Relations and FunctionsClass 11 Mathematics NCERT Solutions

Given: The ordered pairs are equal: $(\frac{x}{3}+1, y-\frac{2}{3})=(\frac{5}{3}, \frac{1}{3})$.

To Find: The values of $x$ and $y$.

Solution: According to the property of equality of ordered pairs, the corresponding elements must be equal.

Equating the first elements: $\frac{x}{3} + 1 = \frac{5}{3}$ $\frac{x}{3} = \frac{5}{3} - 1$ $\frac{x}{3} = \frac{5-3}{3}$ $\frac{x}{3} = \frac{2}{3}$ Multiplying both sides by 3, we get: $x = 2$

Equating the second elements: $y - \frac{2}{3} = \frac{1}{3}$ $y = \frac{1}{3} + \frac{2}{3}$ $y = \frac{1+2}{3}$ $y = \frac{3}{3}$ $y = 1$

Final Answer: The values are $x = 2$ and $y = 1$.

Given: Number of elements in set A, $n(A) = 3$. Set B = {3, 4, 5}.

To Find: The number of elements in the Cartesian product A × B, i.e., $n(A \times B)$.

Solution: First, we find the number of elements in set B. Since B = {3, 4, 5}, the number of elements in B is $n(B) = 3$.

The number of elements in the Cartesian product of two sets is the product of the number of elements in each set. Formula: $n(A \times B) = n(A) \times n(B)$.

Substituting the given values: $n(A \times B) = 3 \times 3$ $n(A \times B) = 9$

Final Answer: The number of elements in (A × B) is 9.

Given: Set G = {7, 8} Set H = {5, 4, 2}

To Find: The Cartesian products G × H and H × G.

Solution: The Cartesian product G × H is the set of all ordered pairs $(g, h)$ such that $g \in G$ and $h \in H$. $G \times H = \{(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)\}$

The Cartesian product H × G is the set of all ordered pairs $(h, g)$ such that $h \in H$ and $g \in G$. $H \times G = \{(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)\}$

Final Answer: $G \times H = \{(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)\}$ $H \times G = \{(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)\}$

Answer: False.

Reason: We have P = {m, n} and Q = {n, m}. The Cartesian product P × Q is the set of all ordered pairs $(p, q)$ where $p \in P$ and $q \in Q$. $P \times Q = \{(m, n), (m, m), (n, n), (n, m)\}$ The given set P × Q = {(m, n), (n, m)} is incomplete.

Correct Statement: If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (m, m), (n, n), (n, m)}.

Answer: True.

Reason: This is the definition of the Cartesian product of two non-empty sets. Since both A and B are non-empty, their Cartesian product A × B will also be non-empty and will consist of ordered pairs $(x, y)$ where $x \in A$ and $y \in B$.

Answer: True.

Reason: The intersection of any set B with the empty set φ is the empty set itself. So, $B \cap \phi = \phi$. Now, we need to find A × (B ∩ φ), which is A × φ. The Cartesian product of any set with the empty set is the empty set. Therefore, $A \times \phi = \phi$. The statement is correct.

Given: Set A = {-1, 1}.

To Find: The Cartesian product A × A × A.

Solution: First, we find the Cartesian product A × A. $A \times A = \{(-1, -1), (-1, 1), (1, -1), (1, 1)\}$

Next, we find A × (A × A). This is A × A × A, which is the set of all ordered triplets $(a, b, c)$ where $a, b, c \in A$. We pair each element of A with each ordered pair in A × A.

For the element -1 in A: $(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1)$

For the element 1 in A: $(1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)$

Combining these, we get A × A × A.

Final Answer: $A \times A \times A = \{(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)\}$

Given: The Cartesian product A × B = {(a, x), (a, y), (b, x), (b, y)}.

To Find: The sets A and B.

Solution: By the definition of the Cartesian product, the set A is the set of all first elements in the ordered pairs of A × B, and the set B is the set of all second elements.

The first elements in the ordered pairs are 'a' and 'b'. Therefore, $A = \{a, b\}$.

The second elements in the ordered pairs are 'x' and 'y'. Therefore, $B = \{x, y\}$.

Final Answer: $A = \{a, b\}$ $B = \{x, y\}$

Given: A = {1, 2} B = {1, 2, 3, 4} C = {5, 6} D = {5, 6, 7, 8}

LHS: A × (B ∩ C) First, find B ∩ C. $B \cap C = \{1, 2, 3, 4\} \cap \{5, 6\} = \phi$ Now, find A × (B ∩ C). $A \times (B \cap C) = A \times \phi = \phi$ So, LHS = φ.

RHS: (A × B) ∩ (A × C) First, find A × B. $A \times B = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)\}$ Next, find A × C. $A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\}$ Now, find the intersection of these two sets. $(A \times B) \cap (A \times C) = \phi$ (since there are no common ordered pairs). So, RHS = φ.

Since LHS = RHS = φ, the statement is verified.

First, we find the set A × C. $A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\}$

Next, we find the set B × D. $B \times D = \{1, 2, 3, 4\} \times \{5, 6, 7, 8\}$ $B \times D = \{(1, 5), (1, 6), (1, 7), (1, 8),$ $(2, 5), (2, 6), (2, 7), (2, 8),$ $(3, 5), (3, 6), (3, 7), (3, 8),$ $(4, 5), (4, 6), (4, 7), (4, 8)\}$

To verify if A × C is a subset of B × D, we check if every element of A × C is also an element of B × D.

Since all elements of A × C are present in B × D, A × C is a subset of B × D.

Given: Set A = {1, 2} Set B = {3, 4}

1. Write A × B: The Cartesian product A × B is the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$. $A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$

2. How many subsets will A × B have? First, find the number of elements in A × B. $n(A \times B) = n(A) \times n(B) = 2 \times 2 = 4$ The number of subsets of a set with $n$ elements is $2^n$. So, the number of subsets of A × B is $2^{n(A \times B)} = 2^4 = 16$.

3. List the subsets: The subsets of A × B are:

Final Answer: $A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$. A × B will have 16 subsets. The list of subsets is provided above.

Given: $n(A) = 3$ $n(B) = 2$ The ordered pairs (x, 1), (y, 2), (z, 1) are elements of A × B. $x, y, z$ are distinct elements.

To Find: The sets A and B.

Solution: By the definition of the Cartesian product A × B, if an ordered pair $(a, b)$ is in A × B, then $a \in A$ and $b \in B$.

From the given ordered pairs:

From these, we can determine the elements of sets A and B. The set A is the set of all first elements. $A = \{x, y, z\}$

The set B is the set of all second elements. $B = \{1, 2\}$

Now we check if this matches the given conditions. $n(A) = 3$, which is correct since $x, y, z$ are distinct. $n(B) = 2$, which is correct.

Final Answer: $A = \{x, y, z\}$ $B = \{1, 2\}$

Given: $n(A \times A) = 9$. The ordered pairs (-1, 0) and (0, 1) are elements of A × A.

To Find: The set A and the remaining elements of A × A.

Solution: 1. Find the set A: We know that $n(A \times A) = n(A) \times n(A) = (n(A))^2$. Given $n(A \times A) = 9$, so $(n(A))^2 = 9$, which means $n(A) = 3$.

If an ordered pair $(a, b) \in A \times A$, then both $a \in A$ and $b \in A$.

Combining these, the elements of A are -1, 0, and 1. Since $n(A)=3$, we have found all the elements. Therefore, $A = \{-1, 0, 1\}$.

2. Find all elements of A × A: Now we find the full Cartesian product A × A. $A \times A = \{-1, 0, 1\} \times \{-1, 0, 1\}$ $A \times A = \{(-1, -1), (-1, 0), (-1, 1),$ $(0, -1), (0, 0), (0, 1),$ $(1, -1), (1, 0), (1, 1)\}$

3. Find the remaining elements of A × A: We are given two elements: (-1, 0) and (0, 1). The remaining elements are all the elements in A × A except for these two.

Remaining elements = $A \times A - \{(-1, 0), (0, 1)\}$ Remaining elements = $\{(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)\}$

Final Answer: The set A is $\{-1, 0, 1\}$. The remaining elements of A × A are $\{(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)\}$.

Given: Set A = {1, 2, 3, ..., 14}. Relation R from A to A is defined as R = {(x, y): 3x - y = 0, where x, y ∈ A}.

To Find: The domain, codomain, and range of R.

Solution: The relation R is defined by the equation $3x - y = 0$, which can be rewritten as $y = 3x$. We need to find pairs $(x, y)$ that satisfy this condition, where both $x$ and $y$ are in set A.

Let's find the pairs by substituting values of $x$ from A:

So, the relation R in roster form is: $R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$

Now we can determine the domain, codomain, and range.

Final Answer: Domain = {1, 2, 3, 4} Codomain = {1, 2, 3, ..., 14} Range = {3, 6, 9, 12}

Given: A relation R on the set N of natural numbers. R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈ N}.

Solution: The condition for $x$ is that it is a natural number less than 4. The natural numbers are 1, 2, 3, ... So, the possible values for $x$ are 1, 2, and 3.

The relation is defined by $y = x + 5$. We find the corresponding $y$ values for each $x$:

1. Roster Form: The relation R in roster form is: $R = \{(1, 6), (2, 7), (3, 8)\}$

Final Answer: Roster form: $R = \{(1, 6), (2, 7), (3, 8)\}$ Domain = {1, 2, 3} Range = {6, 7, 8}

Given: Set A = {1, 2, 3, 5} Set B = {4, 6, 9} Relation R from A to B is defined as R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. The difference can be considered as $|x - y|$.

To Find: The relation R in roster form.

Solution: We need to check each pair $(x, y)$ where $x \in A$ and $y \in B$ to see if their difference is odd.

Combining all the pairs that satisfy the condition, we get the relation R in roster form.

Final Answer: $R = \{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\}$

Given: The relation R is defined as R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.

To Find: The domain and range of R.

Solution: First, let's write the relation R in roster form by substituting the possible values of $x$.

So, the relation in roster form is: $R = \{(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)\}$

Now, we determine the domain and range.

Final Answer: Domain = {0, 1, 2, 3, 4, 5} Range = {5, 6, 7, 8, 9, 10}

Given: The relation R = {(x, x³): x is a prime number less than 10}.

To Find: The relation R in roster form.

Solution: First, we need to identify the prime numbers less than 10. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The prime numbers less than 10 are 2, 3, 5, and 7.

So, the possible values for $x$ are 2, 3, 5, 7.

Now, we find the corresponding second element, $x^3$, for each value of $x$.

Combining these pairs gives the relation R in roster form.

Final Answer: $R = \{(2, 8), (3, 27), (5, 125), (7, 343)\}$

Given: Set A = {x, y, z} Set B = {1, 2}

To Find: The number of relations from A to B.

Formula: The total number of relations from a set A to a set B is the number of possible subsets of the Cartesian product A × B. If $n(A) = p$ and $n(B) = q$, then $n(A \times B) = pq$. The number of subsets of A × B is $2^{pq}$.

Solution: First, find the number of elements in sets A and B. $n(A) = 3$ $n(B) = 2$

Next, find the number of elements in the Cartesian product A × B. $n(A \times B) = n(A) \times n(B) = 3 \times 2 = 6$

The number of relations from A to B is the number of possible subsets of A × B. Number of relations = $2^{n(A \times B)} = 2^6$.

Calculating $2^6$: $2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$

Final Answer: The number of relations from A to B is 64.

Given: The relation R is defined on the set of integers Z. R = {(a, b): a, b ∈ Z, a - b is an integer}.

To Find: The domain and range of R.

Solution: The condition for an ordered pair (a, b) to be in R is that both a and b are integers, and their difference, $a - b$, is also an integer.

We know that the difference between any two integers is always an integer. This means that for any integer 'a' and any integer 'b', the pair (a, b) will always satisfy the condition $a - b \in Z$.

Therefore, the relation R consists of all possible ordered pairs of integers. This means R is the Cartesian product Z × Z. $R = Z \times Z = \{(a, b): a \in Z, b \in Z\}$

Domain of R: The domain is the set of all possible first elements 'a'. Since 'a' can be any integer, the domain is the set of all integers, Z. Domain of R = Z.

Range of R: The range is the set of all possible second elements 'b'. Since 'b' can be any integer, the range is the set of all integers, Z. Range of R = Z.

Final Answer: The domain of R is Z (the set of all integers). The range of R is Z (the set of all integers).

Answer: This relation is a function.

Reason: A relation is a function if every element in the domain has one and only one image. Here, each distinct first element (2, 5, 8, 11, 14, 17) has a unique image, which is 1. It is permissible for multiple elements in the domain to have the same image.

Answer: This relation is a function.

Reason: Every first element in the ordered pairs is distinct, and each has exactly one image. For example, 2 has only image 1, 4 has only image 2, and so on.

Answer: This relation is not a function.

Reason: The definition of a function requires that no two distinct ordered pairs have the same first element. In this relation, the first element '1' is associated with two different images, 3 and 5. This violates the condition for a relation to be a function.

Domain: The modulus function $|x|$ is defined for all real numbers. Therefore, the function $f(x) = -|x|$ is also defined for all real numbers. Domain = R (the set of all real numbers).

Range: We know that for any real number $x$, $|x| \geq 0$. Multiplying by -1 reverses the inequality sign: $-|x| \leq 0$ This means that the value of $f(x)$ is always less than or equal to 0. Range = $(-\infty, 0]$.

Domain: The function involves a square root, so the expression inside the square root must be non-negative. $9 - x^2 \geq 0$ $9 \geq x^2$ $x^2 \leq 9$ Taking the square root of both sides: $|x| \leq 3$ This means $-3 \leq x \leq 3$. Domain = [-3, 3].

Range: Let $y = f(x) = \sqrt{9-x^2}$. By definition, the square root function gives a non-negative result, so $y \geq 0$. To find the maximum value of y, we look at the expression $9-x^2$. Since $x^2 \geq 0$, the maximum value of $9-x^2$ occurs when $x^2$ is minimum, i.e., $x^2=0$ (or $x=0$). When $x=0$, $y = \sqrt{9-0} = 3$. The minimum value of $y$ occurs when $9-x^2$ is minimum. This happens when $x^2$ is maximum, i.e., $x^2=9$ (or $x=\pm 3$). When $x=\pm 3$, $y = \sqrt{9-9} = 0$. So, the value of y ranges from 0 to 3. Range = [0, 3].

Given: The relation $f(x)=\begin{cases}x^{2}, 0 \leq x \leq 3 \ 3 x, 3 \leq x \leq 10\end{cases}$

Proof: The relation $f$ is defined from $x=0$ to $x=10$. The definition changes at $x=3$. For a relation to be a function, every element in its domain must have a unique image.

We need to check the point where the definition changes, which is $x=3$.

Since both definitions give the same value at $x=3$, the element 3 has a unique image, which is 9. Therefore, for every $x$ in the domain $[0, 10]$, there is a unique image.

Conclusion: $f$ is a function.

Given: The relation $g(x)=\begin{cases}x^{2}, 0 \leq x \leq 2 \ 3 x, 2 \leq x \leq 10\end{cases}$

Proof: The relation $g$ is defined from $x=0$ to $x=10$. The definition changes at $x=2$. We check the image of $x=2$ using both rules.

At $x=2$, the relation $g$ yields two different images, 4 and 6. This means the ordered pairs (2, 4) and (2, 6) are both in the relation $g$. Since the element 2 in the domain does not have a unique image, the relation $g$ violates the definition of a function.

Conclusion: $g$ is not a function.

Given: The function $f(x) = x^2$.

To Find: The value of the expression $\frac{f(1.1)-f(1)}{(1.1-1)}$.

Solution: First, we evaluate the function at the required points.

Now, we substitute these values into the given expression. $\frac{f(1.1)-f(1)}{(1.1-1)} = \frac{1.21 - 1}{1.1 - 1}$

Simplify the numerator and the denominator. $= \frac{0.21}{0.1}$

To divide by 0.1, we can multiply the numerator and denominator by 10. $= \frac{0.21 \times 10}{0.1 \times 10} = \frac{2.1}{1} = 2.1$

Final Answer: The value of the expression is 2.1.

Given: The function $f(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$.

To Find: The domain of the function $f(x)$.

Solution: The function $f(x)$ is a rational function. A rational function is defined for all real numbers except for those values of $x$ that make the denominator equal to zero.

So, we must find the values of $x$ for which the denominator is zero. Set the denominator to zero: $x^2 - 8x + 12 = 0$

We can solve this quadratic equation by factoring. We need two numbers that multiply to 12 and add up to -8. These numbers are -6 and -2. $(x - 6)(x - 2) = 0$

This gives two possible solutions: $x - 6 = 0 \implies x = 6$ $x - 2 = 0 \implies x = 2$

The function is not defined at $x=2$ and $x=6$. Therefore, the domain of the function is the set of all real numbers except 2 and 6.

Final Answer: The domain of $f$ is $R - \{2, 6\}$.

Given: The real function $f(x)=\sqrt{(x-1)}$.

To Find: The domain and the range of the function $f(x)$.

Solution: 1. Domain: The function involves a square root. For the function to be defined for real numbers, the expression inside the square root must be non-negative. $x - 1 \geq 0$ $x \geq 1$ So, the domain of the function is all real numbers greater than or equal to 1. Domain = $[1, \infty)$.

2. Range: Let $y = f(x) = \sqrt{x-1}$. By the definition of the principal square root, the output value $y$ must be non-negative. So, $y \geq 0$.

To see if $y$ can take any non-negative value, we can express $x$ in terms of $y$. $y = \sqrt{x-1}$ $y^2 = x-1$ $x = y^2 + 1$ Since $y$ can be any non-negative number ($y \geq 0$), we can always find a corresponding value of $x$ in the domain $[1, \infty)$. For example, if we want $y=100$, then $x=100^2+1=10001$, which is in the domain. This means the range of the function is all non-negative real numbers. Range = $[0, \infty)$.

Final Answer: Domain = $[1, \infty)$ Range = $[0, \infty)$

Given: The real function $f(x)=|x-1|$.

To Find: The domain and the range of the function $f(x)$.

Solution: 1. Domain: The modulus function (absolute value) is defined for any real number input. The expression $x-1$ is a real number for any real number $x$. Therefore, the function $f(x)=|x-1|$ is defined for all real numbers. Domain = R.

2. Range: The modulus function always returns a non-negative value. For any real number $z$, $|z| \geq 0$. In this case, $z = x-1$. So, $|x-1| \geq 0$. This means $f(x) \geq 0$.

The minimum value of $f(x)$ is 0, which occurs when $x-1=0$, i.e., $x=1$. For any other value of $x$, $f(x)$ will be a positive number. There is no upper limit to the value of $|x-1|$. Therefore, the range of the function is all non-negative real numbers. Range = $[0, \infty)$.

Final Answer: Domain = R (the set of all real numbers) Range = $[0, \infty)$

Given: The function $f(x) = \frac{x^2}{1+x^2}$, where the domain is R.

To Find: The range of the function $f$.

Solution: Let $y = f(x)$, so $y = \frac{x^2}{1+x^2}$.