Samacheer Kalvi Class 11 Maths Solutions Chapter 1 Sets, Relations and Functions Exercise 1.3

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets, Relations and Functions Ex 1.3

 

Question 1. Suppose that 120 students are studying in 4 sections of eleventh standard in a school. Let A denote the set of students and B denote the set of the sections. Define a relation from A to B as "x related to y if the student x belongs to the section y". Is this relation a function? What can you say about the inverse relation? Explain your answer.
Answer: Let A be the set of 120 students, and B be the set of 4 sections. The relation is defined as a student `x` belonging to a section `y`.
₁ B₂ B₃ B₄
**Is this relation a function?** Yes, this relation is a function. This is because every student in set A belongs to exactly one section in set B. A student cannot belong to more than one section at the same time. Also, even if many students belong to the same section, each student still maps to only one section. **What can you say about the inverse relation?** The inverse relation from B to A is generally not a function. This is because:
(i) It is not a one-to-one function: Many students can be in the same section, meaning one section in B would map to multiple students in A.
(ii) It is not an onto function: It is possible that some sections in B might not have any students from A, meaning they would not have a preimage in A.
Since the original function is not one-to-one and not onto, its inverse relation cannot be considered a function.
In simple words: This relation is a function because each student belongs to only one section. The inverse relation is not a function because many students can be in the same section, and some sections might not have any students.

🎯 Exam Tip: Remember that for a relation to be a function, each element in the domain must map to exactly one element in the codomain. For an inverse relation to be a function, the original function must be both one-to-one and onto.

 

Question 2. Write the values of f at – 4, 1, -2, 7, 0 if
\[ f(x) = \begin{cases} -x + 4 & \text{if } -\infty < x \le -3 \\ x + 4 & \text{if } -3 < x < -2 \\ x^2 - x & \text{if } -2 \le x < 1 \\ x - x^2 & \text{if } 1 \le x < 7 \\ 0 & \text{otherwise} \end{cases} \]Answer: We need to find the value of \( f(x) \) for each given `x` by choosing the correct rule from the piecewise function definition.

When \( x = -4 \):
Since \( -4 \le -3 \), we use \( f(x) = -x + 4 \).
\( f(-4) = -(-4) + 4 = 4 + 4 = 8 \)

When \( x = 1 \):
Since \( 1 \le 1 < 7 \), we use \( f(x) = x - x^2 \).
\( f(1) = 1 - 1^2 = 1 - 1 = 0 \)

When \( x = -2 \):
Since \( -2 \le -2 < 1 \), we use \( f(x) = x^2 - x \).
\( f(-2) = (-2)^2 - (-2) = 4 + 2 = 6 \)

When \( x = 7 \):
Since \( x = 7 \) does not fall into any of the first four defined intervals (the last interval is \( 1 \le x < 7 \), not including 7), we use the "otherwise" rule.
\( f(7) = 0 \)

When \( x = 0 \):
Since \( -2 \le 0 < 1 \), we use \( f(x) = x^2 - x \).
\( f(0) = 0^2 - 0 = 0 \)
In simple words: For each number, we look at which part of the rule it fits into. Then we use that specific math formula to find the answer for that number.

🎯 Exam Tip: Pay close attention to the strict inequalities ( < or > ) versus inclusive inequalities ( ≤ or ≥ ) when determining which rule of a piecewise function to apply for a given x-value. Boundary points are critical.

 

Question 3. Write the values of f at – 3, 5, 2, – 1, 0 if
\[ f(x) = \begin{cases} x^2+x-5 & \text{if } x \in (-\infty, 0) \\ x^2+3x-2 & \text{if } x \in (3, \infty) \\ x^2 & \text{if } x \in [0,2) \\ x^2-3 & \text{otherwise} \end{cases} \]Answer: We will evaluate \( f(x) \) for each specified `x` value using the correct rule from the piecewise function.

When \( x = -3 \):
Since \( -3 \in (-\infty, 0) \), we use \( f(x) = x^2 + x - 5 \).
\( f(-3) = (-3)^2 + (-3) - 5 = 9 - 3 - 5 = 1 \)

When \( x = 5 \):
Since \( 5 \in (3, \infty) \), we use \( f(x) = x^2 + 3x - 2 \).
\( f(5) = 5^2 + 3(5) - 2 = 25 + 15 - 2 = 38 \)

When \( x = 2 \):
The value \( x = 2 \) does not fall into the interval \( (-\infty, 0) \), \( (3, \infty) \), or \( [0,2) \) (because \( [0,2) \) means \( 0 \le x < 2 \)). So, we use the "otherwise" rule.
\( f(2) = x^2 - 3 \).
\( f(2) = 2^2 - 3 = 4 - 3 = 1 \)

When \( x = -1 \):
Since \( -1 \in (-\infty, 0) \), we use \( f(x) = x^2 + x - 5 \).
\( f(-1) = (-1)^2 + (-1) - 5 = 1 - 1 - 5 = -5 \)

When \( x = 0 \):
The value \( x = 0 \) falls into the interval \( [0,2) \). So, we use \( f(x) = x^2 \).
\( f(0) = 0^2 = 0 \)
In simple words: For each number given, we find which rule applies to it. Then, we put the number into the formula for that rule to get the answer.

🎯 Exam Tip: Always double-check the interval boundaries, especially for 'otherwise' conditions. An interval like \( [0,2) \) includes 0 but excludes 2. Carefully determine which condition each input value satisfies.

 

Question 4. State whether the following relations are functions or not. If it is a function, check for one to oneness and ontoness. If it is not a function, state why?
(i) If \( A = \{ a, b, c \} \) and \( f = \{ (a, c), (b, c), (c, b) \} \); (f : A → A)
Answer: Given \( A = \{ a, b, c \} \) and \( f = \{ (a, c), (b, c), (c, b) \} \). The function maps from A to A.

**Is it a function?** Yes, it is a function. Every element in the domain A (a, b, c) has exactly one image in the codomain A.
**Is it one-to-one?** No, it is not a one-to-one function. Both elements 'a' and 'b' from the domain map to the same element 'c' in the codomain.
**Is it onto?** No, it is not an onto function. The element 'a' in the codomain does not have any element in the domain mapping to it (no preimage).

(ii) If \( X = \{ x, y, z \} \) and \( f = \{ (x, y), (x, z), (z, x) \} \); (f: X → X)
Answer: Given \( X = \{ x, y, z \} \) and \( f = \{ (x, y), (x, z), (z, x) \} \). The function maps from X to X.

**Is it a function?** No, it is not a function. The element 'x' in the domain has two different images ('y' and 'z') in the codomain. For a relation to be a function, each domain element must map to only one image.
In simple words: For part (i), it is a function but not one-to-one or onto. For part (ii), it is not a function at all because one input has two outputs.

🎯 Exam Tip: To check if a relation is a function, ensure every element in the domain has exactly one output. For one-to-one, each output must come from only one input. For onto, every element in the codomain must be an output.

 

Question 5. Let \( A = \{1, 2, 3, 4\} \) and \( B = \{ a, b, c, d \} \) Give a function from A → B for each of the following.
(i) neither one-to-one nor onto.
Answer: Let the function be defined as \( f = \{ (1, b), (2, c), (3, d), (4, d) \} \).

This is a function: Every element in A has exactly one image in B.
It is not one-to-one: Both 3 and 4 map to 'd'.
It is not onto: Element 'a' in B does not have a preimage in A.

(ii) not one to one but onto
Answer: Does not exist.

(iii) one – to one but not onto
Answer: Does not exist.

(iv) one to one and onto
Answer: Let the function be defined as \( f = \{ (1, a), (2, b), (3, c), (4, d) \} \).

This is a function: Every element in A has exactly one image in B.
It is one-to-one: Each element in A maps to a unique element in B.
It is onto: Every element in B has a preimage in A.
In simple words: We create different mappings between two sets to show what functions look like when they are or are not one-to-one or onto. Sometimes, a specific type of function might not be possible with the given sets.

🎯 Exam Tip: When constructing functions for specific properties (one-to-one, onto), ensure your mapping clearly demonstrates those properties. A one-to-one and onto function (bijection) means every element in the domain maps uniquely to every element in the codomain.

 

Question 6. Find the domain of \( \frac{1}{1-2 \sin x} \)
Answer: For the expression \( \frac{1}{1-2 \sin x} \) to be defined, the denominator cannot be equal to zero.
So, we set the denominator to zero and find the values of \( x \) that must be excluded from the domain.
\( 1 - 2 \sin x = 0 \)
\( \implies 1 = 2 \sin x \)
\( \implies \sin x = \frac{1}{2} \)
We know that \( \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \). So,
\( \implies \sin x = \sin \left(\frac{\pi}{6}\right) \)
The general solution for \( \sin x = \sin \alpha \) is \( x = n\pi + (-1)^n \alpha \), where \( n \in Z \) (an integer).
Using this formula, the values of \( x \) for which the function is undefined are:
\( x = n\pi + (-1)^n \frac{\pi}{6}, \text{ where } n \in Z \)
Therefore, the domain of \( f(x) \) is all real numbers except these specific values of \( x \).
Domain of \( f(x) \) is \( R - \left\{ x \mid x = n\pi + (-1)^n \frac{\pi}{6}, n \in Z \right\} \)
In simple words: The domain of a fraction is all numbers where the bottom part is not zero. We find when the bottom part is zero by solving the equation. Then, we remove those 'bad' numbers from all possible real numbers to get the domain.

🎯 Exam Tip: When finding the domain of a rational function, always identify values that make the denominator zero. For trigonometric functions, remember the general solutions for sine, cosine, and tangent equations to correctly exclude all such problematic values.

 

Question 7. Find the largest possible domain of the real valued function \( f(x) = \frac{\sqrt{4-x^{2}}}{\sqrt{x^{2}-9}} \)
Answer: For the function \( f(x) \) to be a real-valued function, two conditions must be met:
1. The expression under the square root in the numerator must be non-negative.
\( 4 - x^2 \ge 0 \)
\( \implies x^2 \le 4 \)
\( \implies -2 \le x \le 2 \)

2. The expression under the square root in the denominator must be strictly positive (since it's in the denominator, it cannot be zero).
\( x^2 - 9 > 0 \)
\( \implies x^2 > 9 \)
\( \implies x < -3 \text{ or } x > 3 \)

Now, we need to find the values of \( x \) that satisfy both conditions simultaneously.
The first condition states that \( x \) must be between -2 and 2, including -2 and 2.
The second condition states that \( x \) must be less than -3 or greater than 3.

If we look at these two sets of values on a number line, there is no overlap between the interval \( [-2, 2] \) and the intervals \( (-\infty, -3) \cup (3, \infty) \). This means there is no real number \( x \) that can satisfy both conditions at the same time.
Therefore, for no real values of \( x \), \( f(x) \) is defined.
The largest possible domain of \( f(x) \) is the empty set.
Domain of \( f(x) = \{\} \)
In simple words: We need to find `x` values for which both the top and bottom square roots are valid. The top part needs `x` to be between -2 and 2. The bottom part needs `x` to be less than -3 or greater than 3. These two needs don't overlap, so there are no `x` values that work for both. The function has no domain.

🎯 Exam Tip: When finding the domain of functions involving square roots and fractions, remember two key rules: the expression inside a square root must be non-negative, and the denominator of a fraction must not be zero. Combine these conditions carefully.

 

Question 8. Find the range of the function \( \frac{1}{2 \cos x-1} \)
Answer: Let \( f(x) = \frac{1}{2 \cos x-1} \).
First, we know the range of the cosine function is:
\( -1 \le \cos x \le 1 \)
Multiply by 2:
\( -2 \le 2 \cos x \le 2 \)
Subtract 1 from all parts:
\( -2 - 1 \le 2 \cos x - 1 \le 2 - 1 \)
\( -3 \le 2 \cos x - 1 \le 1 \)
Let \( y = 2 \cos x - 1 \). So, \( y \) can take any value in the interval \( [-3, 1] \).
However, we need to find the range of \( \frac{1}{y} \). We must exclude the case where \( y = 0 \), because division by zero is not allowed.
\( 2 \cos x - 1 = 0 \implies \cos x = \frac{1}{2} \), which is possible for many values of \( x \).
So, \( y \) can be in \( [-3, 1] \) but \( y \ne 0 \).
This means \( y \in [-3, 0) \cup (0, 1] \).
Now, let's find the range of \( \frac{1}{y} \):
1. For \( y \in [-3, 0) \):
As \( y \) approaches 0 from the negative side, \( \frac{1}{y} \) approaches \( -\infty \).
When \( y = -3 \), \( \frac{1}{y} = -\frac{1}{3} \).
So, for this part, the range is \( (-\infty, -\frac{1}{3}] \).

2. For \( y \in (0, 1] \):
As \( y \) approaches 0 from the positive side, \( \frac{1}{y} \) approaches \( \infty \).
When \( y = 1 \), \( \frac{1}{y} = 1 \).
So, for this part, the range is \( [1, \infty) \).

Combining both parts, the range of \( f(x) \) is \( (-\infty, -\frac{1}{3}] \cup [1, \infty) \). This covers all possible output values.
In simple words: We first find the smallest and largest values that the bottom part of the fraction can be. Since the bottom part can be zero, we must split it into two ranges: negative numbers and positive numbers. Then, we find what numbers the fraction becomes for each of these ranges and combine them.

🎯 Exam Tip: To find the range of a function of the form \( \frac{1}{g(x)} \), first determine the range of \( g(x) \). Remember to exclude any values for which \( g(x)=0 \), as these will lead to asymptotes. Then, analyze how \( \frac{1}{y} \) behaves over the resulting intervals for \( y \).

 

Question 9. Show that the relation \( xy = -2 \) is a function for a suitable domain. Find the domain and the range of the function.
Answer: We are given the relation \( xy = -2 \).
To show if it is a function, we express \( y \) in terms of \( x \):
\( y = \frac{-2}{x} \)
For this relation to be a function, for every input \( x \), there must be exactly one output \( y \). In this expression, for any given value of \( x \) (except for when \( x=0 \)), there is only one unique value for \( y \). Therefore, this relation is a function.

**Domain of the function:**
The function \( y = \frac{-2}{x} \) is defined for all real values of \( x \) except where the denominator is zero.
The denominator is \( x \), so \( x \ne 0 \).
Thus, the domain of the function is all real numbers except 0.
Domain: \( (-\infty, 0) \cup (0, \infty) \) or \( R - \{0\} \)

**Range of the function:**
To find the range, we consider what values \( y \) can take. Since \( y = \frac{-2}{x} \), if \( x \) can be any non-zero real number, then \( y \) can also take any non-zero real number.
For example, as \( x \) approaches 0, \( |y| \) approaches infinity. As \( |x| \) approaches infinity, \( y \) approaches 0.
However, \( y \) will never actually be zero, because \( \frac{-2}{x} = 0 \) has no solution.
Thus, the range of the function is all real numbers except 0.
Range: \( R - \{0\} \)
In simple words: The relation `xy = -2` can be written as `y = -2/x`. This is a function because for every number you put in for `x` (except zero), you get only one answer for `y`. The `x` values that can be put into the function are all numbers except zero. The `y` values that come out are also all numbers except zero.

🎯 Exam Tip: To determine if a relation is a function, always try to express \( y \) in terms of \( x \). If for every \( x \) there is a unique \( y \), it's a function. For domain, look for values that make denominators zero or square roots negative. For range, consider what values the function can produce.

 

Question 10. If \( f, g : R \rightarrow R \) are defined by \( f(x) = |x| + x \) and \( g(x) = |x| - x \) find gof and fog.
Answer: First, let's write the definitions of \( f(x) \) and \( g(x) \) as piecewise functions:
\[ f(x) = |x| + x = \begin{cases} x+x=2x & \text{if } x \ge 0 \\ -x+x=0 & \text{if } x < 0 \end{cases} \]
\[ g(x) = |x| - x = \begin{cases} x-x=0 & \text{if } x \ge 0 \\ -x-x=-2x & \text{if } x < 0 \end{cases} \]
**Now, let's find \( fog(x) = f(g(x)) \):**
Case 1: If \( x \ge 0 \)
\( g(x) = 0 \).
So, \( fog(x) = f(0) \).
Since \( 0 \ge 0 \), we use the rule \( f(x) = 2x \) for \( x \ge 0 \).
\( f(0) = 2(0) = 0 \).
Therefore, \( fog(x) = 0 \) for \( x \ge 0 \).

Case 2: If \( x < 0 \)
\( g(x) = -2x \).
Since \( x < 0 \implies -2x > 0 \), the argument of \( f \) (which is \( -2x \)) is positive.
So, we use the rule \( f(x) = 2x \) for \( x \ge 0 \).
\( fog(x) = f(-2x) = 2(-2x) = -4x \).
Therefore, \( fog(x) = -4x \) for \( x < 0 \).

Combining these, \( fog(x) = \begin{cases} 0 & \text{if } x \ge 0 \\ -4x & \text{if } x < 0 \end{cases} \)

**Next, let's find \( gof(x) = g(f(x)) \):**
Case 1: If \( x \ge 0 \)
\( f(x) = 2x \).
Since \( x \ge 0 \implies 2x \ge 0 \), the argument of \( g \) (which is \( 2x \)) is non-negative.
So, we use the rule \( g(x) = 0 \) for \( x \ge 0 \).
\( gof(x) = g(2x) = 0 \).
Therefore, \( gof(x) = 0 \) for \( x \ge 0 \).

Case 2: If \( x < 0 \)
\( f(x) = 0 \).
Since \( 0 \ge 0 \), the argument of \( g \) (which is \( 0 \)) is non-negative.
So, we use the rule \( g(x) = 0 \) for \( x \ge 0 \).
\( gof(x) = g(0) = 0 \).
Therefore, \( gof(x) = 0 \) for \( x < 0 \).

Combining these, \( gof(x) = 0 \) for all \( x \in R \).
In simple words: We first write `f(x)` and `g(x)` using simple rules for positive and negative numbers. Then, to find `fog(x)`, we put `g(x)` into `f(x)`, checking if `g(x)` itself is positive or negative. To find `gof(x)`, we put `f(x)` into `g(x)`, again checking if `f(x)` is positive or negative.

🎯 Exam Tip: When dealing with composite functions involving absolute values or piecewise definitions, always break down the problem into cases based on the conditions of the inner function's argument. Carefully determine which part of the outer function's definition applies to the output of the inner function.

 

Question 11. If f, g, h are real-valued functions defined on R, then prove that \( (f + g) \circ h = f \circ h + g \circ h \) what can you say about \( f \circ (g + h) \)? Justify your answer.
Answer: We need to prove the given identity and discuss the second expression.

**Proof for \( (f + g) \circ h = f \circ h + g \circ h \):**
Let \( x \) be any real number in the domain.
Consider the left-hand side (LHS):
\( [(f + g) \circ h](x) = (f + g)(h(x)) \)
By the definition of addition of functions, \( (f + g)(y) = f(y) + g(y) \).
So, \( (f + g)(h(x)) = f(h(x)) + g(h(x)) \)
By the definition of composite functions, \( f(h(x)) = (f \circ h)(x) \) and \( g(h(x)) = (g \circ h)(x) \).
\( = (f \circ h)(x) + (g \circ h)(x) \)
This is the right-hand side (RHS).
Therefore, \( (f + g) \circ h = f \circ h + g \circ h \) is proven. This shows that function composition distributes over addition from the right side.

**What can we say about \( f \circ (g + h) \)?**
Consider the expression \( f \circ (g + h) \).
\( [f \circ (g + h)](x) = f((g + h)(x)) \)
\( = f(g(x) + h(x)) \)
In general, \( f(g(x) + h(x)) \) is not equal to \( f(g(x)) + f(h(x)) \).
This means that \( f \circ (g + h) \) is generally not equal to \( (f \circ g) + (f \circ h) \).
The property \( f(A + B) = f(A) + f(B) \) is called linearity, and it is not true for all real-valued functions \( f \). It holds only for special types of functions, like linear functions where \( f(x) = mx + c \). If \( f(x) = x^2 \), for example, then \( f(A+B) = (A+B)^2 = A^2 + 2AB + B^2 \), while \( f(A)+f(B) = A^2+B^2 \). These are not equal.
So, composition of functions does not generally distribute over addition from the left side.
In simple words: When we add two functions and then compose with a third, it's the same as composing each of the first two with the third and then adding the results. But if we compose a function with the sum of two others, it's usually not the same as composing it with each one separately and then adding them. It only works that way for special kinds of functions.

🎯 Exam Tip: Understand the difference between right and left distribution of function composition over addition. Composition distributes from the right, but not generally from the left. Provide a counterexample (like \( f(x)=x^2 \)) if asked to justify why it doesn't distribute from the left.

 

Question 12. If \( f : R \rightarrow R \) is defined by \( f(x) = 3x – 5 \), Prove that f is a bijection and find its inverse.
Answer: We are given the function \( f(x) = 3x - 5 \) defined from real numbers \( R \) to real numbers \( R \).

**To prove \( f \) is a bijection, we need to show it is both one-to-one and onto.**

**1. Proof that \( f \) is one-to-one (Injective):**
Assume \( f(x_1) = f(x_2) \) for any \( x_1, x_2 \in R \).
\( 3x_1 - 5 = 3x_2 - 5 \)
Add 5 to both sides:
\( 3x_1 = 3x_2 \)
Divide by 3:
\( x_1 = x_2 \)
Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function \( f \) is one-to-one.

**2. Proof that \( f \) is onto (Surjective):**
For any \( y \in R \) (in the codomain), we need to find an \( x \in R \) (in the domain) such that \( f(x) = y \).
Let \( y = f(x) \).
\( y = 3x - 5 \)
Add 5 to both sides:
\( y + 5 = 3x \)
Divide by 3:
\( x = \frac{y+5}{3} \)
Since \( y \) is a real number, \( y+5 \) is also a real number, and \( \frac{y+5}{3} \) is also a real number. So, for every \( y \) in the codomain, there exists a real number \( x \) in the domain such that \( f(x) = y \).
Therefore, the function \( f \) is onto.

**Conclusion:** Since \( f \) is both one-to-one and onto, it is a bijection.

**To find the inverse of \( f \):**
From the onto proof, we already found the expression for \( x \) in terms of \( y \).
Let \( f^{-1}(y) = x \).
So, \( f^{-1}(y) = \frac{y+5}{3} \)
It is standard practice to express the inverse function in terms of \( x \), so we replace \( y \) with \( x \).
\( f^{-1}(x) = \frac{x+5}{3} \)
We can check this by verifying \( (f \circ f^{-1})(x) = x \) and \( (f^{-1} \circ f)(x) = x \).
\( (f \circ f^{-1})(x) = f\left( \frac{x+5}{3} \right) = 3\left( \frac{x+5}{3} \right) - 5 = (x+5) - 5 = x \)
\( (f^{-1} \circ f)(x) = f^{-1}(3x-5) = \frac{(3x-5)+5}{3} = \frac{3x}{3} = x \)
Both compositions result in the identity function, confirming the inverse.
In simple words: To show a function is special (a bijection), we prove two things: first, that different inputs always give different outputs (one-to-one); and second, that every possible output can be reached (onto). Then, to find the inverse, we swap the input and output variables and solve for the new output.

🎯 Exam Tip: For linear functions like \( f(x) = ax + b \) (where \( a \ne 0 \)), they are always bijections. The process for finding the inverse function involves setting \( y = f(x) \), solving for \( x \) in terms of \( y \), and then replacing \( y \) with \( x \) to write \( f^{-1}(x) \).

 

Question 13. The weight of the muscles of a man is a function of his bodyweight x and can be expressed as \( W ( x ) = 0.35x \). Determine the domain of this function.
Answer: The function given is \( W(x) = 0.35x \), where \( W(x) \) represents the muscle weight and \( x \) represents the total bodyweight.
In a real-world context, bodyweight \( x \) must be a positive value. A person's weight cannot be zero or negative.
Therefore, the variable \( x \) must satisfy \( x > 0 \).
The domain of this function, considering the physical meaning, is the set of all positive real numbers.
Domain is \( (0, \infty) \).
As bodyweight \( x \) increases, the muscle weight \( W(x) \) also increases proportionally.
In simple words: Since body weight must be a positive number, the domain of this function (the `x` values we can use) must also be positive numbers. So, `x` can be any number greater than zero.

🎯 Exam Tip: When a problem involves real-world quantities, always consider the practical limitations. Quantities like length, weight, time, or population cannot be negative, and often cannot be zero, which restricts the function's domain.

 

Question 14. The distance of an object falling is a function of time t and can be expressed as \( s (t) = – 16t^2 \). Graph the function and determine if it is one – to – one.
Answer: The function for the distance of an object falling is given by \( s(t) = -16t^2 \).

**To determine if the function is one-to-one:**
A function is one-to-one if \( s(t_1) = s(t_2) \) implies \( t_1 = t_2 \).
Let \( s(t_1) = s(t_2) \).
\( -16t_1^2 = -16t_2^2 \)
Divide both sides by -16:
\( t_1^2 = t_2^2 \)
Taking the square root of both sides:
\( t_1 = \pm t_2 \)
Since \( t_1 \) can be equal to \( t_2 \) or \( -t_2 \), the function is not one-to-one over all real numbers. For example, \( s(2) = -16(2^2) = -64 \) and \( s(-2) = -16(-2)^2 = -64 \). Here, two different input values (2 and -2) give the same output (-64).
However, in the physical context of an object falling, time \( t \) is typically considered to be non-negative (i.e., \( t \ge 0 \)). If we restrict the domain to \( t \ge 0 \), then \( t_1 = \pm t_2 \) would imply \( t_1 = t_2 \) (since both \( t_1, t_2 \) must be non-negative). In this restricted domain, the function would be one-to-one. The question does not explicitly state the domain, and the general mathematical analysis shows it's not one-to-one over the real numbers.
Therefore, the function \( s(t) = -16t^2 \) is **not one-to-one**.

**Graph the function:**
We can create a table of values to plot the graph. For falling objects, time \( t \) is usually non-negative, but for the general function \( s(t) = -16t^2 \), we can include negative values to see the full shape (a parabola opening downwards).

The graph of \( s(t) = -16t^2 \) is a parabola that opens downwards, with its vertex at the origin (0,0). The time (t-axis) is usually plotted horizontally and distance (s(t)-axis) vertically. This function shows that as time passes, the distance fallen becomes more negative (further downwards). The symmetrical nature of the parabola confirms that it's not one-to-one over its natural domain \( R \).
In simple words: The function for falling distance `s(t) = -16t^2` means distance changes with the square of time. It's not a one-to-one function because different times (like 2 seconds and -2 seconds) can give the same distance value. The graph is a downward-opening curve (parabola).

🎯 Exam Tip: To check if a function is one-to-one, use the horizontal line test on its graph or algebraically assume \( f(x_1) = f(x_2) \) and try to prove \( x_1 = x_2 \). If you get \( x_1 = \pm x_2 \) (or similar multiple solutions), it's not one-to-one.

 

Question 15. The total cost of airfare on a given route is comprised of the base cost C and the fuel Surcharge S in rupee. Both C and S are functions of the mileage m; \( C ( m ) = 0.4 m + 50 \) and \( S ( m ) = 0.03m \). Determine a function for the total cost of a ticket in terms of the mileage and find the airfare for flying 1600 miles.
Answer: We are given:
Base cost function: \( C(m) = 0.4m + 50 \)
Fuel surcharge function: \( S(m) = 0.03m \)
Where \( m \) is the mileage.

**1. Determine a function for the total cost of a ticket in terms of mileage:**
The total cost \( T(m) \) is the sum of the base cost and the fuel surcharge.
\( T(m) = C(m) + S(m) \)
Substitute the given functions:
\( T(m) = (0.4m + 50) + 0.03m \)
Combine the terms with \( m \):
\( T(m) = (0.4 + 0.03)m + 50 \)
\( T(m) = 0.43m + 50 \)
This is the function for the total cost of a ticket.

**2. Find the airfare for flying 1600 miles:**
We need to find \( T(m) \) when \( m = 1600 \).
\( T(1600) = 0.43(1600) + 50 \)
First, calculate \( 0.43 \times 1600 \):
\( 0.43 \times 1600 = 688 \)
Now, add the fixed cost:
\( T(1600) = 688 + 50 \)
\( T(1600) = 738 \)
The airfare for flying 1600 miles is Rs. 738.
In simple words: The total cost is found by adding the base cost and the fuel cost. Both depend on how many miles you fly. Once we have the total cost formula, we just put in 1600 miles to find the ticket price.

🎯 Exam Tip: When combining functions, clearly define the new function's name and its dependence on variables. Always remember to perform the full calculation, including any constant terms, for the final numerical answer.

 

Question 16. A salesperson whose annual earnings can be represented by the function \( A (x) = 30,000 + 0.04 x \), where x is the rupee value of the merchandise, he sells. His son is also in sales and his earnings are represented by the function \( S(x) = 25,000 + 0.05 x \). Find \( (A + S)(x) \) and determine the total family income if they each sell Rs. 1,50,00,000 worth of merchandise.
Answer: We are given the earnings functions for the salesperson and his son:
Salesperson's earnings: \( A(x) = 30,000 + 0.04x \)
Son's earnings: \( S(x) = 25,000 + 0.05x \)
Where \( x \) is the rupee value of merchandise sold.

**1. Find the total family income function \( (A + S)(x) \):**
The total family income is the sum of their individual earnings functions.
\( (A + S)(x) = A(x) + S(x) \)
\( (A + S)(x) = (30,000 + 0.04x) + (25,000 + 0.05x) \)
Combine the constant terms and the terms with \( x \):
\( (A + S)(x) = (30,000 + 25,000) + (0.04x + 0.05x) \)
\( (A + S)(x) = 55,000 + 0.09x \)
This function represents the total family income.

**2. Determine the total family income if they each sell Rs. 1,50,00,000 worth of merchandise:**
Given that the value of merchandise sold \( x = \text{Rs. } 1,50,00,000 \).
Substitute this value into the total family income function:
\( (A + S)(1,50,00,000) = 55,000 + 0.09(1,50,00,000) \)
First, calculate \( 0.09 \times 1,50,00,000 \):
\( 0.09 \times 1,50,00,000 = 13,50,000 \)
Now, add this to the constant income:
\( (A + S)(1,50,00,000) = 55,000 + 13,50,000 \)
\( (A + S)(1,50,00,000) = 14,05,000 \)
The total family income if they each sell Rs. 1,50,00,000 worth of merchandise is Rs. 14,05,000.
In simple words: We add the father's and son's income rules together to get one rule for total family income. Then, we use this rule to calculate the family's total money if they both sell Rs. 1,50,00,000 worth of items.

🎯 Exam Tip: When calculating large numbers with decimals, be careful with place values. It's often helpful to convert percentages (like 0.04 or 0.09) to fractions (4/100 or 9/100) for easier multiplication with large figures.

 

Question 17. The function for exchanging American dollars for Singapore Dollar on a given day is \( f (x) = 1.23x \), where x represents the number of American dollars. On the same day, the function for exchanging Singapore Dollar to Indian Rupee is \( g(y) = 50.50y \), where y represents the number of Singapore dollars. Write a function which will give the exchange rate of American dollars in terms of Indian rupee.
Answer: We are given two exchange rate functions:
1. American dollars to Singapore dollars: \( f(x) = 1.23x \)
Here, \( x \) is the amount in American dollars.

2. Singapore dollars to Indian Rupees: \( g(y) = 50.50y \)
Here, \( y \) is the amount in Singapore dollars.

To find a function that converts American dollars directly to Indian rupees, we need to compose these two functions. This means applying the first function (f) and then applying the second function (g) to the result of the first. This is represented by \( (g \circ f)(x) \).
\( (g \circ f)(x) = g(f(x)) \)
Substitute \( f(x) \) into \( g(y) \):
\( (g \circ f)(x) = g(1.23x) \)
Now, replace \( y \) with \( 1.23x \) in the function \( g(y) = 50.50y \):
\( (g \circ f)(x) = 50.50(1.23x) \)
Multiply the numerical values:
\( 50.50 \times 1.23 = 62.115 \)
So, the combined exchange rate function is:
\( (g \circ f)(x) = 62.115x \)
This function will give the exchange rate of American dollars in terms of Indian rupees.
In simple words: We have a rule to change American dollars to Singapore dollars, and another rule to change Singapore dollars to Indian rupees. To go straight from American dollars to Indian rupees, we just put the first rule into the second rule, creating one new rule.

🎯 Exam Tip: For currency conversion chains, the final function is a composite function. If you convert currency A to B using \( f \) and B to C using \( g \), then converting A to C is \( (g \circ f) \), meaning \( g(f(A)) \).

 

Question 18. The owner of a small restaurant can prepare a particular meal at a cost of Rs. 100. He estimates that if the menu price of the meal is x rupees, then the number of customers who will order that meal at that price in an evening is given by the function \( D (x) = 200 – x \). Express his day revenue total cost and profit on this meal as functions of x.
Answer: Given information:
Cost to prepare one meal = Rs. 100
Menu price of the meal = Rs. \( x \)
Number of customers (demand function): \( D(x) = 200 - x \)

**1. Day Revenue as a function of \( x \):**
Revenue is the total money collected from sales. It is calculated by multiplying the price per meal by the number of meals sold.
Revenue \( R(x) = (\text{Price per meal}) \times (\text{Number of customers}) \)
\( R(x) = x \times (200 - x) \)
\( R(x) = 200x - x^2 \)

**2. Total Cost as a function of \( x \):**
Total cost is the cost to prepare all the meals sold. It is calculated by multiplying the cost per meal by the number of meals prepared (which is equal to the number of customers).
Total Cost \( TC(x) = (\text{Cost per meal}) \times (\text{Number of customers}) \)
\( TC(x) = 100 \times (200 - x) \)
\( TC(x) = 20000 - 100x \)

**3. Profit as a function of \( x \):**
Profit is the money left after subtracting total costs from total revenue.
Profit \( P(x) = \text{Revenue } R(x) - \text{Total Cost } TC(x) \)
\( P(x) = (200x - x^2) - (20000 - 100x) \)
\( P(x) = 200x - x^2 - 20000 + 100x \)
Combine like terms:
\( P(x) = -x^2 + 300x - 20000 \)

Summary of functions:
Day Revenue: \( R(x) = 200x - x^2 \)
Total Cost: \( TC(x) = 20000 - 100x \)
Profit: \( P(x) = -x^2 + 300x - 20000 \)
In simple words: We find three different formulas based on the meal price `x`. Revenue is `x` times how many meals are sold. Total cost is the cost to make one meal times how many are sold. Profit is the money earned (revenue) minus the money spent (total cost).

🎯 Exam Tip: Remember the basic economic definitions: Revenue = Price × Quantity, Total Cost = Cost per unit × Quantity, and Profit = Revenue - Total Cost. Pay attention to how the quantity (number of customers) is itself a function of the price.

 

Question 19. The formula for converting from Fahrenheit to Celsius temperature is \( y = \frac{5x - 160}{9} \). Find the inverse of this function and determine whether the inverse is also a function?
Answer:
Given the function for converting Fahrenheit to Celsius is \( f(x) = y = \frac{5x - 160}{9} \).
To find the inverse function, we need to express \( x \) in terms of \( y \):
\( y = \frac{5x - 160}{9} \)
Now multiply both sides by 9:
\( 9y = 5x - 160 \)
Next, add 160 to both sides:
\( 5x = 9y + 160 \)
Finally, divide by 5 to isolate \( x \):
\( x = \frac{9y + 160}{5} \)
The inverse function, in terms of \( y \), is \( g(y) = \frac{9y + 160}{5} \). When we replace \( y \) with \( x \), the inverse function is \( f^{-1}(x) = \frac{9x + 160}{5} \).
To check if the inverse is also a function, we verify if the original function is a bijection (both one-to-one and onto). A linear function like \( f(x) = ax + b \) (where \( a \neq 0 \)) is always a bijection, so its inverse will also be a function.
We can also verify by finding \( gof(x) \) and \( fog(y) \):
\( gof(x) = g(f(x)) = g\left(\frac{5x - 160}{9}\right) \)
\( \implies g\left(\frac{5x - 160}{9}\right) = \frac{9\left(\frac{5x - 160}{9}\right) + 160}{5} \)
\( \implies = \frac{5x - 160 + 160}{5} \)
\( \implies = \frac{5x}{5} = x \)
And
\( fog(y) = f(g(y)) = f\left(\frac{9y + 160}{5}\right) \)
\( \implies f\left(\frac{9y + 160}{5}\right) = \frac{5\left(\frac{9y + 160}{5}\right) - 160}{9} \)
\( \implies = \frac{9y + 160 - 160}{9} \)
\( \implies = \frac{9y}{9} = y \)
Since \( gof(x) = x \) and \( fog(y) = y \), the function \( f \) is a bijection, which means its inverse \( f^{-1}(x) \) is also a function. The inverse function converts Celsius to Fahrenheit.
In simple words: First, switch the \( x \) and \( y \) in the formula, then solve for \( y \). The new formula you get is the inverse. Since the original temperature conversion is a simple, straight-line relationship, its inverse will also be a proper function.

🎯 Exam Tip: Remember that for a function to have an inverse that is also a function, the original function must be a bijection (one-to-one and onto). Linear functions always satisfy this condition.

 

Question 20. A simple cipher codes a number using the function \( f( x) = 3x - 4 \). Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line \( y = x \) (by drawing the lines).
Answer:
Given the function: \( f(x) = 3x - 4 \)
To find the inverse function, let \( y = f(x) \):
\( y = 3x - 4 \)
Now, we swap \( x \) and \( y \) to find the inverse relationship:
\( x = 3y - 4 \)
Next, solve for \( y \):
\( x + 4 = 3y \)
\( \implies y = \frac{x + 4}{3} \)
So, the inverse function is \( f^{-1}(x) = \frac{x + 4}{3} \).
Since \( f(x) \) is a linear function, it is always one-to-one and onto (a bijection). Therefore, its inverse \( f^{-1}(x) \) is also a function. Linear functions are always invertible because each input maps to a unique output, and every output comes from a unique input.
To verify the symmetrical property about the line \( y = x \), we can plot both functions and the line \( y = x \).
Let's find some points for \( f(x) = 3x - 4 \):

And for the inverse function \( f^{-1}(x) = \frac{x + 4}{3} \):

From the graph, we can see that the lines for \( f(x) = 3x - 4 \) and its inverse \( f^{-1}(x) = \frac{x + 4}{3} \) are mirror images of each other with respect to the line \( y = x \). This shows the symmetrical property of inverse functions.
In simple words: First, switch \( x \) and \( y \) in the formula, then solve for \( y \) again to get the inverse. Since the function is a straight line, its inverse is also a function. If you draw both functions on a graph, along with the line \( y=x \), you'll see they are perfectly mirrored across the \( y=x \) line.

🎯 Exam Tip: When graphing inverse functions, always include the line \( y = x \). The graphs of a function and its inverse are reflections of each other across this line.