GSEB Class 12 Maths Solutions Chapter 1 સંબંધ અને વિધેય Misc. Questions

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

 

Question 1. f : R → R, \(f(x) = 10x + 7\) દ્વારા વ્યાખ્યાયિત છે. એવું વિધેય g : R → R શોધો કે જેથી \(gof = fog = I_R\).

Answer:Let \(f : R \to R\) with \(f(x) = 10x + 7\). We need to find a function \(g : R \to R\) such that \(gof = fog = I_R\). If \(fog = I_R\), it means \(f(g(x)) = I_R(x)\). This implies that \(10g(x) + 7 = x\). So, \(10g(x) = x - 7\). Therefore, \(g(x) = \frac{x-7}{10}\).
In simple words: We are given a function f and need to find another function g such that applying f then g, or g then f, gives us the original input back. We solved for g(x) by setting f(g(x)) equal to x.

🎯 Exam Tip: Remember that \(I_R\) represents the identity function, where \(I_R(x) = x\). When finding the inverse of a function, set \(y = f(x)\) and solve for x in terms of y.

 

Question 2. ધારો કે W એ પૂર્ણસંખ્યાઓનો ગણ છે. \(f : W \to W\), n અયુગ્મ માટે \(f(n) = n - 1\) દ્વારા વ્યાખ્યાયિત છે અને n યુગ્મ માટે \(f(n) = n + 1\) વ્યાખ્યાયિત કરો. સાબિત કરો કે f એ વ્યસ્ત સંપન્ન છે. f નો વ્યસ્ત શોધો.

Answer:Let W be the set of whole numbers. The function \(f : W \to W\) is defined as: \[f(n) = \begin{cases} n - 1, & \text{n અયુગ્મ.} \\ n + 1, & \text{n યુગ્મ.} \end{cases}\] Consider two whole numbers, n and m. **Case 1: n and m are both odd.** If \(f(n) = f(m)\), then \(n - 1 = m - 1\).
\( \implies n = m\) **Case 2: n and m are both even.** If \(f(n) = f(m)\), then \(n + 1 = m + 1\).
\( \implies n = m\) **Case 3: One is odd and the other is even.** Let n be odd and m be even. If \(f(n) = f(m)\), then \(n - 1 = m + 1\). This means \(n = m + 2\). If n is odd, then n-1 is even. If m is even, then m+1 is odd. The range of f for odd n gives even numbers, and for even n gives odd numbers. So, \(f(n) \neq f(m)\) if n is odd and m is even. Thus, if \(f(n) = f(m)\), then n and m must be of the same parity (both odd or both even). In both such situations, \(n = m\). Therefore, function f is one-to-one (એક-એક વિષય). Now, let's check if f is onto (વ્યાપ્ત). Let \(n \in W\) (co-domain). If n is an odd number, we need to find an even number \(m \in W\) (domain) such that \(f(m) = n\). Since m is even, \(f(m) = m + 1\). So, \(m + 1 = n\), which means \(m = n - 1\). Since n is odd, \(n - 1\) is an even whole number, so \(m \in W\). If n is an even number, we need to find an odd number \(m \in W\) (domain) such that \(f(m) = n\). Since m is odd, \(f(m) = m - 1\). So, \(m - 1 = n\), which means \(m = n + 1\). Since n is even, \(n + 1\) is an odd whole number, so \(m \in W\). For every element n in the co-domain W, there exists a pre-image in the domain W. Therefore, f is an onto function (વ્યાપ્ત વિષય). Since f is both one-to-one and onto, it is invertible. To find the inverse, let \(y = f(n)\). If n is odd, \(y = n - 1 \implies n = y + 1\). If n is even, \(y = n + 1 \implies n = y - 1\). So, the inverse function \(f^{-1}(n)\) is defined as: \[f^{-1}(n) = \begin{cases} n + 1, & \text{n અયુગ્મ.} \\ n - 1, & \text{n યુગ્મ.} \end{cases}\]
In simple words: This function swaps even and odd numbers in a specific way. If you give it an odd number, it subtracts one to make it even. If you give it an even number, it adds one to make it odd. We showed it's "one-to-one" because different inputs always give different outputs, and it's "onto" because every number in the output set can be reached. The inverse function simply reverses these steps.

🎯 Exam Tip: When proving a piecewise function is one-to-one, consider all possible parity combinations (odd-odd, even-even, odd-even, even-odd). For proving onto, consider an arbitrary element in the codomain and find its pre-image based on its parity.

 

Question 3. f : R → R, \(f(x) = x^2 - 3x + 2\) દ્વારા વ્યાખ્યાયિત હોય, તો શું \(f(f(x))\) શોધો.

Answer:Given the function \(f : R \to R\) defined by \(f(x) = x^2 - 3x + 2\). We need to find \(f(f(x))\). Substitute \(f(x)\) into \(f(x)\): \(f(f(x)) = f(x^2 - 3x + 2)\) Now, replace \(x\) with \((x^2 - 3x + 2)\) in the definition of \(f(x)\): \(f(x^2 - 3x + 2) = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2\) Expand the terms: \((x^2 - 3x + 2)^2 = (x^2)^2 + (-3x)^2 + (2)^2 + 2(x^2)(-3x) + 2(x^2)(2) + 2(-3x)(2)\) \(= x^4 + 9x^2 + 4 - 6x^3 + 4x^2 - 12x\) \(= x^4 - 6x^3 + 13x^2 - 12x + 4\) Now, distribute the -3: \(-3(x^2 - 3x + 2) = -3x^2 + 9x - 6\) Combine all parts: \(f(f(x)) = (x^4 - 6x^3 + 13x^2 - 12x + 4) + (-3x^2 + 9x - 6) + 2\) Combine like terms: \(= x^4 - 6x^3 + (13x^2 - 3x^2) + (-12x + 9x) + (4 - 6 + 2)\) \(= x^4 - 6x^3 + 10x^2 - 3x + 0\) So, \(f(f(x)) = x^4 - 6x^3 + 10x^2 - 3x\).
In simple words: To find \(f(f(x))\), you take the entire function \(f(x)\) and put it inside itself. This means wherever you see 'x' in the original function's rule, you replace it with the whole expression for \(f(x)\). Then, you simplify the resulting expression by expanding and combining terms.

🎯 Exam Tip: Be careful with algebraic expansions, especially squaring trinomials and distributing negative signs. It's easy to make a mistake when combining many terms.

 

Question 4. સાબિત કરો કે વિધેય \(f : R \to \{x \in R: -1 < x < 1\}\), \(f(x) = \frac{x}{1+|x|}\) એક એક અને વ્યાપ્ત વિધેય છે.

Answer:Let the function be \(f : R \to \{x \in R: -1 < x < 1\}\), defined by \(f(x) = \frac{x}{1+|x|}\). To prove f is one-to-one (એક-એક), let \(x, y \in R\) such that \(f(x) = f(y)\). **Case 1: \(x > 0\) and \(y > 0\).** In this case, \(|x| = x\) and \(|y| = y\). So, \(f(x) = \frac{x}{1+x}\) and \(f(y) = \frac{y}{1+y}\). If \(f(x) = f(y)\), then \(\frac{x}{1+x} = \frac{y}{1+y}\).
\( \implies x(1+y) = y(1+x)\)
\( \implies x + xy = y + xy\)
\( \implies x = y\) **Case 2: \(x < 0\) and \(y < 0\).** In this case, \(|x| = -x\) and \(|y| = -y\). So, \(f(x) = \frac{x}{1-x}\) and \(f(y) = \frac{y}{1-y}\). If \(f(x) = f(y)\), then \(\frac{x}{1-x} = \frac{y}{1-y}\).
\( \implies x(1-y) = y(1-x)\)
\( \implies x - xy = y - xy\)
\( \implies x = y\) **Case 3: One is positive and the other is negative.** Assume \(x > 0\) and \(y < 0\). Then \(f(x) = \frac{x}{1+x}\). Since \(x > 0\), \(1+x > x > 0\), so \(0 < f(x) < 1\). And \(f(y) = \frac{y}{1-y}\). Since \(y < 0\), let \(y = -z\) where \(z > 0\). \(f(y) = \frac{-z}{1-(-z)} = \frac{-z}{1+z}\). Since \(z > 0\), \(1+z > z > 0\), so \(-1 < f(y) < 0\). Therefore, if \(x > 0\) and \(y < 0\), then \(f(x)\) is positive and \(f(y)\) is negative, so \(f(x) \neq f(y)\). Thus, in all cases, if \(f(x) = f(y)\), then \(x = y\). So, f is a one-to-one function (એક-એક વિધેય છે). To prove f is onto (વ્યાપ્ત), let \(y \in R\) such that \(-1 < y < 1\). We need to find \(x \in R\) such that \(f(x) = y\). **Subcase 1: \(0 < y < 1\).** Since \(f(x) = y\) and y is positive, x must be positive. So \(|x| = x\). \(\frac{x}{1+x} = y\)
\( \implies x = y(1+x)\)
\( \implies x = y + xy\)
\( \implies x - xy = y\)
\( \implies x(1 - y) = y\)
\( \implies x = \frac{y}{1-y}\) Since \(0 < y < 1\), \(1-y > 0\), and \(y > 0\), so \(x = \frac{y}{1-y} > 0\). This value of x exists in R. We can check \(f(x) = f\left(\frac{y}{1-y}\right) = \frac{\frac{y}{1-y}}{1+\left|\frac{y}{1-y}\right|}\). Since \(\frac{y}{1-y} > 0\), \(\left|\frac{y}{1-y}\right| = \frac{y}{1-y}\). So, \(f(x) = \frac{\frac{y}{1-y}}{1+\frac{y}{1-y}} = \frac{\frac{y}{1-y}}{\frac{1-y+y}{1-y}} = \frac{\frac{y}{1-y}}{\frac{1}{1-y}} = y\). **Subcase 2: \(-1 < y < 0\).** Since \(f(x) = y\) and y is negative, x must be negative. So \(|x| = -x\). \(\frac{x}{1-x} = y\)
\( \implies x = y(1-x)\)
\( \implies x = y - xy\)
\( \implies x + xy = y\)
\( \implies x(1 + y) = y\)
\( \implies x = \frac{y}{1+y}\) Since \(-1 < y < 0\), \(1+y > 0\), and \(y < 0\), so \(x = \frac{y}{1+y} < 0\). This value of x exists in R. We can check \(f(x) = f\left(\frac{y}{1+y}\right) = \frac{\frac{y}{1+y}}{1+\left|\frac{y}{1+y}\right|}\). Since \(\frac{y}{1+y} < 0\), \(\left|\frac{y}{1+y}\right| = -\frac{y}{1+y}\). So, \(f(x) = \frac{\frac{y}{1+y}}{1-\frac{y}{1+y}} = \frac{\frac{y}{1+y}}{\frac{1+y-y}{1+y}} = \frac{\frac{y}{1+y}}{\frac{1}{1+y}} = y\). **Subcase 3: \(y = 0\).** If \(f(x) = 0\), then \(\frac{x}{1+|x|} = 0 \implies x = 0\). So \(x=0\) exists. Since for every \(y\) in the co-domain \((-1, 1)\), there exists an \(x\) in the domain R such that \(f(x) = y\), the function f is onto. Therefore, f is both one-to-one and onto.
In simple words: This function takes any real number and maps it to a number between -1 and 1. To show it's "one-to-one," we proved that if two different inputs give the same output, then the inputs must have been the same. To show it's "onto," we proved that for any number between -1 and 1, we can always find a real number that maps to it using this function. We did this by considering positive, negative, and zero inputs and outputs separately.

🎯 Exam Tip: When dealing with absolute values in functions, always split the problem into cases based on the sign of the variable inside the absolute value. This is crucial for proving both one-to-one and onto properties.

 

Question 5. સાબિત કરો કે વિધેય \(f : R \to R\), \(f(x) = x^3\) એક-એક છે.

Answer:Let the function be \(f : R \to R\), defined by \(f(x) = x^3\). To prove f is one-to-one (એક-એક), let \(x_1, x_2 \in R\) such that \(f(x_1) = f(x_2)\). This means \(x_1^3 = x_2^3\). Taking the cube root of both sides:
\( \implies x_1 = x_2\) Since \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), the function f is one-to-one.
In simple words: A function is "one-to-one" if every different input always gives a different output. For \(f(x) = x^3\), if two numbers have the same cube, they must be the same number. So, it is a one-to-one function.

🎯 Exam Tip: For proving one-to-one, the standard method is to assume \(f(x_1) = f(x_2)\) and then algebraically show that this implies \(x_1 = x_2\). Be aware of functions like \(f(x) = x^2\) where this wouldn't hold (\(x_1^2 = x_2^2\) implies \(x_1 = \pm x_2\)).

 

Question 6. બે વિધેયો \(f : N \to Z\) અને \(g : Z \to Z\) નાં ઉદાહરણ આપો કે જેથી \(gof\) એક-એક હોય પરંતુ \(g\) એક-એક ન હોય. (સૂચન : \(f(x) = x\) અને \(g(x) = |x|\) નો વિચાર કરો.)

Answer:Let's consider the suggested functions: Let \(f : N \to Z\) be defined by \(f(x) = x\). Let \(g : Z \to Z\) be defined by \(g(x) = |x|\). First, let's check if \(g\) is one-to-one. For \(g : Z \to Z\), consider \(5 \in Z\) and \(-5 \in Z\). We have \(g(5) = |5| = 5\) and \(g(-5) = |-5| = 5\). So, \(g(5) = g(-5)\) but \(5 \neq -5\). Therefore, \(g\) is not a one-to-one function (એક-એક વિષય નથી). Now, let's find the composite function \(gof\). \(gof(x) = g(f(x))\) Substitute \(f(x) = x\): \(gof(x) = g(x) = |x|\) So, \(gof : N \to Z\) is defined by \(gof(x) = |x|\). The domain of \(gof\) is N (natural numbers), which consists only of positive integers. Let \(x_1, x_2 \in N\) such that \(gof(x_1) = gof(x_2)\).
\( \implies |x_1| = |x_2|\) Since \(x_1, x_2 \in N\), both \(x_1\) and \(x_2\) are positive integers. So, \(|x_1| = x_1\) and \(|x_2| = x_2\).
\( \implies x_1 = x_2\) Therefore, \(gof\) is a one-to-one function (એક-એક વિશેષ છે).
In simple words: We need to pick two functions. We chose a simple function \(f(x) = x\) that takes natural numbers to integers. Then, we picked \(g(x) = |x|\) that takes integers to integers. The function \(g\) is not "one-to-one" because both 5 and -5 give the same output (5). But when we combine them as \(gof(x)\), it acts like \(|x|\) but only for natural numbers, which are always positive. So, for natural numbers, \(|x|\) just means \(x\), making \(gof(x) = x\), which is "one-to-one."

🎯 Exam Tip: When constructing examples, test the conditions carefully. For one-to-one, explicitly find two different inputs that give the same output to show it's *not* one-to-one. For composite functions like gof, the domain of f determines the inputs for gof, which can change the properties of the overall function.

 

Question 7. બે વિધેયો \(f : N \to N\) અને \(g : N \to N\) નાં ઉદાહરણ આપો કે જેથી \(gof\) વ્યાપ્ત હોય પરંતુ \(f\) વ્યાપ્ત ન હોય. (સૂચન : \(f(x) = x + 1\) અને \(g(x) = \left\{\begin{array}{cc} x-1 & x>1 \\ 1 & x=1 \end{array}\right.\) નો વિચાર કરો.)

Answer:Let's consider the suggested functions: Let \(f : N \to N\) be defined by \(f(x) = x + 1\). Let \(g : N \to N\) be defined by \(g(x) = \left\{\begin{array}{cc} x-1 & x>1 \\ 1 & x=1 \end{array}\right.\). First, let's check if \(f\) is onto (વ્યાપ્ત). The function is \(f : N \to N\), defined by \(f(x) = x + 1\). Let \(y \in N\) (co-domain). We need to find an \(x \in N\) (domain) such that \(f(x) = y\). So, \(x + 1 = y \implies x = y - 1\). If we take \(y = 1 \in N\), then \(x = 1 - 1 = 0\). However, \(0 \notin N\) (natural numbers start from 1). This means that for the element \(1\) in the co-domain, there is no pre-image in the domain N. Therefore, function \(f\) is not onto (વ્યાપ્ત નથી). Now, let's find the composite function \(gof\). \(gof(x) = g(f(x))\) Substitute \(f(x) = x + 1\): \(gof(x) = g(x + 1)\) Since \(x \in N\), \(x \ge 1\), so \(x + 1 \ge 2\). Thus, \(x+1 > 1\). According to the definition of \(g(x)\), if the input is greater than 1, we subtract 1 from it. So, \(g(x + 1) = (x + 1) - 1 = x\). Therefore, \(gof : N \to N\) is defined by \(gof(x) = x\). Now, let's check if \(gof\) is onto. For \(gof : N \to N\), defined by \(gof(x) = x\). Let \(y \in N\) (co-domain). We need to find an \(x \in N\) (domain) such that \(gof(x) = y\). Since \(gof(x) = x\), we have \(x = y\). For any \(y \in N\), \(x = y\) is also in N. So, every element in the co-domain has a pre-image in the domain. Therefore, \(gof\) is an onto function (વ્યાપ્ત વિષય છે).
In simple words: We chose a function \(f\) that adds 1 to natural numbers. This function is not "onto" because there's no natural number you can put in to get 1 out. Then we chose \(g\), which subtracts 1 if the number is greater than 1, or returns 1 if it is 1. When we combine them, \(gof(x)\) ends up simply returning \(x\) for any natural number input. This combined function is "onto" because every natural number can be reached as an output.

🎯 Exam Tip: To show a function is *not* onto, find a specific element in the co-domain that does not have a pre-image. For composite functions, always determine the explicit rule for \(gof(x)\) or \(fog(x)\) before testing its properties.

 

Question 8. X એ આપેલ અતિ ગણ છે. X ના તમામ ઉપગણોના ગણ P(X) નો વિચાર કરો. P(X) માં સંબંધ R આ પ્રમાણે વ્યાખ્યાયિત છે : P(X) ના ઉપગણો A અને B માટે, \(A \subset B\) તો અને તો જ ARB. R, P(X) પર સામ્ય સંબંધ છે ? તમારા જવાબનું સમર્થન કરો.

Answer:Let X be a given non-empty set. P(X) is the power set of X (the set of all subsets of X). The relation R on P(X) is defined as: for \(A, B \in P(X)\), \(ARB \iff A \subset B\). We need to check if R is an equivalence relation. An equivalence relation must be reflexive, symmetric, and transitive. **1. Reflexivity (સ્વવાચક સંબંધ):** For any set \(A \in P(X)\), we know that every set is a subset of itself, i.e., \(A \subset A\). According to the definition of R, \(A \subset A \implies ARA\). Therefore, R is reflexive (સ્વવાચક સંબંધ છે). **2. Symmetry (સંમિત સંબંધ):** For any \(A, B \in P(X)\), suppose \(ARB\). This means \(A \subset B\). For R to be symmetric, \(BRA\) must also hold, which means \(B \subset A\). However, \(A \subset B\) does not necessarily imply \(B \subset A\). For example, let \(X = \{1, 2, 3\}\). Let \(A = \{1\}\) and \(B = \{1, 2\}\). Then \(A \subset B\), so \(ARB\). But \(B \not\subset A\), so \(BRA\) does not hold. Therefore, R is not symmetric (સંમિત સંબંધ નથી). **3. Transitivity (પરંપરિત સંબંધ):** For any \(A, B, C \in P(X)\), suppose \(ARB\) and \(BRC\). \(ARB \implies A \subset B\). \(BRC \implies B \subset C\). We know that if \(A \subset B\) and \(B \subset C\), then \(A \subset C\). According to the definition of R, \(A \subset C \implies ARC\). Therefore, R is transitive (પરંપરિત સંબંધ છે). Since R is reflexive and transitive but not symmetric, it is not an equivalence relation.
In simple words: This question asks if the "is a subset of" relationship is an "equivalence relation" on a set of all possible subsets. An equivalence relation needs three things: it must be "reflexive" (a set is a subset of itself), "symmetric" (if A is a subset of B, then B must be a subset of A), and "transitive" (if A is a subset of B, and B is a subset of C, then A is a subset of C). We found that it is reflexive and transitive, but not symmetric, because if A is a smaller set inside B, B is not necessarily a smaller set inside A. So, it's not an equivalence relation.

🎯 Exam Tip: To prove a relation is NOT symmetric, a single counterexample is sufficient. Choose simple sets to illustrate the non-symmetry clearly. Remember all three properties must hold for an equivalence relation.

 

Question 9. આપેલ અરિક્ત ગણ X નો ઘાતગણ P(X) છે. દ્વિક્રિયા \( * : P(X) \times P(X) \to P(X)\) એ \(A * B = A \cap B\), \(\forall A, B \in P(X)\) દ્વારા વ્યાખ્યાયિત છે. સાબિત કરો કે X એ આ ક્રિયા માટે તટસ્થ ઘટક છે અને દ્વિક્રિયા \( * \) ને સાપેક્ષ P(X) માં કેવળ X જ વ્યસ્તસંપન્ન છે.

Answer:Let X be a non-empty set. P(X) is the power set of X. The binary operation \( * : P(X) \times P(X) \to P(X)\) is defined by \(A * B = A \cap B\), for all \(A, B \in P(X)\). **1. Proving X is the identity element:** An element \(E \in P(X)\) is an identity element if for every \(A \in P(X)\), \(A * E = E * A = A\). Let's test \(X\) as the identity element. For any \(A \in P(X)\), we need to check if \(A * X = A\) and \(X * A = A\). \(A * X = A \cap X\). Since A is a subset of X (because \(A \in P(X)\)), \(A \cap X = A\). \(X * A = X \cap A\). Similarly, \(X \cap A = A\). So, \(A * X = A = X * A\). Therefore, X is the identity element for the operation \( * \). **2. Proving only X has an inverse:** An element \(A \in P(X)\) has an inverse \(B \in P(X)\) if \(A * B = B * A = X\) (where X is the identity element). We are looking for an inverse for an arbitrary \(A \in P(X)\). Let this inverse be \(A^{-1}\). So, \(A * A^{-1} = X\), which means \(A \cap A^{-1} = X\). For the intersection of two subsets of X to be equal to X itself, both subsets must be X. This means \(A = X\) and \(A^{-1} = X\). If \(A \neq X\), then \(A \cap A^{-1}\) cannot be X, because \(A \cap A^{-1}\) would be a proper subset of X. Therefore, only the element X itself has an inverse, which is X.
In simple words: We have an operation on subsets where combining two sets means finding their common elements (intersection). We showed that the biggest set, X, acts as the "identity element" because when you combine any subset A with X, you get A back. Then, we looked for "inverse elements." For a set A to have an inverse, when you combine A with its inverse, you must get X. The only way the intersection of two sets is X is if both sets are X. So, only X has an inverse, and its inverse is X itself.

🎯 Exam Tip: Understanding the definitions of identity element and inverse element in the context of binary operations is crucial. For sets, remember that \(A \cap X = A\) when A is a subset of X, and \(A \cap B = X\) implies \(A=X\) and \(B=X\) if A and B are subsets of X.

 

Question 10. ગણ \(\{1, 2, 3,......, n\}\) થી \(\{1, 2, 3, ... ..., n\}\) સુધીના તમામ વ્યાપ્ત વિષયોની સંખ્યા શોધો.

Answer:Let the set be \(A = \{1, 2, 3, ... , n\}\). We need to find the number of surjective (onto) functions from A to A. For a function \(f : A \to A\) where the domain and co-domain have the same finite number of elements (n elements in this case), the function is surjective (onto) if and only if it is injective (one-to-one). If a function is both one-to-one and onto, it is a bijection. The number of bijections from a finite set of n elements to itself is given by n!. Therefore, the number of surjective functions from A to A is n!.
In simple words: We are looking for how many ways we can map each number from 1 to n to a unique number from 1 to n, making sure every number in the second set is used. This is just like arranging n distinct items, which is found by calculating n factorial (n!).

🎯 Exam Tip: For functions between finite sets of the same size, the properties "one-to-one," "onto," and "bijective" are equivalent. This simplifies counting problems significantly, as you only need to count permutations.

 

Question 11. S = {a, b, c} અને T = {1, 2, 3} લો. જો અસ્તિત્વ હોય, તો નીચે આપેલાં વિધેયો F : S → T માટે \(F^{-1}\) શોધો. (i) F = {(a, 3), (b, 2), (c, 1)} (ii) F = {(a, 2), (b, 1), (c, 1)}

Answer:Given sets \(S = \{a, b, c\}\) and \(T = \{1, 2, 3\}\). **(i) F = {(a, 3), (b, 2), (c, 1)}** For \(F^{-1}\) to exist, F must be a bijective function (one-to-one and onto). Let's check the properties of F: - Each element in S maps to a unique element in T: a→3, b→2, c→1. So, F is one-to-one. - Every element in T is mapped by an element in S (3, 2, 1 are all used). So, F is onto. Since F is both one-to-one and onto, its inverse \(F^{-1}\) exists. To find \(F^{-1}\), we reverse the order of the pairs: \(F^{-1} = \{(3, a), (2, b), (1, c)\}\).
In simple words: For the first function, each unique letter (a, b, c) goes to a unique number (3, 2, 1), and all numbers are used. This means an inverse exists. To find the inverse, we just swap the letter and number in each pair.**(ii) F = {(a, 2), (b, 1), (c, 1)}** Let's check the properties of F: - Here, \(F(b) = 1\) and \(F(c) = 1\). Two different elements (b and c) in the domain map to the same element (1) in the co-domain. Therefore, F is not a one-to-one function. Since F is not one-to-one, it is not a bijection, and its inverse \(F^{-1}\) does not exist.
In simple words: For the second function, both 'b' and 'c' are mapped to the number '1'. Since two different inputs give the same output, the function is not "one-to-one." For an inverse function to exist, each input must map to a unique output, so this function does not have an inverse.

🎯 Exam Tip: An inverse function exists if and only if the original function is a bijection (both one-to-one and onto). If a function is not one-to-one (meaning two different inputs map to the same output), its inverse cannot be defined.

 

Question 12. \(a * b = |a - b|\) અને \(a \circ b = a\), \(\forall a, b \in R\) દ્વારા વ્યાખ્યાયિત દ્વિક્રિયાઓ \( * : R \times R \to R\) અને \( \circ : R \times R \to R\) નો વિચાર કરો. સાબિત કરો કે કે સમક્રમી છે પરંતુ જૂથના નિયમને અનુસરતી નથી. \( \circ \) જૂથના નિયમને અનુસરે છે પરંતુ સમક્રમી નથી. વધુમાં, બતાવો કે \(\forall a, b, c \in R\), \(a * (b \circ c) = (a * b) \circ (a * c)\). [જો આ સત્ય હોય, તો આપણે દ્વિક્રિયા \( * \) ને દ્વિક્રિયા \( \circ \) પર વિભાજનીય કહીશું.] શું દ્વિક્રિયા \( \circ \) એ દ્વિક્રિયા \( * \) પર વિભાજનીય થશે ? તમારા જવાબની સત્યાર્થતા ચકાસો.

Answer:Given two binary operations on R: \( * : R \times R \to R\) defined by \(a * b = |a - b|\) \( \circ : R \times R \to R\) defined by \(a \circ b = a\) **Part 1: Operation \( * \)** **a) Commutativity (સમક્રમી):** For any \(a, b \in R\), \(a * b = |a - b|\) \(b * a = |b - a|\) Since \(|a - b| = |-(b - a)| = |b - a|\), we have \(a * b = b * a\). Therefore, operation \( * \) is commutative (સમક્રમી ક્રિયા છે). **b) Associativity (જૂથના નિયમનું પાલન):** For any \(a, b, c \in R\), we need to check if \( (a * b) * c = a * (b * c) \). Let's take specific numbers, e.g., \(a=5, b=8, c=10\). \((5 * 8) * 10 = |5 - 8| * 10 = |-3| * 10 = 3 * 10 = |3 - 10| = |-7| = 7\). \(5 * (8 * 10) = 5 * |8 - 10| = 5 * |-2| = 5 * 2 = |5 - 2| = |3| = 3\). Since \((5 * 8) * 10 \neq 5 * (8 * 10)\) (i.e., \(7 \neq 3\)). Therefore, operation \( * \) is not associative (જૂથનાં નિયમનું પાલન કરતું નથી). **Part 2: Operation \( \circ \)** **a) Commutativity (સમક્રમી):** For any \(a, b \in R\), \(a \circ b = a\) \(b \circ a = b\) Since \(a \circ b \neq b \circ a\) (unless \(a=b\)), for example, \(5 \circ 8 = 5\) but \(8 \circ 5 = 8\). Therefore, operation \( \circ \) is not commutative (સમક્રમી નથી). **b) Associativity (જૂથના નિયમનું પાલન):** For any \(a, b, c \in R\), we need to check if \( (a \circ b) \circ c = a \circ (b \circ c) \). \((a \circ b) \circ c = a \circ c\) (because \(a \circ b = a\)) \(= a\) (because \(a \circ c = a\)) \(a \circ (b \circ c) = a \circ b\) (because \(b \circ c = b\)) \(= a\) (because \(a \circ b = a\)) Since \((a \circ b) \circ c = a \circ (b \circ c) = a\). Therefore, operation \( \circ \) is associative (જૂથના નિયમનું પાલન કરે છે). **Part 3: Distributivity of \( * \) over \( \circ \)** We need to check if \(a * (b \circ c) = (a * b) \circ (a * c)\) for all \(a, b, c \in R\). Left Hand Side (LHS): \(a * (b \circ c) = a * b\) (since \(b \circ c = b\)) \(= |a - b|\) Right Hand Side (RHS): \((a * b) \circ (a * c) = |a - b| \circ |a - c|\) According to the definition of \( \circ \), \(X \circ Y = X\). So, \(|a - b| \circ |a - c| = |a - b|\). Since LHS = RHS (\(|a - b| = |a - b|\)), the operation \( * \) is distributive over \( \circ \). **Part 4: Distributivity of \( \circ \) over \( * \)** We need to check if \(a \circ (b * c) = (a \circ b) * (a \circ c)\) for all \(a, b, c \in R\). Left Hand Side (LHS): \(a \circ (b * c) = a \circ |b - c|\) According to the definition of \( \circ \), \(X \circ Y = X\). So, \(a \circ |b - c| = a\). Right Hand Side (RHS): \((a \circ b) * (a \circ c) = a * a\) (since \(a \circ b = a\) and \(a \circ c = a\)) \(= |a - a|\) \(= |0| = 0\). Since LHS \(\neq\) RHS (i.e., \(a \neq 0\) in general, for example if \(a=5\), \(5 \neq 0\)). Therefore, operation \( \circ \) is not distributive over \( * \).
In simple words: We examined two ways to combine numbers: \( * \) (absolute difference) and \( \circ \) (always picking the first number). \( * \) is like regular addition/multiplication in that the order doesn't matter (commutative), but grouping numbers differently changes the result (not associative). \( \circ \) is the opposite: grouping doesn't matter (associative), but order does (not commutative). We then checked if \( * \) "distributes" over \( \circ \), meaning \(a * (b \circ c)\) is the same as \((a * b) \circ (a * c)\), and it was. Finally, we checked if \( \circ \) distributes over \( * \), and it did not.

🎯 Exam Tip: Always verify associativity and commutativity with general elements \(a, b, c\), but a single counterexample with specific numbers is sufficient to prove non-associativity or non-commutativity. Distributivity involves testing two different forms of distribution.

 

Question 13. X એ આપેલ અરિક્ત ગણ છે. X ના તમામ ઉપગણોના ગણ P(X) નો વિચાર કરો. દ્વિક્રિયા \( * : P(X) \times P(X) \to P(X)\) એ \(A * B = (A - B) \cup (B - A)\), \(\forall A, B \in P(X)\) દ્વારા વ્યાખ્યાયિત દ્વિક્રિયા છે. સાબિત કરો કે રિક્ત ગણ \(\Phi\) એ દ્વિક્રિયા \( * \) માટે તટસ્થ ઘટક છે અને P(X) ના તમામ ઘટક A વ્યસ્ત સંપન્ન છે તથા \(A^{-1} = A\). (સૂચના : \((A - \Phi) \cup (\Phi - A) = A\) અને \((A - A) \cup (A - A) = \Phi\))

Answer:Let X be a non-empty set. P(X) is the power set of X. The binary operation \( * : P(X) \times P(X) \to P(X)\) is defined by \(A * B = (A - B) \cup (B - A)\), which is the symmetric difference of sets. **1. Proving \(\Phi\) is the identity element:** An element \(E \in P(X)\) is an identity element if for every \(A \in P(X)\), \(A * E = E * A = A\). Let's test \(\Phi\) (the empty set) as the identity element. For any \(A \in P(X)\): \(A * \Phi = (A - \Phi) \cup (\Phi - A)\) We know that \(A - \Phi = A\) (removing nothing from A leaves A) and \(\Phi - A = \Phi\) (removing elements of A from the empty set leaves the empty set). So, \(A * \Phi = A \cup \Phi = A\). Similarly, \(\Phi * A = (\Phi - A) \cup (A - \Phi)\) \(= \Phi \cup A = A\). Since \(A * \Phi = A = \Phi * A\), \(\Phi\) is the identity element for the operation \( * \). **2. Proving every element A is invertible and \(A^{-1} = A\):** An element \(A \in P(X)\) has an inverse \(A^{-1} \in P(X)\) if \(A * A^{-1} = A^{-1} * A = \Phi\) (where \(\Phi\) is the identity element). Let's try if \(A\) itself is its own inverse, i.e., \(A^{-1} = A\). We need to check if \(A * A = \Phi\). \(A * A = (A - A) \cup (A - A)\) We know that \(A - A = \Phi\) (removing elements of A from A leaves the empty set). So, \(A * A = \Phi \cup \Phi = \Phi\). Since \(A * A = \Phi\), every element \(A \in P(X)\) is its own inverse. Therefore, every element A in P(X) is invertible, and its inverse is \(A^{-1} = A\).
In simple words: This problem uses a special way to combine sets, called symmetric difference, where \(A * B\) means all elements that are in A or B, but not in both. We showed that the empty set (\(\Phi\)) is like the "zero" of this operation, because combining any set A with \(\Phi\) leaves A unchanged. Then, we found that every set is its own inverse, meaning if you combine a set A with itself using this operation, you get the empty set.

🎯 Exam Tip: Symmetric difference \(A \Delta B = (A - B) \cup (B - A)\) is a common set operation. Remember its properties: \(A - \Phi = A\), \(\Phi - A = \Phi\), and \(A - A = \Phi\). These are key to proving identity and inverse elements.

 

Question 14. ગણ \(\{0, 1, 2, 3, 4, 5\}\) પર દ્વિક્રિયા \( * \) નીચે પ્રમાણે વ્યાખ્યાયિત છે : \[a * b = \begin{cases} a+b, & a+b<6 \\ a+b-6, & a+b \geq 6 \end{cases}\] સાબિત કરી કે આ ક્રિયા માટે શૂન્ય એ તટસ્થ ઘટક છે અને આ ગણનો પ્રત્યેક શૂન્યેતર ઘટક વ્યસ્ત સંપન્ન છે. અહીં, \(6 - a\) એ ઘટક \(a\) નો વ્યસ્ત છે.

Answer:Let the set be \(A = \{0, 1, 2, 3, 4, 5\}\). The binary operation \( * \) is defined as: \[a * b = \begin{cases} a+b, & a+b<6 \\ a+b-6, & a+b \geq 6 \end{cases}\] **1. Proving 0 is the identity element:** An element \(e \in A\) is an identity element if for every \(a \in A\), \(a * e = e * a = a\). Let's test \(e = 0\). For any \(a \in A\): \(a * 0\): If \(a+0 < 6\), then \(a * 0 = a+0 = a\). Since \(a \in \{0, 1, 2, 3, 4, 5\}\), \(a+0\) is always less than 6. So, \(a * 0 = a\). \(0 * a\): If \(0+a < 6\), then \(0 * a = 0+a = a\). Thus, \(a * 0 = 0 * a = a\) for all \(a \in A\). Therefore, 0 is the identity element for the operation \( * \). **2. Proving every non-zero element has an inverse and \(a^{-1} = 6 - a\):** An element \(a \in A\) has an inverse \(b \in A\) if \(a * b = b * a = 0\) (where 0 is the identity element). We need to check if \(6 - a\) is the inverse for each non-zero \(a \in A\). Let \(a \in \{1, 2, 3, 4, 5\}\). The proposed inverse is \(b = 6 - a\). Let's calculate \(a * (6 - a)\): \(a * (6 - a) = a + (6 - a)\) or \(a + (6 - a) - 6\). In both cases, \(a + (6 - a) = 6\). Since \(a + (6 - a) = 6\), which is \(\geq 6\), we use the second rule for the operation: \(a * (6 - a) = (a + (6 - a)) - 6 = 6 - 6 = 0\). Let's also check \((6 - a) * a\): \((6 - a) * a = (6 - a) + a - 6 = 6 - 6 = 0\). Since \(a * (6 - a) = 0\) and \((6 - a) * a = 0\), \(6 - a\) is indeed the inverse of \(a\). Let's list the inverses for each non-zero element: - For \(a = 1\), \(a^{-1} = 6 - 1 = 5\). Check: \(1 * 5 = (1+5)-6 = 0\). - For \(a = 2\), \(a^{-1} = 6 - 2 = 4\). Check: \(2 * 4 = (2+4)-6 = 0\). - For \(a = 3\), \(a^{-1} = 6 - 3 = 3\). Check: \(3 * 3 = (3+3)-6 = 0\). - For \(a = 4\), \(a^{-1} = 6 - 4 = 2\). Check: \(4 * 2 = (4+2)-6 = 0\). - For \(a = 5\), \(a^{-1} = 6 - 5 = 1\). Check: \(5 * 1 = (5+1)-6 = 0\). Thus, every non-zero element in A has an inverse, and the inverse of \(a\) is \(6 - a\). To illustrate the operation, let's construct the composition table (Cayley table):

*	0	1	2	3	4	5
0	0	1	2	3	4	5
1	1	2	3	4	5	0
2	2	3	4	5	0	1
3	3	4	5	0	1	2
4	4	5	0	1	2	3
5	5	0	1	2	3	4

From the table, it is clear: - The row/column for 0 matches the elements of A, confirming 0 is the identity element. - For each non-zero element \(a\), we find 0 in its row/column. For example: - \(1 * 5 = 0\) and \(5 * 1 = 0\). So, 5 is the inverse of 1, and 1 is the inverse of 5. - \(2 * 4 = 0\) and \(4 * 2 = 0\). So, 4 is the inverse of 2, and 2 is the inverse of 4. - \(3 * 3 = 0\). So, 3 is its own inverse. This confirms that \(a^{-1} = 6 - a\) for all \(a \in \{1, 2, 3, 4, 5\}\).
In simple words: We have an operation that's like addition but "wraps around" at 6 (like a clock with numbers 0 to 5). We showed that 0 is the "neutral" number (identity element) because adding 0 doesn't change anything. Then, we showed that for any number (except 0), you can find another number that, when combined, gives 0. This "inverse" number is simply 6 minus the original number. We used a table to clearly show these rules.

🎯 Exam Tip: When given an arithmetic operation modulo n, constructing a Cayley table is often helpful to visualize the properties of the operation, such as identity elements and inverses. Remember that \(a + b \pmod{n}\) is equivalent to the given piecewise definition.

 

Question 15. \(A = \{-1, 0, 1, 2\}\), \(B = \{-4, -2, 0, 2\}\) અને વિધેયો \(f, g : A \to B\), \(f(x) = x^2 - x\), \(x \in A\) અને \(g(x) = 2\left|x-\frac{1}{2}\right|-1\), \(x \in A\) દ્વારા વ્યાખ્યાયિત છે. \(f\) અને \(g\) સમાન વિધેયો છે ? તમારા જવાબની યથાર્થતા ચકાસો. (સૂચન : યાદ કરી કે વિધેયો \(f : A \to B\) અને \(g : A \to B\) માટે \(f(a) = g(a)\), \(\forall a \in A\) હોય, તો \(f\) અને \(g\) સમાન વિધેયો કહેવાય છે.)

Answer:Given sets \(A = \{-1, 0, 1, 2\}\) and \(B = \{-4, -2, 0, 2\}\). Functions \(f, g : A \to B\) are defined by: \(f(x) = x^2 - x\) for \(x \in A\) \(g(x) = 2\left|x-\frac{1}{2}\right|-1\) for \(x \in A\) For functions \(f\) and \(g\) to be equal, they must have the same domain and co-domain (which they do), and \(f(x) = g(x)\) for all \(x\) in their common domain A. Let's evaluate both functions for each element in set A: **For \(x = -1\):** \(f(-1) = (-1)^2 - (-1) = 1 + 1 = 2\) \(g(-1) = 2\left|-1-\frac{1}{2}\right|-1 = 2\left|-\frac{3}{2}\right|-1 = 2\left(\frac{3}{2}\right)-1 = 3 - 1 = 2\) So, \(f(-1) = g(-1) = 2\). **For \(x = 0\):** \(f(0) = (0)^2 - 0 = 0\) \(g(0) = 2\left|0-\frac{1}{2}\right|-1 = 2\left|-\frac{1}{2}\right|-1 = 2\left(\frac{1}{2}\right)-1 = 1 - 1 = 0\) So, \(f(0) = g(0) = 0\). **For \(x = 1\):** \(f(1) = (1)^2 - 1 = 1 - 1 = 0\) \(g(1) = 2\left|1-\frac{1}{2}\right|-1 = 2\left|\frac{1}{2}\right|-1 = 2\left(\frac{1}{2}\right)-1 = 1 - 1 = 0\) So, \(f(1) = g(1) = 0\). **For \(x = 2\):** \(f(2) = (2)^2 - 2 = 4 - 2 = 2\) \(g(2) = 2\left|2-\frac{1}{2}\right|-1 = 2\left|\frac{3}{2}\right|-1 = 2\left(\frac{3}{2}\right)-1 = 3 - 1 = 2\) So, \(f(2) = g(2) = 2\). Since \(f(x) = g(x)\) for all \(x \in A\), the functions \(f\) and \(g\) are equal.
In simple words: We are given two functions and asked if they are the same. For them to be the same, they must have the same input set, output set, and for every input, they must give the exact same output. We tested each number from the input set in both functions and found that they always produced the same result. Therefore, the functions are equal.

🎯 Exam Tip: To prove two functions are equal, ensure their domains and co-domains match and then verify that \(f(x) = g(x)\) for *every* element in their shared domain. Be careful with absolute value calculations.

 

Question 16. ગણ \(A = \{1, 2, 3\}\) લો. ઘટક (1, 2) અને (1, 3) સમાવતા હોય અને સ્વવાચક અને સંમિત હોય, પરંતુ પરંપરિત ન હોય તેવા સંબંધોની સંખ્યા ............... છે.

(A) 1
(B) 2
(C) 3
(D) 4
Answer:Given set \(A = \{1, 2, 3\}\). We need a relation R on A that contains (1, 2) and (1, 3), is reflexive, symmetric, but not transitive. **1. Reflexive:** R must contain \((1, 1), (2, 2), (3, 3)\). **2. Contains (1, 2) and (1, 3):** R must contain \((1, 2)\) and \((1, 3)\). **3. Symmetric:** Since R contains \((1, 2)\), it must also contain \((2, 1)\). Since R contains \((1, 3)\), it must also contain \((3, 1)\). So far, R must contain: \(\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)\}\). **4. Not Transitive:** A relation is transitive if for all \((a, b) \in R\) and \((b, c) \in R\), then \((a, c) \in R\). We need to find a pair \((a, c)\) that should be in R but is not, based on existing pairs. Let's check for transitivity with the elements we have: - \((2, 1) \in R\) and \((1, 3) \in R\). For R to be transitive, \((2, 3)\) must be in R. - If we include \((2, 3)\) in R, then for symmetry, \((3, 2)\) must also be in R. Let's consider two cases for R: **Case 1: R does not contain (2, 3).** Consider the relation \(R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)\}\). - Reflexive: Yes, \((1, 1), (2, 2), (3, 3)\) are in \(R_1\). - Symmetric: Yes, for every \((a, b)\), \((b, a)\) is also in \(R_1\). - Not Transitive: - We have \((2, 1) \in R_1\) and \((1, 3) \in R_1\). - If it were transitive, \((2, 3)\) must be in \(R_1\). - However, \((2, 3) \notin R_1\). Thus, \(R_1\) is not transitive. This relation satisfies all the conditions. So, there is at least 1 such relation. **Case 2: R does contain (2, 3).** If \((2, 3) \in R\), then for symmetry, \((3, 2)\) must also be in R. So, the relation would be \(R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}\). Let's check transitivity for \(R_2\): - \((1, 2) \in R_2\) and \((2, 3) \in R_2 \implies (1, 3) \in R_2\) (True). - \((2, 1) \in R_2\) and \((1, 2) \in R_2 \implies (2, 2) \in R_2\) (True). - \((3, 1) \in R_2\) and \((1, 2) \in R_2 \implies (3, 2) \in R_2\) (True). - \((1, 3) \in R_2\) and \((3, 2) \in R_2 \implies (1, 2) \in R_2\) (True). And so on. If \((2, 3)\) and \((3, 2)\) are added, the relation becomes transitive. This is a complete equivalence relation (all 9 pairs for set of 3 elements). Since \(R_2\) is transitive, it does not satisfy the "not transitive" condition. Therefore, only one such relation exists, which is \(R_1\). The number of such relations is 1.
Answer: (A) 1
In simple words: We need to build a relationship among numbers 1, 2, 3 that follows certain rules. It must link each number to itself (reflexive), if 1 is linked to 2, then 2 must be linked to 1 (symmetric), and it must include the links (1,2) and (1,3). The tricky part is that it must NOT be "transitive" (meaning if 1 links to 2, and 2 links to 3, then 1 must link to 3). We found only one way to make such a relationship without adding the (2,3) link, which would make it transitive.

🎯 Exam Tip: Systematically construct the relation by first adding all necessary elements for reflexivity and the given pairs. Then add pairs for symmetry. Finally, test for transitivity and identify missing pairs. If adding a missing pair makes the relation transitive, it doesn't fit the "not transitive" criteria. A single such counterexample proves non-transitivity.

 

Question 17. ગણ \(A = \{1, 2, 3\}\) લો. (1, 2) ને સમાવતા સામ્ય સંબંધોની સંખ્યા ............... છે.

(A) 1
(B) 2
(C) 3
(D) 4
Answer:Given set \(A = \{1, 2, 3\}\). We need to find the number of equivalence relations on A that contain \((1, 2)\). An equivalence relation must be reflexive, symmetric, and transitive. **Step 1: Reflexivity:** Any equivalence relation R on A must contain \((1, 1), (2, 2), (3, 3)\). **Step 2: Including (1, 2) and ensuring symmetry:** Since R must contain \((1, 2)\), and it must be symmetric, R must also contain \((2, 1)\). So, at this point, R must contain at least \(\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}\). **Step 3: Ensuring transitivity for existing elements:** Let's check transitivity for the current set of elements: - \((1, 2) \in R\) and \((2, 1) \in R \implies (1, 1) \in R\) (already present). - \((2, 1) \in R\) and \((1, 2) \in R \implies (2, 2) \in R\) (already present). This relation \(R_0 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}\) is reflexive, symmetric, and transitive. This is an equivalence relation. It partitions A into \(\{\{1, 2\}, \{3\}\}\). **Step 4: Considering other elements that might be included.** If an equivalence relation contains \((1, 2)\), it effectively links 1 and 2, putting them in the same equivalence class. What if we also include \((1, 3)\)? If \((1, 3) \in R\), then due to symmetry, \((3, 1)\) must also be in R. Now, check transitivity: - Since \((1, 2) \in R\) and \((2, 1) \in R\), and \((1, 3) \in R\), and \((3, 1) \in R\). - Consider \((2, 1) \in R\) and \((1, 3) \in R \implies (2, 3) \in R\). - If \((2, 3) \in R\), then due to symmetry, \((3, 2)\) must also be in R. If \((2, 3)\) and \((3, 2)\) are added, the relation becomes: \(R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}\). This is the universal relation \(A \times A\), which is always an equivalence relation. It partitions A into \(\{\{1, 2, 3\}\}\). So we have found two equivalence relations that contain \((1, 2)\): 1. \(R_0 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}\). This corresponds to partitions \(\{\{1, 2\}, \{3\}\}\). 2. \(R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}\). This corresponds to partitions \(\{\{1, 2, 3\}\}\). Any other combination would either violate reflexivity, symmetry, or transitivity, or would be a superset of \(R_0\) that eventually becomes \(R_1\). For instance, if we add \((2, 3)\) but not \((1, 3)\), then for symmetry, \((3, 2)\) must be present. Then, \((1, 2) \in R\) and \((2, 3) \in R \implies (1, 3) \in R\). For symmetry, \((3, 1) \in R\). This leads to \(R_1\) again. Thus, there are 2 such equivalence relations.
Answer: (B) 2
In simple words: We are looking for different ways to group numbers from 1, 2, 3 such that the groups follow the rules of an "equivalence relation," and 1 and 2 must be in the same group. We found two such ways: one where 1 and 2 are grouped together, and 3 is by itself; and another where all three numbers (1, 2, and 3) are grouped together. These are the only two possible valid groupings that include 1 and 2 being related.

🎯 Exam Tip: Equivalence relations are directly related to partitions of a set. When asked to find equivalence relations that contain specific pairs, first ensure reflexivity and symmetry, then use transitivity to deduce all other necessary pairs. Each distinct set of equivalence classes (partition) corresponds to a unique equivalence relation.

 

Question 18. \(f : R \to R\), ચિત્ર વિધેય (Signum Function) લો. \[f(x) = \begin{cases} 1, & x>0 \\ 0, & x=0 \\ -1, & x<0 \end{cases}\] અને \(g : R \to R\), મહત્તમ પૂર્ણાંક વિધેય \(g(x) = [x]\), જ્યાં \([x] = x\) અથવા \(x\) થી નાના પૂર્ણાંકો પૈકી મહત્તમ પૂર્ણાંક દ્વારા વ્યાખ્યાયિત છે, તો \(fog\) અને \(gof\) એ \((0, 1]\) માં એકના એક જ (સમાન) છે?

(A) સમાન
(B) ના
(C) સંદિગ્ધ
(D) સંયોજિત વિધેયનું અસ્તિત્વ નથી.
Answer:Given functions: \(f : R \to R\) is the Signum function: \[f(x) = \begin{cases} 1, & x>0 \\ 0, & x=0 \\ -1, & x<0 \end{cases}\] \(g : R \to R\) is the Greatest Integer function: \(g(x) = [x]\). We need to compare \(fog\) and \(gof\) in the interval \((0, 1]\). **1. Calculate \(gof(x)\) for \(x \in (0, 1]\):** \(gof(x) = g(f(x))\) For \(x \in (0, 1]\), we know that \(x > 0\). According to the definition of \(f(x)\), \(f(x) = 1\) for \(x > 0\). So, \(gof(x) = g(1)\). The greatest integer function \(g(1) = [1] = 1\). Thus, \(gof(x) = 1\) for all \(x \in (0, 1]\). **2. Calculate \(fog(x)\) for \(x \in (0, 1]\):** \(fog(x) = f(g(x)) = f([x])\) For \(x \in (0, 1]\): - If \(x = 1\), then \([x] = [1] = 1\). So, \(f([x]) = f(1)\). Since \(1 > 0\), \(f(1) = 1\). - If \(0 < x < 1\), then \([x] = 0\). So, \(f([x]) = f(0)\). According to the definition of \(f(x)\), \(f(0) = 0\). Therefore, \(fog(x)\) has two different values in the interval \((0, 1]\): \[fog(x) = \begin{cases} 0, & 0Answer: (B) ના
In simple words: We are comparing two combined functions for numbers between 0 and 1 (including 1). The first function, \(gof\), first takes the sign of the number, which is always 1 for positive numbers, then finds the greatest integer of that result, which is \( [1] = 1\). So \(gof\) always gives 1. The second function, \(fog\), first finds the greatest integer of the number. For numbers between 0 and 1 (not including 1), the greatest integer is 0. For 1, it's 1. Then it takes the sign of that result. So, for numbers like 0.5, \(fog\) gives sign(0) = 0, which is different from \(gof\). Thus, they are not the same.

🎯 Exam Tip: When dealing with composite functions involving piecewise definitions or special functions like Signum or Greatest Integer, always determine the value of the inner function first, and then apply the outer function. Pay close attention to the specified interval for evaluation.

 

Question 19. ગણ \(\{a, b\}\) પર દ્વિક્રિયાઓની સંખ્યા

(A) 10
(B) 16
(C) 20
(D) 8
Answer:Let the set be \(A = \{a, b\}\). The number of elements in set A is \(n(A) = 2\). A binary operation on set A is a function from \(A \times A\) to A. First, let's find the number of elements in the Cartesian product \(A \times A\). \(n(A \times A) = n(A) \times n(A) = 2 \times 2 = 4\). The elements of \(A \times A\) are \(\{(a, a), (a, b), (b, a), (b, b)\}\). For each element in \(A \times A\), there are \(n(A)\) possible choices for its image in A. Since there are 4 elements in \(A \times A\), and for each element there are 2 possible images (either 'a' or 'b'), the total number of binary operations is \(n(A)^{n(A \times A)}\). Number of binary operations = \((n(A))^{n(A)^2} = 2^{2^2} = 2^4 = 16\). Therefore, the number of binary operations on the set \(\{a, b\}\) is 16.
Answer: (B) 16
In simple words: A binary operation on a set means you take any two elements from the set, combine them, and get an element from the same set. For a set with two elements, say 'a' and 'b', there are four possible pairs you can combine ((a,a), (a,b), (b,a), (b,b)). For each of these four pairs, you have two choices for the result (either 'a' or 'b'). So, the total number of ways to define this operation is \(2 \times 2 \times 2 \times 2\), which is 16.

🎯 Exam Tip: The number of binary operations on a set with \(n\) elements is \(n^{n^2}\). This formula is derived from the definition of a binary operation as a function from \(A \times A\) to A, where \(n(A \times A) = n^2\).