Samacheer Kalvi Class 10 Maths Solutions Chapter 1 Relations and Functions Exercise 1

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Unit Exercise 1

Question 1. If the ordered pairs \((x^2 – 3x, y^2 + 4y)\) and \((-2, 5)\) are equal, then find x and y.
Answer:
Given that the ordered pairs are equal: \((x^2 – 3x, y^2 + 4y) = (-2, 5)\)
When two ordered pairs are equal, their corresponding components must be equal. So, we set the x-components equal and the y-components equal.

First, for x:
\(x^2 - 3x = -2\)
Move -2 to the left side to form a quadratic equation:
\(x^2 - 3x + 2 = 0\)
Factor the quadratic equation:
\((x - 2)(x - 1) = 0\)
This means either \((x - 2) = 0\)
\( \implies x = 2 \)
Or \((x - 1) = 0\)
\( \implies x = 1 \)
So, the possible values for x are 1 and 2.

Next, for y:
\(y^2 + 4y = 5\)
Move 5 to the left side to form a quadratic equation:
\(y^2 + 4y - 5 = 0\)
Factor the quadratic equation:
\((y + 5)(y - 1) = 0\)
This means either \((y + 5) = 0\)
\( \implies y = -5 \)
Or \((y - 1) = 0\)
\( \implies y = 1 \)
So, the possible values for y are -5 and 1.
The values for x are 1 and 2, and the values for y are -5 and 1.
In simple words: When two pairs are equal, their first parts must match, and their second parts must match. We solve these two separate equations to find the values for x and y.

🎯 Exam Tip: Remember to solve both equations (for x and y) completely, as they often result in quadratic equations with two possible solutions for each variable.

Question 2. The Cartesian product \(A \times A\) has 9 elements among which \((-1, 0)\) and \((0, 1)\) are found. Find the set A and the remaining elements of \(A \times A\).
Answer:
We are given that the Cartesian product \(A \times A\) has 9 elements. This means the number of elements in set A, denoted as \(n(A)\), must be 3, because \(n(A \times A) = n(A) \times n(A)\), so \(n(A)^2 = 9\), which gives \(n(A) = 3\).

We are also given that \((-1, 0)\) and \((0, 1)\) are elements of \(A \times A\).
Since \((-1, 0) \in A \times A\), it means that \(-1 \in A\) and \(0 \in A\).
Since \((0, 1) \in A \times A\), it means that \(0 \in A\) and \(1 \in A\).

Combining these, the elements \(-1, 0, 1\) must all be in set A.
Since we know \(n(A) = 3\), and we have identified three distinct elements \(-1, 0, 1\), the set A must be exactly \(\{-1, 0, 1\}\).

Now we need to find all the elements of \(A \times A\). This involves pairing every element of A with every other element of A:
\(A \times A = \{ (x, y) \mid x \in A, y \in A \}\)
\(A \times A = \{ (-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1) \}\)
These are all the elements of \(A \times A\). The Cartesian product helps to combine elements from two sets into ordered pairs.
In simple words: If \(A \times A\) has 9 items, set A must have 3 items. Since \((-1,0)\) and \((0,1)\) are in \(A \times A\), it means \(-1, 0,\) and \(1\) are all in A. So, set A is \(\{-1, 0, 1\}\). Then we list all possible pairs you can make by picking one number from A, and another number from A.

🎯 Exam Tip: When determining the elements of a set from its Cartesian product, remember that each component of the ordered pair must belong to the respective set. Also, \(n(A \times A) = n(A)^2\).

Question 3. Given that \(f(x) = \left\{\begin{array}{rl} {\sqrt{x-1}} & {x \geq 1} \\ {4} & {x<1} \end{array}\right.\). Find (i) \(f(0)\) (ii) \(f(3)\) (iii) \(f(a + 1)\) in terms of a. (Given that \(a > 0\))
Answer:
The function \(f(x)\) is a piecewise function, meaning it has different rules depending on the value of x.

(i) Find \(f(0)\):
For \(x = 0\), we check the conditions: \(0 \geq 1\) is false, and \(0 < 1\) is true.
So, we use the rule \(f(x) = 4\).
Therefore, \(f(0) = 4\).

(ii) Find \(f(3)\):
For \(x = 3\), we check the conditions: \(3 \geq 1\) is true.
So, we use the rule \(f(x) = \sqrt{x-1}\).
Therefore, \(f(3) = \sqrt{3-1} = \sqrt{2}\).

(iii) Find \(f(a + 1)\) in terms of a (given that \(a > 0\)):
Since \(a > 0\), it means \(a\) is a positive number. If we add 1 to a positive number, \(a+1\) will always be greater than 1. So, \(a+1 \geq 1\) is true.
Therefore, we use the rule \(f(x) = \sqrt{x-1}\).
Substitute \(x = a+1\):
\(f(a+1) = \sqrt{(a+1)-1} = \sqrt{a}\)
So, \(f(a+1) = \sqrt{a}\).
In simple words: For each value, check which rule the function uses. If the number is less than 1, the answer is 4. If the number is 1 or more, you subtract 1 and then find the square root.

🎯 Exam Tip: When evaluating a piecewise function, always first determine which condition the input value satisfies to use the correct function rule. Be careful with inequalities.

Question 4. Let \(A = \{9, 10, 11, 12, 13, 14, 15, 16, 17\}\) and let \(f: A \rightarrow N\) be defined by \(f(n) = \) the highest prime factor of \(n \in A\). Write f as a set of ordered pairs and find the range of f.
Answer:
The function \(f(n)\) takes a number \(n\) from set A and gives its highest prime factor. A prime factor is a prime number that divides a given number evenly.

Let's find the highest prime factor for each number in set A:
1. For \(n = 9\): Prime factors of 9 are 3. The highest prime factor is 3. So, \((9, 3)\).
2. For \(n = 10\): Prime factors of 10 are 2, 5. The highest prime factor is 5. So, \((10, 5)\).
3. For \(n = 11\): Prime factors of 11 are 11 (since 11 is a prime number). The highest prime factor is 11. So, \((11, 11)\).
4. For \(n = 12\): Prime factors of 12 are 2, 3. The highest prime factor is 3. So, \((12, 3)\).
5. For \(n = 13\): Prime factors of 13 are 13 (since 13 is a prime number). The highest prime factor is 13. So, \((13, 13)\).
6. For \(n = 14\): Prime factors of 14 are 2, 7. The highest prime factor is 7. So, \((14, 7)\).
7. For \(n = 15\): Prime factors of 15 are 3, 5. The highest prime factor is 5. So, \((15, 5)\).
8. For \(n = 16\): Prime factors of 16 are 2. The highest prime factor is 2. So, \((16, 2)\).
9. For \(n = 17\): Prime factors of 17 are 17 (since 17 is a prime number). The highest prime factor is 17. So, \((17, 17)\).

The function \(f\) as a set of ordered pairs is:
\(f = \{(9, 3), (10, 5), (11, 11), (12, 3), (13, 13), (14, 7), (15, 5), (16, 2), (17, 17)\}\)

The range of \(f\) is the set of all second components (the outputs) from the ordered pairs. We list only the unique values:
Range of \(f = \{3, 5, 11, 13, 7, 2, 17\}\)
Arranging them in ascending order for clarity:
Range of \(f = \{2, 3, 5, 7, 11, 13, 17\}\)
In simple words: For each number in the list A, find all the prime numbers that can divide it. Then pick the biggest prime number from that list. Write down each original number with its biggest prime factor as a pair. The range is simply all these biggest prime factors collected together.

🎯 Exam Tip: To find the highest prime factor, it's helpful to first perform prime factorization for each number. The range of a function consists only of the unique output values.

Question 5. Find the domain of the function \(f(x) = \sqrt{1 + \sqrt{1 - \sqrt{1 - x^2}}}\).
Answer:
To find the domain of the function, we need to ensure that all parts of the function are well-defined in the real number system. Specifically, the expression under any square root sign must be greater than or equal to zero.

We work from the innermost square root outwards:
1. The innermost expression is \(\sqrt{1 - x^2}\). For this to be a real number, we must have:
\(1 - x^2 \geq 0\)
\(1 \geq x^2\)
\(x^2 \leq 1\)
This inequality holds true for \(-1 \leq x \leq 1\).

2. Next, consider the middle square root: \(\sqrt{1 - \sqrt{1 - x^2}}\). For this to be a real number, we must have:
\(1 - \sqrt{1 - x^2} \geq 0\)
\(1 \geq \sqrt{1 - x^2}\)
Since both sides are non-negative (because \(\sqrt{1 - x^2}\) is non-negative), we can square both sides without changing the inequality direction:
\(1^2 \geq (\sqrt{1 - x^2})^2\)
\(1 \geq 1 - x^2\)
\(0 \geq -x^2\)
\(x^2 \geq 0\)
This condition is always true for any real number \(x\), since the square of any real number is always non-negative.

3. Finally, consider the outermost square root: \(\sqrt{1 + \sqrt{1 - \sqrt{1 - x^2}}}\). For this to be a real number, we must have:
\(1 + \sqrt{1 - \sqrt{1 - x^2}} \geq 0\)
We already know that \(\sqrt{1 - \sqrt{1 - x^2}}\) is a non-negative number (from step 2, provided \(-1 \leq x \leq 1\)). Therefore, \(1\) plus a non-negative number will always be greater than or equal to 1, and thus always greater than or equal to 0.
So, this condition is always satisfied as long as the inner parts are defined.

Combining all these conditions, the most restrictive condition that must be satisfied for the function to be defined is \(-1 \leq x \leq 1\). The domain of the function is all real numbers from -1 to 1, including -1 and 1.
Thus, the domain of \(f(x)\) is \([-1, 1]\).
In simple words: To find where this function works, we must make sure that everything inside each square root is not negative. Starting from the innermost square root, we find that x must be between -1 and 1 (including -1 and 1). All other conditions for the outer square roots are met if this first condition is true.

🎯 Exam Tip: For functions involving multiple nested square roots, always determine the domain by working from the innermost root outwards, making sure each expression under the root is non-negative.

Question 6. If \(f(x) = x^2\), \(g(x) = 3x\) and \(h(x) = x - 2\). Prove that \((fog)oh = fo(goh)\).
Answer:
We need to prove that the composition of functions is associative, meaning \((f \circ g) \circ h = f \circ (g \circ h)\).

First, let's calculate the Left Hand Side (LHS): \((f \circ g) \circ h\)
1. Calculate \((f \circ g)(x)\):
\((f \circ g)(x) = f(g(x))\)
Substitute \(g(x) = 3x\) into \(f(x)\):
\((f \circ g)(x) = f(3x) = (3x)^2 = 9x^2\)

2. Now, calculate \((f \circ g) \circ h(x)\):
\(((f \circ g) \circ h)(x) = (f \circ g)(h(x))\)
Substitute \(h(x) = x - 2\) into the expression for \((f \circ g)(x)\):
\(((f \circ g) \circ h)(x) = 9(x - 2)^2\)
Expand \((x - 2)^2\):
\(((f \circ g) \circ h)(x) = 9(x^2 - 4x + 4)\)
Distribute the 9:
\(((f \circ g) \circ h)(x) = 9x^2 - 36x + 36 \quad \text{--- (1)}\)

Next, let's calculate the Right Hand Side (RHS): \(f \circ (g \circ h)\)
1. Calculate \((g \circ h)(x)\):
\((g \circ h)(x) = g(h(x))\)
Substitute \(h(x) = x - 2\) into \(g(x)\):
\((g \circ h)(x) = g(x - 2) = 3(x - 2)\)
Distribute the 3:
\((g \circ h)(x) = 3x - 6\)

2. Now, calculate \(f \circ (g \circ h)(x)\):
\((f \circ (g \circ h))(x) = f((g \circ h)(x))\)
Substitute \((g \circ h)(x) = 3x - 6\) into \(f(x)\):
\((f \circ (g \circ h))(x) = f(3x - 6) = (3x - 6)^2\)
Expand \((3x - 6)^2\):
\((f \circ (g \circ h))(x) = (3x)^2 - 2(3x)(6) + 6^2\)
\((f \circ (g \circ h))(x) = 9x^2 - 36x + 36 \quad \text{--- (2)}\)

From (1) and (2), we can see that both sides are equal:
\(9x^2 - 36x + 36 = 9x^2 - 36x + 36\)
Therefore, \((f \circ g) \circ h = f \circ (g \circ h)\) is proved. This demonstrates the associative property of function composition.
In simple words: We need to show that combining three functions in two different orders gives the same result. First, we found \((f \circ g)\) and then combined it with \(h\). Next, we found \((g \circ h)\) and then combined it with \(f\). Since both final answers were the same, we proved that the order of grouping doesn't change the outcome.

🎯 Exam Tip: When proving associative properties for functions, clearly calculate each side (LHS and RHS) separately, showing all intermediate steps of function composition and algebraic expansion.

Question 7. Let \(A = \{1, 2\}\) and \(B = \{1, 2, 3, 4\}\), \(C = \{5, 6\}\) and \(D = \{5, 6, 7, 8\}\). Verify whether \(A \times C\) is a subset of \(B \times D\).
Answer:
We are given four sets:
\(A = \{1, 2\}\)
\(B = \{1, 2, 3, 4\}\)
\(C = \{5, 6\}\)
\(D = \{5, 6, 7, 8\}\)

We need to verify if \(A \times C\) is a subset of \(B \times D\). This means every element (ordered pair) in \(A \times C\) must also be present in \(B \times D\).

First, let's find the Cartesian product \(A \times C\):
\(A \times C = \{ (a, c) \mid a \in A, c \in C \}\)
\(A \times C = \{ (1, 5), (1, 6), (2, 5), (2, 6) \}\)

Next, let's find the Cartesian product \(B \times D\):
\(B \times D = \{ (b, d) \mid b \in B, d \in D \}\)
\(B \times D = \{ (1, 5), (1, 6), (1, 7), (1, 8), \)
\( (2, 5), (2, 6), (2, 7), (2, 8), \)
\( (3, 5), (3, 6), (3, 7), (3, 8), \)
\( (4, 5), (4, 6), (4, 7), (4, 8) \}\)

Now, we check if every element of \(A \times C\) is an element of \(B \times D\):
- Is \((1, 5) \in B \times D\)? Yes, it is.
- Is \((1, 6) \in B \times D\)? Yes, it is.
- Is \((2, 5) \in B \times D\)? Yes, it is.
- Is \((2, 6) \in B \times D\)? Yes, it is.

Since all the elements of \(A \times C\) are also found in \(B \times D\), we can conclude that \(A \times C\) is indeed a subset of \(B \times D\). The verification is complete.
In simple words: First, we make all possible pairs from set A and set C. Then, we make all possible pairs from set B and set D. To check if the first group of pairs is a "subset" of the second, we just need to see if every single pair from the first group is also present in the second group.

🎯 Exam Tip: To verify if one Cartesian product is a subset of another, list all elements of both products and then individually check if every element from the first set is contained in the second set.

Question 8. If \(f(x) = \frac{x-1}{x+1}\), \(x \neq 1\). Show that \(f(f(x)) = -\frac{1}{x}\), provided \(x \neq 0\).
Answer:
We are given the function \(f(x) = \frac{x-1}{x+1}\) and we need to find \(f(f(x))\). This means we substitute the entire function \(f(x)\) into itself.

Let's replace \(x\) in \(f(x)\) with \(f(x)\):
\(f(f(x)) = \frac{f(x) - 1}{f(x) + 1}\)

Now, substitute \(f(x) = \frac{x-1}{x+1}\) into this expression:
\(f(f(x)) = \frac{\frac{x-1}{x+1} - 1}{\frac{x-1}{x+1} + 1}\)

To simplify this complex fraction, we first simplify the numerator and the denominator separately:
**Numerator:**
\(\frac{x-1}{x+1} - 1 = \frac{x-1}{x+1} - \frac{x+1}{x+1}\)
\(= \frac{(x-1) - (x+1)}{x+1}\)
\(= \frac{x-1-x-1}{x+1}\)
\(= \frac{-2}{x+1}\)

**Denominator:**
\(\frac{x-1}{x+1} + 1 = \frac{x-1}{x+1} + \frac{x+1}{x+1}\)
\(= \frac{(x-1) + (x+1)}{x+1}\)
\(= \frac{x-1+x+1}{x+1}\)
\(= \frac{2x}{x+1}\)

Now, substitute these simplified expressions back into the fraction:
\(f(f(x)) = \frac{\frac{-2}{x+1}}{\frac{2x}{x+1}}\)

When dividing fractions, we multiply by the reciprocal of the denominator:
\(f(f(x)) = \frac{-2}{x+1} \times \frac{x+1}{2x}\)

We can cancel out the \((x+1)\) terms (assuming \(x+1 \neq 0\), i.e., \(x \neq -1\)):
\(f(f(x)) = \frac{-2}{2x}\)

Finally, simplify the fraction:
\(f(f(x)) = -\frac{1}{x}\)

This result holds true provided \(x \neq 0\) (because \(x\) is in the denominator) and \(x \neq -1\) (from the definition of \(f(x)\)). We have successfully shown that \(f(f(x)) = -\frac{1}{x}\).
In simple words: We take the whole function \(f(x)\) and put it inside itself. Then, we do the math to make the complex fraction simpler. After combining the top and bottom parts and canceling what we can, we get the simpler answer \(-\frac{1}{x}\).

🎯 Exam Tip: When evaluating composite functions like \(f(f(x))\), substitute the inner function completely and then simplify the resulting complex algebraic expression step-by-step, taking care with common denominators and cancellation.

Question 9. The functions f and g are defined by \(f(x) = 6x + 8\); \(g(x) = \frac{x-2}{3}\).
(i) Calculate the value of \(g(g(\frac{1}{2}))\).
(a) Write an expression for \(gf(x)\) in its simplest form.
Answer:
(i) Calculate the value of \(g(g(\frac{1}{2}))\):
We need to apply the function \(g\) twice. First, we find \(g(\frac{1}{2})\):
\(g(x) = \frac{x-2}{3}\)
Substitute \(x = \frac{1}{2}\):
\(g(\frac{1}{2}) = \frac{\frac{1}{2}-2}{3}\)
To subtract in the numerator, find a common denominator:
\(g(\frac{1}{2}) = \frac{\frac{1}{2}-\frac{4}{2}}{3} = \frac{\frac{1-4}{2}}{3} = \frac{\frac{-3}{2}}{3}\)
Divide the fractions by multiplying by the reciprocal:
\(g(\frac{1}{2}) = \frac{-3}{2} \times \frac{1}{3} = \frac{-3}{6} = -\frac{1}{2}\)

Now, we use this result to find \(g(g(\frac{1}{2}))\), which is \(g(-\frac{1}{2})\):
Substitute \(x = -\frac{1}{2}\) into \(g(x) = \frac{x-2}{3}\):
\(g(-\frac{1}{2}) = \frac{-\frac{1}{2}-2}{3}\)
Again, find a common denominator in the numerator:
\(g(-\frac{1}{2}) = \frac{-\frac{1}{2}-\frac{4}{2}}{3} = \frac{\frac{-1-4}{2}}{3} = \frac{\frac{-5}{2}}{3}\)
Divide the fractions:
\(g(-\frac{1}{2}) = \frac{-5}{2} \times \frac{1}{3} = \frac{-5}{6}\)
So, \(g(g(\frac{1}{2})) = -\frac{5}{6}\).

(a) Write an expression for \(gf(x)\) in its simplest form:
The expression \(gf(x)\) means \(g(f(x))\). We substitute \(f(x)\) into \(g(x)\).
\(f(x) = 6x + 8\)
\(g(x) = \frac{x-2}{3}\)
Substitute \(f(x)\) for \(x\) in \(g(x)\):
\(g(f(x)) = g(6x + 8)\)
\(= \frac{(6x + 8) - 2}{3}\)
\(= \frac{6x + 6}{3}\)
We can factor out 6 from the numerator:
\(= \frac{6(x + 1)}{3}\)
Now, simplify the fraction:
\(= 2(x + 1)\)
\(= 2x + 2\)
So, \(gf(x) = 2x + 2\).
In simple words: For part (i), we first put \(\frac{1}{2}\) into the function \(g\), and then we take that answer and put it back into \(g\) again to get the final result. For part (a), we take the entire function \(f(x)\) and put it into \(g(x)\), then simplify the whole expression until it's as simple as possible.

🎯 Exam Tip: When evaluating nested functions, always work from the innermost function outward. For composite functions, substitute the entire inner function expression and then simplify algebraically.

Question 10. Write the domain of the following real functions
(i) \(f(x) = \frac{2x+1}{x-9}\)
(ii) \(p(x) = \frac{-5}{4x^2 +1}\)
(iii) \(g(x) = \sqrt{x-2}\)
(iv) \(h(x) = x + 6\)
Answer:
The domain of a real function is the set of all real numbers for which the function produces a real number output. We need to consider restrictions like division by zero and taking the square root of a negative number.

(i) For \(f(x) = \frac{2x+1}{x-9}\):
This is a rational function. A rational function is undefined when its denominator is zero. So, we must ensure that the denominator is not equal to zero.
\(x - 9 \neq 0\)
\( \implies x \neq 9\)
The domain of \(f(x)\) is all real numbers except 9. We write this as \(\mathbb{R} \setminus \{9\}\) or \((-\infty, 9) \cup (9, \infty)\).

(ii) For \(p(x) = \frac{-5}{4x^2 +1}\):
This is also a rational function. We need to check if the denominator can be zero.
\(4x^2 + 1 = 0\)
\(4x^2 = -1\)
\(x^2 = -\frac{1}{4}\)
The square of a real number cannot be negative. Therefore, there is no real value of \(x\) for which the denominator is zero. This means the denominator is never zero. In fact, \(4x^2+1\) is always greater than or equal to 1.
The function \(p(x)\) is defined for all real numbers. The domain of \(p(x)\) is \(\mathbb{R}\) or \((-\infty, \infty)\).

(iii) For \(g(x) = \sqrt{x-2}\):
This is a square root function. For the output to be a real number, the expression inside the square root (the radicand) must be greater than or equal to zero.
\(x - 2 \geq 0\)
\( \implies x \geq 2\)
The domain of \(g(x)\) is all real numbers greater than or equal to 2. We write this as \([2, \infty)\).

(iv) For \(h(x) = x + 6\):
This is a linear function, which is a type of polynomial function. Polynomial functions are defined for all real numbers because there are no restrictions (like division by zero or square roots of negative numbers).
The domain of \(h(x)\) is all real numbers. We write this as \(\mathbb{R}\) or \((-\infty, \infty)\).
In simple words: For fractions, make sure the bottom part is not zero. For square roots, make sure the number inside is not negative. For simple straight-line functions like \(x+6\), all numbers work.

🎯 Exam Tip: Always check for two main domain restrictions: division by zero (for rational functions) and non-negative arguments under even roots (like square roots). Polynomials usually have a domain of all real numbers.