Maharashtra Board Class 11 Maths Part 1 Chapter 2 Functions Miscellaneous Solutions

Question 1. Which of the following relations are functions? If it is a function determine its domain and range.
(i) \( \{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)\} \)
(ii) \( \{(0, 0), (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3), (16, 4), (16, -4)\} \)
(iii) \( \{(1, 1), (3, 1), (5, 2)\} \)
Answer:
(i) Let \( R_1 = \{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)\} \)
Here, every element in the domain \( \{2, 4, 6, 8, 10, 12, 14\} \) has a unique image in the co-domain. Therefore, this relation is a function. Understanding this mapping helps us analyze real-world inputs and outputs clearly.
Domain = \( \{2, 4, 6, 8, 10, 12, 14\} \)
Range = \( \{1, 2, 3, 4, 5, 6, 7\} \)

(ii) Let \( R_2 = \{(0, 0), (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3), (16, 4), (16, -4)\} \)
Here, the elements \( 1, 4, 9, 16 \) in the domain do not have a unique image. For example, the element \( 1 \) is associated with two different images, \( 1 \) and \( -1 \). Therefore, this relation is not a function.

(iii) Let \( R_3 = \{(1, 1), (3, 1), (5, 2)\} \)
Here, every element in the domain \( \{1, 3, 5\} \) has a unique image in the co-domain. Therefore, this relation is a function.
Domain = \( \{1, 3, 5\} \)
Range = \( \{1, 2\} \)
In simple words: A relation is a function if every input has exactly one output. If any input points to more than one output, like in the second relation, it is not a function.

🎯 Exam Tip: To check if a relation is a function, look at the first numbers in the ordered pairs; if any first number repeats with a different second number, it is not a function.

Question 1. Determine whether the following relations are functions. If they are, find their domain and range.
(i) Relation represented by mapping:

• Domain Set A: {2, 4, 6, 8, 10, 12, 14}
• Codomain Set B: {1, 2, 3, 4, 5, 6, 7}
• Mappings: 2 → 1, 4 → 2, 6 → 3, 8 → 4, 10 → 5, 12 → 6, 14 → 7

(ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3), (16, 4), (16, -4)}
(iii) {(1, 1), (3, 1), (5, 2)}
Answer:
(i) Every element of set A has been assigned a unique element in set B.
\( \therefore \) Given relation is a function.
Domain = {2, 4, 6, 8, 10, 12, 14},
Range = {1, 2, 3, 4, 5, 6, 7}

(ii) Given relation: {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3), (16, 4), (16, -4)}
\( \therefore \) (1, 1), (1, -1) \( \in \) the relation.
\( \dots \) Given relation is not a function.
As element 1 of the domain has not been assigned a unique element of co-domain.

(iii) Given relation: {(1, 1), (3, 1), (5, 2)}
Every element of set A has been assigned a unique element in set B.
\( \therefore \) Given relation is a function.
Domain = {1, 3, 5}, Range = {1, 2}
In simple words: A relation is a function if every input has exactly one output. If an input has more than one output, like 1 going to both 1 and -1, it is not a function.

🎯 Exam Tip: To check if a relation is a function, ensure that no first element in the ordered pairs is repeated with different second elements.

Question 2. A function \( f: \mathbb{R} \rightarrow \mathbb{R} \) defined by \( f(x) = \frac{3x}{5} + 2 \), \( x \in \mathbb{R} \). Show that \( f \) is one-one and onto. Hence, find \( f^{-1} \).
Answer:
Given function: \( f: \mathbb{R} \rightarrow \mathbb{R} \) defined by \( f(x) = \frac{3x}{5} + 2 \)

1. To prove \( f \) is one-one:
Let \( x_1, x_2 \in \mathbb{R} \) such that \( f(x_1) = f(x_2) \).

\( \implies \frac{3x_1}{5} + 2 = \frac{3x_2}{5} + 2 \)

\( \implies \frac{3x_1}{5} = \frac{3x_2}{5} \)

\( \implies x_1 = x_2 \)
Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function \( f \) is one-one.

2. To prove \( f \) is onto:
Let \( y \in \mathbb{R} \) (codomain).
We need to find \( x \in \mathbb{R} \) (domain) such that \( f(x) = y \).

\( \implies \frac{3x}{5} + 2 = y \)

\( \implies \frac{3x}{5} = y - 2 \)

\( \implies 3x = 5(y - 2) \)

\( \implies x = \frac{5(y - 2)}{3} \)
Since \( y \in \mathbb{R} \), \( x = \frac{5(y - 2)}{3} \) is also a real number, so \( x \in \mathbb{R} \). This confirms that the mapping is bijective, meaning it is both injective and surjective.
Thus, for every \( y \in \mathbb{R} \), there exists \( x \in \mathbb{R} \) such that \( f(x) = y \).
Therefore, \( f \) is an onto function.

3. To find \( f^{-1} \):
Since \( f \) is one-one and onto, its inverse exists.
Let \( f(x) = y \)

\( \implies x = f^{-1}(y) \)
From the onto proof, we have:
\( x = \frac{5(y - 2)}{3} \)

\( \implies f^{-1}(y) = \frac{5(y - 2)}{3} \)
Replacing \( y \) with \( x \), we get:
\( f^{-1}(x) = \frac{5(x - 2)}{3} \)
In simple words: A function is one-one if different inputs always give different outputs, and it is onto if every possible output has a matching input. The inverse function simply reverses the process to find the original input from a given output.

🎯 Exam Tip: To prove a function is one-one, always start with \( f(x_1) = f(x_2) \) and show that it leads to \( x_1 = x_2 \). For onto, express \( x \) in terms of \( y \) and verify that \( x \) belongs to the domain.

Question 3. A function \( f \) is defined as follows: \( f(x) = 4x + 5 \), for \( -4 \le x < 0 \). Find the values of \( f(-1) \), \( f(-2) \), \( f(0) \), if they exist.
Answer: Given function is \( f(x) = 4x + 5 \) for \( -4 \le x < 0 \).
For \( f(-1) \): Since \( -1 \) lies in the interval \( [-4, 0) \), we have:
\( f(-1) = 4(-1) + 5 = -4 + 5 = 1 \)
For \( f(-2) \): Since \( -2 \) lies in the interval \( [-4, 0) \), we have:
\( f(-2) = 4(-2) + 5 = -8 + 5 = -3 \)
For \( f(0) \): Since \( x = 0 \) does not belong to the domain of \( f \) (which is \( -4 \le x < 0 \)), the value of \( f(0) \) cannot be determined. Since the value 0 does not lie within the specified interval, we cannot compute its image under this function.
\( \implies x = 0 \notin \) domain of \( f \)
\( \implies f(0) \) does not exist.
In simple words: We find the values of the function by plugging the numbers into the formula, but only if those numbers are between -4 and 0. Since 0 is not in this range, we cannot find a value for f(0).

🎯 Exam Tip: Always check if the given value of x lies within the defined interval (domain) before calculating the function's value.

Question 4. A function \( f \) is defined as follows: \( f(x) = 5 - x \) for \( 0 \le x \le 4 \). Find the value of \( x \) such that \( f(x) = 3 \).
Answer: Given function is \( f(x) = 5 - x \) for \( 0 \le x \le 4 \).
We are given \( f(x) = 3 \).
Substituting the function's expression:
\( 5 - x = 3 \)
\( \implies -x = 3 - 5 \)
\( \implies -x = -2 \)
\( \implies x = 2 \)
Since \( 2 \) lies in the interval \( [0, 4] \), the required value of \( x \) is \( 2 \). This confirms that the solution is valid within the given domain of the function.
In simple words: We set the equation 5 - x equal to 3 and solve for x, which gives us 2. Since 2 is between 0 and 4, it is the correct answer.

🎯 Exam Tip: Always verify that your final value of x falls within the restricted domain given in the question.

Question 5. If \( f(x) = 3x^2 - 5x + 7 \), find \( f(x - 1) \).
Answer: Given, \( f(x) = 3x^2 - 5x + 7 \)
\( \implies f(x - 1) = 3(x - 1)^2 - 5(x - 1) + 7 \)
\( \implies f(x - 1) = 3(x^2 - 2x + 1) - 5(x - 1) + 7 \)
\( \implies f(x - 1) = 3x^2 - 6x + 3 - 5x + 5 + 7 \)
\( \implies f(x - 1) = 3x^2 - 11x + 15 \) This simplified quadratic expression represents the function shifted to the right by one unit.
In simple words: To find \( f(x-1) \), we replace every \( x \) in the original formula with \( (x-1) \) and then simplify the algebra.

🎯 Exam Tip: Be very careful when expanding \( (x-1)^2 \); remember it is \( x^2 - 2x + 1 \), not just \( x^2 - 1 \).

Question 6. If \( f(x) = 3x + a \) and \( f(1) = 7 \), find \( a \) and \( f(4) \).
Answer: Given, \( f(x) = 3x + a \) and \( f(1) = 7 \)
\( \implies 3(1) + a = 7 \)
\( \implies 3 + a = 7 \)
\( \implies a = 7 - 3 = 4 \)
Substituting the value of \( a \) back into the function gives \( f(x) = 3x + 4 \).
\( \implies f(4) = 3(4) + 4 \)
\( \implies f(4) = 12 + 4 = 16 \)
In simple words: First, use the given value \( f(1) = 7 \) to find the missing number \( a \). Once you know \( a = 4 \), you can easily calculate \( f(4) \).

🎯 Exam Tip: Always find the unknown constant first before trying to evaluate the function at any other point.

Question 7. If \( f(x) = ax^2 + bx + 2 \) and \( f(1) = 3 \), \( f(4) = 42 \), find \( a \) and \( b \).
Answer: Given, \( f(x) = ax^2 + bx + 2 \) and \( f(1) = 3 \)
\( \implies a(1)^2 + b(1) + 2 = 3 \)
\( \implies a + b + 2 = 3 \)
\( \implies a + b = 1 \) ......(i)
Also, \( f(4) = 42 \)
\( \implies a(4)^2 + b(4) + 2 = 42 \)
\( \implies 16a + 4b + 2 = 42 \)
\( \implies 16a + 4b = 40 \)
Dividing both sides by 4, we get:
\( 4a + b = 10 \) ......(ii) These two linear equations can be solved simultaneously to find the unique values of \( a \) and \( b \).
In simple words: Use the two given function values to set up two equations with \( a \) and \( b \), which can then be solved together.

🎯 Exam Tip: Simplify your equations by dividing by a common factor (like dividing by 4 here) to make solving simultaneous equations much easier.

Question 8. If \( f(x) = \frac{2x-1}{5x-2}, x \neq \frac{2}{5} \), verify whether \( (fof)(x) = x \)
Answer: We are given the function \( f(x) = \frac{2x-1}{5x-2} \). To find \( (fof)(x) \), we substitute \( f(x) \) into itself:
\( (fof)(x) = f(f(x)) \)
\( = f\left(\frac{2x-1}{5x-2}\right) \)
\( = \frac{2\left(\frac{2x-1}{5x-2}\right) - 1}{5\left(\frac{2x-1}{5x-2}\right) - 2} \)
Multiplying the numerator and the denominator by \( (5x-2) \) to simplify the complex fraction:
\( = \frac{2(2x-1) - 1(5x-2)}{5(2x-1) - 2(5x-2)} \)
\( = \frac{4x - 2 - 5x + 2}{10x - 5 - 10x + 4} \)
\( = \frac{-x}{-1} \)
\( = x \)
Therefore, \( (fof)(x) = x \) is verified.
In simple words: To find \( (fof)(x) \), we plug the function \( f(x) \) back into itself. After simplifying the fraction, all the extra terms cancel out, leaving us with just \( x \).

🎯 Exam Tip: When simplifying complex fractions, multiply the numerator and denominator by the common denominator to clear the fractions quickly and avoid algebraic errors.

Question 9. If \( f(x) = \frac{x+3}{4x-5}, g(x) = \frac{3+5x}{4x-1} \), then verify that \( (fog)(x) = x \).
Answer: We are given \( f(x) = \frac{x+3}{4x-5} \) and \( g(x) = \frac{3+5x}{4x-1} \). To find the composite function \( (fog)(x) \), we substitute \( g(x) \) into \( f(x) \):
\( (fog)(x) = f(g(x)) \)
\( = f\left(\frac{3+5x}{4x-1}\right) \)
\( = \frac{\left(\frac{3+5x}{4x-1}\right) + 3}{4\left(\frac{3+5x}{4x-1}\right) - 5} \)
Multiplying the numerator and the denominator by \( (4x-1) \) to simplify:
\( = \frac{(3+5x) + 3(4x-1)}{4(3+5x) - 5(4x-1)} \)
\( = \frac{3 + 5x + 12x - 3}{12 + 20x - 20x + 5} \)
\( = \frac{17x}{17} \)
\( = x \)
Therefore, \( (fog)(x) = x \) is verified.
In simple words: We put the entire formula of \( g(x) \) into the \( x \) spots of \( f(x) \). When we simplify the math, everything cancels out perfectly to leave just \( x \).

🎯 Exam Tip: Be very careful with negative signs when expanding brackets in the denominator, such as \( -5(4x-1) = -20x + 5 \).