GSEB Class 11 Statistics Solutions Chapter 8 Function

Section - A

Question 1.Which of the following statements is true?
(a) f: {1, 2, 3, 4} → {3, 4, 5}, the rule 'add 2 to the elements of domain' is not a function.
(b) f: A → B, A = {-2, -1, 0, 1, 2}; B = {0, 1, 2, 3, 4}; f(x) = x² is not a function.
(c) g : P → Q, P = {-1, 0, 1}; Q = \( \{-\frac{1}{3}, -1, 3\} \); g (x) = \( \frac{x+2}{x-2} \). then g is called function.
(d) g : {2, 3, 4, 5} → {-1, 0, 1) and g(x) = 4x - 3 is a function.
Answer: (a) f: {1, 2, 3, 4} → {3, 4, 5}, the rule 'add 2 to the elements of domain' is not a function.
In simple words: This statement correctly identifies a scenario where the given rule fails to define a function, as adding 2 to elements of the domain {1, 2, 3, 4} would result in images {3, 4, 5, 6}, and 6 is not in the codomain {3, 4, 5}.

🎯 Exam Tip: Understanding function definitions and identifying conditions that violate them (like images not being in the codomain) is crucial for multiple-choice questions.

Question 2.Which of the following statements is true for the range of function f: A B?
(a) f(A) = {f(x)|x ∈ A}
(b) It is not a co-domain or subset of co-domain.
Answer: (a) f(A) = {f(x)|x ∈ A}
In simple words: The range of a function consists of all actual output values (f(x)) obtained by applying the function to every element (x) in its domain.

🎯 Exam Tip: Clearly differentiate between the codomain (all possible output values) and the range (all actual output values) of a function. The range is always a subset of the codomain.

Question 3.Which of the following statements is true for the relation g : X → Y, X = {- 1, 0}; Y = {2, 4}; g (x) = 4 - 2x ?
(a) g is called a function.
(b) g is not a function.
(c) X is called function.
(d) Y is called function.
Answer: (b) g is not a function
In simple words: For a relation to be a function, every element in the domain must map to exactly one element in the codomain. Here, \(g(-1) = 4 - 2(-1) = 6\), but 6 is not in the codomain Y = {2, 4}, so g is not a function.

🎯 Exam Tip: When evaluating if a relation is a function, always check two conditions: (1) every element in the domain has an image, and (2) each element in the domain has exactly one image within the codomain.

Question 4.What is the type of function f: A → B, wherein, for two different values of domain their functional values are same?
(a) One-one function
(b) Many-one function
(c) One-many function
(d) Many-many function
Answer: (b) Many-one function
In simple words: A many-one function occurs when multiple distinct inputs from the domain produce the same output in the codomain.

🎯 Exam Tip: Remember the definitions: one-one (injective) means distinct inputs have distinct outputs; many-one means multiple inputs can map to the same output. One-many is not a valid function type.

Question 5.What is the type of function f: A → B, where each value of domain A has the same image in set B?
(a) Not a function
(b) One-one function
(c) Constant function
(d) Many-one function
Answer: (c) Constant function
In simple words: A constant function is one where every input from the domain maps to the exact same single output value in the codomain.

🎯 Exam Tip: Constant functions are a specific type of many-one function where all elements of the domain map to *one specific* element in the codomain. This is a common function type to recognize.

Question 6.Which of the following statements is true for a one-one function?
(b) For any two values of the domain, their images are same.
(c) For any two different values of the domain, their images are different.
(d) For each value of the domain, their images are same.
Answer: (c) For any two different values of the domain their images are different.
In simple words: In a one-one function, if you take any two distinct elements from the domain, their corresponding output values in the codomain will always be distinct as well.

🎯 Exam Tip: The core characteristic of a one-one (injective) function is that no two distinct elements of the domain map to the same element of the codomain. This is key for proving injectivity.

Question 7.What is the type of function f: Z – {0} → N and f(x) = x², x ∈ Z-{0}?
(a) One-one function
(b) Many-one function
(c) Constant function
(d) None of the given
Answer: (b) Many-one function
In simple words: For the function \(f(x) = x^2\) with domain Z – {0}, both positive and negative integers map to the same positive square (e.g., \(f(2) = 4\) and \(f(-2) = 4\)), making it a many-one function.

🎯 Exam Tip: When dealing with functions involving squaring, absolute values, or even powers, always check for input pairs (like \(x\) and \(-x\)) that might yield the same output, indicating a many-one function.

Question 8.Which of the following is a sufficient condition for two different functions to be equal?
(a) Domains of both the functions must be same.
(b) Ranges of both the functions must be same.
(c) (a) and (b)
(d) (a) or (b)
Answer: (c) (a) and (b)
In simple words: For two functions to be considered equal, they must have identical domains, identical codomains, and also produce the same output for every input value (meaning their ranges must also be the same).

🎯 Exam Tip: The equality of functions requires three conditions: identical domains, identical codomains, and the same rule mapping elements. Option (c) encapsulates the critical conditions related to the elements themselves.

Section - B

Answer the following questions in one sentence:

Question 1.Give the necessary condition for defining a function.
Answer:The essential requirement for defining a function is that both the domain and the codomain sets must be non-empty.
In simple words: A function needs a set of inputs (domain) and a set of possible outputs (codomain) that aren't empty for it to exist.

🎯 Exam Tip: Emphasize that a function always operates between two specific, non-empty sets, making their existence a fundamental prerequisite.

Question 2.f: A → B, A = {-3, -1, 1, 3}; \(f(x) = x^2\). Is f a function?
Answer:For the given mapping f: A → B, where A = {-3, -1, 1, 3} and B = {1, 0, 9}, the rule is \(f(x) = x^2\). Evaluating the function for each element in A:
\(f(-3) = (-3)^2 = 9\)
\(f(-1) = (-1)^2 = 1\)
\(f(1) = (1)^2 = 1\)
\(f(3) = (3)^2 = 9\) Since every element in A has a unique image in B, f is indeed a function.
In simple words: Yes, f is a function because each input from set A correctly maps to an output in set B according to the rule \(f(x) = x^2\).

🎯 Exam Tip: To verify if a relation is a function, ensure every element in the domain has exactly one image in the codomain. List out the mappings explicitly to confirm.

Question 3.g: N → N, 'subtract 2 from the elements of the domain'. Can this rule be called a function?
Answer:For the rule g: N → N, where the domain N = {1, 2, 3, ...}, and the rule is to 'subtract 2 from the elements of the domain', this relation cannot be considered a function. This is because if we take an element like 1 from the domain, \(g(1) = 1 - 2 = -1\). However, -1 is not an element of the natural numbers (N), which is the specified codomain. Therefore, not all elements in the domain map to an element within the codomain.
In simple words: No, this rule is not a function because subtracting 2 from some natural numbers (like 1) results in values that are not natural numbers, meaning the output isn't in the specified codomain.

🎯 Exam Tip: Always check if the images of all domain elements fall within the specified codomain. If even one image lies outside the codomain, the relation is not a function.

Question 4.Define a one-one function in notations.
Answer:Let f be a function from set A to set B, denoted as f: A → B. This function f is defined as one-one (or injective) if, for any two distinct elements \(x_1\) and \(x_2\) in the domain A, their corresponding images \(f(x_1)\) and \(f(x_2)\) are also distinct. Symbolically: If \(x_1 \neq x_2 \implies f(x_1) \neq f(x_2)\) for all \(x_1, x_2 \in A\). Equivalently, if \(f(x_1) = f(x_2) \implies x_1 = x_2\) for all \(x_1, x_2 \in A\).
In simple words: A one-one function means that every different input always gives a different output.

🎯 Exam Tip: The formal definition of a one-one function (injectivity) is crucial. Mastering its symbolic representation is essential for proving functions are one-one in problems.

Question 5.Define many-one function in notations.
Answer:Let f be a function from set A to set B, denoted as f: A → B. This function f is defined as many-one if there exist at least two distinct elements \(x_1\) and \(x_2\) in the domain A that map to the same image in the codomain B. Symbolically: There exist \(x_1, x_2 \in A\) such that \(x_1 \neq x_2\) but \(f(x_1) = f(x_2)\).
In simple words: A many-one function means that two or more different inputs can lead to the same output.

🎯 Exam Tip: To demonstrate a many-one function, it's sufficient to provide a single example where two different domain elements yield the same image. This contrasts with proving a function is one-one, which requires a general proof.

Question 6.Define constant function in notations.
Answer:Let f be a function from set A to set B, denoted as f: A → B. This function f is defined as a constant function if for all elements \(x_1, x_2, x_3, \ldots\) in the domain A, their images in the codomain B are identical. Symbolically: There exists a fixed element \(c \in B\) such that \(f(x) = c\) for all \(x \in A\). This implies that \(f(x_1) = f(x_2) = f(x_3) = \ldots = c\).
In simple words: A constant function always produces the exact same output value, no matter what input you give it from its domain.

🎯 Exam Tip: Constant functions are a special case of many-one functions where all domain elements map to *only one* specific element in the codomain. They are easily recognizable by their output rule.

Question 7.f: {1, 2, 3} → N, g : {2, 3, 4} → N,
f (x) = 2x + 1 and g (x) = x - 1. Can these two functions f and g be equal functions? Why?
Answer:For two functions to be equal, they must have the same domain, the same codomain, and the same rule of correspondence. Given functions: f: {1, 2, 3} → N with \(f(x) = 2x + 1\) g: {2, 3, 4} → N with \(g(x) = x - 1\) The domains of f and g are \(D_f = \{1, 2, 3\}\) and \(D_g = \{2, 3, 4\}\) respectively. Since \(D_f \neq D_g\), the domains of functions f and g are not identical. Therefore, functions f and g cannot be considered equal functions.
In simple words: No, functions f and g are not equal because their sets of allowed inputs (domains) are different.

🎯 Exam Tip: The first and most critical condition for function equality is identical domains. If domains differ, the functions are unequal, regardless of their rules or codomains.

Question 8.f: N → N, \(f(t) = t^2 + 1\), t ∈ N. Determine the type of function f.
Answer:Given the function f: N → N, defined by \(f(t) = t^2 + 1\), where the domain N = {1, 2, 3, 4, ...}. Let's evaluate the function for different values in the domain: \(f(1) = 1^2 + 1 = 1 + 1 = 2\) \(f(2) = 2^2 + 1 = 4 + 1 = 5\) \(f(3) = 3^2 + 1 = 9 + 1 = 10\) For any two distinct natural numbers \(t_1\) and \(t_2\), if \(t_1 \neq t_2\), then \(t_1^2 \neq t_2^2\), which implies \(t_1^2 + 1 \neq t_2^2 + 1\). Thus, \(f(t_1) \neq f(t_2)\) for \(t_1 \neq t_2\). Therefore, function f is a one-one function.
In simple words: This function is a one-one function because every unique natural number input produces a unique natural number output.

🎯 Exam Tip: When the domain is natural numbers (N) or positive integers, functions like \(x^2+c\) or \(ax+b\) are often one-one, as squaring distinct positive numbers yields distinct results. Always consider the specific domain given.

Question 9.f: N → N, \(f(t) = t^2 + 1\), t ∈ N. Determine the type of function f.
Answer:Given the function f: N → N, defined by \(f(t) = t^2 + 1\), where the domain N = {1, 2, 3, 4, ...} and the codomain is also N. Let's analyze the images for different elements in the domain: \(f(1) = 1^2 + 1 = 2\) \(f(2) = 2^2 + 1 = 5\) \(f(3) = 3^2 + 1 = 10\) For any distinct \(t_1, t_2 \in N\), if \(t_1 \neq t_2\), then \(t_1^2 \neq t_2^2\). Consequently, \(t_1^2 + 1 \neq t_2^2 + 1\). This means that distinct elements in the domain N map to distinct images in the codomain N. Therefore, function f is a one-one function.
In simple words: This function is one-one because each different natural number input always leads to a unique natural number output.

🎯 Exam Tip: Pay close attention to the domain and codomain when classifying function types. For \(f(x) = x^2+1\) with domain N, it is one-one. However, if the domain were Z (integers), it would be many-one (e.g., \(f(-1)=2\) and \(f(1)=2\)).

Question 10.Define a function of real variable.
Answer:A function of a real variable is a function where the domain is a subset of the set of real numbers (R) and the codomain is also the set of real numbers (R) or a subset thereof. Specifically, if a function f maps from A to B, where A is a subset of R (A ⊆ R), then f is called a function of a real variable.
In simple words: A real variable function is one that takes real numbers as inputs and gives real numbers as outputs.

🎯 Exam Tip: The key characteristic of a real variable function is that its domain and codomain are typically subsets of the real numbers, allowing for continuous graphs and calculations.

Section - C

Answer the following questions as required:

Question 1.Give definition of a function.
Answer:If A and B are any two non-empty sets, a function from set A to set B is a rule, relation, or correspondence that associates each element of set A with one and only one element of set B. It is typically denoted by f: A → B.
In simple words: A function is a rule that assigns exactly one output from a set B to every single input from a set A.

🎯 Exam Tip: The two critical components of a function's definition are: (1) every element in the domain must be mapped, and (2) each element in the domain must map to exactly one element in the codomain. Emphasize "one and only one."

Question 2.Define domain and co-domain of a function.
Answer:**Domain of a function:** For a function f: A → B, the set A, from which the input values are taken, is called the domain of the function. It is denoted by \(D_f\). **Co-domain of a function:** For a function f: A → B, the set B, which contains all the possible output values, is called the codomain of the function.
In simple words: The domain is the set of all possible inputs for a function, and the codomain is the set of all possible outputs.

🎯 Exam Tip: Differentiate domain (inputs) and codomain (all potential outputs) clearly. These are fundamental concepts for understanding function behavior and properties.

Question 3.Define range of a function.
Answer:Suppose f is a function from set A to set B (f: A → B). The range of function f is the set of all actual images or functional values that are produced when applying the function to all elements of set A. It is denoted by \(R_f\), and formally, \(R_f = \{f(x) | x \in A\}\). The range of a function can either be equal to the codomain itself or be a proper subset of the codomain.
In simple words: The range of a function is the collection of all the actual output values that the function generates from its inputs.

🎯 Exam Tip: The range is a subset of the codomain. Emphasize that it contains only the *actual* values produced by the function, not all *possible* values in the codomain.

Question 4.g: A → N, A = {x| x ∈ N, 1 < x ≤ 4}; g(x) = x + 1. Find range of function g.
Answer:Given the function g: A → N, where the domain A is defined as \(A = \{x | x \in N, 1 < x \leq 4\}\). Identifying the elements in set A: Since x is a natural number greater than 1 and less than or equal to 4, A = {2, 3, 4}. The rule for the function is \(g(x) = x + 1\). Now, we find the images for each element in the domain: \(g(2) = 2 + 1 = 3\) \(g(3) = 3 + 1 = 4\) \(g(4) = 4 + 1 = 5\) The range of function g, denoted \(R_g\), is the set of all these images. Therefore, \(R_g = \{g(x) | x \in A\} = \{g(2), g(3), g(4)\} = \{3, 4, 5\}\).
In simple words: For the given function, the domain includes 2, 3, and 4. Applying the rule \(g(x) = x+1\) to these gives outputs 3, 4, and 5, which form the range.

🎯 Exam Tip: To find the range, first correctly identify all elements in the domain from the given condition, then apply the function rule to each element to find its image. List all unique images to form the range.

Question 5.k:X → Y, X = {t| t ∈ Z, -3 < t ≤ 3}; Y = {a| a ∈ N, 1 ≤ a≤ 20}; k(t) = t²+2. State the type of function k.
Answer:Given the function k: X → Y, with: Domain \(X = \{t | t \in Z, -3 < t \leq 3\}\). The elements of X are \(\{-2, -1, 0, 1, 2, 3\}\). Codomain \(Y = \{a | a \in N, 1 \leq a \leq 20\}\). The elements of Y are \(\{1, 2, 3, \ldots, 20\}\). The function rule is \(k(t) = t^2 + 2\). Let's evaluate k(t) for each element in the domain X: \(k(-2) = (-2)^2 + 2 = 4 + 2 = 6\) \(k(-1) = (-1)^2 + 2 = 1 + 2 = 3\) \(k(0) = (0)^2 + 2 = 0 + 2 = 2\) \(k(1) = (1)^2 + 2 = 1 + 2 = 3\) \(k(2) = (2)^2 + 2 = 4 + 2 = 6\) \(k(3) = (3)^2 + 2 = 9 + 2 = 11\) Observe the images: \(k(-2) = 6\) and \(k(2) = 6\). Here, two different elements from the domain (-2 and 2) map to the same image (6). \(k(-1) = 3\) and \(k(1) = 3\). Here, two different elements from the domain (-1 and 1) map to the same image (3). Since distinct elements from the domain map to the same image in the codomain, the function k is a many-one function.
In simple words: The function k is a many-one function because different negative and positive inputs (like -2 and 2) result in the same output value.

🎯 Exam Tip: For functions involving \(t^2\) with an integer domain, always check for symmetric inputs like \(t\) and \(-t\). If \(f(t) = f(-t)\) for \(t \neq 0\), it's a many-one function.

Question 6.h: A → B, A = {1, 2, 3}, B = {3, 4, 5, 6, 7, 8}, h (x) = x + 5. State the type of function h.
Answer:Given the function h: A → B, with: Domain A = {1, 2, 3} Codomain B = {3, 4, 5, 6, 7, 8} The function rule is \(h(x) = x + 5\). Let's find the images for each element in the domain A: \(h(1) = 1 + 5 = 6\) \(h(2) = 2 + 5 = 7\) \(h(3) = 3 + 5 = 8\) For the elements 1, 2, 3 in the domain, their images (6, 7, 8) are distinct from each other. Also, all these images are present in the codomain B. Since distinct elements in the domain map to distinct images in the codomain, function h is a one-one function.
In simple words: Function h is a one-one function because each unique input from the domain A produces a unique output in the codomain B.

🎯 Exam Tip: A simple linear function like \(x+c\) is often one-one, especially over a discrete domain. Verify that each input yields a unique, distinct output to confirm injectivity.

Question 7.If P : A → B, P(x) = 2x - 3 and \(R_P = \{-2, -1, 0\}\), then find domain of the function.
Answer:Given the function P: A → B, with the rule \(P(x) = 2x - 3\). The range of the function is given as \(R_P = \{-2, -1, 0\}\). To find the domain A, we need to find the input values of x for which P(x) equals each value in the range. 1. Set \(P(x) = -2\): \(2x - 3 = -2\)
\(2x = -2 + 3\)
\(2x = 1\)
\(x = \frac{1}{2}\) 2. Set \(P(x) = -1\): \(2x - 3 = -1\)
\(2x = -1 + 3\)
\(2x = 2\)
\(x = 1\) 3. Set \(P(x) = 0\): \(2x - 3 = 0\)
\(2x = 3\)
\(x = \frac{3}{2}\) Therefore, the domain of the function P is the set of these x values: \(A = \{\frac{1}{2}, 1, \frac{3}{2}\}\).
In simple words: To find the domain, we set the function's rule equal to each value in the given range and solve for x. This gives us the inputs that produce those outputs.

🎯 Exam Tip: When given the range and the function rule, work backward to find the domain. Solve \(P(x) = y\) for x, where y takes on each value from the range.

Question 8.If \(f(x) = 1 - \frac{1}{1-x^2}\), \(x \in R - \{-1, 1\}\), then find \(f(2) - f(-2)\).
Answer:Given the function \(f(x) = 1 - \frac{1}{1-x^2}\), with domain \(x \in R - \{-1, 1\}\). First, calculate \(f(2)\): \(f(2) = 1 - \frac{1}{1-(2)^2} = 1 - \frac{1}{1-4} = 1 - \frac{1}{-3} = 1 + \frac{1}{3} = \frac{3+1}{3} = \frac{4}{3}\) Next, calculate \(f(-2)\): \(f(-2) = 1 - \frac{1}{1-(-2)^2} = 1 - \frac{1}{1-4} = 1 - \frac{1}{-3} = 1 + \frac{1}{3} = \frac{3+1}{3} = \frac{4}{3}\) Finally, find the difference \(f(2) - f(-2)\): \(f(2) - f(-2) = \frac{4}{3} - \frac{4}{3} = 0\)
In simple words: We calculate the function value for \(x=2\) and \(x=-2\) separately. Since both yield \(4/3\), their difference is 0.

🎯 Exam Tip: Be careful with signs, especially when squaring negative numbers. \( (-x)^2 = x^2 \). This property often leads to functions being many-one, and in this specific case, results in the same functional values for \(f(2)\) and \(f(-2)\).

Question 9.If domain of \(f(x) = \frac{x-3}{x+4}\) is (0, 3, 6) then find its range.
Answer:Given the function \(f(x) = \frac{x-3}{x+4}\) and its domain is \(\{0, 3, 6\}\). To find the range, we evaluate the function at each element in the domain: For \(x = 0\): \(f(0) = \frac{0-3}{0+4} = \frac{-3}{4}\) For \(x = 3\): \(f(3) = \frac{3-3}{3+4} = \frac{0}{7} = 0\) For \(x = 6\): \(f(6) = \frac{6-3}{6+4} = \frac{3}{10}\) The range \(R_f\) is the set of all these functional values. Therefore, \(R_f = \{-\frac{3}{4}, 0, \frac{3}{10}\}\).
In simple words: To find the range, simply substitute each value from the given domain into the function's formula and collect all the resulting output values.

🎯 Exam Tip: Remember that the range is a set of all *actual* output values. Calculate \(f(x)\) for each \(x\) in the domain and list the unique results in a set.

Question 10.If \(f(x) = \frac{x^2(x+1)^2}{4}\) is a real function then find the value of \(f(3) - f(2)\).
Answer:Given the function \(f(x) = \frac{x^2(x+1)^2}{4}\). First, calculate \(f(2)\): \(f(2) = \frac{(2)^2(2+1)^2}{4} = \frac{4(3)^2}{4} = \frac{4 \times 9}{4} = 9\) Next, calculate \(f(3)\): \(f(3) = \frac{(3)^2(3+1)^2}{4} = \frac{9(4)^2}{4} = \frac{9 \times 16}{4} = 9 \times 4 = 36\) Finally, find the difference \(f(3) - f(2)\): \(f(3) - f(2) = 36 - 9 = 27\)
In simple words: We calculate the function's output for \(x=3\) and \(x=2\) separately, then subtract the second result from the first to get the final answer.

🎯 Exam Tip: When evaluating functions, carefully substitute the values and follow the order of operations (parentheses, exponents, multiplication/division, addition/subtraction) to avoid calculation errors.

Question 11.If \(f(x) = x^2 + 2x - 1\), then state the type of function f.
Answer:Given the function f: R → R, defined by \(f(x) = x^2 + 2x - 1\). The domain is R (real numbers), and the codomain is R (real numbers). Let's test for distinct inputs yielding the same output. Consider two inputs \(x_1\) and \(x_2\) such that \(f(x_1) = f(x_2)\). \(x_1^2 + 2x_1 - 1 = x_2^2 + 2x_2 - 1\) \(x_1^2 + 2x_1 = x_2^2 + 2x_2\) \(x_1^2 - x_2^2 + 2x_1 - 2x_2 = 0\) \((x_1 - x_2)(x_1 + x_2) + 2(x_1 - x_2) = 0\) \((x_1 - x_2)(x_1 + x_2 + 2) = 0\) This implies either \(x_1 - x_2 = 0\) (so \(x_1 = x_2\)) or \(x_1 + x_2 + 2 = 0\). If \(x_1 + x_2 + 2 = 0\), then \(x_1 \neq x_2\) is possible while \(f(x_1) = f(x_2)\). For example, let \(x_1 = -2\). Then \(-2 + x_2 + 2 = 0 \implies x_2 = 0\). Let's check: \(f(-2) = (-2)^2 + 2(-2) - 1 = 4 - 4 - 1 = -1\) \(f(0) = (0)^2 + 2(0) - 1 = 0 + 0 - 1 = -1\) Here, for two different elements (-2 and 0) from the domain, their images are the same (-1) in the codomain. Therefore, function f is a many-one function.
In simple words: This is a many-one function because different input values (like -2 and 0) produce the same output value (-1).

🎯 Exam Tip: For quadratic functions \(ax^2+bx+c\), if the domain is R, they are almost always many-one due to their parabolic shape. Look for pairs of inputs that are symmetric around the axis of symmetry, as they will share the same output.

Question 12.If \(f(x) = \frac{2x-4}{x+7}\) is a real function then for which value of x the image is zero?
Answer:Given the real function \(f(x) = \frac{2x-4}{x+7}\). We need to find the value of x for which the image (output) is zero. So, we set \(f(x) = 0\): \( \frac{2x-4}{x+7} = 0 \) For a fraction to be zero, its numerator must be zero, provided the denominator is not zero. \(2x - 4 = 0\) \(2x = 4\) \(x = \frac{4}{2}\) \(x = 2\) We also need to check that the denominator is not zero at \(x = 2\). \(x+7 = 2+7 = 9 \neq 0\). Since the denominator is not zero at \(x = 2\), the image is zero when \(x = 2\).
In simple words: To find when the image is zero, set the numerator of the fraction to zero and solve for x, making sure that this x-value does not make the denominator zero.

🎯 Exam Tip: For rational functions (fractions), the function equals zero only when the numerator is zero, and the value of x must not make the denominator zero. Always verify both conditions.

Question 13.f:Z – {2} → Z, \(f(x) = \frac{x^2+x-6}{x-2}\) State the type of the function.
Answer:Given the function f: Z – {2} → Z, defined by \(f(x) = \frac{x^2+x-6}{x-2}\). First, simplify the expression for \(f(x)\): The numerator \(x^2+x-6\) can be factored as \((x+3)(x-2)\). So, \(f(x) = \frac{(x+3)(x-2)}{x-2}\). Since \(x \in Z - \{2\}\), \(x-2 \neq 0\), so we can cancel out the \((x-2)\) term. Thus, \(f(x) = x+3\) for \(x \in Z - \{2\}\). Let's evaluate the function for different integer values in the domain (excluding 2): Domain = {\(\ldots\), -3, -2, -1, 0, 1, 3, 4, \(\ldots\)} \(f(-3) = -3 + 3 = 0\) \(f(-2) = -2 + 3 = 1\) \(f(-1) = -1 + 3 = 2\) \(f(0) = 0 + 3 = 3\) \(f(1) = 1 + 3 = 4\) \(f(3) = 3 + 3 = 6\) \(f(4) = 4 + 3 = 7\) For any two distinct integers \(x_1, x_2 \in Z - \{2\}\), if \(x_1 \neq x_2\), then \(x_1 + 3 \neq x_2 + 3\). This implies \(f(x_1) \neq f(x_2)\). Therefore, the function f is a one-one function.
In simple words: After simplifying the function, we get \(f(x) = x+3\). Since each distinct integer input yields a unique integer output, this is a one-one function.

🎯 Exam Tip: Always simplify rational functions before determining their type. Factorizing and canceling common terms can reveal a simpler function that is easier to classify, but remember to note the domain restriction from the original function.

Section - D

Answer the following questions as required:

Question 1.For f:A → B, A = {10, 20, 30); B = (18, 48, 98, 128, 148); f(x) = 5x – 2; obtain domain, co-domain and range.
Answer:Given the function f: A → B, with: Domain A = {10, 20, 30} Codomain B = {18, 48, 98, 128, 148} The function rule is \(f(x) = 5x - 2\). **Domain:** From the given information, the domain of function f is \(D_f = \{10, 20, 30\}\). **Co-domain:** From the given information, the codomain of function f is \(B = \{18, 48, 98, 128, 148\}\). **Range:** To find the range, we apply the function rule to each element in the domain: \(f(10) = (5 \times 10) - 2 = 50 - 2 = 48\) \(f(20) = (5 \times 20) - 2 = 100 - 2 = 98\) \(f(30) = (5 \times 30) - 2 = 150 - 2 = 148\) The range of function f, denoted \(R_f\), is the set of these images. Therefore, \(R_f = \{48, 98, 148\}\).
In simple words: The domain and codomain are given directly. The range is found by calculating the output for each input in the domain using the given function rule.

🎯 Exam Tip: Clearly state the domain and codomain as given. For the range, systematically calculate each image and collect them into a set. Ensure all elements are accounted for and no extra elements are included.

Question 2.Obtain domain, co-domain and range for \(f:P \rightarrow Q\), \(P = \{-\frac{1}{2}, 1, \frac{1}{2}, \frac{3}{2}\}\); \(Q = \{-\frac{1}{5}, 1, \frac{1}{3}, 3\}\), \(f(x) = \frac{x}{2-x}\).
Answer:Given the function \(f:P \rightarrow Q\), with: Domain \(P = \{-\frac{1}{2}, 1, \frac{1}{2}, \frac{3}{2}\}\) Codomain \(Q = \{-\frac{1}{5}, 1, \frac{1}{3}, 3\}\) The function rule is \(f(x) = \frac{x}{2-x}\). **Domain:** From the given information, the domain of function f is \(D_f = \{-\frac{1}{2}, 1, \frac{1}{2}, \frac{3}{2}\}\). **Co-domain:** From the given information, the codomain of function f is \(Q = \{-\frac{1}{5}, 1, \frac{1}{3}, 3\}\). **Range:** To find the range, we evaluate the function at each element in the domain P: For \(x = -\frac{1}{2}\): \(f(-\frac{1}{2}) = \frac{-\frac{1}{2}}{2 - (-\frac{1}{2})} = \frac{-\frac{1}{2}}{2 + \frac{1}{2}} = \frac{-\frac{1}{2}}{\frac{4+1}{2}} = \frac{-\frac{1}{2}}{\frac{5}{2}} = -\frac{1}{2} \times \frac{2}{5} = -\frac{1}{5}\) For \(x = 1\): \(f(1) = \frac{1}{2-1} = \frac{1}{1} = 1\) For \(x = \frac{1}{2}\): \(f(\frac{1}{2}) = \frac{\frac{1}{2}}{2 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{4-1}{2}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}\) For \(x = \frac{3}{2}\): \(f(\frac{3}{2}) = \frac{\frac{3}{2}}{2 - \frac{3}{2}} = \frac{\frac{3}{2}}{\frac{4-3}{2}} = \frac{\frac{3}{2}}{\frac{1}{2}} = \frac{3}{2} \times \frac{2}{1} = 3\) The range \(R_f\) is the set of all these functional values. Therefore, \(R_f = \{-\frac{1}{5}, 1, \frac{1}{3}, 3\}\).
In simple words: The domain and codomain are explicitly given. The range is determined by applying the function formula to each input in the domain and collecting all the calculated outputs.

🎯 Exam Tip: When evaluating rational functions with fractional inputs, perform fraction arithmetic carefully. Ensure you simplify complex fractions correctly to find the precise image for each domain element.

Question 3.If \(f:R – \{0\} \rightarrow R\), \(f(x) = \frac{1}{x}(1+\frac{1}{x}) – 1\) then find the value of \(f(-1), f(-2)\) and \(f(\frac{1}{2})\).
Answer:Given the function \(f(x) = \frac{1}{x}(1+\frac{1}{x}) – 1\), with domain \(R – \{0\}\). First, simplify the function expression: \(f(x) = \frac{1}{x} + \frac{1}{x^2} - 1\) Now, let's find the values for \(f(-1)\), \(f(-2)\), and \(f(\frac{1}{2})\): 1. Calculate \(f(-1)\): \(f(-1) = \frac{1}{-1} + \frac{1}{(-1)^2} - 1 = -1 + \frac{1}{1} - 1 = -1 + 1 - 1 = -1\) 2. Calculate \(f(-2)\): \(f(-2) = \frac{1}{-2} + \frac{1}{(-2)^2} - 1 = -\frac{1}{2} + \frac{1}{4} - 1\) \(f(-2) = \frac{-2+1-4}{4} = \frac{-5}{4}\) 3. Calculate \(f(\frac{1}{2})\): \(f(\frac{1}{2}) = \frac{1}{\frac{1}{2}} + \frac{1}{(\frac{1}{2})^2} - 1 = 2 + \frac{1}{\frac{1}{4}} - 1 = 2 + 4 - 1 = 5\) So, \(f(-1) = -1\), \(f(-2) = -\frac{5}{4}\), and \(f(\frac{1}{2}) = 5\).
In simple words: We first simplify the function's formula, then plug in each given x-value to calculate its corresponding output.

🎯 Exam Tip: Always simplify the function expression first if possible, as it can make subsequent evaluations easier and reduce calculation errors. Pay close attention to fraction arithmetic and signs when substituting negative or fractional values.

Question 4.For the function f:A → B, \(f(x) = 4x - 3\); \(R_f = \{9, 13, 17, 25\}\), then find \(D_f\).
Answer:Given the function f: A → B, with the rule \(f(x) = 4x - 3\). The range of the function is given as \(R_f = \{9, 13, 17, 25\}\). To find the domain \(D_f\), we set \(f(x)\) equal to each value in the range and solve for x: 1. Set \(f(x) = 9\): \(4x - 3 = 9\)
\(4x = 9 + 3\)
\(4x = 12\)
\(x = \frac{12}{4} = 3\) 2. Set \(f(x) = 13\): \(4x - 3 = 13\)
\(4x = 13 + 3\)
\(4x = 16\)
\(x = \frac{16}{4} = 4\) 3. Set \(f(x) = 17\): \(4x - 3 = 17\)
\(4x = 17 + 3\)
\(4x = 20\)
\(x = \frac{20}{4} = 5\) 4. Set \(f(x) = 25\): \(4x - 3 = 25\)
\(4x = 25 + 3\)
\(4x = 28\)
\(x = \frac{28}{4} = 7\) Therefore, the domain of the function is \(D_f = \{3, 4, 5, 7\}\).
In simple words: To find the domain, we take each value from the given range, set it equal to the function's rule, and then solve for x. This gives us the input values.

🎯 Exam Tip: Finding the domain from the range involves solving a series of equations. Be methodical and accurate in your algebraic steps for each value in the range.

Question 5.For the real function \(f(x) = 2x^2 – 5x + 4\), find the value of x, for which \(f(3x) – 3f(x) + 5 = 0\).
Answer:Given the real function \(f(x) = 2x^2 – 5x + 4\). We need to solve the equation \(f(3x) – 3f(x) + 5 = 0\). First, find \(f(3x)\) by replacing x with 3x in the function definition: \(f(3x) = 2(3x)^2 - 5(3x) + 4\) \(f(3x) = 2(9x^2) - 15x + 4\) \(f(3x) = 18x^2 - 15x + 4\) Now, substitute \(f(3x)\) and \(f(x)\) into the given equation: \((18x^2 - 15x + 4) - 3(2x^2 - 5x + 4) + 5 = 0\) Expand the term with 3: \(18x^2 - 15x + 4 - (6x^2 - 15x + 12) + 5 = 0\) Distribute the negative sign: \(18x^2 - 15x + 4 - 6x^2 + 15x - 12 + 5 = 0\) Combine like terms: \((18x^2 - 6x^2) + (-15x + 15x) + (4 - 12 + 5) = 0\) \(12x^2 + 0x - 3 = 0\) \(12x^2 - 3 = 0\) Solve for \(x^2\): \(12x^2 = 3\) \(x^2 = \frac{3}{12}\) \(x^2 = \frac{1}{4}\) Take the square root of both sides: \(x = \pm \sqrt{\frac{1}{4}}\) \(x = \pm \frac{1}{2}\) Therefore, the values of x for which the equation holds true are \(x = \frac{1}{2}\) and \(x = -\frac{1}{2}\).
In simple words: We substituted \(3x\) and \(x\) into the function, then simplified the entire equation. Solving the resulting quadratic equation gave us the x-values.

🎯 Exam Tip: Be meticulous with algebraic substitutions and sign distribution, especially when dealing with multiple terms and coefficients. Quadratic equations often yield two solutions, so remember to consider both positive and negative square roots.

Question 6.If f: A → M, A = {x | x ∈ N, 1 ≤ x < 5} and M = {x | x ∈ N, 1 ≤ x ≤ 20} and \(f(x) = x^2 + 1\), then find the range of f.
Answer:Given the function f: A → M, with: Domain \(A = \{x | x \in N, 1 \leq x < 5\}\). The natural numbers in this range are \(A = \{1, 2, 3, 4\}\). Codomain \(M = \{x | x \in N, 1 \leq x \leq 20\}\). This means M = {1, 2, 3, ..., 20}. The function rule is \(f(x) = x^2 + 1\). To find the range of f, we evaluate the function for each element in the domain A: \(f(1) = 1^2 + 1 = 1 + 1 = 2\) \(f(2) = 2^2 + 1 = 4 + 1 = 5\) \(f(3) = 3^2 + 1 = 9 + 1 = 10\) \(f(4) = 4^2 + 1 = 16 + 1 = 17\) All these images {2, 5, 10, 17} are natural numbers and fall within the codomain M. The range of f, denoted \(R_f\), is the set of these images. Therefore, \(R_f = \{2, 5, 10, 17\}\).
In simple words: We first identified the specific numbers in the domain. Then, we applied the function rule to each of these numbers to find their corresponding outputs, which together form the range.

🎯 Exam Tip: Carefully determine the discrete elements of the domain based on the given conditions. Then, systematically calculate the image for each domain element to form the range. Ensure all range elements are indeed within the defined codomain.

Question 7.If \(f(x) = \frac{x^2-4}{x-2}\) where \(x \in Z – \{2\}\) then find the value of \(f(0) + f(1) – f(-2)\).
Answer:Given the function \(f(x) = \frac{x^2-4}{x-2}\) with domain \(x \in Z – \{2\}\). First, simplify the function expression. The numerator \(x^2-4\) is a difference of squares, which can be factored as \((x-2)(x+2)\). So, \(f(x) = \frac{(x-2)(x+2)}{x-2}\). Since \(x \in Z – \{2\}\), \(x-2 \neq 0\), allowing us to cancel the \((x-2)\) terms. Thus, \(f(x) = x+2\) for \(x \in Z – \{2\}\). Now, we calculate \(f(0)\), \(f(1)\), and \(f(-2)\): 1. Calculate \(f(0)\): \(f(0) = 0 + 2 = 2\) 2. Calculate \(f(1)\): \(f(1) = 1 + 2 = 3\) 3. Calculate \(f(-2)\): \(f(-2) = -2 + 2 = 0\) Finally, find the value of \(f(0) + f(1) – f(-2)\): \(f(0) + f(1) – f(-2) = 2 + 3 - 0 = 5\)
In simple words: We simplified the function to \(f(x) = x+2\). Then, we calculated the function's output for \(x=0, 1, \text{ and } -2\) and combined these results as requested.

🎯 Exam Tip: Simplifying rational function expressions is key, but remember to always consider the original domain restrictions. After simplification, evaluate each term carefully before performing the final arithmetic operation.

Question 8. If the domain of a function \( f: A \rightarrow N \cup \{0\} \); \( f(x) = \sqrt{x^2-16} \) is \( A = \{4, 5\} \), then find its range.
Answer:
Given the function \( f(x) = \sqrt{x^2-16} \) and its domain \( A = \{4, 5\} \). To determine the range, we evaluate the function at each element in the domain.
For \( x = 4 \): \( f(4) = \sqrt{4^2-16} = \sqrt{16-16} = \sqrt{0} = 0 \).
For \( x = 5 \): \( f(5) = \sqrt{5^2-16} = \sqrt{25-16} = \sqrt{9} = 3 \).
Thus, the range of the function \( f \) is \( \{0, 3\} \).
In simple words: The range of a function includes all the possible output values you get when you put in the allowed input values from the domain. Here, for the given domain values, the function produces 0 and 3.

🎯 Exam Tip: Remember to calculate the function's value for *each* element in the domain to accurately determine the range. Pay attention to square roots and potential undefined values.

Question 9. If \( f(x) = x^2 \) and \( g(x) = 5x-6 \), where \( x \in \{2, 3, 4\} \), check the equality of the functions.
Answer:
We are given two functions, \( f(x) = x^2 \) and \( g(x) = 5x-6 \), both defined over the domain \( X = \{2, 3, 4\} \). To check if these functions are equal, we must compare their domains and ranges.
The domain for both functions is \( \{2, 3, 4\} \).
Now, let's find the range for \( f(x) \):
\( f(2) = 2^2 = 4 \)
\( f(3) = 3^2 = 9 \)
\( f(4) = 4^2 = 16 \)
So, the range for \( f \) is \( R_f = \{4, 9, 16\} \).
Next, let's find the range for \( g(x) \):
\( g(2) = 5(2)-6 = 10-6 = 4 \)
\( g(3) = 5(3)-6 = 15-6 = 9 \)
\( g(4) = 5(4)-6 = 20-6 = 14 \)
So, the range for \( g \) is \( R_g = \{4, 9, 14\} \).
Since the range of \( f \) (\( \{4, 9, 16\} \)) is not identical to the range of \( g \) (\( \{4, 9, 14\} \)), the functions \( f \) and \( g \) are not equal, even though their domains are the same.
In simple words: Two functions are equal only if they have the exact same domain and produce the exact same output (range) for every corresponding input. Here, the outputs are different for some inputs, so the functions are not equal.

🎯 Exam Tip: For functions to be equal, both their domains and their ranges must be identical. A single different output for a common input means the functions are not equal.

Question 10. If \( k : R \rightarrow R \), \( k(x) = x^2 + 3x - 12 \); then determine the type of the function k.
Answer:
Given the function \( k: R \rightarrow R \) defined by \( k(x) = x^2 + 3x - 12 \). We need to determine its type. Let's evaluate the function for a few integer values from the domain \( R \):
For \( x = -3 \): \( k(-3) = (-3)^2 + 3(-3) - 12 = 9 - 9 - 12 = -12 \)
For \( x = -2 \): \( k(-2) = (-2)^2 + 3(-2) - 12 = 4 - 6 - 12 = -14 \)
For \( x = -1 \): \( k(-1) = (-1)^2 + 3(-1) - 12 = 1 - 3 - 12 = -14 \)
For \( x = 0 \): \( k(0) = 0^2 + 3(0) - 12 = -12 \)
For \( x = 1 \): \( k(1) = 1^2 + 3(1) - 12 = 1 + 3 - 12 = -8 \)
For \( x = 2 \): \( k(2) = 2^2 + 3(2) - 12 = 4 + 6 - 12 = -2 \)
We observe that \( k(-3) = -12 \) and \( k(0) = -12 \). Also, \( k(-2) = -14 \) and \( k(-1) = -14 \).
Since different elements in the domain (e.g., -3 and 0, or -2 and -1) map to the same image in the co-domain, this function is classified as a many-one function.
In simple words: If multiple different input values give you the exact same output value, the function is called a "many-one" function. Here, -3 and 0 both result in -12, and -2 and -1 both result in -14.

🎯 Exam Tip: To classify a function as one-one or many-one, check if any two distinct inputs produce the same output. If they do, it's many-one. If all distinct inputs yield distinct outputs, it's one-one.

Question 11. If \( f(x) = x(3x-2) \), \( g(x) = x^3 \) and \( x \in \{0, 1, 2\} \); then prove that f and g are equal functions.
Answer:
We are given two functions: \( f(x) = x(3x-2) \) and \( g(x) = x^3 \), both defined for the domain \( X = \{0, 1, 2\} \). To prove that \( f \) and \( g \) are equal functions, we must demonstrate that they have the same domain and produce identical outputs for every element in that domain.
First, observe that the domain for both functions is explicitly given as \( X = \{0, 1, 2\} \).
Next, let's calculate the range for \( f(x) \):
For \( x = 0 \): \( f(0) = 0(3 \cdot 0 - 2) = 0(-2) = 0 \)
For \( x = 1 \): \( f(1) = 1(3 \cdot 1 - 2) = 1(1) = 1 \)
For \( x = 2 \): \( f(2) = 2(3 \cdot 2 - 2) = 2(6 - 2) = 2(4) = 8 \)
So, the range of \( f \) is \( R_f = \{0, 1, 8\} \).
Now, let's calculate the range for \( g(x) \):
For \( x = 0 \): \( g(0) = 0^3 = 0 \)
For \( x = 1 \): \( g(1) = 1^3 = 1 \)
For \( x = 2 \): \( g(2) = 2^3 = 8 \)
So, the range of \( g \) is \( R_g = \{0, 1, 8\} \).
Since both functions share the same domain \( \{0, 1, 2\} \) and their ranges are identical \( \{0, 1, 8\} \), we can conclude that \( f \) and \( g \) are equal functions.
In simple words: If two functions always give the same output for the same inputs, and their allowed inputs are the same, then they are considered the same function. This is what we showed for \(f\) and \(g\).

🎯 Exam Tip: To prove two functions are equal, verify that they share the exact same domain and that for every element in that domain, both functions yield the identical output value.

Question 12. If \( f(x) = \frac{2x+3}{5x+2} \), \( x \in R - \{-\frac{2}{5}\} \); then find the value of \( f(2) \cdot f(1) \).
Answer:
Given the function \( f(x) = \frac{2x+3}{5x+2} \), where \( x \in R - \{-\frac{2}{5}\} \). We are required to find the value of the product \( f(2) \cdot f(1) \).
First, calculate the value of \( f(2) \):
\( f(2) = \frac{2(2)+3}{5(2)+2} = \frac{4+3}{10+2} = \frac{7}{12} \)
Next, calculate the value of \( f(1) \):
\( f(1) = \frac{2(1)+3}{5(1)+2} = \frac{2+3}{5+2} = \frac{5}{7} \)
Now, multiply these two values:
\( f(2) \cdot f(1) = \frac{7}{12} \cdot \frac{5}{7} \)
\( = \frac{5}{12} \)
Therefore, the value of \( f(2) \cdot f(1) \) is \( \frac{5}{12} \).
In simple words: To find the product of two function values, first calculate each value separately by plugging the numbers into the function formula, then multiply the results.

🎯 Exam Tip: Always perform calculations carefully step-by-step. If a function involves fractions, simplify them correctly before proceeding with further operations.

Question 13. If \( f(x) = 2x^2 + \frac{1}{x} \), \( x \in R \); obtain the value of \( f(3) + f(-3) \).
Answer:
Given the function \( f(x) = 2x^2 + \frac{1}{x} \), for \( x \in R \). We need to determine the value of \( f(3) + f(-3) \).
First, calculate \( f(3) \):
\( f(3) = 2(3)^2 + \frac{1}{3} = 2(9) + \frac{1}{3} = 18 + \frac{1}{3} \)
Next, calculate \( f(-3) \):
\( f(-3) = 2(-3)^2 + \frac{1}{-3} = 2(9) - \frac{1}{3} = 18 - \frac{1}{3} \)
Now, add these two results:
\( f(3) + f(-3) = \left(18 + \frac{1}{3}\right) + \left(18 - \frac{1}{3}\right) \)
\( = 18 + \frac{1}{3} + 18 - \frac{1}{3} \)
\( = 18 + 18 \)
\( = 36 \)
Therefore, the value of \( f(3) + f(-3) \) is 36.
In simple words: To add function values, calculate each function value separately for the given inputs, then sum the results. Be careful with signs when substituting negative numbers.

🎯 Exam Tip: When dealing with functions involving powers and fractions, pay close attention to the order of operations and the signs of terms, especially when evaluating for negative input values.

Question 14. If \( f(x) = 15x^3 - 4x^2 + x + 10 \), \( x \in R \); obtain the value of \( \frac{f(2)}{f(1)} \).
Answer:
Given the function \( f(x) = 15x^3 - 4x^2 + x + 10 \), for \( x \in R \). We need to find the value of the ratio \( \frac{f(2)}{f(1)} \).
First, calculate the value of \( f(2) \):
\( f(2) = 15(2)^3 - 4(2)^2 + 2 + 10 \)
\( = 15(8) - 4(4) + 2 + 10 \)
\( = 120 - 16 + 2 + 10 \)
\( = 104 + 12 \)
\( = 116 \)
Next, calculate the value of \( f(1) \):
\( f(1) = 15(1)^3 - 4(1)^2 + 1 + 10 \)
\( = 15(1) - 4(1) + 1 + 10 \)
\( = 15 - 4 + 1 + 10 \)
\( = 11 + 11 \)
\( = 22 \)
Now, find the ratio \( \frac{f(2)}{f(1)} \):
\( \frac{f(2)}{f(1)} = \frac{116}{22} \)
\( = \frac{58}{11} \)
Therefore, the value of \( \frac{f(2)}{f(1)} \) is \( \frac{58}{11} \).
In simple words: To find the ratio of two function values, calculate each function value individually and then divide the first result by the second, simplifying the fraction if possible.

🎯 Exam Tip: Be meticulous with arithmetic, especially when dealing with higher powers and multiple terms in a polynomial function. Simplify fractions to their lowest terms.

Question 15. If domain is \( \{500, 1000, 1300, 1400\} \), \( f(x) = \sqrt{5600-4x} \), \( x \in R \); then find the value of \( f(x) \) for \( x = 1000 \). Also, for which value of \( x \), \( f(x) = 20 \)?
Answer:
Given the function \( f(x) = \sqrt{5600-4x} \) with the domain \( \{500, 1000, 1300, 1400\} \). We need to perform two tasks:
1. Find the value of \( f(x) \) when \( x = 1000 \).
2. Determine the value of \( x \) for which \( f(x) = 20 \).

**Part 1: Find \( f(1000) \)**
Substitute \( x = 1000 \) into the function:
\( f(1000) = \sqrt{5600-4(1000)} \)
\( = \sqrt{5600-4000} \)
\( = \sqrt{1600} \)
\( = 40 \)
So, when \( x = 1000 \), \( f(x) = 40 \).

**Part 2: Find \( x \) such that \( f(x) = 20 \)**
Set the function equal to 20:
\( 20 = \sqrt{5600-4x} \)
To solve for \( x \), square both sides of the equation:
\( (20)^2 = (\sqrt{5600-4x})^2 \)
\( 400 = 5600-4x \)
Now, rearrange the equation to isolate \( x \):
\( 4x = 5600-400 \)
\( 4x = 5200 \)
Divide by 4:
\( x = \frac{5200}{4} \)
\( x = 1300 \)
Therefore, \( f(x) = 20 \) when \( x = 1300 \).
In simple words: For the first part, just plug the number 1000 into the formula for \(x\) and calculate the output. For the second part, set the whole function equal to 20, then use algebra to work backward and find the value of \(x\).

🎯 Exam Tip: When solving for \(x\) in an equation involving a square root, remember to square both sides to eliminate the root. Always double-check your arithmetic, especially with large numbers. Also, ensure the obtained \(x\) value is within the given domain if one is specified.