GSEB Class 12 Maths Solutions Chapter 1 સંબંધ અને વિધેય Exercise 1.3

GSEB Solutions Class 12 Maths Chapter 1 संघ अने विधेय Ex 1.3

 

Question 1. धारी के f = {1, 3, 4} → {1, 2, 5} अने g = {1, 2, 5} → {1, 3} \ f = {(1, 2), (3, 5), (4, 1)} अने g = {(1, 3), (3, 3), (5, 1)} द्वारा व्याख्यायित विधेयो छे. gof शोघो.

Answer:Let function \( f \) map from set \( \{1, 3, 4\} \) to \( \{1, 2, 5\} \), defined as \( f = \{(1, 2), (3, 5), (4, 1)\} \). This means: \( f(1) = 2 \) \( f(3) = 5 \) \( f(4) = 1 \) Let function \( g \) map from set \( \{1, 2, 5\} \) to \( \{1, 3\} \), defined as \( g = \{(1, 3), (2, 3), (5, 1)\} \). This means: \( g(1) = 3 \) \( g(2) = 3 \) \( g(5) = 1 \) Now, we need to find the composite function \( gof \). The domain of \( gof \) will be \( \{1, 3, 4\} \) and its codomain will be \( \{1, 3\} \). To find \( gof \): \( (gof)(1) = g(f(1)) \) \( = g(2) \) \( = 3 \) \( (gof)(3) = g(f(3)) \) \( = g(5) \) \( = 1 \) \( (gof)(4) = g(f(4)) \) \( = g(1) \) \( = 3 \) So, the composite function \( gof \) is: \( gof: \{(1, 3), (3, 1), (4, 3)\} \)
In simple words: We are given two functions, \( f \) and \( g \). We first find what \( f \) maps each input to, then use those outputs as inputs for \( g \) to find the final output for the composite function \( gof \).

🎯 Exam Tip: Always clearly state the domain and codomain of the composite function. Carefully follow the mapping for each element to avoid errors in calculating \( gof \) or \( fog \).

 

Question 2. धारी के विधेयो f, g अने h ओ R धी R आपेलां छे. साबित करो के, (f + g)oh = foh + goh

Answer:Given that functions \( f \), \( g \), and \( h \) are defined from \( R \) to \( R \). **(i) To prove: \( (f + g)oh = foh + goh \)** Let's consider the left-hand side (L.H.S.): \( (f + g)oh \) For any \( x \in R \), we can write this as: \( (f + g)oh (x) = (f + g)[h(x)] \) By the definition of function addition: \( = f[h(x)] + g[h(x)] \) This can be rewritten using composition notation: \( = (foh)(x) + (goh)(x) \) And combining the functions: \( = (foh + goh)(x) \) Thus, \( (f + g)oh = foh + goh \). This proves the first part. **(ii) To prove: \( (f \cdot g)oh = (foh) \cdot (goh) \)** Let's consider the left-hand side (L.H.S.): \( (f \cdot g)oh \) For any \( x \in R \), we can write this as: \( (f \cdot g)oh (x) = (f \cdot g)[h(x)] \) By the definition of function multiplication: \( = f[h(x)] \cdot g[h(x)] \) This can be rewritten using composition notation: \( = (foh)(x) \cdot (goh)(x) \) And combining the functions: \( = ((foh) \cdot (goh))(x) \) Thus, \( (f \cdot g)oh = (foh) \cdot (goh) \). This proves the second part.
In simple words: We are showing that function composition with addition and multiplication behaves like distribution. When you add or multiply two functions and then compose them with a third function, it's the same as composing each of the first two functions with the third one separately and then adding or multiplying the results.

🎯 Exam Tip: When proving properties of function composition, it is crucial to apply the definitions of function addition, multiplication, and composition step-by-step. Start by applying the definition to an arbitrary element \( x \) in the domain.

 

Question 3. gof अने fog शोघो :
(i) f(x) = |x| अने g(x) = |5x - 2|
(ii) f(x) = \( 8x^3 \) अने g(x) = \( x^{\frac{1}{3}} \)

Answer:**(i) Given: \( f(x) = |x| \) and \( g(x) = |5x - 2| \)** To find \( gof \): \( gof(x) = g(f(x)) \) Substitute \( f(x) \) into \( g(x) \): \( = g(|x|) \) Apply the definition of \( g \): \( = |5|x| - 2| \) To find \( fog \): \( fog(x) = f(g(x)) \) Substitute \( g(x) \) into \( f(x) \): \( = f(|5x - 2|) \) Apply the definition of \( f \): \( = ||5x - 2|| \) Since the absolute value of an absolute value is just the absolute value itself: \( = |5x - 2| \) **(ii) Given: \( f(x) = 8x^3 \) and \( g(x) = x^{\frac{1}{3}} \)** To find \( gof \): \( gof(x) = g(f(x)) \) Substitute \( f(x) \) into \( g(x) \): \( = g(8x^3) \) Apply the definition of \( g \): \( = (8x^3)^{\frac{1}{3}} \) \( = 8^{\frac{1}{3}} \cdot (x^3)^{\frac{1}{3}} \) \( = 2x \) To find \( fog \): \( fog(x) = f(g(x)) \) Substitute \( g(x) \) into \( f(x) \): \( = f(x^{\frac{1}{3}}) \) Apply the definition of \( f \): \( = 8(x^{\frac{1}{3}})^3 \) \( = 8x \)
In simple words: For composite functions like \( gof \) or \( fog \), you plug one function into the other. For \( gof \), you first calculate \( f(x) \), then use that result as the input for \( g \). For \( fog \), you first calculate \( g(x) \), then use that result as the input for \( f \).

🎯 Exam Tip: Pay close attention to the order of function composition. \( gof(x) \) means \( g(f(x)) \), while \( fog(x) \) means \( f(g(x)) \). The operations are performed from the inside out.

 

Question 4. જો f(x) = \( \frac{4x+3}{6x-4} \), x ≠ \( \frac{2}{3} \), સાબિત કરો કે (fof) (x) = x. f નું પ્રતિવિષય શું છે ?

Answer:Given the function \( f(x) = \frac{4x+3}{6x-4} \), where \( x \neq \frac{2}{3} \). First, let's find \( (fof)(x) \): \( (fof)(x) = f(f(x)) \) Substitute \( f(x) \) into itself: \( = f\left(\frac{4x+3}{6x-4}\right) \) Now, replace \( x \) in \( f(x) \) with \( \frac{4x+3}{6x-4} \): \( = \frac{4\left(\frac{4x+3}{6x-4}\right) + 3}{6\left(\frac{4x+3}{6x-4}\right) - 4} \) To simplify, multiply the numerator and denominator by \( (6x-4) \): \( = \frac{4(4x+3) + 3(6x-4)}{6(4x+3) - 4(6x-4)} \) Expand the terms: \( = \frac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16} \) Combine like terms: \( = \frac{34x}{34} \) \( = x \) Thus, \( (fof)(x) = x \) is proven. Next, we need to find the inverse of \( f \), denoted as \( f^{-1}(x) \). Let \( y = f(x) \). \( y = \frac{4x+3}{6x-4} \) Now, we solve for \( x \) in terms of \( y \): \( y(6x-4) = 4x+3 \) \( 6xy - 4y = 4x+3 \) Move all terms with \( x \) to one side and terms without \( x \) to the other side: \( 6xy - 4x = 4y + 3 \) Factor out \( x \): \( x(6y - 4) = 4y + 3 \) Isolate \( x \): \( x = \frac{4y+3}{6y-4} \) Since \( x = f^{-1}(y) \), we have: \( f^{-1}(y) = \frac{4y+3}{6y-4} \) To express the inverse function in terms of \( x \), replace \( y \) with \( x \): \( f^{-1}(x) = \frac{4x+3}{6x-4} \) Comparing \( f(x) \) and \( f^{-1}(x) \), we see that \( f^{-1}(x) = f(x) \). Therefore, \( f \) નું પ્રતિવિધેય \( f \) પોતે જ છે.
In simple words: We first checked if applying the function \( f \) twice brings us back to the original input \( x \). It does, which means \( f \) is its own inverse. To formally find the inverse, we set \( y = f(x) \) and solved for \( x \) in terms of \( y \), then replaced \( y \) with \( x \) to get the inverse function.

🎯 Exam Tip: When proving \( (fof)(x) = x \), careful algebraic simplification is key. For finding the inverse, the process involves setting \( y = f(x) \), interchanging \( x \) and \( y \), and then solving for the new \( y \). If \( (fof)(x) = x \), then \( f \) is its own inverse.

 

Question 5. નીચે આપેલાં વિષયોનાં પ્રતિવિધેય મળી શકશે ? કારણ સહિત નિર્ણય કરી.
(i) f : {1, 2, 3, 4} → {10}
f: {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4}
g: {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13}
h : {(2, 7), (3, 9), (4, 11), (5, 13)}

Answer:For a function to have an inverse, it must be both one-to-one (injective) and onto (surjective). **(i) Given: \( f : \{1, 2, 3, 4\} \rightarrow \{10\} \)** The function is defined as \( f = \{(1, 10), (2, 10), (3, 10), (4, 10)\} \). From the definition, we observe: \( f(1) = 10 \) \( f(2) = 10 \) \( f(3) = 10 \) \( f(4) = 10 \) Here, multiple distinct elements in the domain (1, 2, 3, 4) map to the same element (10) in the codomain. This means \( f \) is not a one-to-one function.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र समुच्चय A {1, 2, 3, 4} से समुच्चय B {10} तक एक फलन f को दिखाता है। सभी तत्व 1, 2, 3, 4, समुच्चय B के एक ही तत्व 10 से जुड़े हुए हैं। यह दर्शाता है कि फलन एकैकी नहीं है क्योंकि कई इनपुट का एक ही आउटपुट है। Since \( f \) is not one-to-one, it does not have an inverse. Therefore, \( f \) નું પ્રતિવિધેય મળી શકે નહિ. **(ii) Given: \( g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\} \)** The function is defined as \( g = \{(5, 4), (6, 3), (7, 4), (8, 2)\} \). From the definition, we observe: \( g(5) = 4 \) \( g(6) = 3 \) \( g(7) = 4 \) \( g(8) = 2 \) Here, the distinct elements \( 5 \) and \( 7 \) in the domain both map to the same element \( 4 \) in the codomain. This means \( g \) is not a one-to-one function.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र समुच्चय A {5, 6, 7, 8} से समुच्चय B {1, 2, 3, 4} तक एक फलन g को दर्शाता है। इसमें दिखाया गया है कि डोमेन के दो अलग-अलग तत्व, 5 और 7, कोडोमेन के एक ही तत्व 4 पर मैप करते हैं। यह स्पष्ट रूप से दर्शाता है कि फलन एकैकी नहीं है। Since \( g \) is not one-to-one, it does not have an inverse. Therefore, \( g \) નાં પ્રતિવિધેયનું અસ્તિત્વ નથી. **(iii) Given: \( h : \{2, 3, 4, 5\} \rightarrow \{7, 9, 11, 13\} \)** The function is defined as \( h = \{(2, 7), (3, 9), (4, 11), (5, 13)\} \). From the definition, we observe: \( h(2) = 7 \) \( h(3) = 9 \) \( h(4) = 11 \) \( h(5) = 13 \) Here, each distinct element in the domain maps to a unique distinct element in the codomain. This indicates \( h \) is a one-to-one function. Also, the range of \( h \) is \( \{7, 9, 11, 13\} \), which is equal to its codomain. This means \( h \) is an onto function.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र समुच्चय A {2, 3, 4, 5} से समुच्चय B {7, 9, 11, 13} तक एक फलन h को दर्शाता है। इसमें डोमेन के प्रत्येक तत्व को कोडोमेन के एक अद्वितीय तत्व से जोड़ा गया है। इससे पता चलता है कि यह फलन एकैकी और आच्छादक दोनों है। Since \( h \) is both one-to-one and onto, it is an invertible function. Therefore, \( h \) નાં પ્રતિવિધેયનું અસ્તિત્વ છે.
In simple words: For a function to have an inverse, every input must map to a unique output (one-to-one), and every possible output must be reached by some input (onto). If these conditions are not met, the inverse function does not exist.

🎯 Exam Tip: To check for invertibility, first verify if the function is one-to-one (injective) by ensuring no two distinct domain elements map to the same codomain element. Then, check if it is onto (surjective) by ensuring the range equals the codomain. Both conditions must be met for an inverse to exist.

 

Question 6. સાબિત કરો કે f : [-1, 1] → R, f(x) = \( \frac{x}{x+2} \) દ્વારા વ્યાખ્યાયિત વિષેય એક એક છે. વિધેય f : [-1, 1] → f નો વિસ્તાર f(x) = \( \frac{x}{x+2} \), તો f નું પ્રતિવિષય શોધો.
સૂચનઃ f ના વિસ્તારમાં આવેલ y ને સંગત કોઈક x ૯ [-1, 1]
માટે y = f(x) = \( \frac{x}{x+2} \) એટલે કે, x = \( \frac{2y}{1-y} \).

Answer:Given the function \( f : [-1, 1] \rightarrow R \), defined as \( f(x) = \frac{x}{x+2} \). **Part 1: Prove that \( f \) is one-to-one (injective).** Let \( x_1, x_2 \in [-1, 1] \) such that \( f(x_1) = f(x_2) \). \( \frac{x_1}{x_1+2} = \frac{x_2}{x_2+2} \) Cross-multiply: \( x_1(x_2+2) = x_2(x_1+2) \) \( x_1x_2 + 2x_1 = x_1x_2 + 2x_2 \) Subtract \( x_1x_2 \) from both sides: \( 2x_1 = 2x_2 \) Divide by 2: \( x_1 = x_2 \) Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function \( f \) is one-to-one. **Part 2: Find the inverse function \( f^{-1} \).** Let \( y \) be an element in the range of \( f \), so \( y = f(x) \). \( y = \frac{x}{x+2} \), for some \( x \in [-1, 1] \). We need to solve for \( x \) in terms of \( y \): \( y(x+2) = x \) \( yx + 2y = x \) Move all terms with \( x \) to one side: \( 2y = x - yx \) Factor out \( x \): \( 2y = x(1-y) \) Isolate \( x \): \( x = \frac{2y}{1-y} \) The hint suggests this is the case. This expression for \( x \) defines the inverse function \( f^{-1}(y) \). So, \( f^{-1}(y) = \frac{2y}{1-y} \). To express the inverse function in terms of \( x \), we replace \( y \) with \( x \): \( f^{-1}(x) = \frac{2x}{1-x} \) The problem also mentions "f નો વિસ્તાર f(x) = \( \frac{x}{x+2} \), તો f નું પ્રતિવિષય શોધો." This implies we need to find the range of \( f \) as well to define the domain of \( f^{-1} \). For \( x \in [-1, 1] \): If \( x = -1 \), \( f(-1) = \frac{-1}{-1+2} = \frac{-1}{1} = -1 \). If \( x = 1 \), \( f(1) = \frac{1}{1+2} = \frac{1}{3} \). The function \( f(x) = \frac{x}{x+2} \) is an increasing function on \( [-1, 1] \). To see this, consider its derivative \( f'(x) = \frac{(x+2)(1) - x(1)}{(x+2)^2} = \frac{x+2-x}{(x+2)^2} = \frac{2}{(x+2)^2} \). Since \( (x+2)^2 > 0 \) for \( x \neq -2 \), \( f'(x) > 0 \), so \( f(x) \) is strictly increasing. Therefore, the range of \( f \) is \( [-1, \frac{1}{3}] \). The inverse function \( f^{-1}(x) = \frac{2x}{1-x} \) has a domain of \( [-1, \frac{1}{3}] \) and a range of \( [-1, 1] \).
In simple words: We first showed that different inputs always lead to different outputs for function \( f \), meaning it's one-to-one. Then, to find the inverse function, we took \( y = f(x) \) and rearranged the equation to express \( x \) in terms of \( y \). This new expression gives us the inverse function.

🎯 Exam Tip: When proving a function is one-to-one, always start by assuming \( f(x_1) = f(x_2) \) and logically show that \( x_1 = x_2 \). For finding the inverse, algebraic manipulation is critical. Be mindful of domain and range changes when dealing with inverse functions.

 

Question 7. ધારો કે વિધેય f : R → R, f(x) = 4x + 3. સાબિત કરો કે f વ્યસ્તસંપન્ન છે. વિધેય f નું પ્રતિવિધેય શોધો.

Answer:Given the function \( f : R \rightarrow R \), defined as \( f(x) = 4x + 3 \). **Part 1: Prove that \( f \) is one-to-one (injective).** Let \( x_1, x_2 \in R \) such that \( f(x_1) = f(x_2) \). \( 4x_1 + 3 = 4x_2 + 3 \) Subtract 3 from both sides: \( 4x_1 = 4x_2 \) Divide by 4: \( x_1 = x_2 \) Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function \( f \) is one-to-one. **Part 2: Prove that \( f \) is onto (surjective).** Let \( y \in R \) be an arbitrary element in the codomain. We need to find an \( x \in R \) such that \( f(x) = y \). \( y = 4x + 3 \) Solve for \( x \): \( y - 3 = 4x \) \( x = \frac{y-3}{4} \) Since \( y \in R \), \( (y-3)/4 \) is always a real number. Thus, for every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \). So, \( f \) is an onto function. Since \( f \) is both one-to-one and onto, it is invertible (વ્યસ્તસંપન્ન). **Part 3: Find the inverse function \( f^{-1} \).** From Part 2, we found \( x = \frac{y-3}{4} \). This expression for \( x \) is \( f^{-1}(y) \). So, \( f^{-1}(y) = \frac{y-3}{4} \). To express the inverse function in terms of \( x \), replace \( y \) with \( x \): \( f^{-1}(x) = \frac{x-3}{4} \)
In simple words: We showed that function \( f \) is invertible by proving it is both one-to-one (each input has a unique output) and onto (every output can be reached). To find its inverse, we just reversed the function by solving for the input variable in terms of the output variable.

🎯 Exam Tip: For a function defined from R to R, a linear function like \( f(x) = ax + b \) (where \( a \neq 0 \)) is always invertible. To prove it, explicitly show one-to-one and onto properties. The inverse function formula is found by solving \( y = f(x) \) for \( x \).

 

Question 8. વિધેય f : R+ → [4, ∞), f(x) = \( x^2 + 4 \) દ્વારા વ્યાખ્યાયિત છે. સાબિત કરો કે વ્યસ્તસંપન્ન છે અને નું પ્રતિવિષય f-1 એ f-1(y) = \( \sqrt{y-4} \) દ્વારા દર્શાવાય છે. અત્રે, R* એ તમામ અનુત્ર વાસ્તવિક સંખ્યાઓનો ગણ છે.

Answer:Given the function \( f : R^+ \rightarrow [4, \infty) \), defined as \( f(x) = x^2 + 4 \). Here \( R^+ \) denotes the set of all non-negative real numbers. **Part 1: Prove that \( f \) is one-to-one (injective).** Let \( x_1, x_2 \in R^+ \) such that \( f(x_1) = f(x_2) \). \( x_1^2 + 4 = x_2^2 + 4 \) Subtract 4 from both sides: \( x_1^2 = x_2^2 \) Take the square root of both sides: \( \sqrt{x_1^2} = \sqrt{x_2^2} \) Since \( x_1, x_2 \in R^+ \) (meaning they are non-negative), we have: \( x_1 = x_2 \) Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function \( f \) is one-to-one. **Part 2: Prove that \( f \) is onto (surjective).** Let \( y \in [4, \infty) \) be an arbitrary element in the codomain. We need to find an \( x \in R^+ \) such that \( f(x) = y \). \( y = x^2 + 4 \) Solve for \( x \): \( y - 4 = x^2 \) \( x = \sqrt{y-4} \) Since \( y \in [4, \infty) \), \( y-4 \ge 0 \), so \( \sqrt{y-4} \) is a real number. Also, the square root symbol denotes the positive root, so \( x \ge 0 \), which means \( x \in R^+ \). Thus, for every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \). So, \( f \) is an onto function. Since \( f \) is both one-to-one and onto, it is invertible (વ્યસ્તસંપન્ન). **Part 3: Find the inverse function \( f^{-1} \).** From Part 2, we found \( x = \sqrt{y-4} \). This expression for \( x \) is \( f^{-1}(y) \). So, \( f^{-1}(y) = \sqrt{y-4} \). To express the inverse function in terms of \( x \), replace \( y \) with \( x \): \( f^{-1}(x) = \sqrt{x-4} \)
In simple words: We confirmed that this function is invertible because each positive input leads to a unique output, and all outputs greater than or equal to 4 can be reached. The inverse function was found by swapping the input and output variables and solving for the new output.

🎯 Exam Tip: When dealing with functions defined on restricted domains like \( R^+ \), remember to consider the domain constraint when solving for \( x \) (e.g., \( \sqrt{x^2} = |x| \), but for \( x \in R^+ \), it's just \( x \)). This is crucial for proving one-to-one and onto properties correctly.

 

Question 9. વિધેય f : R+ → [5, ∞), f(x) = \( 9x^2 + 6x - 5 \) વ્યાખ્યાયિત છે. સાબિત કરી કે, f વ્યસ્તસંપન્ન છે અને f^{-1}(y)=\(\frac{\sqrt{y+6}-1}{3}\) .

Answer:Given the function \( f : R^+ \rightarrow [5, \infty) \), defined as \( f(x) = 9x^2 + 6x - 5 \). **Part 1: Prove that \( f \) is one-to-one (injective).** Let \( x_1, x_2 \in R^+ \) such that \( f(x_1) = f(x_2) \). \( 9x_1^2 + 6x_1 - 5 = 9x_2^2 + 6x_2 - 5 \) Add 5 to both sides: \( 9x_1^2 + 6x_1 = 9x_2^2 + 6x_2 \) Rearrange the terms: \( 9x_1^2 - 9x_2^2 + 6x_1 - 6x_2 = 0 \) Factor by grouping: \( 9(x_1^2 - x_2^2) + 6(x_1 - x_2) = 0 \) Apply the difference of squares formula \( (x_1^2 - x_2^2) = (x_1 - x_2)(x_1 + x_2) \): \( 9(x_1 - x_2)(x_1 + x_2) + 6(x_1 - x_2) = 0 \) Factor out \( (x_1 - x_2) \): \( (x_1 - x_2)[9(x_1 + x_2) + 6] = 0 \) Since \( x_1, x_2 \in R^+ \), both \( x_1 \) and \( x_2 \) are positive. Therefore, \( (x_1 + x_2) > 0 \), which implies \( 9(x_1 + x_2) + 6 > 6 \). So, \( 9(x_1 + x_2) + 6 \neq 0 \). For the product to be zero, we must have \( (x_1 - x_2) = 0 \). \( x_1 = x_2 \) Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function \( f \) is one-to-one. **Part 2: Prove that \( f \) is onto (surjective).** Let \( y \in [5, \infty) \) be an arbitrary element in the codomain. We need to find an \( x \in R^+ \) such that \( f(x) = y \). \( y = 9x^2 + 6x - 5 \) To solve for \( x \), we can complete the square for the quadratic expression in \( x \): \( y = (9x^2 + 6x + 1) - 1 - 5 \) \( y = (3x + 1)^2 - 6 \) Now, isolate \( (3x + 1)^2 \): \( y + 6 = (3x + 1)^2 \) Take the square root of both sides: \( \sqrt{y+6} = 3x + 1 \) (Since \( x \in R^+ \), \( 3x+1 \) is positive, so we take the positive square root). Isolate \( 3x \): \( \sqrt{y+6} - 1 = 3x \) Isolate \( x \): \( x = \frac{\sqrt{y+6} - 1}{3} \) Since \( y \in [5, \infty) \), \( y+6 \ge 11 \), so \( \sqrt{y+6} \ge \sqrt{11} > 1 \). Therefore, \( \sqrt{y+6} - 1 > 0 \), which means \( x > 0 \). So, \( x \in R^+ \). Thus, for every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \). So, \( f \) is an onto function. Since \( f \) is both one-to-one and onto, it is invertible (વ્યસ્તસંપન્ન). **Part 3: Find the inverse function \( f^{-1} \).** From Part 2, we found \( x = \frac{\sqrt{y+6} - 1}{3} \). This expression for \( x \) is \( f^{-1}(y) \). So, \( f^{-1}(y) = \frac{\sqrt{y+6} - 1}{3} \). This matches the given form.
In simple words: We proved the function \( f \) is invertible. First, we showed that each unique positive input gives a unique output. Then, we showed that every output in the range can be produced by some positive input, which we found by completing the square and solving for \( x \). This formula for \( x \) is the inverse function.

🎯 Exam Tip: For quadratic functions on a restricted domain, completing the square is often the most effective way to find the inverse. Remember to justify taking the positive square root based on the domain of \( x \).

 

Question 10. વિષય f : X → Y વ્યસ્ત સંપન્ન છે. સાબિત કરી કે, f નું પ્રતિવિધેય અનન્ય છે.
(સૂચન : ધારો કે g₁ આને g₂ બંને વિધેય f નાં પ્રતિવિધેયો છે. તેથી પ્રત્યેક y E Y માટે (fog1)(y) = Iy(y) = (fog2)(y). વિધેય f એક-એક છે, તે સત્યનો ઉપયોગ કરો.

Answer:Given that function \( f : X \rightarrow Y \) is invertible. This implies that \( f \) is both one-to-one (injective) and onto (surjective). To prove that the inverse of \( f \) is unique, let's assume, for contradiction, that \( f \) has two inverse functions, say \( g_1: Y \rightarrow X \) and \( g_2: Y \rightarrow X \). By the definition of an inverse function: For \( g_1 \) to be an inverse of \( f \): \( fog_1 = I_Y \) (identity function on Y) and \( g_1of = I_X \) (identity function on X). For \( g_2 \) to be an inverse of \( f \): \( fog_2 = I_Y \) and \( g_2of = I_X \). Now, consider an arbitrary element \( y \in Y \). Since \( fog_1 = I_Y \), we have \( (fog_1)(y) = I_Y(y) = y \). This means \( f(g_1(y)) = y \). Similarly, since \( fog_2 = I_Y \), we have \( (fog_2)(y) = I_Y(y) = y \). This means \( f(g_2(y)) = y \). From these two equations, we can write: \( f(g_1(y)) = f(g_2(y)) \) Since \( f \) is an invertible function, it must be one-to-one. By the definition of a one-to-one function, if \( f(a) = f(b) \), then \( a = b \). Applying this to \( f(g_1(y)) = f(g_2(y)) \), we get: \( g_1(y) = g_2(y) \) Since this holds for every \( y \in Y \), it means the functions \( g_1 \) and \( g_2 \) are identical. Therefore, \( g_1 = g_2 \). This proves that if a function has an inverse, that inverse must be unique.
In simple words: If a function can be reversed, its reverse is always just one specific function, not two or more different ones. We showed this by assuming there could be two different inverse functions and then proving they must actually be the same because the original function is one-to-one.

🎯 Exam Tip: The proof of uniqueness of inverse functions is a standard proof. The key step is to use the definition of a one-to-one function (if \( f(a) = f(b) \), then \( a = b \)) after setting up the equality \( f(g_1(y)) = f(g_2(y)) \).

 

Question 11. ધારો કે f : {1, 2, 3} → {a, b, c} એ f(1) = a, f(2) = b અને f(3) = c દ્વારા આપેલ છે. f-1 શોધી અને સાબિત કરો કે (f-1)-1 = f

Answer:Given the function \( f : \{1, 2, 3\} \rightarrow \{a, b, c\} \), defined as: \( f(1) = a \) \( f(2) = b \) \( f(3) = c \) **Part 1: Determine if \( f \) is invertible and find \( f^{-1} \).** * **One-to-one:** Each element in the domain \( \{1, 2, 3\} \) maps to a unique element in the codomain \( \{a, b, c\} \). So, \( f \) is one-to-one. * **Onto:** The range of \( f \) is \( \{a, b, c\} \), which is equal to its codomain. So, \( f \) is onto. Since \( f \) is both one-to-one and onto, it is invertible, and its inverse \( f^{-1} \) exists. To find \( f^{-1} \), we simply reverse the mapping: If \( f(1) = a \), then \( f^{-1}(a) = 1 \). If \( f(2) = b \), then \( f^{-1}(b) = 2 \). If \( f(3) = c \), then \( f^{-1}(c) = 3 \). So, the inverse function \( f^{-1} : \{a, b, c\} \rightarrow \{1, 2, 3\} \) is given by: \( f^{-1} = \{(a, 1), (b, 2), (c, 3)\} \) **Part 2: Prove that \( (f^{-1})^{-1} = f \).** Now we consider the inverse of \( f^{-1} \), denoted as \( (f^{-1})^{-1} \). The function \( f^{-1} \) is also one-to-one and onto, so its inverse exists. To find \( (f^{-1})^{-1} \), we reverse the mapping of \( f^{-1} \): If \( f^{-1}(a) = 1 \), then \( (f^{-1})^{-1}(1) = a \). If \( f^{-1}(b) = 2 \), then \( (f^{-1})^{-1}(2) = b \). If \( f^{-1}(c) = 3 \), then \( (f^{-1})^{-1}(3) = c \). So, \( (f^{-1})^{-1} : \{1, 2, 3\} \rightarrow \{a, b, c\} \) is given by: \( (f^{-1})^{-1} = \{(1, a), (2, b), (3, c)\} \) By comparing the mappings, we see that \( (f^{-1})^{-1} \) is exactly the same as the original function \( f \). Therefore, \( (f^{-1})^{-1} = f \).
In simple words: We first found the inverse of the function \( f \) by simply reversing its mappings. Then, we found the inverse of that inverse function, which brought us back to the original function \( f \). This shows that if you reverse something twice, you get back to where you started.

🎯 Exam Tip: This property, \( (f^{-1})^{-1} = f \), is fundamental. When asked to prove it for specific functions or generally, clearly define the mappings for \( f \), then \( f^{-1} \), and finally \( (f^{-1})^{-1} \), showing that the initial and final mappings are identical.

 

Question 12. વિધેય f : X – Y એ વ્યસ્તસંપન્ન છે. સાબિત કરો કે f-1 નું પ્રતિવિધેય શું છે, એટલે કે (f-1)-1 = f.

Answer:Given that function \( f : X \rightarrow Y \) is invertible. This means that \( f \) is both one-to-one and onto. By the definition of an invertible function, there exists a unique inverse function \( f^{-1} : Y \rightarrow X \) such that: \( f^{-1}of = I_X \) (the identity function on set \( X \)) and \( fof^{-1} = I_Y \) (the identity function on set \( Y \)) Now, consider the function \( f^{-1} : Y \rightarrow X \). Since \( f \) is one-to-one and onto, its inverse \( f^{-1} \) is also one-to-one and onto. Because \( f^{-1} \) is both one-to-one and onto, it is also an invertible function. This means its inverse, \( (f^{-1})^{-1} \), exists and maps from \( X \) back to \( Y \). Let \( g = f^{-1} \). Then we want to find \( g^{-1} \), which is \( (f^{-1})^{-1} \). By the definition of the inverse of \( g \): \( g^{-1}og = I_Y \) and \( gog^{-1} = I_X \). Substituting \( g = f^{-1} \): \( (f^{-1})^{-1} o f^{-1} = I_Y \) and \( f^{-1} o (f^{-1})^{-1} = I_X \) We know that \( fof^{-1} = I_Y \). Comparing this with \( (f^{-1})^{-1} o f^{-1} = I_Y \), if we consider composition from the right by \( f \): \( ((f^{-1})^{-1} o f^{-1}) o f = I_Y o f \) \( (f^{-1})^{-1} o (f^{-1} o f) = f \) (Associativity of composition) \( (f^{-1})^{-1} o I_X = f \) (Since \( f^{-1} o f = I_X \)) \( (f^{-1})^{-1} = f \) (Since composing with the identity function leaves the function unchanged) Alternatively, we also know that \( f^{-1}of = I_X \). We have \( f^{-1} o (f^{-1})^{-1} = I_X \). Comparing these, if we compose from the left by \( f \): \( f o (f^{-1} o (f^{-1})^{-1}) = f o I_X \) \( (f o f^{-1}) o (f^{-1})^{-1} = f \) \( I_Y o (f^{-1})^{-1} = f \) \( (f^{-1})^{-1} = f \) Both approaches lead to the same conclusion. Therefore, the inverse of \( f^{-1} \) is \( f \), i.e., \( (f^{-1})^{-1} = f \).
In simple words: If a function \( f \) can be reversed, then its reverse function \( f^{-1} \) also has a reverse. This second reversal brings you back to the original function \( f \). It's like undoing an undo operation.

🎯 Exam Tip: This general proof for \( (f^{-1})^{-1} = f \) relies heavily on the definition of an inverse function and the properties of identity functions. Clearly state the identity function for both domain and codomain sets (\( I_X \) and \( I_Y \)) and use the associative property of function composition.

 

Question 13. જો વિષય f : R → R એ f(x) = \( (3 – x^3)^{\frac{1}{3}} \) દ્વારા આપેલ છે, તો (fof)(x) = ............ છે.
(A) \( x^{\frac{1}{3}} \)
(B) \( x^3 \)
(C) x
(D) \( (3 – x^2) \)

Answer:Given the function \( f : R \rightarrow R \) defined as \( f(x) = (3 - x^3)^{\frac{1}{3}} \). We need to find \( (fof)(x) \). \( (fof)(x) = f(f(x)) \) Substitute \( f(x) \) into itself: \( = f\left((3 - x^3)^{\frac{1}{3}}\right) \) Now, replace \( x \) in \( f(x) \) with \( (3 - x^3)^{\frac{1}{3}} \): \( = \left(3 - \left((3 - x^3)^{\frac{1}{3}}\right)^3\right)^{\frac{1}{3}} \) The cube and cube root cancel out for the inner term: \( = (3 - (3 - x^3))^{\frac{1}{3}} \) Distribute the negative sign: \( = (3 - 3 + x^3)^{\frac{1}{3}} \) Simplify: \( = (x^3)^{\frac{1}{3}} \) \( = x \) So, \( (fof)(x) = x \). The correct option is (C).
Answer: (C) x
In simple words: We applied the function \( f \) to itself. By carefully substituting the expression for \( f(x) \) back into \( f \), we found that the cube and cube root operations canceled out, leaving us with just \( x \). This means applying the function twice brings us back to the original input.

🎯 Exam Tip: For composite functions involving powers and roots, remember that \( (a^{1/n})^n = a \) and \( (a^n)^{1/n} = a \). These properties are crucial for simplifying expressions quickly and accurately. Pay attention to the order of operations.

 

Question 14. વિષેય f : R – \( \{-\frac{4}{3}\} \) → R, f(x) = \( \frac{4x}{3x+4} \) વ્યાખ્યાયિત છે. f નું પ્રતિવિધેય, વિધેય g : f નો વિસ્તાર – R - \( \{-\frac{4}{3}\} \) એ દ્વારા મળે છે.
(A) g(y) = \( \frac{3y}{3-4y} \)
(B) g(y) = \( \frac{4y}{4-3y} \)
(C) g(y) = \( \frac{4y}{3-4y} \)
(D) g(y) = \( \frac{3y}{4-3y} \)

Answer:Given the function \( f : R - \{-\frac{4}{3}\} \rightarrow R \), defined as \( f(x) = \frac{4x}{3x+4} \). We need to find its inverse, \( g(y) = f^{-1}(y) \). Let \( y = f(x) \). \( y = \frac{4x}{3x+4} \) Now, we solve for \( x \) in terms of \( y \): \( y(3x+4) = 4x \) \( 3xy + 4y = 4x \) Move all terms with \( x \) to one side: \( 4y = 4x - 3xy \) Factor out \( x \): \( 4y = x(4 - 3y) \) Isolate \( x \): \( x = \frac{4y}{4-3y} \) So, the inverse function \( g(y) \) is: \( g(y) = \frac{4y}{4-3y} \) For this inverse function to be defined, the denominator cannot be zero, so \( 4-3y \neq 0 \), which means \( y \neq \frac{4}{3} \). The range of \( f \) will be \( R - \{\frac{4}{3}\} \). The function \( g \) maps from the range of \( f \) to the domain of \( f \). So, \( g : R - \{\frac{4}{3}\} \rightarrow R - \{-\frac{4}{3}\} \). Comparing our result with the given options, option (B) matches.
Answer: (B) g(y) = \( \frac{4y}{4-3y} \)
In simple words: To find the inverse function, we took the original equation \( y = f(x) \) and solved it to express \( x \) in terms of \( y \). This new expression gives us the inverse function \( g(y) \). We then matched it with the given options.

🎯 Exam Tip: Finding the inverse of a rational function involves careful algebraic manipulation. Cross-multiplication and then isolating the \( x \) term are common steps. Remember to swap \( x \) and \( y \) if you initially use \( f^{-1}(x) \) notation.