Class 12 Maths Relations and Functions Questions with Solutions

Since the school is a boys school, no student of the school can be sister of any student of the school. Hence, $R=\phi$, showing that R is the empty relation. It is also obvious that the difference between heights of any two students of the school has to be less than 3 meters. This shows that $R^{\prime}=A\times A$ is the universal relation.

R is reflexive, since every triangle is congruent to itself. Further, $(T_1, T_2)\in R \Rightarrow T_1$ is congruent to $T_2 \Rightarrow T_2$ is congruent to $T_1 \Rightarrow (T_2, T_1)\in R$. Hence, R is symmetric. Moreover, $(T_1, T_2), (T_2, T_3)\in R \Rightarrow T_1$ is congruent to $T_2$ and $T_2$ is congruent to $T_3 \Rightarrow T_1$ is congruent to $T_3 \Rightarrow (T_1, T_3)\in R$. Therefore, R is an equivalence relation.

R is not reflexive, as a line $L_1$ can not be perpendicular to itself, i.e., $(L_1, L_1)\notin R$. R is symmetric as $(L_1, L_2)\in R \Rightarrow L_1$ is perpendicular to $L_2 \Rightarrow L_2$ is perpendicular to $L_1 \Rightarrow (L_2, L_1)\in R$. R is not transitive. Indeed, if $L_1$ is perpendicular to $L_2$ and $L_2$ is perpendicular to $L_3$, then $L_1$ can never be perpendicular to $L_3$. In fact, $L_1$ is parallel to $L_3$, i.e., $(L_1, L_2)\in R$, $(L_2, L_3)\in R$ but $(L_1, L_3)\notin R$.

R is reflexive, since (1, 1), (2, 2) and (3, 3) lie in R. Also, R is not symmetric, as $(1, 2)\in R$ but $(2, 1)\notin R$. Similarly, R is not transitive, as $(1, 2)\in R$ and $(2, 3)\in R$ but $(1, 3)\notin R$.

R is reflexive, as 2 divides $(a-a)$ for all $a\in Z$. Further, if $(a, b)\in R$, then 2 divides $a-b$. Therefore, 2 divides $b-a$. Hence, $(b, a)\in R$, which shows that R is symmetric. Similarly, if $(a, b)\in R$ and $(b, c)\in R$, then $a-b$ and $b-c$ are divisible by 2. Now, $a-c=(a-b)+(b-c)$ is even, so $(a-c)$ is divisible by 2. This shows that R is transitive. Thus, R is an equivalence relation in Z.

Given any element a in A, both a and a must be either odd or even, so that $(a, a)\in R$. Further, $(a, b)\in R \Rightarrow$ both a and b must be either odd or even $\Rightarrow (b, a)\in R$. Similarly, $(a, b)\in R$ and $(b, c)\in R \Rightarrow$ all elements a, b, c must be either even or odd simultaneously $\Rightarrow (a, c)\in R$. Hence, R is an equivalence relation. Further, all the elements of {1, 3, 5, 7} are related to each other, as all the elements of this subset are odd. Similarly, all the elements of the subset {2, 4, 6} are related to each other, as all of them are even. Also, no element of the subset {1, 3, 5, 7} can be related to any element of {2, 4, 6}, as elements of {1, 3, 5, 7} are odd, while elements of {2, 4, 6} are even.

No two different students of the class can have the same roll number. Therefore, f must be one-one. We can assume without any loss of generality that roll numbers of students are from 1 to 50. This implies that 51 in N is not the roll number of any student of the class, so 51 cannot be the image of any element of X under f. Hence, f is not onto.

The function f is one-one, for $f(x_1)=f(x_2) \Rightarrow 2x_1=2x_2 \Rightarrow x_1=x_2$. Further, f is not onto, as for $1\in N$, there does not exist any x in N such that $f(x)=2x=1$.

f is one-one, as $f(x_1)=f(x_2) \Rightarrow 2x_1=2x_2 \Rightarrow x_1=x_2$. Also, given any real number y in R, there exists $\frac{y}{2}$ in R such that $f(\frac{y}{2})=2(\frac{y}{2})=y$. Hence, f is onto.

f is not one-one, as $f(1)=f(2)=1$. But f is onto, as given any $y\in N$, $y\ne 1$, we can choose x as $y+1$ such that $f(y+1)=y+1-1=y$. Also for $1\in N$, we have $f(1)=1$.

Since $f(-1)=1=f(1)$, f is not one-one. Also, the element -2 in the co-domain R is not the image of any element x in the domain R. Therefore, f is not onto.

Suppose $f(x_1)=f(x_2)$. Note that if $x_1$ is odd and $x_2$ is even, then we will have $x_1+1=x_2-1$, i.e., $x_2-x_1=2$ which is impossible. Similarly, the possibility of $x_1$ being even and $x_2$ being odd can also be ruled out using similar arguments. Therefore, both $x_1$ and $x_2$ must be either odd or even. Suppose both $x_1$ and $x_2$ are odd, then $f(x_1)=f(x_2) \Rightarrow x_1+1=x_2+1 \Rightarrow x_1=x_2$. Similarly, if both are even, $f(x_1)=f(x_2) \Rightarrow x_1-1=x_2-1 \Rightarrow x_1=x_2$. Thus, f is one-one. Also, any odd number $2r+1$ in the co-domain N is the image of $2r+2$ in the domain N, and any even number 2r in the co-domain N is the image of $2r-1$ in the domain N. Thus, f is onto.

Suppose f is not one-one. Then there exist two elements, say 1 and 2 in the domain whose image in the co-domain is the same. Also, the image of 3 under f can be only one element. Therefore, the range set can have at most two elements of the co-domain {1, 2, 3}, showing that f is not onto, which is a contradiction. Hence, f must be one-one.

Since f is one-one, three elements of $\{1, 2, 3\}$ must be mapped to 3 different elements of the co-domain {1, 2, 3} under f. Hence, f has to be onto.

We have $gof(x) = g(f(x)) = g(\cos x) = 3(\cos x)^2 = 3\cos^2 x$. Similarly, $fog(x) = f(g(x)) = f(3x^2) = \cos(3x^2)$. Note that $3\cos^2 x \ne \cos 3x^2$ for $x=0$. Hence, $gof\ne fog$.

Consider an arbitrary element y of Y. By the definition of Y, $y=4x+3$ for some x in the domain N. This shows that $x=\frac{y-3}{4}$. Define $g:Y\rightarrow N$ by $g(y)=\frac{y-3}{4}$. Now, $gof(x) = g(f(x)) = g(4x+3) = \frac{4x+3-3}{4} = x$ and $fog(y) = f(g(y)) = f(\frac{y-3}{4}) = \frac{4(y-3)}{4}+3 = y-3+3=y$. This shows that $gof=I_N$ and $fog=I_Y$, which implies that f is invertible and g is the inverse of f.

Since $R_1$ and $R_2$ are equivalence relations, $(a, a)\in R_1$ and $(a, a)\in R_2$ for all $a\in A$. This implies that $(a, a)\in R_1\cap R_2$, showing $R_1\cap R_2$ is reflexive. Further, $(a, b)\in R_1\cap R_2 \Rightarrow (a, b)\in R_1$ and $(a, b)\in R_2 \Rightarrow (b, a)\in R_1$ and $(b, a)\in R_2 \Rightarrow (b, a)\in R_1\cap R_2$, hence $R_1\cap R_2$ is symmetric. Similarly, $(a, b)\in R_1\cap R_2$ and $(b, c)\in R_1\cap R_2 \Rightarrow (a, c)\in R_1$ and $(a, c)\in R_2 \Rightarrow (a, c)\in R_1\cap R_2$. This shows that $R_1\cap R_2$ is transitive. Thus, $R_1\cap R_2$ is an equivalence relation.

Clearly, $(x, y)R(x, y)$ for all $(x, y)\in A$, since $xy=yx$. This shows that R is reflexive. Further, $(x, y)R(u, v) \Rightarrow xv=yu \Rightarrow uy=vx \Rightarrow (u, v)R(x, y)$, showing R is symmetric. Similarly, $(x, y)R(u, v)$ and $(u, v)R(a, b) \Rightarrow xv=yu$ and $ub=va \Rightarrow xv\frac{b}{u}=yu\frac{b}{u} \Rightarrow xv\frac{b}{v}=yu\frac{a}{u} \Rightarrow xb=ya \Rightarrow (x, y)R(a, b)$. Thus, R is transitive, making it an equivalence relation.

The characteristic of sets {1, 4, 7}, {2, 5, 8} and {3, 6, 9} is that the difference between any two elements of these sets is a multiple of 3. Therefore, $(x, y)\in R_1 \Rightarrow x-y$ is a multiple of $3 \Rightarrow \{x, y\}\subset\{1, 4, 7\}$ or $\{x, y\}\subset\{2, 5, 8\}$ or $\{x, y\}\subset\{3, 6, 9\} \Rightarrow (x, y)\in R_2$. Hence, $R_1\subset R_2$. Similarly, $\{x, y\}\in R_2 \Rightarrow x-y$ is divisible by $3 \Rightarrow (x, y)\in R_1$, showing $R_2\subset R_1$. Thus, $R_1=R_2$.

For every $a\in X$, $(a, a)\in R$ since $f(a)=f(a)$, showing R is reflexive. Similarly, $(a, b)\in R \Rightarrow f(a)=f(b) \Rightarrow f(b)=f(a) \Rightarrow (b, a)\in R$, showing R is symmetric. Further, $(a, b)\in R$ and $(b, c)\in R \Rightarrow f(a)=f(b)$ and $f(b)=f(c) \Rightarrow f(a)=f(c) \Rightarrow (a, c)\in R$, showing R is transitive. Hence, R is an equivalence relation.

One-one functions from {1, 2, 3} to itself are simply permutations of the three symbols 1, 2, 3. Therefore, the total number of one-one maps is equal to the total number of permutations, which is $3! = 6$.

The smallest relation $R_1$ containing (1, 2) and (2, 3) which is reflexive and transitive but not symmetric is $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$. Adding (2, 1) to $R_1$ gives $R_2$, which is reflexive and transitive but not symmetric. Similarly, $R_3$ is obtained by adding (3, 2) to $R_1$. We cannot add both pairs or (3, 1) directly because doing so forces symmetry. Thus, the total number of desired relations is three.

The smallest equivalence relation $R_1$ containing (1, 2) and (2, 1) is $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$. The remaining pairs are (2, 3), (3, 2), (1, 3), and (3, 1). Adding any one of these requires adding its symmetric counterpart and transitively closing it, which immediately expands the relation to the universal relation. Thus, the total number of equivalence relations containing (1, 2) and (2, 1) is two.

Clearly $I_N$ is onto. But $I_N+I_N$ is not onto, because we can find an element (like 3) in the co-domain N such that there does not exist any x in the domain N where $(I_N+I_N)(x)=2x=3$.

For any two distinct elements $x_1$ and $x_2$ in $[0,\frac{\pi}{2}]$, $\sin x_1\ne \sin x_2$ and $\cos x_1\ne \cos x_2$, so both f and g are one-one. However, $(f+g)(0)=\sin 0+\cos 0=1$ and $(f+g)(\frac{\pi}{2})=\sin\frac{\pi}{2}+\cos\frac{\pi}{2}=1$. Since distinct inputs yield the same output, $f+g$ is not one-one.