Samacheer Kalvi Class 10 Maths Solutions Chapter 1 Relations and Functions Exercise 1.6

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.6

Multiple Choice Questions

 

Question 1. If n(A × B) = 6 and A= {1, 3} then n (B) is
(a) 1
(b) 2
(c) 3
(d) 0
Answer: (c) 3
Hint: We know that the number of elements in the Cartesian product of two sets, n(A × B), is equal to the product of the number of elements in each set, n(A) × n(B). Given that n(A × B) = 6 and A = {1, 3}, which means n(A) = 2. So, we can find n(B) by dividing n(A × B) by n(A).
\( n(A \times B) = 6 \)
\( n(A) = 2 \)
\( n(A \times B) = n(A) \times n(B) \)
\( 6 = 2 \times n(B) \)
\( n(B) = \frac { 6 }{ 2 } \)
\( n(B) = 3 \)
This fundamental property of sets helps us count elements in product sets easily.
In simple words: If you multiply the number of things in set A by the number of things in set B, you get the total number of pairs. Here, we know the total pairs (6) and the number of things in A (2), so we divide to find the number of things in B (3).

🎯 Exam Tip: Remember the basic formula \( n(A \times B) = n(A) \times n(B) \) for counting elements in a Cartesian product; it's essential for set theory problems.

 

Question 2. A = {a, b, p}, B = {2, 3}, C = {p, q, r, s} then n[(A U C) × B] is
(a) 8
(b) 20
(c) 12
(d) 16
Answer: (c) 12
Hint: First, find the union of sets A and C (A U C) by combining all unique elements from both sets. Then, count the number of elements in the resulting set n(A U C) and the number of elements in set B, n(B). Finally, multiply these two counts to find n[(A U C) × B]. The union operation combines elements without duplication.
Given: \( A = \{a, b, p\} \)
\( B = \{2, 3\} \)
\( C = \{p, q, r, s\} \)
First, find \( A \cup C \):
\( A \cup C = \{a, b, p, q, r, s\} \)
Now, find the number of elements in \( A \cup C \) and \( B \):
\( n(A \cup C) = 6 \)
\( n(B) = 2 \)
Then, find \( n[(A \cup C) \times B] \):
\( n[(A \cup C) \times B] = n(A \cup C) \times n(B) \)
\( n[(A \cup C) \times B] = 6 \times 2 = 12 \)
The Cartesian product forms ordered pairs from these combined elements.
In simple words: First, mix the items from set A and set C together, making sure not to count any item twice. Then, count how many items are in this new combined set. Next, count the items in set B. Finally, multiply these two counts to get the answer.

🎯 Exam Tip: When finding the union of sets, remember to list each unique element only once. Then, apply the Cartesian product rule correctly.

 

Question 3. If A = {1,2}, B = {1,2, 3, 4}, C = {5,6} and D = {5, 6, 7, 8} then state which of the following statement is true
(a) \( (A \times C) \subset (B \times D) \)
(b) \( (B \times D) \subset (A \times C) \)
(c) \( (A \times B) \subset (A \times D) \)
(d) \( (D \times A) \subset (B \times A) \)
Answer: (a) \( (A \times C) \subset (B \times D) \)
Hint: To determine if one Cartesian product is a subset of another, compare the number of elements in each product. If \( n(X \times Y) \) is less than or equal to \( n(P \times Q) \), it's a potential subset. For \( (A \times C) \subset (B \times D) \), every ordered pair in \( A \times C \) must also be in \( B \times D \). This happens if A is a subset of B and C is a subset of D. We need to check all the options by calculating \( n(A \times B) \), \( n(A \times C) \), etc.
Given: \( A = \{1, 2\} \), \( B = \{1, 2, 3, 4\} \), \( C = \{5, 6\} \), \( D = \{5, 6, 7, 8\} \)
Calculate the number of elements for each Cartesian product:
\( n(A \times C) = n(A) \times n(C) = 2 \times 2 = 4 \)
\( n(B \times D) = n(B) \times n(D) = 4 \times 4 = 16 \)
Since \( n(A \times C) = 4 \) and \( n(B \times D) = 16 \), it's possible for \( (A \times C) \) to be a subset of \( (B \times D) \).
Now, check if \( A \subseteq B \) and \( C \subseteq D \).
Since \( \{1, 2\} \subseteq \{1, 2, 3, 4\} \) is true, \( A \subseteq B \).
Since \( \{5, 6\} \subseteq \{5, 6, 7, 8\} \) is true, \( C \subseteq D \).
Because both conditions are met, \( (A \times C) \) is indeed a subset of \( (B \times D) \). This makes it true that every pair from A and C will be found in the pairs from B and D. Let's briefly check other options:
\( n(A \times B) = 2 \times 4 = 8 \)
\( n(A \times D) = 2 \times 4 = 8 \)
\( n(B \times C) = 4 \times 2 = 8 \)
\( n(C \times D) = 2 \times 4 = 8 \)
Since \( A \not\subseteq B \) or \( A \not\subseteq D \) for other options, or vice-versa, only (a) remains true.
In simple words: We are checking which statement is correct about sets. The first statement says that if you make pairs from A and C, all those pairs can also be found if you make pairs from B and D. This is true because all items in A are also in B, and all items in C are also in D. So, any pair made from A and C will naturally be a pair from B and D.

🎯 Exam Tip: To check if \( (X \times Y) \subset (P \times Q) \), verify two conditions: first, \( n(X \times Y) \le n(P \times Q) \), and second, \( X \subseteq P \) and \( Y \subseteq Q \). Both must be true.

 

Question 4. If there are 1024 relations from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is
(a) 3
(b) 2
(c) 4
(d) 6
Answer: (b) 2
Hint: The total number of relations from set A to set B is given by \( 2^{n(A) \times n(B)} \). We are given the total number of relations as 1024 and the number of elements in set A, n(A). We need to find the number of elements in set B, n(B), by setting up an equation and solving for the exponent. This formula shows how many ways elements can be related between two sets.
Given: Number of relations \( = 1024 \)
Set \( A = \{1, 2, 3, 4, 5\} \implies n(A) = 5 \)
Let \( n(B) = x \)
The total number of relations from A to B is \( 2^{n(A) \times n(B)} \).
So, \( 2^{5 \times x} = 1024 \)
We know that \( 1024 = 2^{10} \)
So, \( 2^{5x} = 2^{10} \)
Since the bases are the same, we can equate the exponents:
\( 5x = 10 \)
\( x = \frac{10}{5} \)
\( x = 2 \)
Thus, the number of elements in set B is 2.
In simple words: The total number of ways to connect items from set A to set B is found by taking 2 and raising it to the power of (number of items in A multiplied by number of items in B). We know the total connections (1024) and the number of items in A (5). We need to find out what number, when multiplied by 5 and then used as an exponent for 2, gives 1024. That number is 2.

🎯 Exam Tip: Remember that the number of relations from set A to set B is always \( 2^{n(A) \times n(B)} \). This is a common formula in set theory.

 

Question 5. The range of the relation R = {(x, x²) a prime number less than 13} is
(a) {2, 3, 5, 7}
(b) {2, 3, 5, 7, 11}
(c) {4, 9, 25, 49, 121}
(d) {1, 4, 9, 25, 49, 121}
Answer: (c) {4, 9, 25, 49, 121}
Hint: First, identify all prime numbers that are less than 13. These numbers will be the 'x' values in our relation. Then, for each prime number 'x', calculate its square 'x²'. The set of all these 'x²' values will form the range of the relation. The range represents all the output values from the relation.
Prime numbers less than 13 are: \( \{2, 3, 5, 7, 11\} \)
The relation R is defined as \( \{(x, x^2)\} \). The range consists of the second elements of these ordered pairs, which are the squares of the prime numbers.
For \( x = 2, x^2 = 2^2 = 4 \)
For \( x = 3, x^2 = 3^2 = 9 \)
For \( x = 5, x^2 = 5^2 = 25 \)
For \( x = 7, x^2 = 7^2 = 49 \)
For \( x = 11, x^2 = 11^2 = 121 \)
Therefore, the range of R is \( \{4, 9, 25, 49, 121\} \). This shows how the range is derived from the domain elements through the given rule.
In simple words: First, list all the special numbers called "prime numbers" that are smaller than 13. Then, for each of these numbers, multiply it by itself (find its square). The list of all these squared numbers is the "range" of the relation.

🎯 Exam Tip: When finding the range, always make sure to correctly identify the domain (input) values first, then apply the given rule to each input to find its corresponding output (range) value.

 

Question 6. If the ordered pairs (a + 2, 4) and (5, 2a + b)are equal then (a, b) is
(a) (2, -2)
(b) (5, 1)
(c) (2, 3)
(d) (3, -2)
Answer: (d) (3, -2)
Hint: When two ordered pairs are equal, their corresponding components must be equal. This means the first component of the first pair must equal the first component of the second pair, and the second component of the first pair must equal the second component of the second pair. Set up two separate equations and solve them to find the values of 'a' and 'b'. This principle is key to understanding coordinate geometry.
Given that the ordered pairs \( (a + 2, 4) \) and \( (5, 2a + b) \) are equal.
Equate the first components:
\( a + 2 = 5 \)
\( a = 5 - 2 \)
\( a = 3 \)
Equate the second components:
\( 4 = 2a + b \)
Now substitute the value of \( a = 3 \) into the second equation:
\( 4 = 2(3) + b \)
\( 4 = 6 + b \)
\( b = 4 - 6 \)
\( b = -2 \)
So, the values are \( a = 3 \) and \( b = -2 \), which means the ordered pair \( (a, b) \) is \( (3, -2) \). This method provides a clear way to find unknown variables in equal ordered pairs.
In simple words: When two sets of numbers in brackets are exactly the same, it means the first number in the first set is equal to the first number in the second set. And the second number in the first set is equal to the second number in the second set. By solving these simple math puzzles, we find the hidden numbers 'a' and 'b'.

🎯 Exam Tip: Always set up two separate equations by equating the x-coordinates and y-coordinates respectively when two ordered pairs are given as equal.

 

Question 7. Let n(A) = m and n(B) = n then the total number of non-empty relations that can be defined from A to B is
(a) mn
(b) nm
(c) \( 2^{mn} - 1 \)
(d) \( 2^{mn} \)
Answer: (c) \( 2^{mn} - 1 \)
The total number of relations from set A to set B is \( 2^{n(A) \times n(B)} \), which is \( 2^{mn} \). This includes the empty relation. Since the question asks for non-empty relations, we subtract 1 (for the empty relation) from the total. A non-empty relation must contain at least one ordered pair.
In simple words: If set A has 'm' items and set B has 'n' items, the total number of all possible ways to connect items from A to B is \( 2^{mn} \). If we only want connections that are *not* empty, we just take away 1 from this total.

🎯 Exam Tip: Distinguish between total relations (\( 2^{mn} \)) and non-empty relations (\( 2^{mn} - 1 \)); the empty set is a valid relation but is excluded when "non-empty" is specified.

 

Question 8. If {(a, 8), (6, b)}represents an identity function, then the value of a and b are respectively
(a) (8, 6)
(b) (8, 8)
(c) (6, 8)
(d) (6, 6)
Answer: (a) (8, 6)
Hint: An identity function maps each element to itself. This means that for any ordered pair (x, y) in an identity function, x must be equal to y. We apply this rule to the given ordered pairs to find the values of 'a' and 'b'. Understanding identity functions is key to many mathematical concepts.
Given that the set of ordered pairs \( \{(a, 8), (6, b)\} \) represents an identity function.
For an identity function, each element maps to itself.
So, for the pair \( (a, 8) \), we must have \( a = 8 \).
And for the pair \( (6, b) \), we must have \( b = 6 \).
Therefore, the values of \( a \) and \( b \) are 8 and 6 respectively. This demonstrates the direct mapping property of an identity function.
In simple words: An "identity function" is like a mirror – whatever number you put in, you get the exact same number out. So, if we have pairs like (a, 8) and (6, b), it means 'a' must be 8 and 'b' must be 6 for the function to be an identity function.

🎯 Exam Tip: Remember that for an identity function, the input always equals the output, i.e., \( f(x) = x \) or for any ordered pair \( (x, y) \), \( x=y \).

 

Question 9. A function f: A → B given by f = {(1, 4), (2, 8),(3,9),(4,10)} is a
(a) Many-one function
(b) Identity function
(c) One-to-one function
(d) Into function
Answer: (c) One-to-one function
Hint: A function is one-to-one if every distinct element in the domain (set A) maps to a distinct element in the codomain (set B). In simpler terms, no two different inputs produce the same output. Examine the given ordered pairs to see if any two different first elements (inputs) share the same second element (output). If all second elements are unique for unique first elements, it's a one-to-one function. This property is crucial for inverse functions.
The given function is \( f = \{(1, 4), (2, 8), (3, 9), (4, 10)\} \).
Let's look at the mapping from set A to set B:

In this function, different elements from set A (the domain) are mapped to different elements in set B (the codomain). For example, 1 goes to 4, 2 goes to 8, 3 goes to 9, and 4 goes to 10. No two domain elements share the same image. Therefore, this is a one-to-one function. Each input has a unique output.
In simple words: A "one-to-one function" is like a special pairing where every different starting number goes to a different ending number. No two different starting numbers ever lead to the same ending number. In this question, each input (1, 2, 3, 4) produces a unique output (4, 8, 9, 10), making it a one-to-one function.

🎯 Exam Tip: To check for a one-to-one function, make sure that for any two distinct elements \( x_1 \ne x_2 \) in the domain, their images are also distinct, i.e., \( f(x_1) \ne f(x_2) \).

 

Question 10. If f (x) = \( 2x^2 \) and g(x) = \( \frac { 1 }{ 3x } \), then fog is
(a) \( \frac{3}{2 x^{2}} \)
(b) \( \frac{2}{3 x^{2}} \)
(c) \( \frac{2}{9 x^{2}} \)
(d) \( \frac{1}{6 x^{2}} \)
Answer: (c) \( \frac{2}{9 x^{2}} \)
Hint: To find the composite function fog (f of g of x), we substitute the entire function g(x) into f(x) wherever 'x' appears in f(x). This means we replace 'x' in \( f(x) = 2x^2 \) with \( \frac{1}{3x} \) and then simplify the expression. Function composition combines two functions into a new one. Ensure you simplify correctly after substitution.
Given: \( f(x) = 2x^2 \) and \( g(x) = \frac{1}{3x} \)
We need to find \( fog(x) \), which is \( f[g(x)] \).
Substitute \( g(x) \) into \( f(x) \):
\( fog(x) = f\left(\frac{1}{3x}\right) \)
Now, replace 'x' in \( f(x) = 2x^2 \) with \( \frac{1}{3x} \):
\( f\left(\frac{1}{3x}\right) = 2 \times \left(\frac{1}{3x}\right)^2 \)
\( = 2 \times \left(\frac{1^2}{(3x)^2}\right) \)
\( = 2 \times \left(\frac{1}{9x^2}\right) \)
\( = \frac{2}{9x^2} \)
Thus, the composite function \( fog(x) \) is \( \frac{2}{9x^2} \). This process shows how one function's output becomes another function's input.
In simple words: To find "fog" (f of g), you take the entire 'g(x)' function and put it into 'f(x)' everywhere you see 'x'. Here, 'g(x)' is \( \frac{1}{3x} \). So, we replace 'x' in \( 2x^2 \) with \( \frac{1}{3x} \) and then do the math to get the final answer.

🎯 Exam Tip: When dealing with function composition \( f(g(x)) \), remember to substitute the entire expression of \( g(x) \) into \( f(x) \) and then simplify carefully, especially with squares or fractions.

 

Question 11. If f: A \( \rightarrow \) B is a bijective function and if n(B) = 7, then n(A) is equal to
(a) 7
(b) 49
(c) 1
(d) 14
Answer: (a) 7
Hint: A bijective function (also known as a one-to-one correspondence) is a function that is both injective (one-to-one) and surjective (onto). A key property of bijective functions between finite sets is that the number of elements in the domain (n(A)) must be equal to the number of elements in the codomain (n(B)). This ensures every element in both sets is uniquely paired. This property is fundamental in set theory and function mapping.
Given: f: A \( \rightarrow \) B is a bijective function.
\( n(B) = 7 \)
For a function to be bijective between two finite sets, the number of elements in the domain must be equal to the number of elements in the codomain.
So, \( n(A) = n(B) \).
Since \( n(B) = 7 \), it means \( n(A) = 7 \).
This equality ensures a perfect pairing of elements between the two sets.
In simple words: A "bijective function" is a special kind of connection between two sets where every item in the first set is linked to exactly one item in the second set, and every item in the second set is linked to exactly one item in the first set. This means both sets must have the same number of items. So, if set B has 7 items, set A must also have 7 items.

🎯 Exam Tip: Always remember that for a bijective function between finite sets, the number of elements in the domain and codomain must be equal. This is a direct definition of bijectivity for finite sets.

 

Question 12. Let f and g be two functions given by f = {(0,1),(2, 0),(3,-4),(4,2),(5,7)} g = {(0,2),(1,0),(2,4),(-4,2),(7,0)} then the range of f og is
(a) {-4,1,0,2,7}
(b) {1,2,3,4,5}
(c) {0,1,2}
(d) None of the options
Answer: (c) {0,1,2}
Hint: To find the range of the composite function fog (f of g), we first need to determine the function fog itself. This involves finding \( g(x) \) for each x in the domain of g, and then using that \( g(x) \) as the input for f. The set of all resulting output values from f will be the range of fog. Make sure that \( g(x) \) is in the domain of f. The range is simply the collection of all possible output values.
Given the functions:
\( f = \{(0,1),(2, 0),(3,-4),(4,2),(5,7)\} \)
\( g = \{(0,2),(1,0),(2,4),(-4,2),(7,0)\} \)
To find \( fog(x) = f[g(x)] \), we need to check values of \( x \) for which \( g(x) \) is defined, and \( g(x) \) is in the domain of \( f \).
The domain of \( g \) is \( \{0, 1, 2, -4, 7\} \).
The domain of \( f \) is \( \{0, 2, 3, 4, 5\} \).
Let's evaluate \( f[g(x)] \) for each \( x \) in the domain of \( g \):
1. For \( x = 0 \): \( g(0) = 2 \). Since 2 is in the domain of \( f \), we find \( f(2) = 0 \). So, \( (0, 0) \) is in fog.
2. For \( x = 1 \): \( g(1) = 0 \). Since 0 is in the domain of \( f \), we find \( f(0) = 1 \). So, \( (1, 1) \) is in fog.
3. For \( x = 2 \): \( g(2) = 4 \). Since 4 is in the domain of \( f \), we find \( f(4) = 2 \). So, \( (2, 2) \) is in fog.
4. For \( x = -4 \): \( g(-4) = 2 \). Since 2 is in the domain of \( f \), we find \( f(2) = 0 \). So, \( (-4, 0) \) is in fog.
5. For \( x = 7 \): \( g(7) = 0 \). Since 0 is in the domain of \( f \), we find \( f(0) = 1 \). So, \( (7, 1) \) is in fog.
The composite function is \( fog = \{(0, 0), (1, 1), (2, 2), (-4, 0), (7, 1)\} \).
The range of \( fog \) is the set of all second elements of these ordered pairs: \( \{0, 1, 2\} \). This example clearly illustrates the step-by-step process of function composition and range identification.
In simple words: To find the range of "f of g," we first find the output of 'g' for each input. Then, we use these outputs as new inputs for 'f' to find the final outputs. The collection of all these final outputs is the range. Here, we calculate \( f(g(x)) \) for each x in the domain of g and collect all the y-values.

🎯 Exam Tip: When determining the domain and range of a composite function \( f(g(x)) \), first ensure that the output of \( g(x) \) falls within the domain of \( f \). The range consists of all possible final output values.

 

Question 13. Let f(x) = \( \sqrt{1+x^{2}} \) then
(a) f(xy) = f(x),f(y)
(b) f(xy) \( \geq \) f(x),f(y)
(c) f(xy) \( \leq \) f(x).f(y)
(d) None of the options
Answer: (c) f(xy) \( \leq \) f(x).f(y)
Hint: This question tests the property of the given function \( f(x) = \sqrt{1+x^{2}} \) with respect to multiplication. We need to compare \( f(xy) \) with \( f(x) \cdot f(y) \). By substituting \( xy \) into \( f(x) \) and comparing it with the product of \( f(x) \) and \( f(y) \), we can determine the correct inequality. This involves algebraic manipulation and understanding of square root properties.
Given: \( f(x) = \sqrt{1+x^{2}} \)
Let's find \( f(xy) \):
\( f(xy) = \sqrt{1+(xy)^2} = \sqrt{1+x^2 y^2} \)
Now, let's find \( f(x) \cdot f(y) \):
\( f(x) \cdot f(y) = \sqrt{1+x^2} \cdot \sqrt{1+y^2} = \sqrt{(1+x^2)(1+y^2)} \)
\( = \sqrt{1+y^2+x^2+x^2 y^2} \)
Comparing \( \sqrt{1+x^2 y^2} \) and \( \sqrt{1+y^2+x^2+x^2 y^2} \).
Since \( x^2 \geq 0 \) and \( y^2 \geq 0 \), it means \( x^2+y^2 \geq 0 \).
Therefore, \( 1+x^2 y^2 \leq 1+y^2+x^2+x^2 y^2 \).
Taking the square root of both sides (since all terms are non-negative):
\( \sqrt{1+x^2 y^2} \leq \sqrt{1+y^2+x^2+x^2 y^2} \)
This implies:
\( f(xy) \leq f(x) \cdot f(y) \). This inequality highlights a specific characteristic of the function when dealing with products.
In simple words: We are comparing what happens when we put 'xy' into the function with what happens when we multiply the results of putting 'x' and 'y' separately into the function. It turns out that putting 'xy' in gives a result that is less than or equal to multiplying the results of 'x' and 'y' separately.

🎯 Exam Tip: When comparing function properties involving products or sums, substitute the expressions carefully and simplify to determine the correct inequality or equality. Pay attention to the domain of variables, especially when dealing with square roots.

 

Question 14. If g= {(1,1),(2,3),(3,5),(4,7)} is a function given by g(x) = \( \alpha x + \beta \) then the values of \( \alpha \) and \( \beta \) are
(a) (-1,2)
(b) (2,-1)
(c) (-1,-2)
(d) (1,2)
Answer: (b) (2, -1)
Hint: Since \( g(x) = \alpha x + \beta \) represents a linear function, we can use any two ordered pairs from the given set of points to form a system of two linear equations with two variables (\( \alpha \) and \( \beta \)). By solving these equations simultaneously, we can find the unique values for \( \alpha \) and \( \beta \). This method is widely used to find the equation of a line passing through given points.
Given the function \( g = \{(1,1), (2,3), (3,5), (4,7)\} \) and \( g(x) = \alpha x + \beta \).
We can use any two points from the function to form equations.
Using the point \( (1,1) \):
\( g(1) = \alpha(1) + \beta \)
\( 1 = \alpha + \beta \) ... (1)
Using the point \( (2,3) \):
\( g(2) = \alpha(2) + \beta \)
\( 3 = 2\alpha + \beta \) ... (2)
Now, we solve these two linear equations simultaneously. Subtract equation (1) from equation (2):
\( (2\alpha + \beta) - (\alpha + \beta) = 3 - 1 \)
\( 2\alpha - \alpha + \beta - \beta = 2 \)
\( \alpha = 2 \)
Substitute the value of \( \alpha = 2 \) into equation (1):
\( 1 = 2 + \beta \)
\( \beta = 1 - 2 \)
\( \beta = -1 \)
So, the values are \( \alpha = 2 \) and \( \beta = -1 \). Thus, \( (\alpha, \beta) \) is \( (2, -1) \). This method is a standard way to determine the parameters of a linear function.
In simple words: We are given some points that follow a rule \( g(x) = \alpha x + \beta \). This rule is like finding a straight line. We pick two of the given points and use them to create two simple math problems (equations). By solving these two problems together, we find the values of \( \alpha \) and \( \beta \) that make the rule work for all the points.

🎯 Exam Tip: When given a set of points and a linear function form like \( y = mx + c \) or \( g(x) = \alpha x + \beta \), use any two distinct points to create a system of two linear equations and solve for the unknown coefficients.

 

Question 15. f(x) = \( (x + 1)^3 - (x - 1)^3 \) represents a function which is
(a) linear
(b) cubic
(c) reciprocal
(d) quadratic
Answer: (d) quadratic
Hint: To determine the type of function, we need to expand the given expression and simplify it. Use the cubic expansion formulas \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) and \( (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \). After expanding and combining like terms, observe the highest power of 'x' remaining in the simplified expression. This highest power dictates the type of polynomial function. This method is crucial for classifying polynomial functions.
Given function: \( f(x) = (x + 1)^3 - (x - 1)^3 \)
Expand the cubic terms using the binomial expansion formulas:
\( (x + 1)^3 = x^3 + 3x^2(1) + 3x(1)^2 + 1^3 = x^3 + 3x^2 + 3x + 1 \)
\( (x - 1)^3 = x^3 - 3x^2(1) + 3x(1)^2 - 1^3 = x^3 - 3x^2 + 3x - 1 \)
Now, substitute these expanded forms back into the function \( f(x) \):
\( f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1) \)
Distribute the negative sign to the second parenthesis:
\( f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 \)
Combine like terms:
\( f(x) = (x^3 - x^3) + (3x^2 + 3x^2) + (3x - 3x) + (1 + 1) \)
\( f(x) = 0 + 6x^2 + 0 + 2 \)
\( f(x) = 6x^2 + 2 \)
The highest power of 'x' in the simplified expression is 2. Therefore, \( f(x) \) represents a quadratic function. This shows how complex expressions can simplify into standard forms.
In simple words: To figure out what kind of function this is, we need to expand both parts with the power of 3 and then subtract them. When we do all the math, we will see that the highest power of 'x' remaining in the final answer is 2. Any function where the highest power of 'x' is 2 is called a "quadratic function."

🎯 Exam Tip: Always expand and simplify complex polynomial expressions to their standard form (e.g., \( ax^2+bx+c \) for quadratic) to correctly identify the type of function based on the highest power of the variable.