Samacheer Kalvi Class 10 Maths Solutions Chapter 1 Relations and Functions Exercise 1.5

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations And Functions Ex 1.5

Question 1. Using the functions f and g given below, find fog and gof Check whether fog = gof.
(i) \( f(x) = x - 6 \), \( g(x) = x^2 \)
Answer:
Given functions are \( f(x) = x - 6 \) and \( g(x) = x^2 \).
First, find \( fog \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(x^2) \) (since \( g(x) = x^2 \))
\( fog(x) = x^2 - 6 \) (substitute \( x^2 \) into \( f(x) \))
Next, find \( gof \):
\( gof(x) = g(f(x)) \)
\( gof(x) = g(x - 6) \) (since \( f(x) = x - 6 \))
\( gof(x) = (x - 6)^2 \) (substitute \( x - 6 \) into \( g(x) \))
\( gof(x) = x^2 - 12x + 36 \) (expand the square)
Comparing \( fog(x) \) and \( gof(x) \):
\( x^2 - 6 \neq x^2 - 12x + 36 \)
So, \( fog \neq gof \). This shows that the order of function composition matters.
In simple words: We found two new functions by combining f and g in two different orders. When we applied f first and then g, we got one result. When we applied g first and then f, we got a different result, meaning they are not the same.

🎯 Exam Tip: When finding composite functions, always remember to substitute the inner function completely into the outer function's variable. Pay close attention to the order of operations.

Question 1. (ii) \( f(x) = \frac{2}{x} \), \( g(x) = 2x^2 - 1 \)
Answer:
Given functions are \( f(x) = \frac{2}{x} \) and \( g(x) = 2x^2 - 1 \).
First, find \( fog \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(2x^2 - 1) \) (substitute \( g(x) \))
\( fog(x) = \frac{2}{2x^2 - 1} \) (substitute \( 2x^2 - 1 \) into \( f(x) \))
Next, find \( gof \):
\( gof(x) = g(f(x)) \)
\( gof(x) = g(\frac{2}{x}) \) (substitute \( f(x) \))
\( gof(x) = 2(\frac{2}{x})^2 - 1 \) (substitute \( \frac{2}{x} \) into \( g(x) \))
\( gof(x) = 2 \times \frac{4}{x^2} - 1 \)
\( gof(x) = \frac{8}{x^2} - 1 \)
Comparing \( fog(x) \) and \( gof(x) \):
\( \frac{2}{2x^2 - 1} \neq \frac{8}{x^2} - 1 \)
So, \( fog \neq gof \). This again shows that the order of combining functions changes the final output.
In simple words: We combined the given functions in two ways: f then g, and g then f. The results were different, which means these two combinations are not the same.

🎯 Exam Tip: When dealing with rational functions in composition, be careful with the domain. Ensure that the denominators do not become zero for the values being substituted.

Question 1. (iii) \( f(x) = \frac{x+6}{3} \), \( g(x) = 3 - x \)
Answer:
Given functions are \( f(x) = \frac{x+6}{3} \) and \( g(x) = 3 - x \).
First, find \( fog \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(3 - x) \) (substitute \( g(x) \))
\( fog(x) = \frac{(3 - x) + 6}{3} \) (substitute \( 3 - x \) into \( f(x) \))
\( fog(x) = \frac{9 - x}{3} \)
Next, find \( gof \):
\( gof(x) = g(f(x)) \)
\( gof(x) = g(\frac{x+6}{3}) \) (substitute \( f(x) \))
\( gof(x) = 3 - (\frac{x+6}{3}) \) (substitute \( \frac{x+6}{3} \) into \( g(x) \))
\( gof(x) = \frac{9 - (x+6)}{3} \)
\( gof(x) = \frac{9 - x - 6}{3} \)
\( gof(x) = \frac{3 - x}{3} \)
Comparing \( fog(x) \) and \( gof(x) \):
\( \frac{9 - x}{3} \neq \frac{3 - x}{3} \)
So, \( fog \neq gof \). This clearly shows that function composition is generally not commutative.
In simple words: When we combined the functions in two different ways, the final mathematical expressions were not the same. This means the order in which you apply the functions matters.

🎯 Exam Tip: Be careful with signs when subtracting composite expressions. A common error is forgetting to distribute the negative sign to all terms within the parentheses.

Question 1. (iv) \( f(x) = 3 + x \), \( g(x) = x - 4 \)
Answer:
Given functions are \( f(x) = 3 + x \) and \( g(x) = x - 4 \).
First, find \( fog \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(x - 4) \) (substitute \( g(x) \))
\( fog(x) = 3 + (x - 4) \) (substitute \( x - 4 \) into \( f(x) \))
\( fog(x) = x - 1 \)
Next, find \( gof \):
\( gof(x) = g(f(x)) \)
\( gof(x) = g(3 + x) \) (substitute \( f(x) \))
\( gof(x) = (3 + x) - 4 \) (substitute \( 3 + x \) into \( g(x) \))
\( gof(x) = x - 1 \)
Comparing \( fog(x) \) and \( gof(x) \):
\( x - 1 = x - 1 \)
So, \( fog = gof \). In this specific case, the order of composition does not affect the final function.
In simple words: For these two functions, combining them in either order gave us the exact same new function. This means the result is the same whether you apply f then g, or g then f.

🎯 Exam Tip: Always simplify the expressions for both \( fog(x) \) and \( gof(x) \) fully before comparing them to see if they are equal. Simplification helps to reveal the true relationship between the functions.

Question 1. (v) \( f(x) = 4x^2 - 1 \), \( g(x) = 1 + x \)
Answer:
Given functions are \( f(x) = 4x^2 - 1 \) and \( g(x) = 1 + x \).
First, find \( fog \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(1 + x) \) (substitute \( g(x) \))
\( fog(x) = 4(1 + x)^2 - 1 \) (substitute \( 1 + x \) into \( f(x) \))
\( fog(x) = 4(1 + 2x + x^2) - 1 \) (expand the square)
\( fog(x) = 4 + 8x + 4x^2 - 1 \)
\( fog(x) = 4x^2 + 8x + 3 \)
Next, find \( gof \):
\( gof(x) = g(f(x)) \)
\( gof(x) = g(4x^2 - 1) \) (substitute \( f(x) \))
\( gof(x) = 1 + (4x^2 - 1) \) (substitute \( 4x^2 - 1 \) into \( g(x) \))
\( gof(x) = 4x^2 \)
Comparing \( fog(x) \) and \( gof(x) \):
\( 4x^2 + 8x + 3 \neq 4x^2 \)
So, \( fog \neq gof \). This further reinforces that function composition is generally not commutative.
In simple words: We combined the two functions in both possible orders. The final expressions we got were different, meaning the sequence in which you apply the functions changes the outcome.

🎯 Exam Tip: Remember the formula for \( (a+b)^2 = a^2 + 2ab + b^2 \) when expanding squared terms in composite functions to avoid calculation errors.

Question 2. Find the value of k, such that fog = gof
(i) \( f(x) = 3x + 2 \), \( g(x) = 6x - k \)
Answer:
Given functions are \( f(x) = 3x + 2 \) and \( g(x) = 6x - k \). We need to find \( k \) such that \( fog = gof \).
First, find \( fog(x) \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(6x - k) \)
\( fog(x) = 3(6x - k) + 2 \)
\( fog(x) = 18x - 3k + 2 \) ... (1)
Next, find \( gof(x) \):
\( gof(x) = g(f(x)) \)
\( gof(x) = g(3x + 2) \)
\( gof(x) = 6(3x + 2) - k \)
\( gof(x) = 18x + 12 - k \) ... (2)
For \( fog = gof \), we set equation (1) equal to equation (2):
\( 18x - 3k + 2 = 18x + 12 - k \)
Subtract \( 18x \) from both sides:
\( -3k + 2 = 12 - k \)
Add \( 3k \) to both sides:
\( 2 = 12 + 2k \)
Subtract \( 12 \) from both sides:
\( -10 = 2k \)
Divide by 2:
\( k = -5 \). This value ensures that the order of function composition does not change the result.
In simple words: We combined the functions in two ways, f then g, and g then f. We set these two results equal to each other to find the special number 'k' that makes them the same. After solving, we found k is -5.

🎯 Exam Tip: When solving for an unknown variable like \( k \), ensure all algebraic steps are performed carefully, especially when isolating the variable. Double-check your distribution and sign changes.

Question 2. (ii) \( f(x) = 2x - k \), \( g(x) = 4x + 5 \)
Answer:
Given functions are \( f(x) = 2x - k \) and \( g(x) = 4x + 5 \). We need to find \( k \) such that \( fog = gof \).
First, find \( fog(x) \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(4x + 5) \)
\( fog(x) = 2(4x + 5) - k \)
\( fog(x) = 8x + 10 - k \) ... (1)
Next, find \( gof(x) \):
\( gof(x) = g(f(x)) \)
\( gof(x) = g(2x - k) \)
\( gof(x) = 4(2x - k) + 5 \)
\( gof(x) = 8x - 4k + 5 \) ... (2)
For \( fog = gof \), we set equation (1) equal to equation (2):
\( 8x + 10 - k = 8x - 4k + 5 \)
Subtract \( 8x \) from both sides:
\( 10 - k = -4k + 5 \)
Add \( 4k \) to both sides:
\( 10 + 3k = 5 \)
Subtract \( 10 \) from both sides:
\( 3k = -5 \)
Divide by 3:
\( k = -\frac{5}{3} \). This fractional value of k also ensures the composite functions are equal.
In simple words: We set the two ways of combining the functions equal to each other to find the unknown value 'k'. Solving the equation, we found that k is minus five-thirds.

🎯 Exam Tip: Be careful with signs when moving terms across the equality sign. It is crucial for isolating the variable correctly and avoiding mistakes in the final answer.

Question 3. If \( f(x) = 2x - 1 \), \( g(x) = \frac{x+1}{2} \), show that fog = gof = x
Answer:
Given functions are \( f(x) = 2x - 1 \) and \( g(x) = \frac{x+1}{2} \). We need to show that \( fog = gof = x \).
First, find \( fog(x) \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(\frac{x+1}{2}) \) (substitute \( g(x) \))
\( fog(x) = 2(\frac{x+1}{2}) - 1 \) (substitute \( \frac{x+1}{2} \) into \( f(x) \))
\( fog(x) = (x+1) - 1 \)
\( fog(x) = x \)
Next, find \( gof(x) \):
\( gof(x) = g(f(x)) \)
\( gof(x) = g(2x - 1) \) (substitute \( f(x) \))
\( gof(x) = \frac{(2x - 1) + 1}{2} \) (substitute \( 2x - 1 \) into \( g(x) \))
\( gof(x) = \frac{2x}{2} \)
\( gof(x) = x \)
Since both \( fog(x) \) and \( gof(x) \) simplify to \( x \), we have shown that \( fog = gof = x \). This means that \( f \) and \( g \) are inverse functions of each other.
In simple words: When we combined these two functions in any order, the result was simply 'x'. This proves they are inverses, meaning one function undoes what the other one does.

🎯 Exam Tip: When asked to 'show that', ensure your steps clearly lead to the stated conclusion. Each algebraic manipulation should be precise and easy to follow for the examiner.

Question 4. (i) If \( f(x) = x^2 - 1 \), \( g(x) = x - 2 \) find a, if gof(a) = 1.
Answer:
Given functions are \( f(x) = x^2 - 1 \) and \( g(x) = x - 2 \). We are also given \( gof(a) = 1 \).
First, find the general composite function \( gof(x) \):
\( gof(x) = g(f(x)) \)
\( gof(x) = g(x^2 - 1) \) (substitute \( f(x) \))
\( gof(x) = (x^2 - 1) - 2 \) (substitute \( x^2 - 1 \) into \( g(x) \))
\( gof(x) = x^2 - 3 \)
Now, substitute \( x = a \) into \( gof(x) \) and set it equal to 1, as given:
\( gof(a) = a^2 - 3 \)
Since \( gof(a) = 1 \):
\( a^2 - 3 = 1 \)
Add 3 to both sides:
\( a^2 = 4 \)
Take the square root of both sides:
\( a = \pm 2 \). This means there are two possible values for 'a' that satisfy the condition.
In simple words: We first found a new function by combining g and f. Then, we put 'a' into this new function and set the result to 1. By solving the equation, we found that 'a' can be either 2 or -2.

🎯 Exam Tip: Remember that when taking the square root of a number to solve an equation, both positive and negative roots must be considered for the variable.

Question 4. (ii) Find k, if \( f(k) = 2k - 1 \) and fof(k) = 5.
Answer:
Given \( f(k) = 2k - 1 \) and \( fof(k) = 5 \). We need to find the value of \( k \).
We know that \( fof(k) = f(f(k)) \).
Substitute \( f(k) = 2k - 1 \) into the expression:
\( fof(k) = f(2k - 1) \)
Now, use the definition of \( f(x) \), replacing \( x \) with \( (2k - 1) \). Since \( f(x) = 2x - 1 \) (using the structure from \( f(k) \)), then:
\( f(2k - 1) = 2(2k - 1) - 1 \)
\( fof(k) = 4k - 2 - 1 \)
\( fof(k) = 4k - 3 \)
We are given that \( fof(k) = 5 \). So, set the expression equal to 5:
\( 4k - 3 = 5 \)
Add 3 to both sides:
\( 4k = 8 \)
Divide by 4:
\( k = 2 \). This value of k satisfies the condition for the self-composition of the function.
In simple words: We were given a function and told that applying it twice to 'k' gives the number 5. We used the function's rule to apply it twice and then solved for 'k', finding that k is 2.

🎯 Exam Tip: When evaluating \( fof(k) \), remember that you are applying the function \( f \) to the result of \( f(k) \). Substitute the inner function completely before applying the outer function.

Question 5. Let A,B,C \( \subseteq N \) and a function \( f: A \rightarrow B \) be defined by \( f(x) = 2x + 1 \) and \( g: B \rightarrow C \) be defined by \( g(x) = x^2 \). Find the range of fog and gof.
Answer:
Given functions are \( f(x) = 2x + 1 \) and \( g(x) = x^2 \), where the domain for \( x \) is natural numbers (\( N \)).
First, find \( fog(x) \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(x^2) \) (substitute \( g(x) \))
\( fog(x) = 2(x^2) + 1 \) (substitute \( x^2 \) into \( f(x) \))
\( fog(x) = 2x^2 + 1 \)
Since \( x \in N \), \( x^2 \) will also be a natural number (1, 4, 9, ...). Then \( 2x^2 \) will be an even natural number (2, 8, 18, ...). Adding 1 means \( 2x^2 + 1 \) will be an odd natural number (3, 9, 19, ...).
The range of \( fog \) is \( \{y \mid y = 2x^2 + 1, x \in N\} \).
Next, find \( gof(x) \):
\( gof(x) = g(f(x)) \)
\( gof(x) = g(2x + 1) \) (substitute \( f(x) \))
\( gof(x) = (2x + 1)^2 \) (substitute \( 2x + 1 \) into \( g(x) \))
Since \( x \in N \), \( 2x \) will be an even natural number. Adding 1 means \( 2x + 1 \) will be an odd natural number (3, 5, 7, ...). Squaring an odd number results in another odd number (9, 25, 49, ...).
The range of \( gof \) is \( \{y \mid y = (2x + 1)^2, x \in N\} \). These ranges describe all possible output values of the combined functions.
In simple words: We found the rules for two new combined functions. The 'range' tells us all the possible numbers these new functions can produce when we use natural numbers as inputs. For \( fog \), the outputs are numbers like 3, 9, 19. For \( gof \), the outputs are numbers like 9, 25, 49.

🎯 Exam Tip: To find the range of a composite function, first determine the composite function itself, then consider the domain of the input variable (in this case, natural numbers) to identify the set of all possible output values. Describing the properties of the output numbers (e.g., odd, even, perfect squares) helps define the range clearly.

Question 6. If \( f(x) = x^2 - 1 \). Find (i) fof(x), (ii) fofof(x).
Answer:
Given function is \( f(x) = x^2 - 1 \).
(i) Find \( fof(x) \):
\( fof(x) = f(f(x)) \)
\( fof(x) = f(x^2 - 1) \) (substitute \( f(x) \))
\( fof(x) = (x^2 - 1)^2 - 1 \) (substitute \( x^2 - 1 \) into \( f(x) \))
\( fof(x) = (x^4 - 2x^2 + 1) - 1 \) (expand the square)
\( fof(x) = x^4 - 2x^2 \). This is the result of composing the function with itself once.
(ii) Find \( fofof(x) \):
\( fofof(x) = f(fof(x)) \)
We already found \( fof(x) = x^4 - 2x^2 \). So, substitute this into \( f \):
\( fofof(x) = f(x^4 - 2x^2) \)
\( fofof(x) = (x^4 - 2x^2)^2 - 1 \) (substitute \( x^4 - 2x^2 \) into \( f(x) \))
\( fofof(x) = (x^4)^2 - 2(x^4)(2x^2) + (2x^2)^2 - 1 \) (expand the square \( (a-b)^2 \))
\( fofof(x) = x^8 - 4x^6 + 4x^4 - 1 \). This is the result of composing the function with itself twice.
In simple words: We took a function and combined it with itself once to get \( fof(x) \). Then we combined it with itself again (so three times in total) to get \( fofof(x) \). This involves substituting the output of one step back into the original function.

🎯 Exam Tip: When finding higher-order compositions like \( fofof(x) \), it's often easiest to calculate \( fof(x) \) first, and then use that result to find \( f(fof(x)) \), simplifying the process and reducing potential errors.

Question 7. If \( f : R \rightarrow R \) and \( g : R \rightarrow R \) are defined by \( f(x) = x^5 \) and \( g(x) = x^4 \) then check if f, g are one - one and fog is one - one?
Answer:
Given functions are \( f(x) = x^5 \) and \( g(x) = x^4 \), both mapping from real numbers to real numbers.
Let's check if \( f(x) \) is one-to-one:
If \( f(x_1) = f(x_2) \), then \( x_1^5 = x_2^5 \). Taking the fifth root of both sides gives \( x_1 = x_2 \).
Therefore, \( f(x) = x^5 \) is a one-to-one function. Each input has a unique output.
Let's check if \( g(x) \) is one-to-one:
If \( g(x_1) = g(x_2) \), then \( x_1^4 = x_2^4 \). This implies \( x_1 = \pm x_2 \). For example, \( g(1) = 1^4 = 1 \) and \( g(-1) = (-1)^4 = 1 \). Here, different inputs (1 and -1) give the same output (1).
Therefore, \( g(x) = x^4 \) is NOT a one-to-one function.
Now, let's find \( fog(x) \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(x^4) \) (substitute \( g(x) \))
\( fog(x) = (x^4)^5 \) (substitute \( x^4 \) into \( f(x) \))
\( fog(x) = x^{20} \)
Let's check if \( fog(x) = x^{20} \) is one-to-one:
If \( fog(x_1) = fog(x_2) \), then \( x_1^{20} = x_2^{20} \). This implies \( x_1 = \pm x_2 \). For example, \( fog(1) = 1^{20} = 1 \) and \( fog(-1) = (-1)^{20} = 1 \).
Therefore, \( fog(x) = x^{20} \) is also NOT a one-to-one function. For a function to be one-to-one, distinct inputs must always lead to distinct outputs.
In simple words: We checked if each function gives a unique answer for every unique input. Function f is unique, but function g is not (for example, 1 and -1 both give the same answer). When we combined them into fog, that new function was also not unique in its outputs.

🎯 Exam Tip: To check if a function is one-to-one, assume \( f(x_1) = f(x_2) \) and try to prove \( x_1 = x_2 \). If \( x_1 = \pm x_2 \) (or other multiple possibilities) is the result, the function is not one-to-one. For power functions like \( x^n \), if \( n \) is odd, it's one-to-one. If \( n \) is even, it's not one-to-one over real numbers.

Question 8. Consider the functions \( f(x), g(x), h(x) \) as given below. Show that \( (fog)oh = fo(goh) \) in each case.
(i) \( f(x) = x - 1 \), \( g(x) = 3x + 1 \) and \( h(x) = x^2 \)
Answer:
Given functions: \( f(x) = x - 1 \), \( g(x) = 3x + 1 \), \( h(x) = x^2 \). We need to show \( (fog)oh = fo(goh) \).
First, let's calculate the Left Hand Side (LHS): \( (fog)oh \).
Step 1: Find \( fog(x) \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(3x + 1) \)
\( fog(x) = (3x + 1) - 1 \)
\( fog(x) = 3x \)
Step 2: Find \( (fog)oh(x) \):
\( (fog)oh(x) = (fog)(h(x)) \)
\( (fog)oh(x) = (fog)(x^2) \) (substitute \( h(x) \))
\( (fog)oh(x) = 3(x^2) \) (substitute \( x^2 \) into \( fog(x) \))
\( (fog)oh(x) = 3x^2 \) ... (1)
Next, let's calculate the Right Hand Side (RHS): \( fo(goh) \).
Step 1: Find \( goh(x) \):
\( goh(x) = g(h(x)) \)
\( goh(x) = g(x^2) \)
\( goh(x) = 3(x^2) + 1 \)
\( goh(x) = 3x^2 + 1 \)
Step 2: Find \( fo(goh)(x) \):
\( fo(goh)(x) = f(goh(x)) \)
\( fo(goh)(x) = f(3x^2 + 1) \) (substitute \( goh(x) \))
\( fo(goh)(x) = (3x^2 + 1) - 1 \) (substitute \( 3x^2 + 1 \) into \( f(x) \))
\( fo(goh)(x) = 3x^2 \) ... (2)
Comparing (1) and (2), we see that \( (fog)oh(x) = fo(goh)(x) \). Both sides are equal to \( 3x^2 \). This demonstrates the associative property of function composition.
In simple words: We showed that when combining three functions, it doesn't matter if you combine the first two then combine with the third, or combine the last two then combine with the first. The final result is the same.

🎯 Exam Tip: The associative property of function composition \( (f \circ g) \circ h = f \circ (g \circ h) \) is a fundamental concept. Always calculate the inner composite function first, then substitute its result into the outer function.

Question 8. (ii) \( f(x) = x^2 \), \( g(x) = 2x \) and \( h(x) = x + 4 \)
Answer:
Given functions: \( f(x) = x^2 \), \( g(x) = 2x \), \( h(x) = x + 4 \). We need to show \( (fog)oh = fo(goh) \).
First, let's calculate the Left Hand Side (LHS): \( (fog)oh \).
Step 1: Find \( fog(x) \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(2x) \)
\( fog(x) = (2x)^2 \)
\( fog(x) = 4x^2 \)
Step 2: Find \( (fog)oh(x) \):
\( (fog)oh(x) = (fog)(h(x)) \)
\( (fog)oh(x) = (fog)(x + 4) \) (substitute \( h(x) \))
\( (fog)oh(x) = 4(x + 4)^2 \) (substitute \( x + 4 \) into \( fog(x) \))
\( (fog)oh(x) = 4(x^2 + 8x + 16) \) (expand the square)
\( (fog)oh(x) = 4x^2 + 32x + 64 \) ... (1)
Next, let's calculate the Right Hand Side (RHS): \( fo(goh) \).
Step 1: Find \( goh(x) \):
\( goh(x) = g(h(x)) \)
\( goh(x) = g(x + 4) \)
\( goh(x) = 2(x + 4) \)
\( goh(x) = 2x + 8 \)
Step 2: Find \( fo(goh)(x) \):
\( fo(goh)(x) = f(goh(x)) \)
\( fo(goh)(x) = f(2x + 8) \) (substitute \( goh(x) \))
\( fo(goh)(x) = (2x + 8)^2 \) (substitute \( 2x + 8 \) into \( f(x) \))
\( fo(goh)(x) = (2x)^2 + 2(2x)(8) + 8^2 \)
\( fo(goh)(x) = 4x^2 + 32x + 64 \) ... (2)
Comparing (1) and (2), we see that \( (fog)oh(x) = fo(goh)(x) \). Both sides are equal to \( 4x^2 + 32x + 64 \). This further confirms the associative property for these functions.
In simple words: We calculated the combination of three functions in two different ways. Both ways gave us the exact same complex function, proving that the order of grouping functions does not change the final combined function.

🎯 Exam Tip: Ensure that when squaring a binomial like \( (2x+8)^2 \), you correctly apply the formula \( (a+b)^2 = a^2 + 2ab + b^2 \) to all terms, not just \( a^2+b^2 \). Careful expansion is key to accuracy.

Question 8. (iii) \( f(x) = x - 4 \), \( g(x) = x^2 \) and \( h(x) = 3x - 5 \)
Answer:
Given functions: \( f(x) = x - 4 \), \( g(x) = x^2 \), \( h(x) = 3x - 5 \). We need to show \( (fog)oh = fo(goh) \).
First, let's calculate the Left Hand Side (LHS): \( (fog)oh \).
Step 1: Find \( fog(x) \):
\( fog(x) = f(g(x)) \)
\( fog(x) = f(x^2) \)
\( fog(x) = x^2 - 4 \)
Step 2: Find \( (fog)oh(x) \):
\( (fog)oh(x) = (fog)(h(x)) \)
\( (fog)oh(x) = (fog)(3x - 5) \) (substitute \( h(x) \))
\( (fog)oh(x) = (3x - 5)^2 - 4 \) (substitute \( 3x - 5 \) into \( fog(x) \))
\( (fog)oh(x) = (9x^2 - 30x + 25) - 4 \) (expand the square)
\( (fog)oh(x) = 9x^2 - 30x + 21 \) ... (1)
Next, let's calculate the Right Hand Side (RHS): \( fo(goh) \).
Step 1: Find \( goh(x) \):
\( goh(x) = g(h(x)) \)
\( goh(x) = g(3x - 5) \)
\( goh(x) = (3x - 5)^2 \)
\( goh(x) = 9x^2 - 30x + 25 \)
Step 2: Find \( fo(goh)(x) \):
\( fo(goh)(x) = f(goh(x)) \)
\( fo(goh)(x) = f(9x^2 - 30x + 25) \) (substitute \( goh(x) \))
\( fo(goh)(x) = (9x^2 - 30x + 25) - 4 \) (substitute \( 9x^2 - 30x + 25 \) into \( f(x) \))
\( fo(goh)(x) = 9x^2 - 30x + 21 \) ... (2)
Comparing (1) and (2), we see that \( (fog)oh(x) = fo(goh)(x) \). Both sides are equal to \( 9x^2 - 30x + 21 \). This final example confirms the associative property for function composition universally.
In simple words: This exercise confirms that when you combine three functions, how you group them does not change the final combined function. The result is the same regardless of the initial pairing.

🎯 Exam Tip: Always follow the order of operations carefully, especially when dealing with nested functions. Expanding squared binomials accurately is crucial to avoid common errors.

Question 9. Let \( f = \{(-1, 3), (0, -1), (2, -9)\} \) be a linear function from \( Z \) into \( Z \). Find \( f(x) \).
Answer:
A linear function can be written in the form \( f(x) = ax + b \).
We are given three points that the function passes through: \( (-1, 3) \), \( (0, -1) \), and \( (2, -9) \). We can use any two points to find \( a \) and \( b \). Using \( (0, -1) \) is the easiest.
Using the point \( (0, -1) \):
Substitute \( x = 0 \) and \( f(x) = -1 \) into \( f(x) = ax + b \):
\( f(0) = a(0) + b \)
\( -1 = 0 + b \)
\( b = -1 \)
Now that we have \( b = -1 \), use another point, for example \( (-1, 3) \):
Substitute \( x = -1 \) and \( f(x) = 3 \) into \( f(x) = ax + b \):
\( f(-1) = a(-1) + b \)
\( 3 = -a + b \)
Substitute \( b = -1 \) into this equation:
\( 3 = -a + (-1) \)
\( 3 = -a - 1 \)
Add 1 to both sides:
\( 4 = -a \)
Multiply by -1:
\( a = -4 \)
So, the linear function is \( f(x) = ax + b \), which becomes \( f(x) = -4x - 1 \). This function describes the relationship between the given input and output values.
In simple words: We know a linear function looks like \( f(x) = ax + b \). By using the given points, especially the one where x is 0, we found the values for 'a' and 'b'. This helped us write the exact rule for the function.

🎯 Exam Tip: When finding the equation of a linear function from points, always use the point where \( x=0 \) (the y-intercept) first to directly find the value of \( b \). This simplifies the process and avoids solving a system of two equations.

Question 10. In electrical circuit theory, a circuit \( C(t) \) is called a linear circuit if it satisfies the superposition principle given by \( C(at_1 + bt_2) = aC(t_1) + bC(t_2) \), where a,b are constants. Show that the circuit \( C(t) = 3t \) is linear.
Answer:
Given the function \( C(t) = 3t \). We need to show that it satisfies the superposition principle, which defines a linear circuit.
The superposition principle states: \( C(at_1 + bt_2) = aC(t_1) + bC(t_2) \).
Let's evaluate the Left Hand Side (LHS) of the principle:
\( C(at_1 + bt_2) = 3(at_1 + bt_2) \) (substituting \( at_1 + bt_2 \) into \( C(t) = 3t \))
\( C(at_1 + bt_2) = 3at_1 + 3bt_2 \)
Now, let's evaluate the Right Hand Side (RHS) of the principle:
We know that \( C(t_1) = 3t_1 \) and \( C(t_2) = 3t_2 \).
So, \( aC(t_1) + bC(t_2) = a(3t_1) + b(3t_2) \)
\( aC(t_1) + bC(t_2) = 3at_1 + 3bt_2 \)
Comparing the LHS and RHS, we see that \( 3at_1 + 3bt_2 = 3at_1 + 3bt_2 \).
Since \( C(at_1 + bt_2) = aC(t_1) + bC(t_2) \) is satisfied, the function \( C(t) = 3t \) obeys the superposition principle. Therefore, \( C(t) = 3t \) represents a linear circuit. Linear systems are fundamental in many areas of engineering and physics because they are easy to analyze.
In simple words: A circuit is 'linear' if it follows a special rule about adding things up. We checked if the given circuit \( C(t) = 3t \) follows this rule. Since it does, we proved it is a linear circuit.

🎯 Exam Tip: When proving linearity using the superposition principle, ensure you clearly calculate both sides of the equation \( C(at_1 + bt_2) = aC(t_1) + bC(t_2) \) independently and then show that they are equal. Do not skip any substitution steps.