\displaystyle \textbf{Question 1. }\text{The number of all one-one functions from set }A=\{1,2,3\}\text{ to itself is}
\displaystyle \text{(a) }2\qquad \text{(b) }6
\displaystyle \text{(c) }3\qquad \text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \text{Number of one-one functions from }A\text{ to }A=3!=6
\displaystyle \therefore \text{Correct option is (b).}
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\displaystyle \textbf{Question 2. }\text{The inverse of the function }f:R\rightarrow \{a\in R:a<1\},\text{ defined by } \\ f(a)=\frac{e^{a}-e^{-a}}{e^{a}+e^{-a}},\text{ is}
\displaystyle \text{(a) }\frac{1}{2}\log\frac{1+a}{1-a}\qquad \text{(b) }\frac{1}{2}\log\frac{2+a}{2-a}
\displaystyle \text{(c) }\frac{1}{2}\log\frac{1-a}{1+a}\qquad \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}
\displaystyle y=\frac{e^{2x}-1}{e^{2x}+1}
\displaystyle y(e^{2x}+1)=e^{2x}-1
\displaystyle e^{2x}(1-y)=1+y
\displaystyle e^{2x}=\frac{1+y}{1-y}
\displaystyle 2x=\log\frac{1+y}{1-y}
\displaystyle x=\frac{1}{2}\log\frac{1+y}{1-y}
\displaystyle \therefore f^{-1}(a)=\frac{1}{2}\log\frac{1+a}{1-a}
\displaystyle \therefore \text{Correct option is (a).}
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\displaystyle \textbf{Question 3. }\text{Let }R\text{ be a relation from }R\text{ to }R\text{ the set of real numbers defined by } \\ R=\{(x,y):x,y\in R\text{ and }x-y+\sqrt{3}\text{ is an}
\displaystyle \text{irrational number}\}.\text{ Then, }R\text{ is}
\displaystyle \text{(a) reflexive}\qquad \text{(b) transitive}
\displaystyle \text{(c) symmetric}\qquad \text{(d) an equivalence relation}
\displaystyle \text{Answer:}
\displaystyle \text{For any }x\in R,\ x-x+\sqrt{3}=\sqrt{3},\text{ which is irrational.}
\displaystyle \therefore (x,x)\in R
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 4. }\text{Let }A=\{0,1,2,3\}\text{ and define a relation }R\text{ on }A\text{ as}
\displaystyle R=\{(0,0),(0,1),(0,3),(1,0),(1,1),(2,2),(3,0),(3,3)\}.\text{ Is }R\text{ reflexive, symmetric} \\ \text{and transitive?}
\displaystyle \text{Answer:}
\displaystyle \text{For reflexivity, }(0,0),(1,1),(2,2),(3,3)\text{ should belong to }R.
\displaystyle \text{Here, all these ordered pairs belong to }R.
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Now, }(0,1)\in R\text{ and }(1,0)\in R.
\displaystyle (0,3)\in R\text{ and }(3,0)\in R.
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{But }(1,0)\in R\text{ and }(0,3)\in R,\text{ while }(1,3)\notin R.
\displaystyle \therefore R\text{ is not transitive.}
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\displaystyle \textbf{Question 5. }\text{For real numbers }x\text{ and }y,\text{ define }xRy\text{ if and only if } \\ x-y+\sqrt{2}\text{ is an irrational number. Is }R\text{ transitive?}   \text{Explain your answer.}
\displaystyle \text{Answer:}
\displaystyle \text{No, }R\text{ is not transitive.}
\displaystyle \text{Take }x=0,\ y=\sqrt{2}-\sqrt{3},\ z=\sqrt{2}.
\displaystyle x-y+\sqrt{2}=0-(\sqrt{2}-\sqrt{3})+\sqrt{2}=\sqrt{3},\text{ which is irrational.}
\displaystyle \therefore xRy
\displaystyle y-z+\sqrt{2}=(\sqrt{2}-\sqrt{3})-\sqrt{2}+\sqrt{2}=\sqrt{2}-\sqrt{3},\text{ which is irrational.}
\displaystyle \therefore yRz
\displaystyle \text{But }x-z+\sqrt{2}=0-\sqrt{2}+\sqrt{2}=0,\text{ which is rational.}
\displaystyle \therefore xRz\text{ is not true.}
\displaystyle \therefore R\text{ is not transitive.}
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\displaystyle \textbf{Question 6. }\text{Let }A=\{a,b,c\}\text{ and the relation }R\text{ be defined on }A\text{ as } \\ R=\{(a,a),(b,c),(a,b)\}.\text{ Then, write minimum}   \text{number of ordered pairs to be added in } \\ R\text{ to make }R\text{ reflexive and transitive.}
\displaystyle \text{Answer:}
\displaystyle \text{For reflexivity, }(b,b)\text{ and }(c,c)\text{ must be added.}
\displaystyle \text{Also, }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \therefore \text{For transitivity, }(a,c)\text{ must be added.}
\displaystyle \therefore \text{Minimum ordered pairs to be added are }(b,b),(c,c),(a,c).
\displaystyle \therefore \text{Minimum number of ordered pairs}=3.
\\

\displaystyle \textbf{Question 7. }\text{State whether the function }f:R\rightarrow R\text{ defined by } \\ f(x)=3-4x\text{ is onto or not.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }y\in R.
\displaystyle y=3-4x
\displaystyle \Rightarrow x=\frac{3-y}{4}
\displaystyle \text{Since }x\in R\text{ for every }y\in R,\text{ every element of co-domain has a pre-image.}
\displaystyle \therefore f\text{ is onto.}
\\

\displaystyle \textbf{Question 8. }\text{Let }f:R\rightarrow R\text{ be defined by }f(x)=x^{2}+1.\text{ Find pre-images of } \\ 17\text{ and }-3.
\displaystyle \text{Answer:}
\displaystyle f(x)=17
\displaystyle \Rightarrow x^{2}+1=17
\displaystyle \Rightarrow x^{2}=16
\displaystyle \Rightarrow x=\pm4
\displaystyle \therefore \text{Pre-images of }17\text{ are }4\text{ and }-4.
\displaystyle \text{Now, }f(x)=-3
\displaystyle \Rightarrow x^{2}+1=-3
\displaystyle \Rightarrow x^{2}=-4
\displaystyle \text{This is not possible for }x\in R.
\displaystyle \therefore -3\text{ has no pre-image in }R.
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\displaystyle \textbf{Question 9. }\text{Show that the function }f:R\rightarrow R,\ f(x)=x^{4}\text{ is many-one and into.}
\displaystyle \text{Answer:}
\displaystyle f(1)=1^{4}=1
\displaystyle f(-1)=(-1)^{4}=1
\displaystyle \text{Here, }1\neq-1\text{ but }f(1)=f(-1).
\displaystyle \therefore f\text{ is many-one.}
\displaystyle \text{Also, }f(x)=x^{4}\geq0\text{ for all }x\in R.
\displaystyle \therefore \text{Range}(f)=[0,\infty)
\displaystyle \text{But co-domain is }R.
\displaystyle \therefore \text{Range}(f)\neq R
\displaystyle \therefore f\text{ is into.}
\\

\displaystyle \textbf{Question 10. }\text{Check the injectivity of the function }f:R\rightarrow R\text{ given by } \\ f(x)=x^{3}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x_{1})=f(x_{2}).
\displaystyle \Rightarrow x_{1}^{3}=x_{2}^{3}
\displaystyle \Rightarrow x_{1}=x_{2}
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \therefore f\text{ is injective.}
\\

\displaystyle \textbf{Question 4. }\text{State whether the function }f:R\rightarrow R,\text{ defined by } \\ f(x)=3-4x\text{ is onto or not.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }y\in R.
\displaystyle \text{Then }y=3-4x
\displaystyle \Rightarrow x=\frac{3-y}{4}
\displaystyle \text{Since }x\in R\text{ for every }y\in R,\text{ there exists a pre-image of every }y.
\displaystyle \therefore f\text{ is onto.}
\\

\displaystyle \textbf{Question 11. }\text{Show that the function }f:N\rightarrow N,\text{ given by } \\ f(x)=2x\text{ is one-one but not onto.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x_{1})=f(x_{2}).
\displaystyle \Rightarrow 2x_{1}=2x_{2}
\displaystyle \Rightarrow x_{1}=x_{2}
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Now, Range}(f)=\{2,4,6,8,\ldots\}
\displaystyle \text{Since odd natural numbers do not have pre-images in }N,
\displaystyle \text{Range}(f)\neq N
\displaystyle \therefore f\text{ is not onto.}
\displaystyle \therefore f(x)=2x\text{ is one-one but not onto.}
\\

\displaystyle \textbf{Question 12. }\text{Consider }f:\{1,2,3\}\rightarrow\{a,b,c\}\text{ given by }f(1)=a, \\ f(2)=b,\ f(3)=c.\text{ Find }f^{-1}.   \text{Show that }(f^{-1})^{-1}=f.
\displaystyle \text{Answer:}
\displaystyle f^{-1}:\{a,b,c\}\rightarrow\{1,2,3\}
\displaystyle f^{-1}(a)=1,\quad f^{-1}(b)=2,\quad f^{-1}(c)=3
\displaystyle \text{Hence, }f^{-1}=\{(a,1),(b,2),(c,3)\}
\displaystyle \text{Now, }(f^{-1})^{-1}(1)=a,\quad (f^{-1})^{-1}(2)=b,\quad (f^{-1})^{-1}(3)=c
\displaystyle \therefore (f^{-1})^{-1}=\{(1,a),(2,b),(3,c)\}=f
\displaystyle \therefore (f^{-1})^{-1}=f.
\\

\displaystyle \textbf{Question 13. }\text{Show that the relation }R\text{ in the set }A\text{ of real numbers defined as }
\displaystyle R=\{(a,b):a\leq b\}\text{ is reflexive and }\text{transitive but not symmetric.}
\displaystyle \text{Answer:}
\displaystyle \text{For any }a\in A,\ a\leq a.
\displaystyle \therefore (a,a)\in R
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Now, let }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \therefore a\leq b\text{ and }b\leq c
\displaystyle \Rightarrow a\leq c
\displaystyle \therefore (a,c)\in R
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Also, }(2,3)\in R\text{ since }2\leq3.
\displaystyle \text{But }(3,2)\notin R\text{ since }3\nleq2.
\displaystyle \therefore R\text{ is not symmetric.}
\\

\displaystyle \textbf{Question 14. }\text{Let a relation }R\text{ on the set }A\text{ of real numbers be defined as } \\ (a,b)\in R\Rightarrow 1+ab>0,\ \forall\ a,b\in A.   \text{Show that }R\text{ is reflexive and symmetric but not} \\ \text{transitive.}
\displaystyle \text{Answer:}
\displaystyle \text{For any }a\in A,\ 1+a^{2}>0.
\displaystyle \therefore (a,a)\in R
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Now, let }(a,b)\in R.
\displaystyle \therefore 1+ab>0
\displaystyle \Rightarrow 1+ba>0
\displaystyle \therefore (b,a)\in R
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{Now, }(2,0)\in R\text{ since }1+2\cdot0=1>0.
\displaystyle \text{Also, }(0,-2)\in R\text{ since }1+0\cdot(-2)=1>0.
\displaystyle \text{But }(2,-2)\notin R\text{ since }1+2(-2)=-3<0.
\displaystyle \therefore R\text{ is not transitive.}
\\

\displaystyle \textbf{Question 15. }\text{Show that relation }R\text{ in the set of real numbers, defined as }
\displaystyle R=\{(a,b):a\leq b^{2}\}\text{ is neither reflexive nor} \text{symmetric nor transitive.}
\displaystyle \text{Answer:}
\displaystyle \text{For reflexivity, we need }a\leq a^{2}\text{ for every }a\in R.
\displaystyle \text{But for }a=\frac{1}{2},\ \frac{1}{2}\nleq\frac{1}{4}.
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Now, }(1,2)\in R\text{ since }1\leq2^{2}.
\displaystyle \text{But }(2,1)\notin R\text{ since }2\nleq1^{2}.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Also, }(8,3)\in R\text{ since }8\leq3^{2}.
\displaystyle \text{and }(3,2)\in R\text{ since }3\leq2^{2}.
\displaystyle \text{But }(8,2)\notin R\text{ since }8\nleq2^{2}.
\displaystyle \therefore R\text{ is not transitive.}
\\

\displaystyle \textbf{Question 16. }\text{If }A=\{1,2,3,4\},\text{ define relations on }A\text{ which have properties of being}
\displaystyle \text{(i) reflexive, transitive but not symmetric}
\displaystyle \text{(ii) symmetric but neither reflexive nor transitive.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }R_{1}=\{(1,1),(2,2),(3,3),(4,4),(1,2)\}.
\displaystyle \text{Then }R_{1}\text{ is reflexive and transitive but not symmetric, since }(1,2)\in R_{1}\text{ but }(2,1)\notin R_{1}.
\displaystyle \text{(ii) Let }R_{2}=\{(1,2),(2,1)\}.
\displaystyle \text{Then }R_{2}\text{ is symmetric, but not reflexive since }(1,1),(2,2),(3,3),(4,4)\notin R_{2}.
\displaystyle \text{Also, }(1,2)\in R_{2}\text{ and }(2,1)\in R_{2},\text{ but }(1,1)\notin R_{2}.
\displaystyle \therefore R_{2}\text{ is not transitive.}
\\

\displaystyle \textbf{Question 17. }\text{If }f:R\rightarrow R\text{ is the function, defined by } f(x)=4x^{3}+7, \\ \text{ then show that }f\text{ is a bijection.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x_{1})=f(x_{2}).
\displaystyle \Rightarrow 4x_{1}^{3}+7=4x_{2}^{3}+7
\displaystyle \Rightarrow x_{1}^{3}=x_{2}^{3}
\displaystyle \Rightarrow x_{1}=x_{2}
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Now, let }y\in R.
\displaystyle y=4x^{3}+7
\displaystyle \Rightarrow x=\sqrt[3]{\frac{y-7}{4}}
\displaystyle \text{Since }x\in R\text{ for every }y\in R,\text{ every element of co-domain has a pre-image.}
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is bijective.}
\\

\displaystyle \textbf{Question 18. }\text{Let }N\text{ be the set of all natural numbers and let }R\text{ be a relation in }N,\text{ defined by}
\displaystyle R=\{(a,b):a\text{ is a multiple of }b\}.\text{ Show that }R\text{ is reflexive and transitive but not symmetric.}
\displaystyle \text{Answer:}
\displaystyle \text{For any }a\in N,\ a\text{ is a multiple of }a.
\displaystyle \therefore (a,a)\in R
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Now, let }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \therefore a\text{ is a multiple of }b\text{ and }b\text{ is a multiple of }c.
\displaystyle \therefore a\text{ is a multiple of }c.
\displaystyle \therefore (a,c)\in R
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Also, }(6,3)\in R\text{ since }6\text{ is a multiple of }3.
\displaystyle \text{But }(3,6)\notin R\text{ since }3\text{ is not a multiple of }6.
\displaystyle \therefore R\text{ is not symmetric.}
\\

\displaystyle \textbf{Question 19. }\text{Show that the relation }R\text{ in the set }R\text{ of real numbers, defined as }
\displaystyle R=\{(a,b):a\leq b^{2}\}\text{ is neither reflexive nor}\text{symmetric nor transitive.}
\displaystyle \text{Answer:}
\displaystyle \text{For reflexivity, we need }a\leq a^{2}\text{ for every }a\in R.
\displaystyle \text{But for }a=\frac{1}{2},\ \frac{1}{2}\nleq \frac{1}{4}.
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Now, }(2,2)\in R\text{ since }2\leq 2^{2}.
\displaystyle \text{But to show not symmetric, take }(2,1).
\displaystyle (2,1)\notin R\text{ since }2\nleq 1^{2}.
\displaystyle \text{Instead, }(1,2)\in R\text{ since }1\leq 2^{2},\text{ but }(2,1)\notin R.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Now, }(8,3)\in R\text{ since }8\leq 3^{2}.
\displaystyle \text{Also, }(3,2)\in R\text{ since }3\leq 2^{2}.
\displaystyle \text{But }(8,2)\notin R\text{ since }8\nleq 2^{2}.
\displaystyle \therefore R\text{ is not transitive.}
\displaystyle \therefore R\text{ is neither reflexive nor symmetric nor transitive.}
\\

\displaystyle \textbf{Question 20. }\text{Consider }f:R^{+}\rightarrow[-9,\infty)\text{ given by } f(x)=5x^{2}+6x-9.
\displaystyle \text{ Prove that }f\text{ is invertible with } f^{-1}(y)=\left(\frac{\sqrt{54+5y}-3}{5}\right). \\ \text{ [where, }R^{+}\text{ is the set of all non-negative real numbers].}
\displaystyle \text{Answer:}
\displaystyle f(x)=5x^{2}+6x-9
\displaystyle \text{Let }y=5x^{2}+6x-9
\displaystyle 5x^{2}+6x-(9+y)=0
\displaystyle \text{Using quadratic formula,}
\displaystyle x=\frac{-6\pm\sqrt{36+20(9+y)}}{10}
\displaystyle x=\frac{-6\pm\sqrt{216+20y}}{10}
\displaystyle x=\frac{-6\pm2\sqrt{54+5y}}{10}
\displaystyle x=\frac{-3\pm\sqrt{54+5y}}{5}
\displaystyle \text{Since }x\in R^{+},\text{ we take the positive value.}
\displaystyle x=\frac{\sqrt{54+5y}-3}{5}
\displaystyle \therefore f^{-1}(y)=\frac{\sqrt{54+5y}-3}{5}
\displaystyle \text{Also, }f'(x)=10x+6>0\text{ for all }x\in R^{+}.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{For }x\in R^{+},\ f(0)=-9\text{ and }f(x)\rightarrow\infty\text{ as }x\rightarrow\infty.
\displaystyle \therefore \text{Range of }f=[-9,\infty),\text{ which is the co-domain.}
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is invertible.}
\\

\displaystyle \textbf{Question 21. }\text{Show that the relation 'is similar to' on the set of all triangles in} \\ \text{a plane is an equivalence relation.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }R\text{ be the relation 'is similar to' on the set of all triangles in a plane.}
\displaystyle \text{Every triangle is similar to itself.}
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{If triangle }T_{1}\text{ is similar to triangle }T_{2},\text{ then }T_{2}\text{ is similar to }T_{1}.
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{If }T_{1}\text{ is similar to }T_{2}\text{ and }T_{2}\text{ is similar to }T_{3},\text{ then }T_{1}\text{ is similar to }T_{3}.
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \therefore R\text{ is an equivalence relation.}
\\

\displaystyle \textbf{Question 22. }\text{Show that the function }
\displaystyle f:R\rightarrow\{x\in R:-1<x<1\}\text{ defined by }f(x)=\frac{x}{1+|x|},\ x\in R \\ \text{ is one-one and onto} \text{function.}
\displaystyle \text{Answer:}
\displaystyle f(x)=\frac{x}{1+|x|}
\displaystyle \text{For }x\geq0,\ f(x)=\frac{x}{1+x}
\displaystyle \text{For }x<0,\ f(x)=\frac{x}{1-x}
\displaystyle \text{Both branches are strictly increasing. Also, }f(x)<0\text{ for }x<0\text{ and }f(x)\geq0\text{ for }x\geq0.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Let }y\in(-1,1).
\displaystyle \text{If }0\leq y<1,\text{ take }x=\frac{y}{1-y}.
\displaystyle \text{Then }f(x)=\frac{x}{1+x}=y.
\displaystyle \text{If }-1<y<0,\text{ take }x=\frac{y}{1+y}.
\displaystyle \text{Then }f(x)=\frac{x}{1-x}=y.
\displaystyle \therefore \text{every }y\in(-1,1)\text{ has a pre-image in }R.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is one-one and onto.}
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\displaystyle \textbf{Question 23. }\text{Given, a function defined by }f(x)=\sqrt{4-x^{2}};\ 0\leq x\leq2,\ 
\displaystyle 0\leq f(x)\leq2. \ \text{ Show that }f\text{ is bijective }  \text{function.}
\displaystyle \text{Answer:}
\displaystyle f(x)=\sqrt{4-x^{2}},\quad 0\leq x\leq2
\displaystyle \text{If }f(x_{1})=f(x_{2}),\text{ then}
\displaystyle \sqrt{4-x_{1}^{2}}=\sqrt{4-x_{2}^{2}}
\displaystyle \Rightarrow x_{1}^{2}=x_{2}^{2}
\displaystyle \Rightarrow x_{1}=x_{2}\quad \text{as }x_{1},x_{2}\in[0,2]
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Also, for every }y\in[0,2],\text{ take }x=\sqrt{4-y^{2}}.
\displaystyle \text{Then }x\in[0,2]\text{ and }f(x)=y.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is bijective.}
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