OP Malhotra Class 11 Maths Solutions Chapter 2 Relations and Functions Exercise 2 (F)

Question 1. The range and domain of function \( f(x) = \frac { 3 }{ x } + 1 \) are subsets of A and B respectively, where \( A = \left\{-\frac{1}{2}, 0, \frac{2}{3}, \frac{6}{7}, 1\right\} \) and \( B = \left\{-5,0,4 \frac{1}{2}, 5,5 \frac{1}{2}\right\} \). List the elements of the function as ordered pairs.
Answer: To find the elements of the function as ordered pairs, we need to apply the function \( f(x) = \frac{3}{x} + 1 \) to each value in the domain A. We will only include pairs where the result belongs to set B. Also, remember that division by zero is not allowed, so \( x \neq 0 \).
Let's calculate for each element in A:
For \( x = -\frac{1}{2} \):
\( f\left(-\frac{1}{2}\right) = \frac{3}{-\frac{1}{2}} + 1 = -6 + 1 = -5 \)
Since \( -5 \in B \), the ordered pair is \( \left(-\frac{1}{2},-5\right) \).

For \( x = 0 \):
\( f(0) = \frac{3}{0} + 1 \)
This value does not exist because we cannot divide by zero. So, 0 is not in the domain of f.

For \( x = \frac{2}{3} \):
\( f\left(\frac{2}{3}\right) = \frac{3}{\frac{2}{3}} + 1 = \frac{9}{2} + 1 = 4.5 + 1 = 5.5 \)
Since \( 5.5 \in B \) (as \( 5\frac{1}{2} = 5.5 \)), the ordered pair is \( \left(\frac{2}{3},5.5\right) \).

For \( x = \frac{6}{7} \):
\( f\left(\frac{6}{7}\right) = \frac{3}{\frac{6}{7}} + 1 = \frac{21}{6} + 1 = \frac{7}{2} + 1 = 3.5 + 1 = 4.5 \)
Since \( 4.5 \in B \) (as \( 4\frac{1}{2} = 4.5 \)), the ordered pair is \( \left(\frac{6}{7},4.5\right) \).

For \( x = 1 \):
\( f(1) = \frac{3}{1} + 1 = 3 + 1 = 4 \)
Since \( 4 \notin B \) (the closest values are 0, 4.5, 5.5), this ordered pair is not included.

So, the elements of the function as ordered pairs are: \( \left\{\left(-\frac{1}{2},-5\right),\left(\frac{2}{3},5.5\right),\left(\frac{6}{7},4.5\right)\right\} \).
In simple words: We put each number from set A into the function \( f(x) \). If the answer we get is also found in set B, then we write that pair of numbers together. We can't use 0 from set A because we can't divide by zero.

🎯 Exam Tip: Always check all elements in the given domain. Pay close attention to restrictions like division by zero or values not present in the codomain set (B in this case), as these will exclude certain pairs from the function's final list.

Question 2. A = \( \left\{- 2, - 1, 1, 2\right\} \) and \( f = \left\{\left(x, \frac{1}{x}\right), x \in A\right\} \).
(i) List the domain of f
(ii) List the range of f
(iii) Is f a function ?
Answer: The function \( f \) is defined by \( f(x) = \frac{1}{x} \) for all \( x \) in set A.

(i) The domain of \( f \) is the set of all input values \( x \) for which the function is defined. From the definition \( f = \left\{\left(x, \frac{1}{x}\right), x \in A\right\} \), it clearly states that \( x \) belongs to set A. Also, since none of the values in A are zero, \( \frac{1}{x} \) is defined for all \( x \in A \).
Therefore, the domain of \( f \) is \( \left\{- 2, - 1, 1, 2\right\} \).

(ii) To find the range of \( f \), we need to calculate \( f(x) \) for each element in the domain:
For \( x = -2 \): \( f(-2) = \frac{1}{-2} = -\frac{1}{2} \)
For \( x = -1 \): \( f(-1) = \frac{1}{-1} = -1 \)
For \( x = 1 \): \( f(1) = \frac{1}{1} = 1 \)
For \( x = 2 \): \( f(2) = \frac{1}{2} \)
Thus, the range of \( f \) is \( \left\{- 1, 1, -\frac{1}{2}, \frac{1}{2}\right\} \).

(iii) A function is a relation where each element in the domain has exactly one corresponding element in the range (a unique image). In other words, no two distinct ordered pairs can have the same first component. In this case, each value in set A (the domain) produces a single, unique output value.
For example, -2 gives only \( -\frac{1}{2} \), -1 gives only -1, and so on. This means that \( f \) represents a function.
In simple words: The domain is all the numbers we can put into the function, which is set A itself. The range is all the answers we get out when we use those numbers. Since each input number gives only one output answer, this is a function.

🎯 Exam Tip: To determine if a relation is a function, always check if every domain element maps to exactly one range element. If any domain element maps to more than one range element, it is not a function.

Question 3. f : A\( \to \) highest prime factor of x.
(i) Find the range of f when the domain is {12, 13, 14, 15, 16, 17}.
(ii) State a domain of five integers for which the range is (3).
(iii) A set of positive integers is called S. What can be said about these integers if f(S) = S?
Answer: The function \( f(x) \) gives the highest prime factor of \( x \). A prime factor is a prime number that divides a given number. For example, the prime factors of 12 are 2 and 3.

(i) We need to find the highest prime factor for each number in the domain {12, 13, 14, 15, 16, 17}:
For 12: Factors are 1, 2, 3, 4, 6, 12. Prime factors are 2, 3. The highest prime factor is 3.
\( \implies \) So, \( f(12) = 3 \).
For 13: Factors are 1, 13. Prime factors are 13. The highest prime factor is 13.
\( \implies \) So, \( f(13) = 13 \).
For 14: Factors are 1, 2, 7, 14. Prime factors are 2, 7. The highest prime factor is 7.
\( \implies \) So, \( f(14) = 7 \).
For 15: Factors are 1, 3, 5, 15. Prime factors are 3, 5. The highest prime factor is 5.
\( \implies \) So, \( f(15) = 5 \).
For 16: Factors are 1, 2, 4, 8, 16. Prime factors are 2. The highest prime factor is 2.
\( \implies \) So, \( f(16) = 2 \).
For 17: Factors are 1, 17. Prime factors are 17. The highest prime factor is 17.
\( \implies \) So, \( f(17) = 17 \).
The range of \( f \) is the set of all unique output values: \( R_f = \{2, 3, 5, 7, 13, 17\} \).

(ii) We need a domain of five integers for which the highest prime factor of each is 3. This means we need numbers whose largest prime factor is 3.
Examples include: 3 (highest prime factor is 3), 6 (prime factors 2, 3; highest is 3), 9 (prime factor 3; highest is 3), 12 (prime factors 2, 3; highest is 3), 18 (prime factors 2, 3; highest is 3).
Therefore, a required domain \( D_f = \{3, 6, 9, 12, 18\} \).

(iii) If \( f(S) = S \), it means that the set of highest prime factors of all numbers in S is equal to S itself. This condition is only possible if S is the set of all prime numbers.
For any prime number \( p \), its only prime factor is \( p \) itself, so the highest prime factor of \( p \) is \( p \). Thus, if S contains only prime numbers, \( f(S) \) would be S. If S contained a composite number, say 4, then \( f(4) = 2 \), and S would not equal \( f(S) \) unless 2 was already in S and 4 was somehow mapped back to 4 as a prime. This is a special property of prime numbers.
In simple words: First, we find the biggest prime number that divides each number in the list for part (i). For part (ii), we think of numbers where the largest prime number that divides them is 3. For part (iii), if a set of numbers always gives back itself when we find the biggest prime factor, then that set must be all the prime numbers.

🎯 Exam Tip: When dealing with prime factors, remember that 1 is not a prime number. For a number to be a prime number itself, it must have exactly two distinct positive divisors: 1 and itself.

Question 4. A function f is defined on the set of integers as follows :
\( f(x) = \left\{\begin{aligned} 1+x ; & 1 \leq x<2 \\ 2 x-1 ; & 2 \leq x<4 \\ 3 x-10 ; & 4 \leq x<6 \end{aligned}\right\} \)
(i) Find the domain of the function
(ii) Find the range of the function
(iii) Find the value of f(4).
Answer: The function \( f(x) \) is defined differently based on the value of \( x \). It is specified for integer values of \( x \). To find the domain, we look at the intervals where \( x \) is defined.

(i) The function is defined for integers \( x \) such that:
\( 1 \leq x < 2 \implies x = 1 \)
\( 2 \leq x < 4 \implies x = 2, 3 \)
\( 4 \leq x < 6 \implies x = 4, 5 \)
Combining these, the domain of \( f \) consists of all integers from 1 up to 5.
Thus, the domain \( D_f = \{1, 2, 3, 4, 5\} \).

(ii) To find the range, we evaluate \( f(x) \) for each value in the domain:
For \( x = 1 \) (using \( 1+x \)): \( f(1) = 1 + 1 = 2 \)
For \( x = 2 \) (using \( 2x-1 \)): \( f(2) = 2(2) - 1 = 4 - 1 = 3 \)
For \( x = 3 \) (using \( 2x-1 \)): \( f(3) = 2(3) - 1 = 6 - 1 = 5 \)
For \( x = 4 \) (using \( 3x-10 \)): \( f(4) = 3(4) - 10 = 12 - 10 = 2 \)
For \( x = 5 \) (using \( 3x-10 \)): \( f(5) = 3(5) - 10 = 15 - 10 = 5 \)
The unique values obtained are 2, 3, and 5. So, the range of \( f \) is \( R_f = \{2, 3, 5\} \).

(iii) The value of \( f(4) \) was calculated in part (ii). For \( x = 4 \), we use the rule \( 3x-10 \).
\( f(4) = 3(4) - 10 = 12 - 10 = 2 \).
In simple words: For part (i), we list all the whole numbers that the function uses, based on the rules given. For part (ii), we put each of those numbers into the right rule and list all the answers we get. For part (iii), we just find the answer when we put in the number 4.

🎯 Exam Tip: When working with piecewise functions, always identify the correct sub-function based on the input value's interval. Mistakes often occur by using the wrong rule for a given \( x \).

Question 5. Let f be a function whose domain is the set of all real number. If \( f(x) = | x | – x \), what is range of f?
Answer: The function is defined as \( f(x) = |x| - x \). The domain is all real numbers. To find the range, we consider two cases for \( x \), depending on the definition of the absolute value \( |x| \). The absolute value \( |x| \) means the positive value of \( x \), so it's \( x \) if \( x \ge 0 \) and \( -x \) if \( x < 0 \).

Case 1: When \( x \ge 0 \)
If \( x \) is zero or a positive number, then \( |x| = x \).
\( f(x) = x - x = 0 \).
So, for all non-negative real numbers, the function's value is 0.

Case 2: When \( x < 0 \)
If \( x \) is a negative number, then \( |x| = -x \).
\( f(x) = -x - x = -2x \).
Since \( x \) is a negative number (e.g., -1, -2, -3...), then \( -x \) will be a positive number (e.g., 1, 2, 3...).
Therefore, \( -2x \) will be a positive number (e.g., 2, 4, 6...).
As \( x \) approaches \( -\infty \), \( -2x \) approaches \( +\infty \). Also, as \( x \) approaches 0 from the negative side, \( -2x \) approaches 0.

Combining both cases: The function output is either 0 (for \( x \ge 0 \)) or a positive real number (for \( x < 0 \)).
Therefore, the range of \( f \) is \( [0, \infty) \). This means the function can produce any non-negative real number.
In simple words: We look at the function in two parts: when x is zero or positive, and when x is negative. If x is positive or zero, the answer is always 0. If x is negative, the answer is a positive number. So, the function can give any answer from 0 upwards.

🎯 Exam Tip: When dealing with absolute value functions, always break the problem into cases based on where the expression inside the absolute value changes its sign. This helps to accurately determine the function's behavior across its domain.

Question 6. Write the domain of the following real functions
(i) \( \sqrt{9-x^2} \)
(ii) \( \sqrt{1-2 x-3 x^2} \)
(iii) \( 10^x \)
(iv) \( \frac{1}{\sqrt{x^2-7}} \)
(v) \( \log (2 – 3x) \)
(vi) \( \log(\sqrt{x-4}+\sqrt{6-x}) \)
(vii) \( \left[\log _{10}\left(\frac{5 x-x^2}{4}\right)\right]^{\frac{1}{2}} \)
(viii) \( \sin ^{-1}\left[\log _2\left(\frac{x}{2}\right)\right] \)
Answer: The domain of a real function is the set of all real input values for which the function gives a real output. This often means avoiding square roots of negative numbers, logarithms of non-positive numbers, and division by zero. Understanding these basic restrictions is key to finding the domain.

(i) Given \( f(x) = \sqrt{9-x^2} \)
For \( f(x) \) to be a real number, the expression under the square root must be non-negative.
\( 9-x^2 \ge 0 \)
\( \implies x^2 \le 9 \)
\( \implies -3 \le x \le 3 \)
The domain \( D_f = [-3, 3] \).

(ii) Given \( f(x) = \sqrt{1-2 x-3 x^2} \)
For \( f(x) \) to be a real number, the expression under the square root must be non-negative.
\( 1-2x-3x^2 \ge 0 \)
Multiply by -1 and reverse the inequality sign:
\( \implies 3x^2+2x-1 \le 0 \)
Factorize the quadratic expression:
\( (3x-1)(x+1) \le 0 \)
The critical points are \( x = -1 \) and \( x = \frac{1}{3} \). For the product to be less than or equal to zero, \( x \) must be between these two values.
\( \implies -1 \le x \le \frac{1}{3} \)
The domain \( D_f = \left[-1, \frac{1}{3}\right] \).

(iii) Given \( f(x) = 10^x \)
This is an exponential function. For any real number \( x \), \( 10^x \) is always a real number.
The domain \( D_f = R \) (all real numbers).

(iv) Given \( f(x) = \frac{1}{\sqrt{x^2-7}} \)
For \( f(x) \) to be a real number, two conditions must be met:
1. The expression under the square root must be non-negative: \( x^2-7 \ge 0 \)
2. The denominator cannot be zero: \( \sqrt{x^2-7} \neq 0 \implies x^2-7 \neq 0 \)
Combining these, we need \( x^2-7 > 0 \).
\( \implies x^2 > 7 \)
\( \implies |x| > \sqrt{7} \)
This means \( x > \sqrt{7} \) or \( x < -\sqrt{7} \).
The domain \( D_f = (-\infty, -\sqrt{7}) \cup (\sqrt{7}, \infty) \).

(v) Given \( f(x) = \log (2 – 3x) \)
For the logarithm of a real number to be defined, its argument must be strictly positive.
\( 2-3x > 0 \)
\( \implies 2 > 3x \)
\( \implies 3x < 2 \)
\( \implies x < \frac{2}{3} \)
The domain \( D_f = \left(-\infty, \frac{2}{3}\right) \).

(vi) Given \( f(x) = \log(\sqrt{x-4}+\sqrt{6-x}) \)
For \( f(x) \) to be a real number, several conditions must be met:
1. The expressions under the square roots must be non-negative:
\( x-4 \ge 0 \implies x \ge 4 \)
\( 6-x \ge 0 \implies x \le 6 \)
Combining these, \( 4 \le x \le 6 \).
2. The argument of the logarithm must be strictly positive:
\( \sqrt{x-4}+\sqrt{6-x} > 0 \)
For \( x \) in the interval \( [4, 6] \), both \( \sqrt{x-4} \) and \( \sqrt{6-x} \) are non-negative. Their sum will be greater than zero unless both are zero simultaneously. This only happens if \( x-4=0 \) AND \( 6-x=0 \), which means \( x=4 \) AND \( x=6 \), which is impossible for a single \( x \). So, the sum is always positive for \( x \in [4, 6] \).
The domain \( D_f = [4, 6] \).

(vii) Given \( f(x) = \left[\log _{10}\left(\frac{5 x-x^2}{4}\right)\right]^{\frac{1}{2}} \)
For \( f(x) \) to be a real number, two main conditions apply:
1. The expression under the square root must be non-negative:
\( \log _{10}\left(\frac{5 x-x^2}{4}\right) \ge 0 \)
\( \implies \frac{5 x-x^2}{4} \ge 10^0 \)
\( \implies \frac{5 x-x^2}{4} \ge 1 \)
\( \implies 5x - x^2 \ge 4 \)
\( \implies x^2 - 5x + 4 \le 0 \)
Factorize: \( (x-1)(x-4) \le 0 \)
This inequality holds when \( 1 \le x \le 4 \).
2. The argument of the logarithm must be strictly positive:
\( \frac{5 x-x^2}{4} > 0 \)
\( \implies 5x - x^2 > 0 \)
\( \implies x(5-x) > 0 \)
This inequality holds when \( 0 < x < 5 \).
Combining both conditions, we need \( x \) to be in \( [1, 4] \) AND \( (0, 5) \). The intersection of these intervals is \( [1, 4] \).
The domain \( D_f = [1, 4] \).

(viii) Given \( f(x) = \sin ^{-1}\left[\log _2\left(\frac{x}{2}\right)\right] \)
For \( f(x) \) to be a real number, two main conditions apply:
1. The argument of \( \sin^{-1} \) must be in the interval \( [-1, 1] \):
\( -1 \le \log_2\left(\frac{x}{2}\right) \le 1 \)
To remove the logarithm base 2, we raise 2 to the power of each part of the inequality. Since the base 2 is greater than 1, the inequality direction does not change.
\( \implies 2^{-1} \le \frac{x}{2} \le 2^1 \)
\( \implies \frac{1}{2} \le \frac{x}{2} \le 2 \)
Multiply all parts by 2:
\( \implies 1 \le x \le 4 \).
2. The argument of the logarithm must be strictly positive:
\( \frac{x}{2} > 0 \)
\( \implies x > 0 \).
Combining both conditions, we need \( x \) to be in \( [1, 4] \) AND \( (0, \infty) \). The intersection of these is \( [1, 4] \).
The domain \( D_f = [1, 4] \).
In simple words: For each function, we find the numbers that make sense to put in. We avoid taking square roots of negative numbers, logarithms of zero or negative numbers, and dividing by zero. For inverse sine, the input must be between -1 and 1. We solve these rules to find the allowed range of x values for each function.

🎯 Exam Tip: When determining domains, always consider all potential restrictions simultaneously: non-negative values under even roots, positive values for logarithms, non-zero denominators, and specific input ranges for inverse trigonometric functions. The domain is the intersection of all these conditions.

Question 7. Find the range of each of the following function
(i) \( |x-3| \)
(ii) \( \sqrt{x-5} \)
(iii) \( \cos\left(\frac { x }{ 3 }\right) \)
(iv) \( \frac{x+2}{|x+2|} \)
(v) \( \sec \left(\frac{\pi}{4} \cos ^2 x\right) \), \( – \infty(vi) \( \frac{x^2+x+2}{x^2+x+1} \)
(vii) \( \frac{x}{1+x^2} \)
(viii) \( f(x) = \frac{3}{2-x^2} \)
Answer: The range of a function is the set of all possible output values it can produce. To find the range, we often analyze the function's behavior across its domain, considering minimum and maximum values or transformations.

(i) Given \( f(x) = |x-3| \)
The absolute value of any real number is always non-negative. This means \( |x-3| \ge 0 \) for all real \( x \). The expression \( x-3 \) can take any real value, so \( |x-3| \) can take any non-negative value. The smallest value is 0 (when \( x=3 \)).
The range \( R_f = [0, \infty) \).

(ii) Given \( f(x) = \sqrt{x-5} \)
First, find the domain: For \( \sqrt{x-5} \) to be real, \( x-5 \ge 0 \implies x \ge 5 \). So, the domain is \( [5, \infty) \).
For any \( x \) in this domain, \( x-5 \) will be non-negative. The square root of a non-negative number is always non-negative. The smallest value is \( \sqrt{0} = 0 \) (when \( x=5 \)). As \( x \) increases, \( \sqrt{x-5} \) also increases without bound.
The range \( R_f = [0, \infty) \).

(iii) Given \( f(x) = \cos\left(\frac{x}{3}\right) \)
The domain of \( \cos(z) \) is all real numbers. Since \( \frac{x}{3} \) can take any real value, the input to the cosine function covers its entire domain.
The cosine function, \( \cos(z) \), always produces values between -1 and 1, inclusive.
So, \( -1 \le \cos\left(\frac{x}{3}\right) \le 1 \).
The range \( R_f = [-1, 1] \).

(iv) Given \( f(x) = \frac{x+2}{|x+2|} \)
The function is defined only when the denominator is not zero, so \( x+2 \ne 0 \implies x \ne -2 \). The domain is \( R - \{-2\} \).
We consider two cases based on the value of \( x+2 \):
Case 1: When \( x+2 > 0 \implies x > -2 \)
In this case, \( |x+2| = x+2 \).
\( f(x) = \frac{x+2}{x+2} = 1 \).
Case 2: When \( x+2 < 0 \implies x < -2 \)
In this case, \( |x+2| = -(x+2) \).
\( f(x) = \frac{x+2}{-(x+2)} = -1 \).
So, the function can only output values of 1 or -1.
The range \( R_f = \{-1, 1\} \).

(v) Given \( f(x) = \sec \left(\frac{\pi}{4} \cos ^2 x\right) \)
The domain is all real numbers. We know that for any real \( x \), \( 0 \le \cos^2 x \le 1 \).
Multiplying by \( \frac{\pi}{4} \), we get \( 0 \le \frac{\pi}{4} \cos^2 x \le \frac{\pi}{4} \).
Let \( \theta = \frac{\pi}{4} \cos^2 x \). So \( \theta \) lies in the interval \( \left[0, \frac{\pi}{4}\right] \).
Now we need to find the range of \( \sec(\theta) \) for \( \theta \in \left[0, \frac{\pi}{4}\right] \).
We know that \( \cos(0) = 1 \) and \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \).
So, \( \sec(0) = \frac{1}{\cos(0)} = 1 \).
And \( \sec\left(\frac{\pi}{4}\right) = \frac{1}{\cos\left(\frac{\pi}{4}\right)} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \).
Since \( \cos(\theta) \) is decreasing on \( \left[0, \frac{\pi}{4}\right] \), \( \sec(\theta) \) is increasing on this interval.
Therefore, the range of \( f(x) \) is \( [1, \sqrt{2}] \).
The range \( R_f = [1, \sqrt{2}] \).

(vi) Given \( f(x) = \frac{x^2+x+2}{x^2+x+1} \)
Let \( y = f(x) \). We can rewrite the function as:
\( y = \frac{(x^2+x+1)+1}{x^2+x+1} = 1 + \frac{1}{x^2+x+1} \).
First, consider the denominator \( x^2+x+1 \). This is a quadratic expression with discriminant \( \Delta = 1^2 - 4(1)(1) = 1-4 = -3 \). Since \( \Delta < 0 \) and the coefficient of \( x^2 \) is positive (1), \( x^2+x+1 \) is always positive for all real \( x \).
To find its minimum value, use the vertex formula \( x = -\frac{b}{2a} = -\frac{1}{2(1)} = -\frac{1}{2} \).
Minimum value of \( x^2+x+1 = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1-2+4}{4} = \frac{3}{4} \).
So, \( x^2+x+1 \in \left[\frac{3}{4}, \infty\right) \).
Now consider \( \frac{1}{x^2+x+1} \). When the denominator is \( \frac{3}{4} \), the fraction is \( \frac{1}{3/4} = \frac{4}{3} \). As the denominator increases, the fraction decreases towards 0.
So, \( \frac{1}{x^2+x+1} \in \left(0, \frac{4}{3}\right] \).
Finally, \( y = 1 + \frac{1}{x^2+x+1} \). Adding 1 to the interval:
\( y \in \left(1+0, 1+\frac{4}{3}\right] \)
\( y \in \left(1, \frac{3+4}{3}\right] \)
\( y \in \left(1, \frac{7}{3}\right] \).
The range \( R_f = \left(1, \frac{7}{3}\right] \).

(vii) Given \( f(x) = \frac{x}{1+x^2} \)
Let \( y = \frac{x}{1+x^2} \). We want to find the possible values of \( y \).
\( y(1+x^2) = x \)
\( y + yx^2 = x \)
\( yx^2 - x + y = 0 \)
This is a quadratic equation in \( x \). For \( x \) to be a real number, the discriminant \( D \) must be greater than or equal to 0.
\( D = b^2 - 4ac = (-1)^2 - 4(y)(y) \)
\( \implies 1 - 4y^2 \ge 0 \)
\( \implies 1 \ge 4y^2 \)
\( \implies 4y^2 \le 1 \)
\( \implies y^2 \le \frac{1}{4} \)
\( \implies |y| \le \frac{1}{2} \)
\( \implies -\frac{1}{2} \le y \le \frac{1}{2} \).
The range \( R_f = \left[-\frac{1}{2}, \frac{1}{2}\right] \).

(viii) Given \( f(x) = \frac{3}{2-x^2} \)
First, find the domain: \( 2-x^2 \ne 0 \implies x^2 \ne 2 \implies x \ne \pm\sqrt{2} \). The domain is \( R - \{-\sqrt{2}, \sqrt{2}\} \).
Let \( y = \frac{3}{2-x^2} \). We want to find the possible values of \( y \).
\( 2-x^2 = \frac{3}{y} \)
\( \implies x^2 = 2 - \frac{3}{y} \)
For \( x \) to be a real number, \( x^2 \) must be non-negative. So, \( 2 - \frac{3}{y} \ge 0 \).
Also, from the domain, \( y \ne 0 \).
\( \implies \frac{2y-3}{y} \ge 0 \)
To solve this inequality, we consider the signs of the numerator and denominator:
Case A: Numerator \( 2y-3 \ge 0 \implies y \ge \frac{3}{2} \) AND Denominator \( y > 0 \). The intersection is \( y \ge \frac{3}{2} \).
Case B: Numerator \( 2y-3 \le 0 \implies y \le \frac{3}{2} \) AND Denominator \( y < 0 \). The intersection is \( y < 0 \).
Combining these, the range is \( (-\infty, 0) \cup \left[\frac{3}{2}, \infty\right) \).
The range \( R_f = (-\infty, 0) \cup \left[\frac{3}{2}, \infty\right) \).
In simple words: For each function, we try to see what kinds of answers we can get out. For absolute values and square roots, answers are usually positive or zero. For cosine, answers are between -1 and 1. For functions with \( x \) in the denominator, we check what happens when \( x \) is very big or very small, or when the denominator is close to zero. Sometimes, we turn the function into an equation for \( x \) and use the rule that \( x \) must be a real number.

🎯 Exam Tip: Finding the range can be more complex than finding the domain. Techniques include: using known ranges of basic functions (like sin/cos), considering limiting values, rewriting the function to isolate the constant term (as in part vi), or using the discriminant for quadratic equations (as in part vii and viii) by setting \( y=f(x) \) and solving for \( x \).

Question 8. Find the domain and range of the function \( \frac{x^2-4}{x-2} \).
Answer: Given the function \( f(x) = \frac{x^2-4}{x-2} \).

Domain:
For \( f(x) \) to be a real number, the denominator cannot be equal to zero.
\( x-2 \ne 0 \)
\( \implies x \ne 2 \)
So, the domain of the function includes all real numbers except 2.
The domain \( D_f = R - \{2\} \).

Range:
We can simplify the function by factoring the numerator:
\( f(x) = \frac{(x-2)(x+2)}{x-2} \)
Since we know from the domain that \( x \ne 2 \), we can safely cancel the \( (x-2) \) terms.
\( \implies f(x) = x+2 \), but with the restriction that \( x \ne 2 \).
Now, if \( x \ne 2 \), then \( x+2 \) cannot be equal to \( 2+2=4 \).
So, \( f(x) \) can take any real value except 4.
The range \( R_f = R - \{4\} \).
In simple words: First, for the domain, we just make sure we don't divide by zero, so x cannot be 2. Then, we simplify the function by canceling out \( (x-2) \) from the top and bottom. This leaves us with \( x+2 \). But since x cannot be 2, the answer \( x+2 \) can never be 4. So, the function can give any answer except 4.

🎯 Exam Tip: Always simplify rational functions (fractions with polynomials) *after* determining the initial domain. The simplified form helps find the range, but the domain restriction from the original denominator must still be applied to the simplified function's output.

Question 9. If the domain of the function \( \frac{|x|}{x} \) be [3, 7], then its range is
(a) [- 1, 1]
(b) [- 1, 1]
(c) {1}
(d) {-1}
Answer: (c) {1}
Given the function \( f(x) = \frac{|x|}{x} \).
The domain of the function is given as \( [3, 7] \). This means that \( x \) can be any real number from 3 to 7, including 3 and 7.
For any \( x \) in the interval \( [3, 7] \), \( x \) is always a positive number. When \( x \) is positive, the absolute value of \( x \), denoted as \( |x| \), is simply equal to \( x \).
So, for \( x \in [3, 7] \), we can replace \( |x| \) with \( x \).
\( f(x) = \frac{x}{x} \)
Since \( x \) is never zero in the given domain \( [3, 7] \), we can safely divide \( x \) by \( x \).
\( f(x) = 1 \).
This means that for every value of \( x \) in the domain \( [3, 7] \), the function's output is always 1.
Therefore, the range of the function is just the set containing the number 1.
The range \( R_f = \{1\} \).
In simple words: The numbers we can use are all positive, from 3 to 7. When we have a positive number, \( |x| \) is the same as \( x \). So, the function becomes \( x \) divided by \( x \), which is always 1. So, the only answer we ever get is 1.

🎯 Exam Tip: When evaluating functions involving absolute values, always consider the sign of the expression inside the absolute value. If the domain restricts the input to be consistently positive or negative, the absolute value function simplifies greatly, making the range easier to find.

Question 1. The range and domain of function \( f(x) = \frac { 3 }{ x } + 1 \) are subsets of A and B respectively, where \( A = \left\{-\frac{1}{2}, 0, \frac{2}{3}, \frac{6}{7}, 1\right\} \) and \( B = \left\{-5,0,4 \frac{1}{2}, 5,5 \frac{1}{2}\right\} \). List the elements of the function as ordered pairs.
Answer: We are given the function \( f(x) = \frac { 3 }{ x } + 1 \). We need to find the output for each element in set A and check if it belongs to set B.
For \( x = -\frac{1}{2} \):
\( f\left(-\frac{1}{2}\right) = \frac{3}{-\frac{1}{2}} + 1 = -6 + 1 = -5 \)
Since \( -5 \in B \), the ordered pair is \( \left(-\frac{1}{2}, -5\right) \).
For \( x = 0 \):
\( f(0) = \frac{3}{0} + 1 \). Division by zero is undefined, so \( f(0) \) does not exist. This means 0 is not in the domain of this function, even if it's in set A.
For \( x = \frac{2}{3} \):
\( f\left(\frac{2}{3}\right) = \frac{3}{\frac{2}{3}} + 1 = \frac{9}{2} + 1 = 4.5 + 1 = 5.5 \)
Since \( 5.5 \in B \) (as \( 5 \frac{1}{2} = 5.5 \)), the ordered pair is \( \left(\frac{2}{3}, 5.5\right) \).
For \( x = \frac{6}{7} \):
\( f\left(\frac{6}{7}\right) = \frac{3}{\frac{6}{7}} + 1 = \frac{3 \times 7}{6} + 1 = \frac{21}{6} + 1 = 3.5 + 1 = 4.5 \)
Since \( 4.5 \in B \) (as \( 4 \frac{1}{2} = 4.5 \)), the ordered pair is \( \left(\frac{6}{7}, 4.5\right) \).
For \( x = 1 \):
\( f(1) = \frac{3}{1} + 1 = 3 + 1 = 4 \)
Since \( 4 \notin B \) (B contains 0, 4.5, 5.5), 1 is not part of the domain for which the range is B.
Therefore, the elements of the function as ordered pairs are:
\( f = \left\{\left(-\frac{1}{2}, -5\right), \left(\frac{2}{3}, 5.5\right), \left(\frac{6}{7}, 4.5\right)\right\} \).
In simple words: We took each number from set A, put it into the function \( f(x) = \frac{3}{x} + 1 \), and checked if the answer was in set B. If it was, we wrote down the pair of numbers. We couldn't use 0 because you can't divide by zero, and 1 didn't work because its answer wasn't in set B.

🎯 Exam Tip: When given a specific domain and range, always verify that the function output for each domain element actually falls within the specified range to form valid ordered pairs.

Question 2. \( A = \{- 2, − 1, 1, 2\} \) and \( f = \left\{\left(x, \frac{1}{x}\right), x \in A\right\} \).
(i) List the domain of f
(ii) List the range of f
(iii) Is f a function ?
Answer: The function \( f \) is defined as \( f(x) = \frac{1}{x} \) for \( x \in A = \{- 2, -1, 1, 2\} \).
Let's find the value of \( f(x) \) for each element in A:
For \( x = -2 \): \( f(-2) = -\frac{1}{2} \)
For \( x = -1 \): \( f(-1) = -1 \)
For \( x = 1 \): \( f(1) = 1 \)
For \( x = 2 \): \( f(2) = \frac{1}{2} \)
So, the function can be written as a set of ordered pairs: \( f = \left\{(-2, -\frac{1}{2}), (-1, -1), (1, 1), (2, \frac{1}{2})\right\} \).

(i) **Domain of f:** The domain is the set of all first elements in the ordered pairs. In this case, it is the set A itself, as \( \frac{1}{x} \) is defined for all non-zero values in A.
Therefore, Domain (f) = \( \{-2, -1, 1, 2\} \).

(ii) **Range of f:** The range is the set of all second elements (output values) in the ordered pairs.
Therefore, Range (f) = \( \left\{-\frac{1}{2}, -1, 1, \frac{1}{2}\right\} \).

(iii) **Is f a function?** A relation is a function if each element in the domain has exactly one unique image (output) in the range. Looking at our ordered pairs, each element from A (the domain) appears only once as the first component, and it maps to a unique value. For instance, -2 maps only to \( -\frac{1}{2} \), -1 maps only to -1, and so on. This ensures a clear mapping.
Therefore, yes, \( f \) represents a function.
In simple words: We calculated the output for each number in set A using the rule \( f(x) = \frac{1}{x} \). The domain is all the input numbers, and the range is all the output numbers. Since each input number gives only one output number, this is indeed a function.

🎯 Exam Tip: To determine if a relation is a function, always check if every element in the domain maps to *exactly one* element in the range. If any domain element maps to more than one range element, it is not a function.

Question 3. \( f : A \rightarrow \) highest prime factor of \( x \).
(i) Find the range of \( f \) when the domain is \( \{12, 13, 14, 15, 16, 17\} \).
(ii) State a domain of five integers for which the range is \( \{3\} \).
(iii) A set of positive integers is called S. What can be said about these integers if \( f(S) = S \)?
Answer: The function \( f(x) \) gives the highest prime factor of \( x \).

(i) **Range of \( f \) for domain \( \{12, 13, 14, 15, 16, 17\} \):**
For \( x = 12 \): Factors are 1, 2, 3, 4, 6, 12. Prime factors are 2, 3. Highest prime factor is 3.
\( \implies 3 \in R_f \)
For \( x = 13 \): Factors are 1, 13. Prime factor is 13. Highest prime factor is 13.
\( \implies 13 \in R_f \)
For \( x = 14 \): Factors are 1, 2, 7, 14. Prime factors are 2, 7. Highest prime factor is 7.
\( \implies 7 \in R_f \)
For \( x = 15 \): Factors are 1, 3, 5, 15. Prime factors are 3, 5. Highest prime factor is 5.
\( \implies 5 \in R_f \)
For \( x = 16 \): Factors are 1, 2, 4, 8, 16. Prime factor is 2. Highest prime factor is 2.
\( \implies 2 \in R_f \)
For \( x = 17 \): Factors are 1, 17. Prime factor is 17. Highest prime factor is 17.
\( \implies 17 \in R_f \)
Therefore, the range of \( f \) is \( \{2, 3, 5, 7, 13, 17\} \).

(ii) **Domain for which the range is \( \{3\} \):** We need to find five integers whose highest prime factor is 3. This means we are looking for numbers that are products of primes, and the largest prime factor in them is 3. For example, 3 itself has a highest prime factor of 3. Also, numbers like \( 3 \times 1 = 3 \), \( 3 \times 2 = 6 \) (highest prime factor is 3, not 2), \( 3 \times 3 = 9 \), \( 3 \times 2 \times 2 = 12 \) (highest prime factor is 3), \( 3 \times 5 = 15 \) (highest prime factor is 5, not 3). So we need to ensure the highest prime factor is 3. Examples of numbers whose highest prime factor is 3 are 3, 6, 9, 12, 18 (highest prime factor of 15 is 5, so 15 is excluded).
Therefore, a required domain \( D_f = \{3, 6, 9, 12, 18\} \).

(iii) **If \( f(S) = S \) for a set S of positive integers:** If the set S contains only prime numbers, then for any prime number \( p \in S \), its highest prime factor is \( p \) itself. So \( f(p) = p \). This means if S is the set of all prime numbers, then \( f(S) \) would indeed be S. Every prime number's highest prime factor is itself. This ensures that applying the function to the set gives the set back. S must be the set of all prime numbers.
In simple words: For the first part, we found the biggest prime number that can divide each number in the list. For the second part, we looked for numbers whose biggest prime factor is 3. For the third part, if a set of numbers gives itself back when you find the highest prime factor of each, that set must be made up only of prime numbers. This is because a prime number's biggest prime factor is just the number itself.

🎯 Exam Tip: Clearly list the prime factors for each number to correctly identify the "highest" one. For part (ii), be careful to pick numbers whose *highest* prime factor is exactly 3, not a larger prime.

Question 4. A function \( f \) is defined on the set of integers as follows :
\( f(x) = \left\{\begin{aligned} 1+x ; & 1 \leq x<2 \\ 2 x-1 ; & 2 \leq x<4 \\ 3 x-10 ; & 4 \leq x<6 \end{aligned}\right\} \)
(i) Find the domain of the function
(ii) Find the range of the function
(iii) Find the value of \( f(4) \).
Answer: The function \( f(x) \) is defined for integers based on specific conditions:
\( f(x) = \left\{\begin{aligned} 1+x ; & 1 \leq x<2 \\ 2 x-1 ; & 2 \leq x<4 \\ 3 x-10 ; & 4 \leq x<6 \end{aligned}\right\} \)
Since \( f \) is defined on the set of integers, we look for integer values that satisfy the conditions.

(i) **Domain of the function:** The conditions specify the range of \( x \) values. The first rule applies for \( x=1 \). The second rule applies for \( x=2, 3 \). The third rule applies for \( x=4, 5 \). The condition \( x<6 \) means 6 is not included. So, the integers for which the function is defined are 1, 2, 3, 4, 5.
Therefore, the domain \( D_f = \{1, 2, 3, 4, 5\} \).

(ii) **Range of the function:** We calculate \( f(x) \) for each value in the domain:
When \( x = 1 \): \( f(1) = 1 + 1 = 2 \)
When \( x = 2 \): \( f(2) = (2 \times 2) - 1 = 4 - 1 = 3 \)
When \( x = 3 \): \( f(3) = (2 \times 3) - 1 = 6 - 1 = 5 \)
When \( x = 4 \): \( f(4) = (3 \times 4) - 10 = 12 - 10 = 2 \)
When \( x = 5 \): \( f(5) = (3 \times 5) - 10 = 15 - 10 = 5 \)
The set of all unique output values is the range.
Therefore, the range \( R_f = \{2, 3, 5\} \).

(iii) **Value of \( f(4) \):** From our calculations above, when \( x=4 \), the third rule \( f(x) = 3x - 10 \) applies.
\( f(4) = (3 \times 4) - 10 = 12 - 10 = 2 \).
In simple words: The function has different rules for different input numbers. For part (i), we found all the integer numbers that fit into any of the rules. For part (ii), we put each of those numbers into the correct rule and wrote down all the answers we got. For part (iii), we just used the rule that applied to the number 4 to find its specific answer.

🎯 Exam Tip: For piecewise functions, carefully identify which rule applies to each input value in the domain. Misapplying a rule for a specific \( x \) value is a common mistake.

Question 5. Let \( f \) be a function whose domain is the set of all real numbers. If \( f(x) = | x | – x \), what is range of \( f \)?
Answer: The function is given as \( f(x) = |x| - x \). We need to find its range when the domain is all real numbers.
We can consider two cases for \( x \):
**Case 1: \( x \geq 0 \)**
If \( x \) is a non-negative number, then \( |x| = x \).
So, \( f(x) = x - x = 0 \).
**Case 2: \( x < 0 \)**
If \( x \) is a negative number, then \( |x| = -x \).
So, \( f(x) = -x - x = -2x \).
Since \( x < 0 \), it means \( -x > 0 \). Therefore, \( -2x \) will always be a positive number. For example, if \( x = -1 \), \( f(x) = -2(-1) = 2 \). If \( x = -5 \), \( f(x) = -2(-5) = 10 \).
Combining both cases: If \( x \geq 0 \), \( f(x) = 0 \). If \( x < 0 \), \( f(x) \) is a positive real number (ranging from just above 0 to infinity). This function always produces non-negative values. The smallest value it produces is 0.
Therefore, the range of \( f \) is \( [0, \infty) \).
In simple words: The function \( f(x) \) takes a number, finds its absolute value, and then subtracts the original number. If the number is zero or positive, the answer is always zero. If the number is negative, the answer will always be a positive number. So, the function will only give answers that are zero or greater than zero, going all the way up to infinity.

🎯 Exam Tip: When dealing with absolute value functions, always split the problem into cases based on whether the expression inside the absolute value is positive or negative. This helps correctly evaluate the function for its entire domain.

Question 6. Write the domain of the following real functions
(i) \( \sqrt{9-x^2} \)
(ii) \( \sqrt{1-2 x-3 x^2} \)
(iii) \( 10^x \)
(iv) \( \frac{1}{\sqrt{x^2-7}} \)
(v) \( \log (2 – 3x) \)
(vi) \( \log(\sqrt{x-4}+\sqrt{6-x}) \)
(vii) \( \left[\log _{10}\left(\frac{5 x-x^2}{4}\right)\right]^{\frac{1}{2}} \)
(viii) \( \sin ^{-1}\left[\log _2\left(\frac{x}{2}\right)\right] \)
Answer: For a function to be a real function, the expressions must result in real numbers.

(i) **For \( f(x) = \sqrt{9-x^2} \):**
For \( f(x) \) to be a real number, the term inside the square root must be non-negative.
\( 9 - x^2 \geq 0 \)
\( 9 \geq x^2 \)
\( x^2 \leq 9 \)
This means that \( |x| \leq 3 \), so \( -3 \leq x \leq 3 \).
Thus, the domain \( D_f = [-3, 3] \).

(ii) **For \( f(x) = \sqrt{1-2 x-3 x^2} \):**
For \( f(x) \) to be a real number, the term inside the square root must be non-negative.
\( 1 - 2x - 3x^2 \geq 0 \)
Multiply by -1 and reverse the inequality sign:
\( 3x^2 + 2x - 1 \leq 0 \)
Factor the quadratic expression:
\( 3x^2 + 3x - x - 1 \leq 0 \)
\( 3x(x+1) - 1(x+1) \leq 0 \)
\( (3x-1)(x+1) \leq 0 \)
For this inequality to hold, \( (3x-1) \) and \( (x+1) \) must have opposite signs, or one of them must be zero. This happens when \( -1 \leq x \leq \frac{1}{3} \). We can visualize this on a number line or test values. The roots are -1 and \( \frac{1}{3} \), and the parabola \( y = 3x^2 + 2x - 1 \) opens upwards, so it is below the x-axis between its roots.
Thus, the domain \( D_f = \left[-1, \frac{1}{3}\right] \).

(iii) **For \( f(x) = 10^x \):**
This is an exponential function. For any real number \( x \), \( 10^x \) is always a real number. Exponential functions are defined for all real numbers.
Thus, the domain \( D_f = \mathbb{R} \) (all real numbers).

(iv) **For \( f(x) = \frac{1}{\sqrt{x^2-7}} \):**
For \( f(x) \) to be a real number, two conditions must be met: the term under the square root must be positive (it cannot be zero because it's in the denominator), and the square root itself must exist.
\( x^2 - 7 > 0 \)
\( x^2 > 7 \)
This means \( |x| > \sqrt{7} \). So, \( x > \sqrt{7} \) or \( x < -\sqrt{7} \).
Thus, the domain \( D_f = (-\infty, -\sqrt{7}) \cup (\sqrt{7}, \infty) \).

(v) **For \( f(x) = \log (2 – 3x) \):**
For a logarithm to be defined, its argument (the term inside the logarithm) must be strictly positive.
\( 2 - 3x > 0 \)
\( 2 > 3x \)
\( \frac{2}{3} > x \)
\( x < \frac{2}{3} \)
Thus, the domain \( D_f = \left(-\infty, \frac{2}{3}\right) \).

(vi) **For \( f(x) = \log(\sqrt{x-4}+\sqrt{6-x}) \):**
For \( f(x) \) to be a real number, several conditions must be met:
1. The terms inside the square roots must be non-negative:
\( x-4 \geq 0 \implies x \geq 4 \)
\( 6-x \geq 0 \implies 6 \geq x \)
Combining these, we get \( 4 \leq x \leq 6 \).
2. The argument of the logarithm must be strictly positive:
\( \sqrt{x-4}+\sqrt{6-x} > 0 \)
Since square roots always produce non-negative values, their sum will be positive unless both are zero. If \( \sqrt{x-4}=0 \) and \( \sqrt{6-x}=0 \), then \( x=4 \) and \( x=6 \), which is not possible simultaneously. So, as long as \( x \) is between 4 and 6 (inclusive), the sum of the square roots will be positive. For instance, if \( x=4 \), \( \sqrt{0} + \sqrt{2} = \sqrt{2} > 0 \). If \( x=6 \), \( \sqrt{2} + \sqrt{0} = \sqrt{2} > 0 \). If \( x=5 \), \( \sqrt{1} + \sqrt{1} = 2 > 0 \).
So, the condition \( 4 \leq x \leq 6 \) ensures the logarithm's argument is positive.
Thus, the domain \( D_f = [4, 6] \).

(vii) **For \( f(x) = \left[\log _{10}\left(\frac{5 x-x^2}{4}\right)\right]^{\frac{1}{2}} \):**
For \( f(x) \) to be a real number, the term inside the outer square root must be non-negative. Also, the logarithm's argument must be strictly positive.
1. Argument of the logarithm must be positive:
\( \frac{5x - x^2}{4} > 0 \)
Since the denominator 4 is positive, we need \( 5x - x^2 > 0 \).
\( x(5 - x) > 0 \)
This inequality holds when \( 0 < x < 5 \).
2. The value of the logarithm must be non-negative:
\( \log _{10}\left(\frac{5x - x^2}{4}\right) \geq 0 \)
This means \( \frac{5x - x^2}{4} \geq 10^0 \)
\( \frac{5x - x^2}{4} \geq 1 \)
\( 5x - x^2 \geq 4 \)
\( 0 \geq x^2 - 5x + 4 \)
\( x^2 - 5x + 4 \leq 0 \)
Factor this quadratic: \( (x-1)(x-4) \leq 0 \)
This inequality holds when \( 1 \leq x \leq 4 \).
We need both conditions to be true. The intersection of \( (0, 5) \) and \( [1, 4] \) is \( [1, 4] \).
Thus, the domain \( D_f = [1, 4] \).

(viii) **For \( f(x) = \sin ^{-1}\left[\log _2\left(\frac{x}{2}\right)\right] \):**
For \( \sin^{-1}(y) \) to be defined, its argument \( y \) must be in the range \( [-1, 1] \).
So, we need \( -1 \leq \log _2\left(\frac{x}{2}\right) \leq 1 \).
Also, for \( \log_2 \) to be defined, its argument \( \frac{x}{2} \) must be strictly positive:
\( \frac{x}{2} > 0 \implies x > 0 \).
Now, let's solve the inequality \( -1 \leq \log _2\left(\frac{x}{2}\right) \leq 1 \).
We can rewrite this in exponential form:
\( 2^{-1} \leq \frac{x}{2} \leq 2^1 \)
\( \frac{1}{2} \leq \frac{x}{2} \leq 2 \)
Multiply all parts by 2:
\( 1 \leq x \leq 4 \)
Combining this with \( x > 0 \), the overall domain is \( [1, 4] \).
Thus, the domain \( D_f = [1, 4] \).
In simple words: For each function, we had to find the values of \( x \) that would make the function give a real number answer. For square roots, the inside part can't be negative. For fractions, the bottom part can't be zero. For logarithms, the inside part must be positive. For inverse sine, the inside part must be between -1 and 1. We solved these conditions for each problem to find the set of allowed \( x \) values.

🎯 Exam Tip: When finding domains, consider all restrictions simultaneously: denominators cannot be zero, square roots must have non-negative arguments, logarithms must have positive arguments, and inverse trigonometric functions have specific argument ranges. Always find the intersection of all conditions.

Question 7. Find the range of each of the following function
(i) \( |x-3| \)
(ii) \( \sqrt{x-5} \)
(iii) \( \cos(\frac { x }{ 3 }) \)
(iv) \( \frac{x+2}{|x+2|} \)
(v) \( \sec \left(\frac{\pi}{4} \cos ^2 x\right) \), \( -\infty < x < \infty \)
(vi) \( \frac{x^2+x+2}{x^2+x+1} \)
(vii) \( \frac{x}{1+x^2} \)
(viii) \( f(x) = \frac{3}{2-x^2} \)
Answer: We need to determine the set of all possible output values (the range) for each function.

(i) **For \( y = f(x) = |x-3| \):**
The absolute value of any real number is always non-negative. This means \( |x-3| \geq 0 \) for all real values of \( x \). The minimum value is 0, which occurs when \( x-3=0 \), so \( x=3 \). The value can go up to infinity.
The domain of \( f(x) = |x-3| \) is all real numbers, \( D_f = \mathbb{R} \).
Therefore, the range \( R_f = [0, \infty) \).

(ii) **For \( y = f(x) = \sqrt{x-5} \):**
First, find the domain. For \( \sqrt{x-5} \) to be a real number, \( x-5 \geq 0 \), so \( x \geq 5 \). The domain is \( D_f = [5, \infty) \).
The square root symbol \( \sqrt{} \) always gives a non-negative value. The smallest value occurs when \( x=5 \), so \( \sqrt{5-5} = \sqrt{0} = 0 \). As \( x \) increases, \( \sqrt{x-5} \) also increases towards infinity.
Therefore, the range \( R_f = [0, \infty) \).

(iii) **For \( y = f(x) = \cos(\frac { x }{ 3 }) \):**
The cosine function is defined for all real numbers. Its domain is \( D_f = \mathbb{R} \).
The range of the cosine function, regardless of its argument (like \( \frac{x}{3} \)), is always between -1 and 1, inclusive.
Therefore, the range \( R_f = [-1, 1] \).

(iv) **For \( f(x) = \frac{x+2}{|x+2|} \):**
First, find the domain. The denominator \( |x+2| \) cannot be zero, so \( x+2 \neq 0 \), which means \( x \neq -2 \). The domain is \( D_f = \mathbb{R} - \{-2\} \).
We can consider two cases:
**Case 1: \( x+2 > 0 \)** (i.e., \( x > -2 \))
In this case, \( |x+2| = x+2 \). So, \( f(x) = \frac{x+2}{x+2} = 1 \).
**Case 2: \( x+2 < 0 \)** (i.e., \( x < -2 \))
In this case, \( |x+2| = -(x+2) \). So, \( f(x) = \frac{x+2}{-(x+2)} = -1 \).
Thus, the function can only take on values 1 or -1.
Therefore, the range \( R_f = \{-1, 1\} \).

(v) **For \( y = f(x) = \sec \left(\frac{\pi}{4} \cos ^2 x\right) \):**
We know that \( 0 \leq \cos^2 x \leq 1 \) for all real \( x \). This is because \( -1 \leq \cos x \leq 1 \), so squaring it makes it non-negative.
Multiply by \( \frac{\pi}{4} \): \( 0 \leq \frac{\pi}{4} \cos^2 x \leq \frac{\pi}{4} \).
Let \( \theta = \frac{\pi}{4} \cos^2 x \). So, \( \theta \in \left[0, \frac{\pi}{4}\right] \).
Now we need to find the range of \( \sec \theta \) for \( \theta \in \left[0, \frac{\pi}{4}\right] \).
The secant function is \( \frac{1}{\cos \theta} \). For \( \theta \in \left[0, \frac{\pi}{4}\right] \), \( \cos \theta \) is positive and decreasing from \( \cos(0)=1 \) to \( \cos(\frac{\pi}{4})=\frac{1}{\sqrt{2}} \).
Therefore, \( \sec \theta \) will be increasing from \( \sec(0) = \frac{1}{1} = 1 \) to \( \sec(\frac{\pi}{4}) = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \).
Therefore, the range \( R_f = [1, \sqrt{2}] \).

(vi) **For \( y = f(x) = \frac{x^2+x+2}{x^2+x+1} \):**
Let \( y = \frac{x^2+x+2}{x^2+x+1} \). We can rewrite this as:
\( y = \frac{(x^2+x+1) + 1}{x^2+x+1} = 1 + \frac{1}{x^2+x+1} \)
First, consider the denominator \( x^2+x+1 \). We can complete the square:
\( x^2+x+1 = \left(x+\frac{1}{2}\right)^2 - \frac{1}{4} + 1 = \left(x+\frac{1}{2}\right)^2 + \frac{3}{4} \)
Since \( \left(x+\frac{1}{2}\right)^2 \geq 0 \), it means \( x^2+x+1 \geq \frac{3}{4} \).
Also, \( x^2+x+1 \) has a minimum value of \( \frac{3}{4} \) (when \( x = -\frac{1}{2} \)) and can go to \( \infty \). It is always positive, so the domain is \( \mathbb{R} \).
Now consider \( \frac{1}{x^2+x+1} \):
Since \( x^2+x+1 \geq \frac{3}{4} \), then \( 0 < \frac{1}{x^2+x+1} \leq \frac{1}{\frac{3}{4}} \)
So, \( 0 < \frac{1}{x^2+x+1} \leq \frac{4}{3} \).
Now add 1 to all parts to find the range of \( y = 1 + \frac{1}{x^2+x+1} \):
\( 1 + 0 < 1 + \frac{1}{x^2+x+1} \leq 1 + \frac{4}{3} \)
\( 1 < y \leq \frac{7}{3} \)
Therefore, the range \( R_f = \left(1, \frac{7}{3}\right] \).

(vii) **For \( y = f(x) = \frac{x}{1+x^2} \):**
The domain of this function is all real numbers because the denominator \( 1+x^2 \) is always positive and never zero.
Let \( y = \frac{x}{1+x^2} \). We want to find the possible values of \( y \).
\( y(1+x^2) = x \)
\( y + yx^2 = x \)
\( yx^2 - x + y = 0 \)
This is a quadratic equation in \( x \). For \( x \) to be a real number, the discriminant must be non-negative. (If \( y=0 \), then \( -x=0 \implies x=0 \), so \( y=0 \) is possible).
Discriminant \( D = (-1)^2 - 4(y)(y) = 1 - 4y^2 \).
For real \( x \), \( D \geq 0 \):
\( 1 - 4y^2 \geq 0 \)
\( 1 \geq 4y^2 \)
\( 4y^2 \leq 1 \)
\( y^2 \leq \frac{1}{4} \)
This means \( |y| \leq \frac{1}{2} \), so \( -\frac{1}{2} \leq y \leq \frac{1}{2} \).
Therefore, the range \( R_f = \left[-\frac{1}{2}, \frac{1}{2}\right] \).

(viii) **For \( f(x) = \frac{3}{2-x^2} \):**
First, find the domain. The denominator cannot be zero:
\( 2-x^2 \neq 0 \implies x^2 \neq 2 \implies x \neq \pm \sqrt{2} \).
The domain is \( D_f = \mathbb{R} - \{-\sqrt{2}, \sqrt{2}\} \).
Let \( y = \frac{3}{2-x^2} \). We want to find the range of \( y \).
\( 2-x^2 = \frac{3}{y} \)
\( x^2 = 2 - \frac{3}{y} \)
For \( x^2 \) to be a real number, it must be non-negative. Also, \( y \) cannot be 0 (as \( 3/0 \) is undefined).
\( x^2 \geq 0 \)
\( 2 - \frac{3}{y} \geq 0 \)
To solve this inequality, combine terms:
\( \frac{2y - 3}{y} \geq 0 \)
This inequality holds if \( y > 0 \) and \( 2y-3 \geq 0 \implies y \geq \frac{3}{2} \). (So \( y \geq \frac{3}{2} \)).
Or if \( y < 0 \) and \( 2y-3 \leq 0 \implies y \leq \frac{3}{2} \). (So \( y < 0 \)).
Combining these, we get \( y \in (-\infty, 0) \cup \left[\frac{3}{2}, \infty\right) \).
Therefore, the range \( R_f = (-\infty, 0) \cup \left[\frac{3}{2}, \infty\right) \).
In simple words: For each function, we looked at what values the output could possibly be. For absolute values and square roots, the answer is usually zero or positive. For cosine, the answer is always between -1 and 1. For fractions with absolute values, the answer is often just -1 or 1. For other complex fractions, we sometimes rewrite the function or use properties of quadratic equations to find the lowest and highest possible output values.

🎯 Exam Tip: Finding the range often involves algebraic manipulation to express \( x \) in terms of \( y \) and then applying domain restrictions to \( y \), or analyzing the behavior of the function by understanding its component parts (e.g., bounds of \( \cos x \), non-negativity of square roots).

Question 8. Find the domain and range of the function \( \frac{x^2-4}{x-2} \).
Answer: The given function is \( f(x) = \frac{x^2-4}{x-2} \).

**Domain:**
For a rational function, the denominator cannot be zero.
\( x-2 \neq 0 \)
\( x \neq 2 \)
So, the domain \( D_f \) is all real numbers except 2.
Therefore, \( D_f = \mathbb{R} - \{2\} \).

**Range:**
We can simplify the function by factoring the numerator:
\( f(x) = \frac{(x-2)(x+2)}{x-2} \)
Since \( x \neq 2 \) (from the domain), we can cancel out the \( (x-2) \) term.
So, \( f(x) = x+2 \), but with the restriction that \( x \neq 2 \).
Now, if \( f(x) = x+2 \) and \( x \neq 2 \), then we need to find what value \( f(x) \) cannot take.
If \( x=2 \) were allowed, \( f(x) \) would be \( 2+2 = 4 \).
Because \( x \) cannot be 2, \( f(x) \) cannot be 4.
So, the range is all real numbers except 4.
Therefore, \( R_f = \mathbb{R} - \{4\} \).
In simple words: First, we found the domain by making sure the bottom part of the fraction is not zero, so \( x \) cannot be 2. Next, we simplified the fraction by canceling common terms. This showed that the function is like \( x+2 \), but since \( x \) can't be 2, the function can't give an answer of \( 2+2=4 \). So, the range includes all numbers except 4.

🎯 Exam Tip: When a rational function can be simplified by canceling terms, remember to apply the original domain restrictions to the simplified form. The "hole" in the graph at the canceled point affects the range.

Question 9. If the domain of the function \( \frac{|x|}{x} \) be \( [3, 7] \), then its range is
(a) \( [- 1, 1] \)
(b) \( [- 1, 1] \)
(c) \( \{1\} \)
(d) \( \{-1\} \)
Answer: (c) {1}
The given function is \( f(x) = \frac{|x|}{x} \). The domain of the function is specified as \( [3, 7] \).
This means that \( 3 \leq x \leq 7 \).
For any \( x \) in this domain, \( x \) is a positive number. Specifically, \( x > 0 \).
When \( x > 0 \), the absolute value of \( x \), i.e., \( |x| \), is simply equal to \( x \).
So, for the given domain, \( f(x) = \frac{x}{x} \).
Since \( x \neq 0 \) in the domain \( [3, 7] \), we can simplify \( \frac{x}{x} \) to 1.
Therefore, for all \( x \in [3, 7] \), \( f(x) = 1 \).
The function only takes on one value, 1.
The range of the function is \( \{1\} \).
In simple words: The function divides the absolute value of a number by the number itself. Since the given numbers are all positive (from 3 to 7), the absolute value is just the number itself. So, for all these numbers, the function always gives 1.

🎯 Exam Tip: Pay close attention to the given domain. If the domain consists of only positive numbers, then \( |x| \) can be directly replaced by \( x \), simplifying the function significantly.