GSEB Class 12 Maths Solutions Chapter 1 સંબંધ અને વિધેય Exercise 1.2

GSEB Solutions Class 12 Maths Chapter 1 સંબંધ અને વિધેય Ex 1.2

Question 1. ધારો કે R* તમામ શૂન્યેતર વાસ્તવિક સંખ્યાઓનો ગણ છે. સાબિત કરો કે, વિધેય \( f : R* \to R* \), \( f(x) = \frac{1}{x} \) વડે વ્યાખ્યાયિત વિધેય એક એક અને વ્યાપ્ત છે. જો પ્રદેશ R* ના બદલે N લેવામાં આવે અને સહપ્રદેશ R* જ રહે તો શું આ પરિણામ સત્ય રહેશે ?
Answer:Let's define the function \( f: R* \to R* \) as \( f(x) = \frac{1}{x} \). Consider two elements \( x_1, x_2 \in R* \). If \( f(x_1) = f(x_2) \), then \( \frac{1}{x_1} = \frac{1}{x_2} \), which means \( x_1 = x_2 \). Thus, the function \( f \) is one-to-one (એક-એક). Now, let's check if it is onto (વ્યાપ્ત). Let \( y \in R* \) and \( y = f(x) \).
\( \implies y = \frac{1}{x} \)
\( \implies x = \frac{1}{y} \) We can see that for every \( y \in R* \), there exists an \( x = \frac{1}{y} \in R* \) such that \( f(x) = f\left(\frac{1}{y}\right) = \frac{1}{\frac{1}{y}} = y \). Therefore, the function \( f \) is onto (વ્યાપ્ત). Now, consider the case where the domain of function \( f \) is N (natural numbers) instead of R*, and the codomain remains R*. In this case, the result is not true. This means \( f(x) = \frac{1}{x} \) is neither one-to-one nor onto. This is because in N, reciprocals of numbers are not always found within N itself. For example, for \( x=2 \in N \), \( f(2) = \frac{1}{2} \notin N \) if N were the codomain. If codomain is R*, then it's one-to-one but not onto from N to R* (e.g., \( y = \frac{1}{2} \) has no preimage if x needs to be in N). Also, numbers like 2 in R* won't have a preimage from N.In simple words: The function \( f(x) = \frac{1}{x} \) maps distinct numbers to distinct numbers and covers all possible output values when operating on non-zero real numbers. However, if the input is restricted to natural numbers, it won't cover all real number outputs, and some outputs might not have a natural number input.

🎯 Exam Tip: When proving one-to-one, show that \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). For onto, show that for every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \). Pay close attention to the domain and codomain sets, as they are crucial for determining these properties.

Question 2. નીચે આપેલ વિષયો એક-એક અથવા વ્યાપ્ત અથવા બંને ગુણધર્મ ધરાવતા વિધેયો છે કે નહિ તે ચકાસો :
(i) \( f : N \to N, f(x) = x^2 \)
(ii) \( f : Z \to Z, f(x) = x^2 \)
(iii) \( f : R \to R, f(x) = x^2 \)
(iv) \( f : N \to N, f(x) = x^3 \)
(v) \( f : Z \to Z, f(x) = x^3 \)
Answer:(i) \( f : N \to N, f(x) = x^2 \) Let \( x_1, x_2 \in N \). If \( f(x_1) = f(x_2) \), then \( x_1^2 = x_2^2 \). Since \( x_1, x_2 \in N \) (natural numbers, which are positive), \( x_1 = x_2 \) must be true. So, function \( f \) is one-to-one (એક-એક). Now, let's check if it is onto (વ્યાપ્ત). Consider \( 3 \in N \) (codomain). There is no natural number \( x \) in the domain such that \( f(x) = 3 \), because \( x^2 = 3 \implies x = \sqrt{3} \), and \( \sqrt{3} \notin N \). So, some elements in the codomain N do not have a preimage in the domain N. Therefore, function \( f \) is not onto (વ્યાપ્ત નથી).In simple words: For natural numbers, different inputs give different square outputs, so it's one-to-one. But not all natural numbers (like 3) are perfect squares, so it's not onto.

🎯 Exam Tip: When working with natural numbers (N), remember they are positive integers. For one-to-one, ensure \( x_1^2 = x_2^2 \) implies \( x_1 = x_2 \) *only* because negative roots are excluded. For onto, check if non-perfect squares in the codomain have preimages.

(ii) \( f : Z \to Z, f(x) = x^2 \) Let's check if it is one-to-one. The set of integers \( Z = \{..., -3, -2, -1, 0, 1, 2, 3, ...\} \). Consider \( f(1) = (1)^2 = 1 \) and \( f(-1) = (-1)^2 = 1 \). Here, \( f(1) = f(-1) \) but \( 1 \neq -1 \). So, function \( f \) is not one-to-one (એક-એક વિધેય નથી). Now, let's check if it is onto (વ્યાપ્ત). Consider \( 3 \in Z \) (codomain). There is no integer \( x \) in the domain such that \( f(x) = 3 \), because \( x^2 = 3 \implies x = \sqrt{3} \), and \( \sqrt{3} \notin Z \). So, some elements in the codomain Z do not have a preimage in the domain Z. Therefore, function \( f \) is not onto (વ્યાપ્ત વિધેય નથી).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फलन \( f: Z \to Z, f(x) = x^2 \) के लिए है। यह दिखाता है कि -1 और 1 दोनों का प्रतिबिंब 1 है, जिसका अर्थ है कि फलन एक-एक नहीं है। यह भी दर्शाता है कि 0 का प्रतिबिंब 0 है, लेकिन अन्य ऋणात्मक पूर्णांकों का कोई पूर्व-प्रतिबिंब नहीं है, इसलिए यह आच्छादक भी नहीं है।In simple words: For integers, both 1 and -1 give the same square (1), so it's not one-to-one. Also, numbers like 2 or 3 in the codomain are not perfect squares of any integer, so it's not onto.

🎯 Exam Tip: When the domain includes negative numbers (like Z or R), it's easy to find counterexamples for one-to-one functions like \( f(x) = x^2 \) by picking \( x \) and \( -x \). For onto, check if non-perfect squares (or numbers with non-integer roots) exist in the codomain.

(iii) \( f : R \to R, f(x) = x^2 \) Let's check if it is one-to-one. Consider \( f(2) = (2)^2 = 4 \) and \( f(-2) = (-2)^2 = 4 \). Here, \( f(2) = f(-2) \) but \( 2 \neq -2 \). So, function \( f \) is not one-to-one (એક-એક વિષય નથી). Now, let's check if it is onto (વ્યાપ્ત). Consider \( -2 \in R \) (codomain). There is no real number \( x \) in the domain such that \( f(x) = -2 \), because \( x^2 = -2 \implies x = \sqrt{-2} \), which is not a real number. So, negative real numbers in the codomain do not have a preimage in the domain R. Therefore, function \( f \) is not onto (વ્યાપ્ત વિષય નથી).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फलन \( f: R \to R, f(x) = x^2 \) के लिए है। यह दिखाता है कि -2 और 2 दोनों का प्रतिबिंब 4 है, जिसका अर्थ है कि फलन एक-एक नहीं है। यह भी दर्शाता है कि codomain में सभी नकारात्मक वास्तविक संख्याओं (जैसे -2) का कोई पूर्व-प्रतिबिंब नहीं है, इसलिए यह आच्छादक भी नहीं है।In simple words: For real numbers, numbers like 2 and -2 both give 4 when squared, so it's not one-to-one. Negative numbers in the output cannot be formed by squaring any real number, so it's not onto.

🎯 Exam Tip: For \( f(x) = x^2 \) with a real domain, remember that squaring always yields a non-negative result. This immediately implies it's not onto for any codomain containing negative numbers. The existence of both positive and negative roots (e.g., for 4, both 2 and -2) shows it's not one-to-one.

(iv) \( f : N \to N, f(x) = x^3 \) Let \( x_1, x_2 \in N \). If \( f(x_1) = f(x_2) \), then \( x_1^3 = x_2^3 \). Since \( x_1, x_2 \in N \), this means \( x_1 = x_2 \). So, function \( f \) is one-to-one (એક-એક વિષય છે). Now, let's check if it is onto (વ્યાપ્ત). Consider \( 2 \in N \) (codomain). There is no natural number \( x \) in the domain such that \( f(x) = 2 \), because \( x^3 = 2 \implies x = \sqrt[3]{2} \), and \( \sqrt[3]{2} \notin N \). So, some elements in the codomain N do not have a preimage in the domain N. Therefore, function \( f \) is not onto (વ્યાપ્ત વિષય નથી).In simple words: For natural numbers, different numbers always have different cubes, making it one-to-one. However, not every natural number is a perfect cube (like 2), so it's not onto.

🎯 Exam Tip: Cubing a number retains its sign and unique value. Thus, \( x^3 \) is generally one-to-one for N, Z, or R. For onto property, specifically for N, check if all natural numbers are perfect cubes. They are not.

(v) \( f : Z \to Z, f(x) = x^3 \) Let \( x_1, x_2 \in Z \). If \( f(x_1) = f(x_2) \), then \( x_1^3 = x_2^3 \). Since \( x_1, x_2 \in Z \), this means \( x_1 = x_2 \). So, function \( f \) is one-to-one (એક-એક વિષય છે). Now, let's check if it is onto (વ્યાપ્ત). Consider \( 5 \in Z \) (codomain). There is no integer \( x \) in the domain such that \( f(x) = 5 \), because \( x^3 = 5 \implies x = \sqrt[3]{5} \), and \( \sqrt[3]{5} \notin Z \). So, some elements in the codomain Z do not have a preimage in the domain Z. Therefore, function \( f \) is not onto (વ્યાપ્ત વિધેય નથી).In simple words: For integers, different inputs lead to different cubed outputs, so it's one-to-one. But many integers (like 5) are not perfect cubes of any integer, so it's not onto.

🎯 Exam Tip: Similar to the previous part, cubing preserves uniqueness for integers, ensuring one-to-one. However, for onto, verify if every integer is a perfect cube of another integer. It's not, making it not onto.

Question 3. \( f: R \to R, f(x) = [x] \) દ્વારા વ્યાખ્યાયિત મહત્તમ પૂર્ણાંક વિષય (Greatest integer function) એક-એક પણ નથી અને વ્યાપ્ત પણ નથી. અહીં \( [x] \), એ \( x \) થી નાના અથવા \( x \) ને સમાન તમામ પૂર્ણાંકોમાં મહત્તમ પૂર્ણાંક દર્શાવે છે, બીજા શબ્દોમાં \( x \) થી અધિક નહિ તેવા પૂર્ણાંકોમાં સૌથી મોટો પૂર્ણાંક \( x \) છે.
Answer:Let the function be \( f : R \to R, f(x) = [x] \). Here, \( [x] \) represents the greatest integer less than or equal to \( x \). It is clear that: If \( 1 \le x < 2 \), then \( f(x) = 1 \). If \( 2 \le x < 3 \), then \( f(x) = 2 \). For example, \( f(1.5) = 1 \), \( f(1.8) = 1 \), \( f(1.9) = 1 \). Since \( f(1.5) = f(1.8) = 1 \) but \( 1.5 \neq 1.8 \), the function \( f \) is not one-to-one (એક-એક વિષય નથી). Now, let's check if it is onto (વ્યાપ્ત). The range ( વિસ્તાર ) of \( f \) is the set of all integers \( Z \), because \( [x] \) always gives an integer. The codomain of \( f \) is \( R \) (real numbers). Since the range \( Z \) is a proper subset of the codomain \( R \), the function is not onto. For example, consider \( 1.5 \in R \) (codomain). There is no real number \( x \) such that \( f(x) = 1.5 \), because \( [x] \) can only produce integer values. Therefore, function \( f \) is not onto (વ્યાપ્ત વિષય નથી). Hence, the function \( f \) is neither one-to-one nor onto.In simple words: The greatest integer function gives the same output (like 1) for many different inputs (like 1.5, 1.8), so it's not one-to-one. Also, its output can only be whole numbers, so it cannot cover all real numbers as outputs, meaning it's not onto.

🎯 Exam Tip: For the greatest integer function, remember its step-like behavior. This directly leads to multiple inputs mapping to the same output, proving it's not one-to-one. Its range being only integers immediately tells you it's not onto for a real number codomain.

Question 4. સાબિત કરો કે માનાંક વિષય \( f : R \to R, f(x) = |x| \) દ્વારા વ્યાખ્યાયિત વિધેય એક-એક પણ નથી અને વ્યાપ્ત પણ નથી. જો \( x \) ધન અથવા શૂન્ય (અનૃણ) હોય, તો \( |x| = x \) અને \( x \) ઋણ હોય, તો \( |x| = -x \).
Answer:Let the function be \( f : R \to R, f(x) = |x| \). The absolute value function is defined as: \[ f(x) = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases} \] Let's check if it is one-to-one. Consider \( 2 > 0 \implies f(2) = |2| = 2 \). Consider \( -2 < 0 \implies f(-2) = |-2| = -(-2) = 2 \). Here, \( f(2) = f(-2) \) but \( 2 \neq -2 \). So, function \( f \) is not one-to-one (એક-એક વિષય નથી). Now, let's check if it is onto (વ્યાપ્ત). The absolute value function always returns a non-negative value (i.e., \( |x| \ge 0 \)). Consider any negative real number, for example, \( -5 \in R \) (codomain). There is no real number \( x \) such that \( f(x) = -5 \), because \( |x| \) cannot be negative. So, negative real numbers in the codomain do not have a preimage in the domain R. Therefore, function \( f \) is not onto (વ્યાપ્ત વિષય નથી). Hence, the function \( f \) is neither one-to-one nor onto.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फलन \( f: R \to R, f(x) = |x| \) के लिए है। यह दर्शाता है कि 2 और -2 दोनों का प्रतिबिंब 2 है, इसलिए यह एक-एक नहीं है। साथ ही, codomain में सभी नकारात्मक वास्तविक संख्याओं (जैसे -2, -5) का कोई पूर्व-प्रतिबिंब नहीं है, इसलिए यह आच्छादक भी नहीं है।In simple words: The absolute value function gives the same positive output for a number and its negative counterpart (e.g., \( |2|=2 \) and \( |-2|=2 \)), so it's not one-to-one. Also, it can never produce a negative output, so it can't cover all real numbers, meaning it's not onto.

🎯 Exam Tip: For the absolute value function, its inherent property of mapping both positive and negative inputs to the same positive output is key to proving it's not one-to-one. Its range being only non-negative numbers instantly shows it's not onto for a codomain including negative values.

Question 5. સાબિત કરો કે ચિહન વિષય (Signum Function) \( f : R \to R \), \[ f(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases} \] દ્વારા વ્યાખ્યાયિત વિધેય એક એક નથી અને વ્યાપ્ત પણ નથી.
Answer:Let the function be \( f : R \to R, f(x) = \text{sgn}(x) \), defined as: \[ f(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases} \] Let's check if it is one-to-one. It is clear that for any \( x_1, x_2 > 0 \), \( f(x_1) = 1 \) and \( f(x_2) = 1 \). For example, \( f(5) = 1 \) and \( f(8) = 1 \), but \( 5 \neq 8 \). Similarly, for any \( x_1, x_2 < 0 \), \( f(x_1) = -1 \) and \( f(x_2) = -1 \). For example, \( f(-2) = -1 \) and \( f(-3) = -1 \), but \( -2 \neq -3 \). Since different inputs can map to the same output (e.g., all positive numbers map to 1), function \( f \) is not one-to-one (એક-એક વિષય નથી). Now, let's check if it is onto (વ્યાપ્ત). The range ( વિસ્તાર ) of \( f \) is \( \{-1, 0, 1\} \). The codomain of \( f \) is \( R \) (real numbers). Since the range \( \{-1, 0, 1\} \) is a proper subset of the codomain \( R \), the function is not onto. For example, consider \( 2 \in R \) (codomain). There is no real number \( x \) such that \( f(x) = 2 \), because \( f(x) \) can only be \( -1, 0 \), or \( 1 \). Therefore, function \( f \) is not onto (વ્યાપ્ત વિષય નથી). Hence, the function \( f \) is neither one-to-one nor onto.In simple words: The signum function gives only three possible outputs (-1, 0, or 1). Since many different positive numbers all output 1, it's not one-to-one. And because it only outputs -1, 0, or 1, it cannot produce all other real numbers as outputs, so it's not onto.

🎯 Exam Tip: The signum function is a classic example of a function that is neither one-to-one nor onto when mapped from R to R. Its piecewise definition and limited range (only -1, 0, 1) are key aspects to highlight in your explanation.

Question 6. A = {1, 2, 3}, B = {4, 5, 6, 7} છે અને વિધેય \( f : A \to B, f = \{(1, 4), (2, 5), (3, 6)\} \) દ્વારા વ્યાખ્યાયિત છે. સાબિત કરો કે હું એક-એક છે.
Answer:Given sets \( A = \{1, 2, 3\} \) and \( B = \{4, 5, 6, 7\} \). The function \( f : A \to B \) is defined as \( f = \{(1, 4), (2, 5), (3, 6)\} \). From the definition of \( f \):
\( f(1) = 4 \)
\( f(2) = 5 \)
\( f(3) = 6 \) We can see that all distinct elements in set A (1, 2, 3) map to distinct elements in set B (4, 5, 6). Since no two distinct elements of A have the same image in B, the function \( f \) is one-to-one (એક-એક વિષય છે). Let's check if it is onto. The range of \( f \) is \( \{4, 5, 6\} \). The codomain is \( B = \{4, 5, 6, 7\} \). Since \( 7 \in B \) does not have a preimage in A, the function \( f \) is not onto.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फलन \( f: A \to B \) को दर्शाता है जहाँ \( A=\{1,2,3\} \) और \( B=\{4,5,6,7\} \)। फलन 1 को 4 से, 2 को 5 से और 3 को 6 से जोड़ता है। चूंकि A के प्रत्येक अद्वितीय तत्व का B में एक अद्वितीय प्रतिबिंब है, यह एक-एक फलन है। लेकिन B में 7 का कोई पूर्व-प्रतिबिंब नहीं है, इसलिए यह आच्छादक नहीं है।In simple words: Each unique number from set A (1, 2, 3) goes to a unique number in set B (4, 5, 6). So, it's one-to-one. However, the number 7 in set B is not mapped to by any number from set A, so it's not onto.

🎯 Exam Tip: For finite sets, to prove one-to-one, explicitly list the mappings and verify that no two different domain elements map to the same codomain element. For onto, compare the range with the codomain; if they are not identical, it's not onto.

Question 7. નીચે આપેલ પ્રત્યેક પ્રશ્નમાં આપેલાં વિધેય એક એક છે કે નહિ, વ્યાપ્ત છે કે નહિ અથવા એક-એક અને વ્યાપ્ત છે કે નહિ તે નક્કી કરો. તમારા જવાબનું સમર્થન કરોઃ
(i) \( f : R \to R \) એ \( f(x) = 3 - 4x \) દ્વારા વ્યાખ્યાયિત વિધેય છે.
Answer:The function is \( f : R \to R, f(x) = 3 - 4x \). Let's check if it is one-to-one. Let \( x_1, x_2 \in R \). If \( f(x_1) = f(x_2) \), then \( 3 - 4x_1 = 3 - 4x_2 \).
\( \implies -4x_1 = -4x_2 \)
\( \implies x_1 = x_2 \) So, function \( f \) is one-to-one (એક-એક વિધેય છે). Now, let's check if it is onto (વ્યાપ્ત). Let \( y \in R \) (codomain). We want to find an \( x \in R \) such that \( f(x) = y \).
\( \implies 3 - 4x = y \)
\( \implies 4x = 3 - y \)
\( \implies x = \frac{3 - y}{4} \) For every real number \( y \), \( x = \frac{3 - y}{4} \) is also a real number. So, for every \( y \in R \), there exists an \( x = \frac{3 - y}{4} \in R \) such that \( f(x) = f\left(\frac{3 - y}{4}\right) = 3 - 4\left(\frac{3 - y}{4}\right) = 3 - (3 - y) = y \). Therefore, function \( f \) is onto (વ્યાપ્ત વિધેય છે). Hence, the function \( f \) is both one-to-one and onto.In simple words: This linear function gives a unique output for every unique input, so it's one-to-one. Also, for any desired output number, you can always find a real number input that produces it, meaning it's onto.

🎯 Exam Tip: Linear functions \( f(x) = ax + b \) (where \( a \neq 0 \)) are generally one-to-one and onto for real numbers. To prove one-to-one, set \( f(x_1) = f(x_2) \) and solve for \( x_1 = x_2 \). To prove onto, set \( f(x) = y \) and solve for \( x \) in terms of \( y \), then confirm that this \( x \) is always in the domain for any \( y \) in the codomain.

(ii) \( f : R \to R \) એ \( f(x) = 1 + x^2 \) દ્વારા વ્યાખ્યાયિત વિધેય છે.
Answer:The function is \( f : R \to R, f(x) = 1 + x^2 \). Let's check if it is one-to-one. Let \( x_1, x_2 \in R \). If \( f(x_1) = f(x_2) \), then \( 1 + x_1^2 = 1 + x_2^2 \).
\( \implies x_1^2 = x_2^2 \)
\( \implies x_1 = \pm x_2 \) For example, \( f(2) = 1 + (2)^2 = 5 \) and \( f(-2) = 1 + (-2)^2 = 5 \). Here, \( f(2) = f(-2) \) but \( 2 \neq -2 \). So, function \( f \) is not one-to-one (એક-એક વિષય નથી). Now, let's check if it is onto (વ્યાપ્ત). Since \( x^2 \ge 0 \) for all \( x \in R \), it means \( 1 + x^2 \ge 1 \). So, the range of \( f \) is \( [1, \infty) \). The codomain of \( f \) is \( R \). Since the range \( [1, \infty) \) is a proper subset of the codomain \( R \), the function is not onto. For example, consider a negative number \( y = -5 \in R \) (codomain). There is no real number \( x \) such that \( f(x) = -5 \), because \( 1 + x^2 = -5 \implies x^2 = -6 \), which has no real solution for \( x \). Therefore, function \( f \) is not onto (વ્યાપ્ત વિષય નથી). Hence, the function \( f \) is neither one-to-one nor onto.In simple words: Since both positive and negative numbers (like 2 and -2) give the same output (5) when plugged into \( 1+x^2 \), it's not one-to-one. Also, the output of \( 1+x^2 \) can never be less than 1, so it cannot produce all real numbers (like negative numbers), meaning it's not onto.

🎯 Exam Tip: For functions involving \( x^2 \) over real numbers, always check for \( f(x) = f(-x) \) for one-to-one property. For onto, evaluate the minimum or maximum value of the function (in this case, \( 1+x^2 \ge 1 \)) to see if it covers the entire codomain.

Question 8. A અને B આપેલ ગણ છે. સાબિત કરો કે \( f : A \times B \to B \times A, f((a, b)) = (b, a) \) દ્વારા વ્યાખ્યાયિત વિધેય એક એક અને વ્યાપ્ત છે.
Answer:The function is \( f : A \times B \to B \times A \) defined as \( f((a, b)) = (b, a) \). Let's check if it is one-to-one. Let \( (a_1, b_1), (a_2, b_2) \in A \times B \), where \( a_1, a_2 \in A \) and \( b_1, b_2 \in B \). If \( f((a_1, b_1)) = f((a_2, b_2)) \), then \( (b_1, a_1) = (b_2, a_2) \). By the definition of equality of ordered pairs, this implies \( b_1 = b_2 \) and \( a_1 = a_2 \).
\( \implies (a_1, b_1) = (a_2, b_2) \) So, function \( f \) is one-to-one (એક-એક વિષય છે). Now, let's check if it is onto (વ્યાપ્ત). Let \( (b, a) \in B \times A \) be any arbitrary element in the codomain, where \( b \in B \) and \( a \in A \). We need to find an element \( (x, y) \in A \times B \) such that \( f((x, y)) = (b, a) \). From the definition of \( f \), \( f((x, y)) = (y, x) \). So, we need \( (y, x) = (b, a) \), which means \( y = b \) and \( x = a \). Since \( a \in A \) and \( b \in B \), the pair \( (a, b) \in A \times B \) is in the domain. Therefore, for every \( (b, a) \in B \times A \), there exists an element \( (a, b) \in A \times B \) such that \( f((a, b)) = (b, a) \). Thus, function \( f \) is onto (વ્યાપ્ત વિષય છે). Hence, the function \( f \) is both one-to-one and onto.In simple words: This function just swaps the order of elements in a pair. Since different input pairs will always result in different output pairs, it's one-to-one. And because every possible output pair (like (b,a)) can be formed by reversing an input pair (like (a,b)), it's onto.

🎯 Exam Tip: Functions that permute the order of elements in ordered pairs or tuples are often bijective (one-to-one and onto). For one-to-one, assume the outputs are equal and show the inputs must be equal. For onto, take an arbitrary element from the codomain and show how to construct its preimage in the domain.

Question 9. \( f : N \to N \) વિધેય \( f(n) = \begin{cases} \frac{n+1}{2}, & \text{જો } n \text{ અયુગ્મ હોય} \\ \frac{n}{2}, & \text{જો } n \text{ યુગ્મ હોય} \end{cases}, \forall n \in N \) દ્વારા વ્યાખ્યાયિત વિધેય આપેલ છે. વિધેય એક-એક છે કે નહિ તથા વ્યાપ્ત છે કે નહિ તે નિશ્ચિત કરો, તમારા જવાબનું સમર્થન કરો.
Answer:The function is \( f : N \to N \) defined as: \[ f(n) = \begin{cases} \frac{n+1}{2}, & \text{જો } n \text{ અયુગ્મ હોય} \\ \frac{n}{2}, & \text{જો } n \text{ યુગ્મ હોય} \end{cases} \] Let's check if it is one-to-one. Consider \( n = 1 \) (odd): \( f(1) = \frac{1+1}{2} = 1 \). Consider \( n = 2 \) (even): \( f(2) = \frac{2}{2} = 1 \). Here, \( f(1) = f(2) = 1 \), but \( 1 \neq 2 \). So, function \( f \) is not one-to-one (એક-એક વિષય નથી). Now, let's check if it is onto (વ્યાપ્ત). Let \( m \in N \) (codomain). We need to find an \( n \in N \) such that \( f(n) = m \). Case 1: If \( n \) is odd, \( f(n) = \frac{n+1}{2} = m \implies n+1 = 2m \implies n = 2m - 1 \). For any \( m \in N \), \( 2m - 1 \) is always an odd natural number. For example, if \( m=1 \), \( n = 2(1)-1 = 1 \). \( f(1) = \frac{1+1}{2} = 1 \). If \( m=2 \), \( n = 2(2)-1 = 3 \). \( f(3) = \frac{3+1}{2} = 2 \). Case 2: If \( n \) is even, \( f(n) = \frac{n}{2} = m \implies n = 2m \). For any \( m \in N \), \( 2m \) is always an even natural number. For example, if \( m=1 \), \( n = 2(1) = 2 \). \( f(2) = \frac{2}{2} = 1 \). If \( m=2 \), \( n = 2(2) = 4 \). \( f(4) = \frac{4}{2} = 2 \). Since for every natural number \( m \) in the codomain, there exists an \( n \in N \) (either \( 2m-1 \) or \( 2m \)) such that \( f(n) = m \), the function \( f \) is onto (વ્યાપ્ત વિધેય છે). Hence, the function \( f \) is onto but not one-to-one.In simple words: This function gives the same output for consecutive natural numbers (like 1 and 2 both giving 1), so it's not one-to-one. However, every natural number in the output can be reached from an input (either an odd number \( 2m-1 \) or an even number \( 2m \)), making it onto.

🎯 Exam Tip: For piecewise functions, carefully test values for each case to check the one-to-one property. For onto, consider an arbitrary element in the codomain and try to find a preimage by solving for \( n \) in both parts of the function definition.

Question 10. \( A = R - \{3\} \) અને \( B = R - \{1\} \) છે. \( f(x) = \frac{x-2}{x-3} \) દ્વારા વ્યાખ્યાયિત વિધેય \( f : A \to B \) નો વિચાર કરો. શું \( f \) એક-એક અને વ્યાપ્ત છે ? તમારા જવાબનું સમર્થન કરો.
Answer:The function is \( f : A \to B \), where \( A = R - \{3\} \) and \( B = R - \{1\} \), defined as \( f(x) = \frac{x-2}{x-3} \). Let's check if it is one-to-one. Let \( x_1, x_2 \in A \). If \( f(x_1) = f(x_2) \), then \( \frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3} \).
\( \implies (x_1-2)(x_2-3) = (x_2-2)(x_1-3) \)
\( \implies x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6 \) Subtracting \( x_1x_2 + 6 \) from both sides:
\( \implies -3x_1 - 2x_2 = -3x_2 - 2x_1 \)
\( \implies -3x_1 + 2x_1 = -3x_2 + 2x_2 \)
\( \implies -x_1 = -x_2 \)
\( \implies x_1 = x_2 \) So, function \( f \) is one-to-one (એક-એક વિષય છે). Now, let's check if it is onto (વ્યાપ્ત). Let \( y \in B \) (codomain). We want to find an \( x \in A \) such that \( f(x) = y \).
\( \implies \frac{x-2}{x-3} = y \)
\( \implies x-2 = y(x-3) \)
\( \implies x-2 = yx - 3y \)
\( \implies x - yx = 2 - 3y \)
\( \implies x(1-y) = 2 - 3y \)
\( \implies x = \frac{2 - 3y}{1 - y} \) Since \( y \in B = R - \{1\} \), this means \( y \neq 1 \), so the denominator \( 1-y \neq 0 \). Thus, \( x \) is always a real number. We also need to check if \( x = 3 \) (the excluded value from A). If \( x = 3 \), then \( 3 = \frac{2 - 3y}{1 - y} \)
\( \implies 3(1 - y) = 2 - 3y \)
\( \implies 3 - 3y = 2 - 3y \)
\( \implies 3 = 2 \), which is false. This means that \( x \) can never be 3. So, for every \( y \in B \), there exists an \( x = \frac{2 - 3y}{1 - y} \in A \) such that \( f(x) = y \). Therefore, function \( f \) is onto (વ્યાપ્ત વિષય છે). Hence, the function \( f \) is both one-to-one and onto.In simple words: This function maps different input values to different output values, so it's one-to-one. Also, for every possible output value in the codomain, there's a valid input value in the domain that produces it, meaning it's onto.

🎯 Exam Tip: For rational functions, to prove one-to-one, cross-multiply and simplify to show \( x_1 = x_2 \). For onto, set \( f(x) = y \) and solve for \( x \) in terms of \( y \). Make sure the resulting \( x \) is always in the domain for any \( y \) in the codomain, specifically checking for values that might make the denominator zero or fall outside the domain restrictions.

પ્રશ્નો 11 તથા 12 માં વિધાન સાચું બને તે રીતે આપેલ વિકલ્પોમાંથી યોગ્ય વિકલ્પ પસંદ કરો :

Question 11. \( f : R \to R, f(x) = x^4 \) દ્વારા વ્યાખ્યાયિત વિધેય છે.
(A) \( f \) એક-એક અને વ્યાપ્ત છે.
(B) \( f \) અનેક-એક અને વ્યાપ્ત છે.
(C) \( f \) એક-એક છે પરંતુ વ્યાપ્ત નથી.
(D) \( f \) એક-એક પણ નથી અને વ્યાપ્ત પણ નથી.
Answer:The function is \( f : R \to R, f(x) = x^4 \). Let's check if it is one-to-one. Let \( x_1, x_2 \in R \). If \( f(x_1) = f(x_2) \), then \( x_1^4 = x_2^4 \).
\( \implies x_1^4 - x_2^4 = 0 \)
\( \implies (x_1^2 - x_2^2)(x_1^2 + x_2^2) = 0 \)
\( \implies (x_1 - x_2)(x_1 + x_2)(x_1^2 + x_2^2) = 0 \) This implies \( x_1 = x_2 \) or \( x_1 = -x_2 \) or \( x_1^2 + x_2^2 = 0 \) (which means \( x_1=0 \) and \( x_2=0 \)). For example, \( f(2) = (2)^4 = 16 \) and \( f(-2) = (-2)^4 = 16 \). Here, \( f(2) = f(-2) \) but \( 2 \neq -2 \). So, function \( f \) is not one-to-one (એક-એક વિષય નથી). Now, let's check if it is onto (વ્યાપ્ત). Since \( x^4 \ge 0 \) for all \( x \in R \), the range of \( f \) is \( [0, \infty) \). The codomain of \( f \) is \( R \). Since the range \( [0, \infty) \) is a proper subset of the codomain \( R \), the function is not onto. For example, consider a negative number \( y = -1 \in R \) (codomain). There is no real number \( x \) such that \( f(x) = -1 \), because \( x^4 = -1 \) has no real solution for \( x \). Therefore, function \( f \) is not onto (વ્યાપ્ત વિષય નથી). Hence, the function \( f \) is neither one-to-one nor onto. The correct option is (D).
Answer: (D) f એક-એક પણ નથી અને વ્યાપ્ત પણ નથી.In simple words: The function \( f(x)=x^4 \) gives the same output for positive and negative inputs (like 2 and -2 both give 16), so it's not one-to-one. Also, it can only produce non-negative numbers, so it can't cover all real numbers (like negative numbers), meaning it's not onto.

🎯 Exam Tip: For functions like \( x^4 \), remember that even powers result in non-negative values and lose the sign information, implying not one-to-one (for R or Z) and not onto (if the codomain includes negative numbers). Always look for easy counterexamples involving positive and negative inputs.

Question 12. વિધેય \( f : R \to R, f(x) = 3x \) દ્વારા વ્યાખ્યાયિત છે.
(A) \( f \) એક-એક અને વ્યાપ્ત છે.
(B) \( f \) અનેક-એક અને વ્યાપ્ત છે.
(C) \( f \) એક-એક છે પરંતુ વ્યાપ્ત નથી.
(D) \( f \) એક-એક પણ નથી અને વ્યાપ્ત પણ નથી.
Answer:The function is \( f : R \to R, f(x) = 3x \). Let's check if it is one-to-one. Let \( x_1, x_2 \in R \). If \( f(x_1) = f(x_2) \), then \( 3x_1 = 3x_2 \).
\( \implies x_1 = x_2 \) So, function \( f \) is one-to-one (એક-એક વિષય છે). Now, let's check if it is onto (વ્યાપ્ત). Let \( y \in R \) (codomain). We want to find an \( x \in R \) such that \( f(x) = y \).
\( \implies 3x = y \)
\( \implies x = \frac{y}{3} \) For every real number \( y \), \( x = \frac{y}{3} \) is also a real number. So, for every \( y \in R \), there exists an \( x = \frac{y}{3} \in R \) such that \( f(x) = f\left(\frac{y}{3}\right) = 3\left(\frac{y}{3}\right) = y \). Therefore, function \( f \) is onto (વ્યાપ્ત વિષય છે). Hence, the function \( f \) is both one-to-one and onto. The correct option is (A).
Answer: (A) f એક-એક અને વ્યાપ્ત છે.In simple words: This function scales the input by 3. Different inputs always give different scaled outputs, so it's one-to-one. Also, for any desired output, you can always find a real number input (by dividing by 3) that produces it, making it onto.

🎯 Exam Tip: Similar to \( f(x) = ax+b \), linear functions of the form \( f(x) = ax \) (where \( a \neq 0 \)) are typically bijective (one-to-one and onto) for real numbers. The steps for proving one-to-one and onto are straightforward algebraic manipulations.