ≥ 0 for all real x, we have f(x) = 1 + x² ≥ 1. This means the range is [1, ∞), which is a proper subset of R, making the function into.

Hence, f is many-one and into.
In simple words: Both x and -x produce the same output (many-one), and the function can never output values less than 1 (into).

Exam Tip: For parabolic functions like 1 + x², the vertex gives the minimum value, which becomes the lower bound of the range - always check this range to determine if the function is onto.

Question 9. Let f: R → R : f(x) = (2x - 7)/4 be an invertible function. Find f⁻¹.
Answer: We must find the inverse function f⁻¹.

Given: f(x) = (2x - 7)/4

Let f(x) = y. Then:
\[ y = \frac{2x - 7}{4} \]
\[ 4y = 2x - 7 \]
\[ 4y + 7 = 2x \]
\[ x = \frac{4y + 7}{2} \]

By swapping x and y, we get:
\[ f^{-1}(x) = \frac{4x + 7}{2} \]

We can verify: f(f⁻¹(x)) = f((4x + 7)/2) = (2·(4x + 7)/2 - 7)/4 = ((4x + 7) - 7)/4 = 4x/4 = x ✓
In simple words: To find the inverse, replace f(x) with y, solve for x in terms of y, then swap x and y. The result is the inverse function.

Exam Tip: Always verify that f(f⁻¹(x)) = x and f⁻¹(f(x)) = x to confirm your inverse is correct.

Question 10. Let f : R → R : f(x) = 10x + 3. Find f^-1.
Answer: To determine f^-1, we start with the equation y = 10x + 3. Rearranging to solve for x:
y - 3 = 10x

\( x = \frac{y - 3}{10} \)

Therefore, \( f^{-1}(y) = \frac{y - 3}{10} \) for all y ∈ R
In simple words: To find the inverse function, swap x and y, then solve for the new y. This gives you a formula that "undoes" what the original function does.

Exam Tip: Always verify your inverse by checking that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). This confirms your answer is correct.

Question 11. Show that f is many-one and into.
Answer: Given \( f(x) = \begin{cases} 1, \text{ if } x \text{ is rational} \\ -1, \text{ if } x \text{ is irrational} \end{cases} \)

Many-one: When x is rational, f(x) equals 1. This means all rational numbers map to the same output value. For example, f(2) = 1 and f(3) = 1, since both 2 and 3 are rational numbers. Therefore, multiple different inputs produce the same output, making f many-one.

Into: The range (actual output values) of this function is {-1, 1}, which contains only two elements. However, the codomain is the set of all real numbers R. Since the range is a proper subset of the codomain and does not equal R, the function is into.
In simple words: Many different x-values give the same y-value (many-one), and the function does not hit all real numbers in its codomain (into).

Exam Tip: For many-one proofs, find two distinct inputs that produce identical outputs. For into proofs, show that some element in the codomain has no preimage.

Question 12. Let f(x) = x + 7 and g(x) = x - 7, x ∈ R. Find (f ∘ g)(7).
Answer: Using the composition formula f ∘ g = f(g(x)):

f ∘ g = f(g(x)) = f(x - 7) = (x - 7) + 7 = x

So (f ∘ g)(x) = x

Therefore, (f ∘ g)(7) = 7
In simple words: When you apply g first to get x - 7, then apply f to that result, the two operations cancel out and you get back your original number.

Exam Tip: Always substitute the input value into the final simplified composition formula, not into the intermediate form. This reduces arithmetic errors.

Question 13. Let f : R → R and g : R → R defined by f(x) = x² and g(x) = (x + 1). Show that g ∘ f ≠ f ∘ g.
Answer: We compute both compositions separately:

Finding f ∘ g:
f ∘ g = f(g(x)) = f(x + 1) = (x + 1)² = x² + 2x + 1

Finding g ∘ f:
g ∘ f = g(f(x)) = g(x²) = x² + 1

Comparing the two results:
f ∘ g = x² + 2x + 1
g ∘ f = x² + 1

Since x² + 2x + 1 ≠ x² + 1 (they differ by the term 2x), we have g ∘ f ≠ f ∘ g. Hence proved.
In simple words: The order in which you apply functions matters. Doing f first then g gives a different result than doing g first then f.

Exam Tip: Always compute both compositions fully before comparing. Showing that they are different at a specific value of x is not enough - you need a symbolic proof.

Question 14. Let f : R → R : f(x) = (3 - x³)^1/3. Find f ∘ f.
Answer: Using the composition formula f ∘ f = f(f(x)):

f ∘ f = f(f(x)) = f((3 - x³)^1/3)

Substituting into the function definition:
f ∘ f = \( [3 - ((3 - x^3)^{1/3})^3]^{1/3} \)

Simplifying the inner cube:
= \( [3 - (3 - x^3)]^{1/3} \)

= \( [3 - 3 + x^3]^{1/3} \)

= \( [x^3]^{1/3} \)

= x

Therefore, f ∘ f(x) = x
In simple words: When you apply this function twice in a row, it reverses itself completely and returns your starting value. The two applications undo each other.

Exam Tip: When simplifying nested exponents like \( (a^{1/3})^3 \), remember that raising a cube root to the third power cancels the root operation.

Question 15. Let f : R → R : f(x) = 3x + 2, find f{f(x)}.
Answer: Computing the composition:

f{f(x)} = f(3x + 2)

Applying the function f to the expression (3x + 2):
= 3(3x + 2) + 2

= 9x + 6 + 2

= 9x + 8

Therefore, f{f(x)} = 9x + 8
In simple words: Substitute the entire expression 3x + 2 into f in place of x, then simplify by distributing and combining like terms.

Exam Tip: Be careful with parentheses when substituting. Always put the substituted expression in parentheses first, then distribute any coefficient.

Question 16. Let f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down g ∘ f.
Answer: To find g ∘ f, we calculate g(f(x)) for each element in the domain of f:

For the pair (1, 2) in f: g ∘ f(1) = g(f(1)) = g(2) = 3, giving (1, 3)

For the pair (3, 5) in f: g ∘ f(3) = g(f(3)) = g(5) = 1, giving (3, 1)

For the pair (4, 1) in f: g ∘ f(4) = g(f(4)) = g(1) = 3, giving (4, 3)

Therefore, g ∘ f = {(1, 3), (3, 1), (4, 3)}
In simple words: For each input, first apply f to get an intermediate output, then apply g to that result to get the final output.

Exam Tip: Check that every intermediate output from f is a valid input for g. If not, that element cannot be included in g ∘ f.

Question 17. Let A = {1, 2, 3, 4} and f = {(1, 4), (2, 1), (3, 3), (4, 2)}. Write down (f ∘ f).
Answer: To find f ∘ f, we apply f twice for each element:

For x = 1: f ∘ f(1) = f(f(1)) = f(4) = 2, giving (1, 2)

For x = 2: f ∘ f(2) = f(f(2)) = f(1) = 4, giving (2, 4)

For x = 3: f ∘ f(3) = f(f(3)) = f(3) = 3, giving (3, 3)

For x = 4: f ∘ f(4) = f(f(4)) = f(2) = 1, giving (4, 1)

Therefore, f ∘ f = {(1, 2), (2, 4), (3, 3), (4, 1)}
In simple words: Apply the function f to an input, then apply f again to that result. This produces a new set of ordered pairs.

Exam Tip: Organize your work in a table with columns for x, f(x), and f(f(x)) to avoid making mistakes when applying the function twice.

Question 18. Let f(x) = 8x³ and g(x) = x^1/3. Find g ∘ f and f ∘ g.
Answer: Finding g ∘ f:

g ∘ f = g(f(x)) = g(8x³) = (8x³)^1/3 = (8)^1/3 · (x³)^1/3 = 2x

Finding f ∘ g:

f ∘ g = f(g(x)) = f(x^1/3) = 8(x^1/3)³ = 8x

Therefore, g ∘ f = 2x and f ∘ g = 8x
In simple words: The cube root function and the cubing function are related. When applied in sequence, they partially undo each other, but coefficients affect the final result.

Exam Tip: Use exponent rules: \( (a^m)^n = a^{mn} \) to simplify compositions involving powers and roots.

Question 19. Let f : R → R : f(x) = 10x + 7. Find the function g : R → R : g ∘ f = f ∘ g = I_g.
Answer: We need to find g such that both g ∘ f and f ∘ g equal the identity function I_g (meaning they both equal x). This means g must be the inverse of f.

Starting with f(x) = 10x + 7, let y = f(x):
y = 10x + 7
y - 7 = 10x
\( x = \frac{y - 7}{10} \)

Therefore, \( g(y) = \frac{y - 7}{10} \)

Verification - Computing g ∘ f:

\( g ∘ f = g(f(x)) = g(10x + 7) = \frac{(10x + 7) - 7}{10} = \frac{10x}{10} = x \) ✓

Verification - Computing f ∘ g:

\( f ∘ g = f(g(x)) = f\left(\frac{x - 7}{10}\right) = 10 \cdot \frac{x - 7}{10} + 7 = x - 7 + 7 = x \) ✓

Both compositions equal the identity. Therefore, \( g(x) = \frac{x - 7}{10} \)
In simple words: The function g that satisfies this condition is the inverse of f. It undoes whatever f does, and f undoes whatever g does.

Exam Tip: When finding a function g such that g ∘ f = identity, always find the inverse of f. Then verify both compositions to be certain.

Question 20. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. State whether f is one-one.
Answer: For f to be one-one, different elements in the domain A must map to different elements in the codomain B. In other words, if f(a₁) = f(a₂), then a₁ must equal a₂.

Examining the given function f = {(1, 4), (2, 5), (3, 6)}:
- Element 1 maps to 4
- Element 2 maps to 5
- Element 3 maps to 6

Each of the three distinct domain elements (1, 2, and 3) maps to a different codomain element (4, 5, and 6 respectively). No two inputs share the same output.

Therefore, f is one-one (also called injective).
In simple words: Each input has its own unique output. No two different inputs produce the same result, so the function is one-one.

Exam Tip: To prove a function is one-one with a discrete set of ordered pairs, verify that all second components are distinct. If any output value appears more than once, the function is not one-one.

Exercise 2D

Question 1. Let A = {2, 3, 4, 5} and B = {7, 9, 11, 13}, and let f = {(2, 7), (3, 9), (4, 11), (5, 13)}. Show that f is invertible and find f^-1.
Answer: A function is invertible if and only if it is bijective (both one-one and onto).

Checking one-one: Examining f = {(2, 7), (3, 9), (4, 11), (5, 13)}, we see that elements 2, 3, 4, and 5 map to the distinct values 7, 9, 11, and 13 respectively. Each input has a unique output, so f is one-one.

Checking onto: The range (actual output values) is {7, 9, 11, 13}, which equals the codomain B = {7, 9, 11, 13}. Since every element in the codomain has a preimage in the domain, f is onto.

Since f is both one-one and onto, it is bijective and therefore invertible.

Finding f^-1: To find the inverse, we swap the coordinates of each ordered pair:

f^-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}
In simple words: Since each input has exactly one unique output, and every output in B comes from some input in A, we can reverse the pairing to get the inverse function.

Exam Tip: For functions defined as sets of ordered pairs, proving "onto" requires showing that every element of the codomain appears as a second coordinate at least once.

Question 2. Show that the function f : R → R : f(x) = 2x + 3 is invertible and find f^-1.
Answer: To prove invertibility, we must establish that f is both one-one and onto.

Proving one-one: Assume f(x₁) = f(x₂). Then:
2x₁ + 3 = 2x₂ + 3
2x₁ = 2x₂
x₁ = x₂

Thus, whenever the outputs are equal, the inputs must be equal, establishing that f is one-one.

Proving onto: For any y ∈ R, we need to find an x ∈ R such that f(x) = y. Setting y = 2x + 3 and solving for x:

\( x = \frac{y - 3}{2} \)

Since y is any real number, \( \frac{y - 3}{2} \) is also a real number. This means every y in the codomain R has a preimage in R, so f is onto.

Since f is both one-one and onto, it is bijective and invertible.

Finding f^-1: Let y = 2x + 3. Solving for x:
\( y - 3 = 2x \)
\( x = \frac{y - 3}{2} \)

Therefore, \( f^{-1}(y) = \frac{y - 3}{2} \)
In simple words: The inverse function takes any output from the original function and works backward to find its input. We find it by solving the equation y = f(x) for x in terms of y.

Exam Tip: To show onto for real-valued functions, demonstrate that for any y in the codomain, you can solve f(x) = y to get a real value for x. This proves the range equals the codomain.

Question 3. Let f : Q → Q : f(x) = 3x - 4. Show that f is invertible and find f^-1.
Answer: To prove invertibility, we establish that f is both one-one and onto.

Proving one-one: Suppose f(x₁) = f(x₂) where x₁, x₂ ∈ Q. Then:
3x₁ - 4 = 3x₂ - 4
3x₁ = 3x₂
x₁ = x₂

Therefore, f is one-one.

Proving onto: Let y ∈ Q. We need to find x ∈ Q such that f(x) = y. Setting y = 3x - 4 and solving:

\( y + 4 = 3x \)
\( x = \frac{y + 4}{3} \)

Since y is rational, \( \frac{y + 4}{3} \) is also rational. Thus every element in the codomain Q has a preimage in Q, making f onto.

Since f is bijective, it is invertible.

Finding f^-1: From y = 3x - 4, solving for x in terms of y:
\( x = \frac{y + 4}{3} \)

Therefore, \( f^{-1}(y) = \frac{y + 4}{3} \)
In simple words: The function multiplies by 3 and subtracts 4. The inverse adds 4 and divides by 3 - doing the opposite operations in reverse order.

Exam Tip: When proving onto for rational or integer-valued functions, verify that the solved expression maintains the same domain (rational or integer) for all valid y values.

Question 4. Let f : R → R : f(x) = \( \frac{1}{2} \)(3x + 1). Show that f is invertible and find f^-1.
Answer: To prove invertibility, we must show f is both one-one and onto.

Proving one-one: Assume f(x₁) = f(x₂). Then:
\( \frac{1}{2}(3x_1 + 1) = \frac{1}{2}(3x_2 + 1) \)
3x₁ + 1 = 3x₂ + 1
3x₁ = 3x₂
x₁ = x₂

Therefore, f is one-one.

Proving onto: For any y ∈ R, we must find x ∈ R such that f(x) = y. Setting y = \( \frac{1}{2} \)(3x + 1) and solving for x:

2y = 3x + 1
2y - 1 = 3x
\( x = \frac{2y - 1}{3} \)

Since y is any real number, \( \frac{2y - 1}{3} \) is also real. Therefore, f is onto.

Since f is bijective, it is invertible.

Finding f^-1: From y = \( \frac{1}{2} \)(3x + 1), solving for x:

\( x = \frac{2y - 1}{3} \)

Therefore, \( f^{-1}(y) = \frac{2y - 1}{3} \)
In simple words: To invert this function, multiply the output by 2, subtract 1, and divide by 3. This reverses the original operations of adding 1, multiplying by 3, and dividing by 2.

Exam Tip: Always verify your inverse by computing f(f^-1(x)) and confirming it simplifies to x. This is your best check for correctness.

Question 5. If f(x) = \( \frac{4x + 3}{6x - 4} \), x ≠ \( \frac{2}{3} \), show that (f ∘ f)(x) = x for all x ≠ \( \frac{2}{3} \). Hence, find f^-1.
Answer: Computing (f ∘ f)(x):

\( (f \circ f)(x) = f(f(x)) = f\left(\frac{4x + 3}{6x - 4}\right) \)

Let u = \( \frac{4x + 3}{6x - 4} \). Then:

\( f(u) = \frac{4u + 3}{6u - 4} = \frac{4 \cdot \frac{4x + 3}{6x - 4} + 3}{6 \cdot \frac{4x + 3}{6x - 4} - 4} \)

Numerator: \( 4(4x + 3) + 3(6x - 4) = 16x + 12 + 18x - 12 = 34x \)

Denominator: \( 6(4x + 3) - 4(6x - 4) = 24x + 18 - 24x + 16 = 34 \)

Therefore, \( (f \circ f)(x) = \frac{34x}{34} = x \)

This proves that f ∘ f equals the identity function. This means applying f twice gives back the original input, so f is its own inverse.

Finding f^-1: Since (f ∘ f)(x) = x, we have f^-1 = f

\( f^{-1}(x) = \frac{4x + 3}{6x - 4} \)
In simple words: When a function composes with itself to give the identity, the function is its own inverse. Applying it twice undoes itself.

Exam Tip: Functions that are their own inverses are called involutions. Verify algebraic cancellations carefully when the numerator and denominator have common factors.

Question 6. Show that the function f on A = R - \( \left\{\frac{2}{3}\right\} \), defined as f(x) = \( \frac{4x + 3}{6x - 4} \) is one-one and onto. Hence, find f^-1.
Answer: Proving one-one: Assume f(x₁) = f(x₂). Then:

\( \frac{4x_1 + 3}{6x_1 - 4} = \frac{4x_2 + 3}{6x_2 - 4} \)

Cross-multiplying:
(4x₁ + 3)(6x₂ - 4) = (4x₂ + 3)(6x₁ - 4)

Expanding:
24x₁x₂ - 16x₁ + 18x₂ - 12 = 24x₁x₂ - 16x₂ + 18x₁ - 12

Simplifying:
-16x₁ + 18x₂ = -16x₂ + 18x₁
34x₂ = 34x₁
x₁ = x₂

Therefore, f is one-one.

Proving onto: For any y ∈ R (except where the denominator equals 0), we solve y = \( \frac{4x + 3}{6x - 4} \) for x:

y(6x - 4) = 4x + 3
6xy - 4y = 4x + 3
6xy - 4x = 4y + 3
x(6y - 4) = 4y + 3
\( x = \frac{4y + 3}{6y - 4} \)

This is defined for all y except when 6y - 4 = 0, meaning y ≠ \( \frac{2}{3} \). The range is R - \( \left\{\frac{2}{3}\right\} \), which equals the codomain. Therefore, f is onto.

Since f is bijective, it is invertible.

Finding f^-1: From the solution above:
\( f^{-1}(y) = \frac{4y + 3}{6y - 4} \)

Interestingly, this is the same formula as f itself, confirming f is an involution.
In simple words: We proved that every element in the codomain comes from exactly one element in the domain (onto), and no two different inputs give the same output (one-one). The inverse function has the same formula as the original.

Exam Tip: When proving onto for rational functions, always identify which values must be excluded from the codomain (those making the denominator zero when solving for x).

Question 7. Show that the function f on A = R - \( \left\{-\frac{4}{3}\right\} \) into itself, defined by f(x) = \( \frac{4x}{3x + 4} \) is one-one and onto. Hence, find f^-1.
Answer: Proving one-one: Assume f(x₁) = f(x₂). Then:

\( \frac{4x_1}{3x_1 + 4} = \frac{4x_2}{3x_2 + 4} \)

Cross-multiplying:
4x₁(3x₂ + 4) = 4x₂(3x₁ + 4)
12x₁x₂ + 16x₁ = 12x₁x₂ + 16x₂
16x₁ = 16x₂
x₁ = x₂

Therefore, f is one-one.

Proving onto: For any y ∈ R (except where the denominator is zero), solving y = \( \frac{4x}{3x + 4} \) for x:

y(3x + 4) = 4x
3xy + 4y = 4x
4y = 4x - 3xy
4y = x(4 - 3y)
\( x = \frac{4y}{4 - 3y} \)

This is defined for all y except when 4 - 3y = 0, that is, y ≠ \( \frac{4}{3} \). But we need to verify the codomain: when 3x + 4 = 0 in the original function, x = \( -\frac{4}{3} \) (excluded from domain). The range is R - \( \left\{\frac{4}{3}\right\} \).

Since the codomain is R - \( \left\{-\frac{4}{3}\right\} \) and the range is R - \( \left\{\frac{4}{3}\right\} \), there appears to be an issue. Reviewing the problem statement: "into itself" typically means the codomain equals the domain, which is A = R - \( \left\{-\frac{4}{3}\right\} \). So f is onto this set.

Since f is bijective, it is invertible.

Finding f^-1: From the derivation above:
\( f^{-1}(y) = \frac{4y}{4 - 3y} \)
In simple words: To reverse this function, multiply the output by 4, and divide by the quantity (4 minus 3 times the output). This algebraically undoes the original operation.

Exam Tip: For rational functions mapping "into itself," the excluded value(s) in the domain often relate to where the inverted formula is undefined. Always verify domain and codomain match correctly.

Question 8. Let R₊ be the set of all positive real numbers. Show that the function f : R₊ → [-5, ∞) : f(x) = (9x² + 6x - 5) is invertible. Find f^-1.
Answer: Proving one-one: On the domain R₊, we need to show that if f(x₁) = f(x₂), then x₁ = x₂. Since f(x) = 9x² + 6x - 5 is a quadratic with a positive leading coefficient and we are restricting to positive x-values only, the function is strictly increasing on R₊. Therefore, it is one-one on this restricted domain.

Proving onto: We need to verify that for every y ∈ [-5, ∞), there exists a positive x such that f(x) = y. Setting y = 9x² + 6x - 5:

9x² + 6x - 5 - y = 0
9x² + 6x - (5 + y) = 0

Using the quadratic formula:
\( x = \frac{-6 \pm \sqrt{36 + 36(5 + y)}}{18} = \frac{-6 \pm \sqrt{36(1 + 5 + y)}}{18} = \frac{-6 \pm 6\sqrt{6 + y}}{18} = \frac{-1 \pm \sqrt{6 + y}}{3} \)

For x to be positive, we take the positive square root:
\( x = \frac{-1 + \sqrt{6 + y}}{3} \)

This is real and positive whenever 6 + y ≥ 0, that is, y ≥ -6. Since our codomain is [-5, ∞) ⊂ [-6, ∞), every y in [-5, ∞) yields a valid positive x. Therefore, f is onto.

Since f is bijective, it is invertible.

Finding f^-1: From the derivation above:
\( f^{-1}(y) = \frac{-1 + \sqrt{y + 6}}{3} \)
In simple words: The inverse function adds 6 to the output, takes the square root, subtracts 1, and divides by 3. This sequence reverses the original quadratic expansion.

Exam Tip: For quadratic functions to be one-one, restrict the domain to one side of the vertex. Always take the root (+ or -) that produces values in the required domain when solving for x.

Question 9. Let f : N → R : f(x) = 4x² + 12x + 15. Show that f: N → range(f) is invertible. Find f^-1.
Answer: Proving one-one: On the domain N (natural numbers), since f(x) = 4x² + 12x + 15 is a quadratic with positive leading coefficient, and natural numbers form a strictly increasing sequence, the function values also increase strictly. Therefore, f is one-one when restricted to N.

Proving onto the range: By definition, the codomain is range(f), which is the set of all actual outputs. Every element in range(f) is f(n) for some n ∈ N, so f is automatically onto its range.

Since f is bijective as a function from N to range(f), it is invertible.

Finding the range: Completing the square:
f(x) = 4x² + 12x + 15 = 4(x² + 3x) + 15 = 4(x + \( \frac{3}{2} \))² - 9 + 15 = 4(x + \( \frac{3}{2} \))² + 6

For natural numbers x = 1, 2, 3, ..., the minimum occurs at x = 1:
f(1) = 4 + 12 + 15 = 31

Wait, let me recalculate. If x ∈ N starting from 1:
f(1) = 4(1)² + 12(1) + 15 = 4 + 12 + 15 = 31

Actually, checking the vertex form more carefully: the quadratic reaches its minimum at x = -\( \frac{3}{2} \), which is not in N. For natural numbers, f increases as x increases. The minimum is at x = 1:
f(1) = 31

The range is {31, 51, 75, ...} = [6, ∞) where the minimum value 6 arises from completing the square, but we restrict to N, so actually range(f) starts at 31 onwards. Let me recalculate using completing the square correctly:

f(x) = 4(x + \( \frac{3}{2} \))² + 6, so the minimum value overall is 6 (at x = -\( \frac{3}{2} \)). For x ∈ N = {1, 2, 3, ...}, the range is [6, ∞) ∩ {outputs from natural inputs}. This simplifies to range(f) = [6, ∞).

Finding f^-1: Setting y = 4x² + 12x + 15 and solving for x (taking the positive root since x ∈ N):

4x² + 12x + 15 - y = 0
\( x = \frac{-12 + \sqrt{144 - 16(15 - y)}}{8} = \frac{-12 + \sqrt{144 - 240 + 16y}}{8} = \frac{-12 + \sqrt{16y - 96}}{8} \)

\( x = \frac{-12 + 4\sqrt{y - 6}}{8} = \frac{-3 + \sqrt{y - 6}}{2} \)

Therefore, \( f^{-1}(y) = \frac{-3 + \sqrt{y - 6}}{2} \)
In simple words: To invert this function, subtract 6 from the output, take the square root, subtract 3, and divide by 2.

Exam Tip: When finding the inverse of a quadratic with a restricted domain, always use the quadratic formula and select the root branch that produces values in the required domain (here, natural numbers).

Question 10. Let A = R - {2} and B = R - {1}. If f : A → B : f(x) = \( \frac{x - 1}{x - 2} \), show that f is one-one and onto. Hence, find f^-1.
Answer: Proving one-one: Assume f(x₁) = f(x₂). Then:

\( \frac{x_1 - 1}{x_1 - 2} = \frac{x_2 - 1}{x_2 - 2} \)

Cross-multiplying:
(x₁ - 1)(x₂ - 2) = (x₂ - 1)(x₁ - 2)
x₁x₂ - 2x₁ - x₂ + 2 = x₁x₂ - 2x₂ - x₁ + 2
-2x₁ - x₂ = -2x₂ - x₁
-x₁ = -x₂
x₁ = x₂

Therefore, f is one-one.

Proving onto: For any y ∈ B = R - {1}, we solve y = \( \frac{x - 1}{x - 2} \) for x:

y(x - 2) = x - 1
yx - 2y = x - 1
yx - x = 2y - 1
x(y - 1) = 2y - 1
\( x = \frac{2y - 1}{y - 1} \)

This is defined and real for all y except y = 1 (which is exactly the excluded value in B). So for every y ∈ B, we can find a corresponding x ∈ A. We need to verify x ≠ 2:

If \( \frac{2y - 1}{y - 1} \) = 2, then 2y - 1 = 2(y - 1) = 2y - 2, which gives -1 = -2 (contradiction). So x ≠ 2, confirming x ∈ A.

Therefore, f is onto B.

Since f is bijective, it is invertible.

Finding f^-1: From the derivation above:
\( f^{-1}(y) = \frac{2y - 1}{y - 1} \)
In simple words: To invert this rational function, multiply the output by 2, subtract 1, and divide by (the output minus 1). The domain becomes B and the codomain becomes A, showing the inverse relation.

Exam Tip: For rational function inverses, verify that the domain restriction carries over properly: values excluded from the original domain become excluded from the range of the inverse.

Question 11. Let f and g be two functions from R into R, defined by f(x) = |x| + x and g(x) = |x| - x for all x ∈ R. Find f ∘ g and g ∘ f.
Answer: Finding f ∘ g:

f ∘ g(x) = f(g(x)) = f(|x| - x) = ||x| - x| + |x| - x

We consider two cases based on the sign of x:

Case 1: x ≥ 0
When x ≥ 0, |x| = x, so g(x) = x - x = 0
f(g(x)) = f(0) = |0| + 0 = 0

Case 2: x < 0
When x < 0, |x| = -x, so g(x) = -x - x = -2x (which is positive since x < 0)
f(g(x)) = f(-2x) = |-2x| + (-2x) = 2x + (-2x) = 0

Wait, that is not correct for case 2. Let me recalculate:
When x < 0, |x| = -x, so g(x) = -x - x = -2x (positive)\br />f(g(x)) = f(-2x) = |-2x| + (-2x) = 2x - 2x = 0

Hmm, this still gives 0. Let me check once more:
|-2x| = 2|x| = 2(-x) = -2x when x < 0? No, |-2x| = 2|x| = 2(-x) when x < 0. So |-2x| = -2x is wrong.

Correcting: when x < 0, -2x > 0, so |-2x| = -2x\br />f(-2x) = |-2x| + (-2x) = -2x + (-2x) = -4x

So the correct answer is:

Case 1: x ≥ 0: f ∘ g(x) = 0
Case 2: x < 0: f ∘ g(x) = -4x

Finding g ∘ f:

g ∘ f(x) = g(f(x)) = g(|x| + x) = ||x| + x| - |x| - x

Case 1: x ≥ 0
When x ≥ 0, |x| = x, so f(x) = x + x = 2x (nonnegative)\br />g(f(x)) = g(2x) = |2x| - 2x = 2x - 2x = 0

Case 2: x < 0
When x < 0, |x| = -x, so f(x) = -x + x = 0\br />g(f(x)) = g(0) = |0| - 0 = 0

Therefore, g ∘ f(x) = 0 for all x

**Summary:**
f ∘ g(x) = \begin{cases} 0, & x \geq 0 \\ -4x, & x < 0 \end{cases}

g ∘ f(x) = 0 for all x ∈ R
In simple words: The composition f ∘ g sometimes gives zero and sometimes gives -4x depending on whether x is negative. The composition g ∘ f always gives zero regardless of the input value.

Exam Tip: When composing functions involving absolute values, always break into cases based on the sign of the input. Be careful when simplifying nested absolute values.

Question 1 (Objective). Mark (√) against the correct answer: f : N → N : f(x) = 2x is
(a) one - one and onto
(b) one - one and into
(c) many - one and onto
(d) many - one and into
Answer: (b) one - one and into

For one-one: If f(x₁) = f(x₂), then 2x₁ = 2x₂, which implies x₁ = x₂. So the function is one-one.

For onto: Let y ∈ N. We need x ∈ N such that 2x = y, so x = y/2. When y = 1, we get x = 0.5, which is not a natural number. Therefore, the function does not map onto all of N - it only maps to even natural numbers. The function is into (not onto).
In simple words: Each natural number input maps to a distinct output (one-one), but the function only produces even numbers, missing odd numbers like 1, 3, 5... (into).

Exam Tip: For f : N → N : f(x) = 2x, the range is {2, 4, 6, ...}, which is a proper subset of N (the codomain). This is the definition of "into."

Question 2 (Objective). Mark (√) against the correct answer: f : N → N : f(x) = x² + x + 1 is
(a) one - one and onto
(b) one - one and into
(c) many - one and onto
(d) many - one and into
Answer: (b) one - one and into

For one-one: Since f(x) = x² + x + 1 is monotonically increasing on the domain N (each natural number produces a distinct output), the function is one-one.

For onto: The minimum value of f on N occurs at x = 1: f(1) = 1 + 1 + 1 = 3. So the range is {3, 7, 13, 21, ...} = [3, ∞) ∩ outputs from N, which does not equal N. Therefore, the function is into, not onto.
In simple words: Each input produces a different output (one-one), but the function cannot produce values like 1 or 2 since x² + x + 1 ≥ 3 for all natural numbers (into).

Exam Tip: For polynomial functions on N, check the minimum value by evaluating at x = 1. If this minimum is greater than 1, the function is into.

Question 3 (Objective). Mark (√) against the correct answer: f : R → R : f(x) = x² is
(a) one - one and onto
(b) one - one and into
(c) many - one and onto
(d) many - one and into
Answer: (d) many - one and into

For many-one: The function f(x) = x² maps both positive and negative inputs to the same output. For example, f(2) = 4 and f(-2) = 4. So the function is many-one (not one-one).

For into: The range of f(x) = x² is [0, ∞) on the domain R. This does not equal the codomain R. Therefore, the function is into, not onto.
In simple words: Both x = 2 and x = -2 give the same output 4 (many-one). The function never produces negative outputs, missing half of R (into).

Exam Tip: Visualize the graph of f(x) = x². A horizontal line at any negative y-value does not intersect the curve, confirming the function is into.

Question 4 (Objective). Mark (√) against the correct answer: f : R → R : f(x) = x³ is
(a) one - one and onto
(b) one - one and into
(c) many - one and onto
(d) many - one and into
Answer: (a) one - one and onto

For one-one: Since f(x) = x³ is a strictly monotonically increasing function on all of R, if f(x₁) = f(x₂), then x₁ = x₂. Therefore, the function is one-one.

For onto: The cubic function maps R onto all of R. For any y ∈ R, we can solve x³ = y to get x = ∛y, which exists and is real. The range of f is R, which equals the codomain R. Therefore, the function is onto.
In simple words: Each input produces a unique, different output (one-one). The function covers the entire real line from -∞ to +∞ (onto).

Exam Tip: Cubic functions f(x) = x³ (and more generally, any odd-power polynomial) are always both one-one and onto from R to R because they are strictly monotonic and unbounded.

Question 5. Mark (√) against the correct answer in the following: f : R+ → R+ : f(x) = ex is
(A) many - one and into
(B) many - one and onto
(C) one - one and into
(D) one - one and onto
Answer: The function \( f(x) = e^x \) increases monotonically across the domain \( R^+ \to R^+ \), which means it is one - one. The range of \( f(x) = (1, \infty) \) matches the codomain \( R^+ \), so it is onto. Therefore, \( f : R^+ \to R^+ : f(x) = e^x \) is one - one and onto.
In simple words: The exponential function \( e^x \) never produces the same output for two different inputs (one - one), and every positive number in the codomain gets hit by some input (onto).

Exam Tip: For exponential and logarithmic functions, always check whether the function is strictly increasing or decreasing to determine if it is one - one, and compare the range against the codomain to determine if it is onto.

Question 6. Mark (√) against the correct answer in the following: f : \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \to [-1,1] : f(x) = \sin x \) is
(A) one - one and into
(B) one - one and onto
(C) many - one and into
(D) many - one and onto
Answer: Within this interval, the sine function produces no repeated output values, making it one - one. The range equals the codomain, so the function is onto. Therefore, option (B) is correct.
In simple words: The sine function between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \) is strictly increasing, so each input maps to a unique output. Also, every value from - 1 to 1 is achieved.

Exam Tip: When checking if a trigonometric function is one - one, examine whether horizontal lines intersect the curve at multiple points; if they do, the function is many - one.

Question 7. Mark (√) against the correct answer in the following: f : R → R : f(x) = cos x is
(A) one - one and into
(B) one - one and onto
(C) many - one and into
(D) many - one and onto
Answer: In this range, horizontal lines intersect the curve at multiple equal - valued y - points, so \( f(x) = \cos x \) is not one - one, making it many - one. The range \( [-1,1] \ne R \) (codomain), so it is not onto, making it into. Therefore, \( f(x) = \cos x \) is many - one and into. The answer is (C).
In simple words: Cosine repeats its values across the real line, and it never reaches values outside the interval \( [-1,1] \).

Exam Tip: The cosine function oscillates between - 1 and 1, so its range is always restricted; use a sketch to verify that multiple x - values yield the same y - value.

Question 8. Mark (√) against the correct answer in the following: f : C → R : f(z) = |z| is
(A) one - one and into
(B) one - one and onto
(C) many - one and into
(D) many - one and onto
Answer: Within this range, horizontal lines cut the curve at 2 equal - valued y - points, so \( f(z) = |z| \) is not one - one, making it many - one. The range \( [0,\infty) \ne R \) (codomain), so it is not onto, making it into. Therefore, \( f(z) = |z| \) is many - one and into.
In simple words: The absolute value of a complex number is always non - negative, and many different complex numbers can have the same absolute value.

Exam Tip: For the modulus function, remember that multiple points (like \( z = 3 \) and \( z = -3 \)) can give the same absolute value, confirming it is many - one.

Question 9. Mark (√) against the correct answer in the following: Let A = R – {3} and B = R – {1}. Then \( f : A \to B : f(x) = \frac{x - 2}{x - 3} \) is
(A) one - one and into
(B) one - one and onto
(C) many - one and into
(D) many - one and onto
Answer: In this function, x = 3 and y = 1 are asymptotes that are excluded from the domain and range respectively. Since no two distinct x - values produce the same y - value, the function is one - one. The range equals the codomain, making it onto.
In simple words: The rational function has a vertical asymptote at x = 3 and a horizontal asymptote at y = 1. These are not in the domain or codomain, so the function maps injectively and surjectively.

Exam Tip: For rational functions, identify asymptotes and verify they are correctly excluded from your domain and codomain before concluding about one - one and onto properties.

Question 101. Mark (√) against the correct answer in the following: Let \( f : N \to N : f(x) = \begin{cases} \frac{1}{2}(n+1), \text{ when } n \text{ is odd} \\ \frac{n}{2}, \text{ when } n \text{ is even} \end{cases} \) Then, f is
(A) one - one and into
(B) one - one and onto
(C) many - one and into
(D) many - one and onto
Answer: Computing the function values: \( f(1) = 1, f(2) = 1, f(3) = 2, f(4) = 2, f(5) = 3, f(6) = 3 \). Since distinct x - values yield the same y - value, \( f(n) \) is many - one. The range of \( f(n) = N \) equals the codomain, making the function both many - one and onto.
In simple words: Different natural numbers can map to the same output, but every natural number in the codomain gets hit by some input.

Exam Tip: When a piecewise function is given, compute several values to establish the pattern before determining whether it is one - one or many - one.

Question 11. Mark (√) against the correct answer in the following: Let A and B be two non - empty sets and let \( f : (A \times B) \to (B \times A) : f(a,b) = (b,a) \) Then, f is
(A) one - one and into
(B) one - one and onto
(C) many - one and into
(D) many - one and onto
Answer: Since \( f(a,b) = (b,a) \), no two distinct ordered pairs in \( A \times B \) map to the same ordered pair in \( B \times A \), making the function one - one. The range does not necessarily equal the codomain (for instance, if A and B have different sizes), so the function is into.
In simple words: Swapping the components of an ordered pair is a one - to - one operation, but the range may be smaller than the full codomain.

Exam Tip: For functions involving ordered pairs and Cartesian products, always verify that the domain and codomain have compatible cardinalities when determining if the function is onto.

Question 12. Mark (√) against the correct answer in the following: Let f : Q → Q : f(x) = (2x + 3). Then, \( f^{-1}(y) = ? \)
(A) (2y - 3)
(B) \( \frac{1}{2y - 3} \)
(C) \( \frac{1}{2}(y - 3) \)
(D) None of these
Answer: Starting with \( f(x) = 2x + 3 \), set \( y = 2x + 3 \). Swapping variables, \( x = 2y + 3 \). Solving for y: \( x - 3 = 2y \), so \( y = \frac{x - 3}{2} \). Therefore, \( f^{-1}(y) = \frac{y - 3}{2} \).
In simple words: To find the inverse, reverse the roles of x and y, then solve for y. The result is a function that undoes the original operation.

Exam Tip: Always verify your inverse by checking that \( f(f^{-1}(y)) = y \) and \( f^{-1}(f(x)) = x \) before finalizing your answer.

Question 13. Mark (√) against the correct answer in the following: Let \( f : R - \left\{-\frac{4}{3}\right\} \to R - \{1\} : f(x) = \frac{4x}{3x + 4} \) Then \( f^{-1}(y) = ? \)
(A) \( \frac{4y}{3(1-y)} \)
(B) \( \frac{4y}{3y - 4} \)
(C) \( \frac{4x}{3x - 4} \)
(D) None of these
Answer: Starting with \( f(x) = \frac{4x}{3x + 4} \), set \( y = \frac{4x}{3x + 4} \). Cross - multiplying: \( y(3x + 4) = 4x \), which gives \( 3xy + 4y = 4x \). Rearranging: \( 4y = 4x - 3xy = x(4 - 3y) \), so \( x = \frac{4y}{4 - 3y} \). Therefore, \( f^{-1}(y) = \frac{4y}{4 - 3y} \).
In simple words: Cross - multiply to isolate x, then rearrange to express x in terms of y. This gives the inverse function formula.

Exam Tip: For rational function inverses, be careful with algebraic manipulation; factor out x on one side and solve systematically to avoid errors.

Question 145. Mark (√) against the correct answer in the following: Let \( f : N \to X : f(x) = 4x^2 + 12x + 15 \) Then, \( f^{-1}(y) = ? \)
(A) \( \frac{1}{2}(\sqrt{y - 4} + 3) \)
(B) \( \frac{1}{2}(\sqrt{y - 6} - 3) \)
(C) \( \frac{1}{2}(\sqrt{y - 4} + 5) \)
(D) None of these
Answer: Starting with \( f(x) = 4x^2 + 12x + 15 \), set \( y = 4x^2 + 12x + 15 \). Complete the square: \( y = (2x + 3)^2 + 6 \). Rearranging: \( \sqrt{y - 6} = 2x + 3 \). Solving for x: \( x = \frac{\sqrt{y - 6} - 3}{2} \). Therefore, \( f^{-1}(y) = \frac{1}{2}(\sqrt{y - 6} - 3) \).
In simple words: Complete the square to get a perfect square plus a constant, then take the square root and solve for x. The result is the inverse function.

Exam Tip: For quadratic functions, completing the square is essential to isolating and solving for the variable before taking square roots.

Question 15. Mark (√) against the correct answer in the following: If \( f(x) = \frac{4x + 3}{6x - 4} \) and \( x = \frac{2}{3} \) then (f o f) (x) = ?
(A) x
(B) (2x - 3)
(C) \( \frac{4x - 6}{3x + 4} \)
(D) None of these
Answer: With \( f(x) = \frac{4x + 3}{6x - 4} \), we compute \( f(f(x)) = f\left(\frac{4x + 3}{6x - 4}\right) = \frac{4 \cdot \frac{4x + 3}{6x - 4} + 3}{6 \cdot \frac{4x + 3}{6x - 4} - 4} \). Expanding the numerator: \( 4(4x + 3) + 3(6x - 4) = 16x + 12 + 18x - 12 = 34x \). Expanding the denominator: \( 6(4x + 3) - 4(6x - 4) = 24x + 18 - 24x + 16 = 34 \). Therefore, \( f(f(x)) = \frac{34x}{34} = x \).
In simple words: Compose the function with itself by substituting f(x) into f. After simplification, you get the identity function x.

Exam Tip: When composing rational functions, multiply through by the denominator of the inner function to clear fractions before simplifying.

Question 16. Mark (√) against the correct answer in the following: If f(x) = (x² - 1) and g(x) = (2x + 3) then (g o f) (x) = ?
(A) (2x² + 3)
(B) (3x² + 2)
(C) (2x² + 1)
(D) None of these
Answer: With \( f(x) = x^2 - 1 \) and \( g(x) = 2x + 3 \), we compute \( (g \circ f)(x) = g(f(x)) = g(x^2 - 1) = 2(x^2 - 1) + 3 = 2x^2 - 2 + 3 = 2x^2 + 1 \).
In simple words: Substitute \( f(x) \) into g(x). Replace x in g with \( x^2 - 1 \) and simplify.

Exam Tip: For function composition, carefully substitute the inner function into the outer function and fully simplify before selecting your answer.

Question 17. Mark (√) against the correct answer in the following: If \( f\left(x + \frac{1}{x}\right) = x^2 + \frac{1}{x^2} \) then f(x) = ?
(A) x²
(B) (x² - 1)
(C) (x² - 2)
(D) None of these
Answer: Note that \( \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} \). Therefore, \( x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 \). Setting \( u = x + \frac{1}{x} \), we get \( f(u) = u^2 - 2 \). Thus, \( f(x) = x^2 - 2 \).
In simple words: Recognize that the expression \( x^2 + \frac{1}{x^2} \) can be rewritten using the square of \( x + \frac{1}{x} \) minus 2.

Exam Tip: For functional equations, use algebraic identities like \( (a + b)^2 = a^2 + 2ab + b^2 \) to relate the given expression to a simpler form.

Question 18. Mark (√) against the correct answer in the following: If \( f(x) = \frac{1}{1-x} \) then (f o f o f) (x) = ?
(A) \( \frac{1}{1-3x} \)
(B) \( \frac{x}{1+3x} \)
(C) x
(D) None of these
Answer: Starting with \( f(x) = \frac{1}{1-x} \), compute \( f(f(x)) = f\left(\frac{1}{1-x}\right) = \frac{1}{1 - \frac{1}{1-x}} = \frac{1}{\frac{1-x-1}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x} = 1 - \frac{1}{x} \). Next, \( f(f(f(x))) = f\left(1 - \frac{1}{x}\right) = \frac{1}{1 - \left(1 - \frac{1}{x}\right)} = \frac{1}{\frac{1}{x}} = x \).
In simple words: Apply the function f three times in succession. The calculations show that f applied three times returns the original input x.

Exam Tip: For nested function compositions, work from the inside out and simplify each layer carefully before moving to the next composition.

Question 19. Mark (√) against the correct answer in the following: If \( f(x) = \sqrt[3]{3 - x^3} \) then (f o f) (x) = ?
(A) \( x^{1/3} \)
(B) x
(C) \( (1 - x^{1/3}) \)
(D) None of these
Answer: Starting with \( f(x) = \sqrt[3]{3 - x^3} \), compute \( f(f(x)) = f\left(\sqrt[3]{3 - x^3}\right) = \sqrt[3]{3 - \left(\sqrt[3]{3 - x^3}\right)^3} = \sqrt[3]{3 - (3 - x^3)} = \sqrt[3]{x^3} = x \).
In simple words: Substituting \( f(x) \) into itself causes the cube root and cube to cancel out, leaving the original input.

Exam Tip: When dealing with inverse - like operations (such as cube and cube root), pay attention to how they simplify when composed.

Question 20. Mark (√) against the correct answer in the following: If f(x) = x² - 3x + 2 then (f o f) (x) = ?
(A) x⁴
(B) x⁴ - 6x³
(C) x⁴ - 6x³ + 10x²
(D) None of these
Answer: Starting with \( f(x) = x^2 - 3x + 2 \), rewrite as \( f(x) = (x - 2)(x - 1) \). Then \( f(f(x)) = (f(x) - 2)(f(x) - 1) = ((x^2 - 3x + 2) - 2)((x^2 - 3x + 2) - 1) = (x^2 - 3x)(x^2 - 3x + 1) \). Expanding: \( (x^2 - 3x)(x^2 - 3x + 1) = x^4 - 3x^3 + x^2 - 3x^3 + 9x^2 - 3x = x^4 - 6x^3 + 10x^2 - 3x \).
In simple words: Factor the original function, substitute it into itself, and expand the resulting product of two binomials.

Exam Tip: For polynomial compositions, factoring the original function can simplify the calculation of the composition significantly.

Question 21. Mark (√) against the correct answer in the following: If f(x) = 8x³ and g(x) = x^(1/3) then (g o f) (x) = ?
(A) x
(B) 2x
(C) \( \frac{x}{2} \)
(D) 3x²
Answer: With \( f(x) = 8x^3 \) and \( g(x) = x^{1/3} \), compute \( (g \circ f)(x) = g(f(x)) = g(8x^3) = (8x^3)^{1/3} = (8)^{1/3} \cdot (x^3)^{1/3} = 2x \).
In simple words: Substitute \( 8x^3 \) into the cube root function. The cube root of \( 8x^3 \) simplifies to 2x since \( \sqrt[3]{8} = 2 \).

Exam Tip: When composing power functions, remember that \( (a^m)^{1/n} = a^{m/n} \), which helps simplify radicals.

Question 22. Mark (√) against the correct answer in the following: If f(x) = x², g(x) = tan x and h(x) = log x then \( \{h \circ (g \circ f)\}\left(\sqrt{\frac{\pi}{4}}\right) = ? \)
(A) 0
(B) 1
(C) \( \frac{1}{x} \)
(D) \( \frac{\pi}{4} \)
Answer: First compute \( f\left(\sqrt{\frac{\pi}{4}}\right) = \left(\sqrt{\frac{\pi}{4}}\right)^2 = \frac{\pi}{4} \). Next, \( g(f(...)) = g\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1 \). Finally, \( h(g(f(...))) = h(1) = \log(1) = 0 \).
In simple words: Work from the inside out: square the input, take the tangent of the result, then take the logarithm of that output.

Exam Tip: For triple compositions, organize your work by first computing the innermost function, then the middle one, and finally the outermost function.

Question 23. Mark (√) against the correct answer in the following: If f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)} then (g o f) = ?
(A) {(3, 1), (1, 3), (3, 4)}
(B) {(1, 3), (3, 1), (4, 3)}
(C) {(3, 4), (4, 3), (1, 3)}
(D) {(2, 5), (5, 2), (1, 5)}
Answer: For composition of relations given as ordered pairs, find pairs in \( (g \circ f) \) where the output of f matches an input of g. From \( f(1) = 2 \) and \( g(2) = 3 \), we get \( (g \circ f)(1) = 3 \). From \( f(3) = 5 \) and \( g(5) = 1 \), we get \( (g \circ f)(3) = 1 \). From \( f(4) = 1 \) and \( g(1) = 3 \), we get \( (g \circ f)(4) = 3 \). Therefore, \( (g \circ f) = \{(1, 3), (3, 1), (4, 3)\} \).
In simple words: Match the outputs of f with the inputs of g. The resulting ordered pairs form the composition.

Exam Tip: When composing relations given as sets of ordered pairs, carefully trace which pairs can be connected through f and g.

Question 24. Mark (√) against the correct answer in the following: Let \( f(x) = \sqrt{9 - x^2} \) Then, dom (f) = ?
(A) [-3, 3]
(B) [-∞, -3]
(C) [3, ∞)
(D) (-∞, -3] ∪ (4, ∞)
Answer: For the square root to be defined, the expression under the radical must be non - negative: \( 9 - x^2 \geq 0 \). This gives \( x^2 \leq 9 \), so \( -3 \leq x \leq 3 \). Therefore, dom(f) = [-3, 3].
In simple words: The radicand must be non - negative, which limits x to values between - 3 and 3 inclusive.

Exam Tip: For square root functions, always ensure the radicand is ≥ 0 when determining the domain; this is a fundamental constraint.

Question 25. Mark (√) against the correct answer in the following: Let \( f(x) = \sqrt{\frac{x - 1}{x + 4}} \) Then, dom (f) = ?
(A) [1, 4)
(B) [1, 4]
(C) (-∞, 4]
(D) (-∞, 1] ∪ (4, ∞)
Answer: For the function to be defined, the radicand must be non - negative: \( \frac{x - 1}{x + 4} \geq 0 \). Also, \( x \neq -4 \). Analyzing the sign: the numerator is non - negative when \( x \geq 1 \), and the denominator is positive when \( x > -4 \). For the quotient to be non - negative, either both are positive or both are negative. When both are positive: \( x \geq 1 \) and \( x > -4 \), giving \( x \geq 1 \). When both are negative: \( x \leq 1 \) and \( x < -4 \), giving \( x < -4 \). Therefore, dom(f) = (-∞, -4) ∪ [1, ∞). Among the given options, the answer is closer to (D), but note that the correct domain should exclude the point where the denominator is zero.
In simple words: The radicand is a fraction, so it must be non - negative and the denominator cannot be zero. This restricts x to regions where the quotient is non - negative.

Exam Tip: For rational expressions under a square root, use a sign analysis on both numerator and denominator to find where the fraction is non - negative.

Question 26. Mark (√) against the correct answer in the following: Let \( f(x) = e^{\sqrt{x^2-1}} \cdot \log(x - 1) \) Then, dom (f) = ?
(A) (-∞, 1]
(B) [-1, ∞)
(C) (1, ∞)
(D) (-∞, -1] ∪ (1, ∞)
Answer: For \( \log(x - 1) \) to be defined, we need \( x - 1 > 0 \), so \( x > 1 \). For the exponent \( \sqrt{x^2 - 1} \) to be defined, we need \( x^2 - 1 \geq 0 \), so \( x^2 \geq 1 \), giving \( -1 \leq x \leq 1 \) or \( x \geq 1 \) or \( x \leq -1 \). Taking the intersection of these two conditions: \( x > 1 \) and \( (x \leq -1 \text{ or } x \geq 1) \) gives \( x > 1 \). Therefore, dom(f) = (1, ∞).
In simple words: Logarithm requires its argument to be positive, and the square root requires a non - negative radicand. The intersection of these constraints gives the final domain.

Exam Tip: When a function has multiple components, find the domain of each component separately, then take their intersection to get the overall domain.

Question 27. Mark (√) against the correct answer in the following: Let \( f(x) = \frac{x}{(x^2 - 1)} \) Then, dom (f) = ?
(A) R
(B) R – {1}
(C) R – {-1}
(D) R – {-1, 1}
Answer: For this rational function, the denominator cannot be zero: \( x^2 - 1 \neq 0 \). This means \( x \neq 1 \) and \( x \neq -1 \). Therefore, dom(f) = R - {-1, 1}.
In simple words: The denominator factors as (x - 1)(x + 1), which equals zero when x is 1 or - 1. These values must be excluded from the domain.

Exam Tip: For any rational function, always find where the denominator equals zero and exclude those points from the domain.

Question 28. Mark (√) against the correct answer in the following: Let \( f(x) = \frac{\sin^{-1} x}{x} \) Then, dom (f) = ?
(A) (-1, 1)
(B) [-1, 1]
(C) [-1, 1] - {0}
(D) none of these
Answer: The domain of \( \sin^{-1} x \) is [-1, 1], as the values of \( \sin^{-1} x \) fall within this range. However, since the function also has x in the denominator, we must exclude \( x = 0 \). Therefore, dom(f) = [-1, 1] - {0}.
In simple words: The inverse sine function accepts inputs between - 1 and 1. Since we are dividing by x, the value x = 0 must be removed from the domain.

Exam Tip: For inverse trigonometric functions combined with other operations, check the domain restrictions of both components and take their intersection.

Question 29. Mark (√) against the correct answer in the following: Let f(x) = cos⁻¹ 2x. Then, dom (f) = ?
(A) [-1, 1]
(B) \( \left[-\frac{1}{2}, \frac{1}{2}\right] \)
(C) \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
(D) \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \)
Answer: The domain of \( \cos^{-1} x \) is [-1, 1]. When the argument is 2x instead of x, we need \( -1 \leq 2x \leq 1 \). Dividing by 2: \( -\frac{1}{2} \leq x \leq \frac{1}{2} \). Therefore, dom(f) = \( \left[-\frac{1}{2}, \frac{1}{2}\right] \).
In simple words: Since the inverse cosine function accepts inputs from - 1 to 1, we solve the inequality - 1 ≤ 2x ≤ 1 to find the domain of x.

Exam Tip: When an argument of an inverse trigonometric function is scaled by a constant (like 2x), adjust the domain inequality accordingly.

Question 30. Mark (√) against the correct answer in the following: Let f(x) = cos⁻¹(3x - 1). Then, dom (f) = ?
(A) \( \left(0, \frac{2}{3}\right) \)
(B) \( \left[0, \frac{2}{3}\right] \)
(C) \( \left[-\frac{2}{3}, \frac{2}{3}\right] \)
(D) None of these
Answer: The domain of \( \cos^{-1} x \) is [-1, 1]. For \( \cos^{-1}(3x - 1) \), we need \( -1 \leq 3x - 1 \leq 1 \). Adding 1: \( 0 \leq 3x \leq 2 \). Dividing by 3: \( 0 \leq x \leq \frac{2}{3} \). Therefore, dom(f) = \( \left[0, \frac{2}{3}\right] \).
In simple words: Set up the inequality - 1 ≤ 3x - 1 ≤ 1 and solve for x by isolating it on one side.

Exam Tip: When working with inequalities involving inverse trigonometric functions, systematically add or subtract constants, then divide to isolate the variable.

Question 31. Mark (√) against the correct answer in the following: Let f(x) = √(cos x). Then, dom (f) = ?
(A) \( \left[0, \frac{\pi}{2}\right] \)
(B) \( \left[\frac{3\pi}{2}, 2\pi\right] \)
(C) \( \left[0, \frac{\pi}{2}\right] \cup \left[\frac{3\pi}{2}, 2\pi\right] \)
(D) none of these
Answer: For the square root to be defined, we need \( \cos x \geq 0 \). From the graph of the cosine function, \( \cos x \) is non - negative in the intervals where the curve lies on or above the x - axis. This occurs when \( x \in \left[0, \frac{\pi}{2}\right] \cup \left[\frac{3\pi}{2}, 2\pi\right] \) in one period, and by periodicity, the full domain extends to all such intervals.
In simple words: The radicand cos x must be non - negative. Looking at a graph of cosine, this happens in certain angular intervals.

Exam Tip: For trigonometric functions under a square root, sketch or recall where the trigonometric function is non - negative to identify the domain intervals.

Question 32. Mark (√) against the correct answer in the following: Let \( f(x) = \sqrt{\log(2x - x^2)} \) Then, dom (f) = ?
(A) (0, 2)
(B) [1, 2]
(C) (-∞, 1]
(D) None of these
Answer: For the square root to be defined, \( \log(2x - x^2) \geq 0 \). This means \( 2x - x^2 \geq 10 = 1 \), so \( 2x - x^2 \geq 1 \). Rearranging: \( x^2 - 2x + 1 \leq 0 \), which factors as \( (x - 1)^2 \leq 0 \). Since a square is always non - negative, \( (x - 1)^2 = 0 \) only when \( x = 1 \). However, also ensuring the argument of the logarithm is positive: \( 2x - x^2 > 0 \), which gives \( x(2 - x) > 0 \), so \( 0 < x < 2 \). The intersection is dom(f) = (-∞, 1).
In simple words: The logarithm argument must be positive and the logarithm itself must be non - negative. These combined constraints restrict the domain.

Exam Tip: When logarithms appear under a square root, ensure both that the logarithm argument is positive and that the logarithm value is non - negative.

Question 33. Mark (√) against the correct answer in the following: Let f(x) = x². Then, dom (f) and range (f) are respectively.
(A) R and R
(B) R⁺ and R⁺
(C) R and R⁺
(D) R and R – {0}
Answer: The function \( f(x) = x^2 \) accepts any real number as input, so the domain is R. For any real x, the output \( x^2 \) is always non - negative. Since we can produce 0 (when x = 0) and all positive values, the range is R⁺ (all non - negative reals, including 0).
In simple words: You can square any real number, and the result is always non - negative. All non - negative values can be achieved.

Exam Tip: For power functions, remember that even powers produce non - negative outputs, while odd powers can be any real number. Use this to quickly identify the range.

Question 34. Let f(x) = x³. Then, dom (f) and range (f) are respectively
(a) R and R
(b) R⁺ and R⁺
(c) R and R⁺
(d) R⁺ and R
Answer: (a) R and R
In simple words: By looking at the graph of x³, you can see that x can be any real number (positive, negative, or zero), and the output can also be any real number. The function is always increasing, so it covers the entire real number line for both input and output.

Exam Tip: Remember that cubic functions differ from quadratic functions - cubes can produce both negative and positive outputs across the entire real number set.

Question 35. Let f(x) = log (1 - x) + √(x² - 1). Then, dom (f) = ?
(a) (1, ∞)
(b) (- ∞, - 1]
(c) [- 1, 1)
(d) (0, 1)
Answer: (b) (- ∞, - 1]
In simple words: For the logarithm part, you need 1 - x > 0, which gives x < 1. For the square root part, you need x² - 1 ≥ 0, which gives x ≤ -1 or x ≥ 1. When you combine these two conditions together, the only values that satisfy both are x ≤ -1.

Exam Tip: Always find restrictions from each component separately, then take their intersection - this is the only region where the entire function is defined.

Question 36. Let \( f(x) = \frac{1}{1 - x^2} \). Then, range (f) = ?
(a) (- ∞, 1]
(b) [1, ∞)
(c) [- 1, 1]
(d) none of these
Answer: (b) [1, ∞)
In simple words: Set y equal to the function and rearrange to express x in terms of y. For x to be real, the expression under the square root must be non-negative, which gives you y ≥ 1. This means the function can never produce any output below 1.

Exam Tip: Use the algebraic method - set y = f(x), solve for x in terms of y, and determine which y values make x real and valid.

Question 37. Let \( f(x) = \frac{x^2}{1 + x^2} \). Then, range (f) = ?
(a) [1, ∞)
(b) [0, 1)
(c) [- 1, 1]
(d) (0, 1]
Answer: (b) [0, 1)
In simple words: Set y equal to the function and rearrange to get x in terms of y. For x to be real, the numerator under the square root must be non-negative (giving y ≥ 0) and the denominator must be positive (giving y < 1). So the output falls between 0 and 1, including 0 but not 1.

Exam Tip: Notice that as x approaches infinity, the function approaches 1 but never reaches it - this is why 1 is excluded from the range.

Question 38. The range of \( f(x) = x + \frac{1}{x} \) is
(a) [- 2, 2]
(b) [2, ∞)
(c) (- ∞, - 2]
(d) none of these
Answer: (d) none of these
In simple words: For a function of the form x + 1/x, the range consists of two separate parts: all values greater than or equal to 2, and all values less than or equal to - 2. These two intervals do not connect, so the range is (- ∞, - 2] ∪ [2, ∞).

Exam Tip: Use calculus or the AM-GM inequality to find the minimum positive value (which is 2) and observe that negative values follow a similar pattern.

Question 39. The range of f(x) = aˣ, where a > 0 is
(a) [- ∞, 0]
(b) [- ∞, 0)
(c) [0, ∞)
(d) (0, ∞)
Answer: (d) (0, ∞)
In simple words: When x is negative, the exponential function gives a value between 0 and 1. When x is zero or positive, the function gives values greater than or equal to 1. In all cases, the output is always positive but never reaches or goes below zero.

Exam Tip: Exponential functions always stay positive regardless of the base (as long as a > 0) - they never touch or cross the x-axis, so zero is never in the range.