Maharashtra Board Class 11 Maths Part 1 Chapter 2 Functions 2.1 Solutions

Question 1. Check if the following relations are functions.

🎯 Exam Tip: To determine if a relation is a function, verify that each element in the domain is mapped to exactly one unique element in the codomain.

Question 1. Check whether the following relations represented by mapping diagrams are functions. Give reasons for your answer.

• Diagram (a): Set A = {2, 1, 0, -1, -2}, Set B = {-3, 2, 1, 5, 6, -1}. Mappings: 2 → -3, 1 → 2, 0 → 1, -1 → 5, -2 → -1.
• Diagram (b): Set A = {p, q, r, s}, Set B = {a, b, c, d, e}. Mappings: p → b, q → d, r → a, r → c, s → e.
• Diagram (c): Set A = {3, -2, 1, 0, 2, 4}, Set B = {9, 7, -6, 3, 2}. Mappings: 3 → 9, -2 → 7, 1 → -6, 0 → 3, 2 → 2 (element 4 has no mapping).
Answer:
(a) Yes Reason: Every element of set A has been assigned a unique element in set B. A function requires every input to have exactly one output.
(b) No Reason: An element of set A has been assigned more than one element from set B.
(c) No Reason: Not every element of set A has been assigned an image from set B.
In simple words: For a relation to be a function, every single member in the first group must have exactly one partner in the second group. If a member has more than one partner, or no partner at all, it is not a function.

🎯 Exam Tip: Remember the two golden rules for a function: every element in the domain must have an image, and that image must be unique. Look out for unmapped elements or elements with multiple outgoing arrows.

Question 2. Which sets of ordered pairs represent functions from \( A = \{1, 2, 3, 4\} \) to \( B = \{-1, 0, 1, 2, 3\} \)? Justify.
(i) \( \{(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)\} \)
(ii) \( \{(1, 2), (2, -1), (3, 1), (4, 3)\} \)
(iii) \( \{(1, 3), (4, 1), (2, 2)\} \)
(iv) \( \{(1, 1), (2, 1), (3, 1), (4, 1)\} \)
Answer:
(i) \( \{(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)\} \) does not represent a function. Reason: \( (2, -1) \) and \( (2, 2) \) show that element \( 2 \in A \) has been assigned two different images -1 and 2 from set B, which violates the definition of a function.
(ii) \( \{(1, 2), (2, -1), (3, 1), (4, 3)\} \) represents a function. Reason: Every element of set A has a unique image in set B.
(iii) \( \{(1, 3), (4, 1), (2, 2)\} \) does not represent a function. Reason: \( 3 \in A \) does not have an image in set B.
(iv) \( \{(1, 1), (2, 1), (3, 1), (4, 1)\} \) represents a function. Reason: Every element of set A has been assigned a unique image in set B.
In simple words: For a relation to be a function, every input from the first set must connect to exactly one output in the second set. If an input has more than one output, or no output at all, it is not a function.

🎯 Exam Tip: To quickly check if a set of ordered pairs is a function, look at the first numbers (x-coordinates). If any first number repeats with a different second number, or if any element from the domain is missing, it is not a function.

Question 3. If \( f(m) = m^2 - 3m + 1 \), find
(i) \( f(0) \)
(ii) \( f(-3) \)
(iii) \( f\left(\frac{1}{2}\right) \)
(iv) \( f(x + 1) \)
(v) \( f(-x) \)
Answer:
Given: \( f(m) = m^2 - 3m + 1 \)
(i) \( f(0) = 0^2 - 3(0) + 1 = 1 \)
(ii) \( f(-3) = (-3)^2 - 3(-3) + 1 = 9 + 9 + 1 = 19 \)
(iii) \( f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{3}{2} + 1 = \frac{1 - 6 + 4}{4} = -\frac{1}{4} \)
(iv) \( f(x + 1) = (x + 1)^2 - 3(x + 1) + 1 = x^2 + 2x + 1 - 3x - 3 + 1 = x^2 - x - 1 \)
(v) \( f(-x) = (-x)^2 - 3(-x) + 1 = x^2 + 3x + 1 \). This shows how the function behaves when the input variable is negated.
In simple words: To find the value of a function, simply replace the variable 'm' with the given number or expression in the brackets and simplify the math.

🎯 Exam Tip: Be very careful with negative signs when substituting values, especially when squaring a negative number like \( (-3)^2 \), which always becomes positive.

Question 3. (Continued) Evaluate the function for the given values:
(iii) \( f\left(\frac{1}{2}\right) \)
(iv) \( f(x + 1) \)
(v) \( f(-x) \)
Answer:
(iii) \( f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 1 \)
\( = \frac{1}{4} - \frac{3}{2} + 1 \)
\( = \frac{1-6+4}{4} \)
\( = -\frac{1}{4} \)

(iv) \( f(x + 1) = (x + 1)^2 - 3(x + 1) + 1 \)
\( = x^2 + 2x + 1 - 3x - 3 + 1 \)
\( = x^2 - x - 1 \)

(v) \( f(-x) = (-x)^2 - 3(-x) + 1 = x^2 + 3x + 1 \). These steps show how the function behaves under different inputs.
In simple words: We substitute different values or expressions like \( \frac{1}{2} \), \( x+1 \), and \( -x \) into the original function to find their simplified final forms.

🎯 Exam Tip: When substituting expressions like \( x+1 \) into a function, always use parentheses to avoid sign errors when expanding terms.

Question 4. Find \( x \), if \( g(x) = 0 \) where
(i) \( g(x) = \frac{5x-6}{7} \)
(ii) \( g(x) = \frac{18-2x^2}{7} \)
(iii) \( g(x) = 6x^2 + x - 2 \)
Answer:
(i) \( g(x) = \frac{5x-6}{7} \)
Given \( g(x) = 0 \)
\( \implies \frac{5x-6}{7} = 0 \)
\( \implies 5x - 6 = 0 \)
\( \implies 5x = 6 \)
\( \implies x = \frac{6}{5} \)

(ii) \( g(x) = \frac{18-2x^2}{7} \)
Given \( g(x) = 0 \)
\( \implies \frac{18-2x^2}{7} = 0 \)
\( \implies 18 - 2x^2 = 0 \)
\( \implies 2x^2 = 18 \)
\( \implies x^2 = 9 \)
\( \implies x = \pm 3 \)

(iii) \( g(x) = 6x^2 + x - 2 \)
Given \( g(x) = 0 \)
\( \implies 6x^2 + x - 2 = 0 \)
\( \implies 6x^2 + 4x - 3x - 2 = 0 \)
\( \implies 2x(3x + 2) - 1(3x + 2) = 0 \)
\( \implies (3x + 2)(2x - 1) = 0 \)
\( \implies 3x + 2 = 0 \) or \( 2x - 1 = 0 \)
\( \implies x = -\frac{2}{3} \) or \( x = \frac{1}{2} \). These values represent the roots or zeroes of the given functions.
In simple words: To find \( x \) when \( g(x) = 0 \), we set each given equation equal to zero and solve for \( x \) using basic algebra and factoring.

🎯 Exam Tip: Remember that taking the square root of both sides of an equation like \( x^2 = 9 \) always yields both positive and negative solutions, \( \pm 3 \).

Question 5. Find x, if f(x) = g(x) where f(x) = x⁴ + 2x², g(x) = 11x².
Answer: Given:
\( f(x) = x^4 + 2x^2 \)
\( g(x) = 11x^2 \)
Since \( f(x) = g(x) \):
\( \therefore x^4 + 2x^2 = 11x^2 \)
\( \therefore x^4 - 9x^2 = 0 \)
\( \dots \)
\( \therefore x^2(x^2 - 9) = 0 \)
\( \therefore x^2 = 0 \) or \( x^2 - 9 = 0 \)
\( \therefore x = 0 \) or \( x^2 = 9 \)
\( \therefore x = 0 \) or \( x = \pm 3 \)
This gives us three distinct real values for x that satisfy the given equation.
In simple words: We set the two equations equal to each other and solve for x by moving all terms to one side and factoring. This helps us find the values of x where both functions produce the exact same output.

🎯 Exam Tip: Always remember to write \( \pm \) when taking the square root of a positive number to ensure you do not lose any valid solutions.

Question 6. If f(x) = \( \begin{cases} x^2 + 3, & x \le 2 \\ 5x + 7, & x > 2 \end{cases} \), then find
(i) f(3)
(ii) f(2)
(iii) f(0)
Answer:
(i) For \( f(3) \), since \( 3 > 2 \), we use the formula \( f(x) = 5x + 7 \).
\( \therefore f(3) = 5(3) + 7 = 15 + 7 = 22 \)
(ii) For \( f(2) \), since \( 2 \le 2 \), we use the formula \( f(x) = x^2 + 3 \).
\( \dots \)
\( \therefore f(2) = 2^2 + 3 = 4 + 3 = 7 \)
(iii) For \( f(0) \), since \( 0 \le 2 \), we use the formula \( f(x) = x^2 + 3 \).
\( \dots \)
\( \therefore f(0) = 0^2 + 3 = 3 \)
Choosing the correct piece of the function based on the value of x is essential for piecewise functions.
In simple words: A piecewise function has different rules depending on the value of x. We look at our input number, find which rule's condition it fits, and then plug it into that specific formula.

🎯 Exam Tip: Pay close attention to the boundary values like \( x = 2 \) and check which interval contains the "or equal to" (\( \le \) or \( \ge \)) sign.

Question 7. If f(x) = \( \begin{cases} 4x - 2, & x \le -3 \\ 5, & -3 < x < 3 \\ x^2, & x \ge 3 \end{cases} \), then find
(i) f(-4)
(ii) f(-3)
(iii) f(1)
Answer:
(i) For \( f(-4) \), since \( -4 \le -3 \), we use the formula \( f(x) = 4x - 2 \).
\( \therefore f(-4) = 4(-4) - 2 = -16 - 2 = -18 \)
(ii) For \( f(-3) \), since \( -3 \le -3 \), we use the formula \( f(x) = 4x - 2 \).
\( \dots \)
\( \dots \)
\( \therefore f(-3) = 4(-3) - 2 = -12 - 2 = -14 \)
(iii) For \( f(1) \), since \( -3 < 1 < 3 \), we use the constant value \( f(x) = 5 \).
\( \therefore f(1) = 5 \)
Evaluating each part carefully ensures we apply the correct interval rule for every given input.
In simple words: We check which of the three conditions our input x satisfies, and then use that specific part of the function to get the answer.

🎯 Exam Tip: For constant intervals like \( -3 < x < 3 \), the output is always the constant value itself without needing any calculation.

Question 7. Find the values of the piecewise function for the following:
(i) \( f(-4) \)
(ii) \( f(-3) \)
(iii) \( f(1) \)
(iv) \( f(5) \)
Answer:
The given piecewise function is:
\( f(x) = 4x - 2 \), for \( x \le -3 \)
\( f(x) = 5 \), for \( -3 < x < 3 \)
\( f(x) = x^2 \), for \( x \ge 3 \)

(i) For \( x = -4 \) (since \( -4 \le -3 \)):
\( f(-4) = 4(-4) - 2 = -16 - 2 = -18 \)

(ii) For \( x = -3 \) (since \( -3 \le -3 \)):
\( f(-3) = 4(-3) - 2 = -12 - 2 = -14 \)

(iii) For \( x = 1 \) (since \( -3 < 1 < 3 \)):
\( f(1) = 5 \)

(iv) For \( x = 5 \) (since \( 5 \ge 3 \)):
\( f(5) = 5^2 = 25 \)
In simple words: To find the value of a piecewise function, look at the given intervals to see which formula fits the input value of x, then calculate using that formula.

🎯 Exam Tip: Always check which interval the given x-value falls into before choosing the formula to substitute the value.

Question 8. If \( f(x) = 3x + 5 \), \( g(x) = 6x - 1 \), then find
(i) \( (f + g)(x) \)
(ii) \( (f - g)(2) \)
(iii) \( (fg)(3) \)
(iv) \( \left(\frac{f}{g}\right)(x) \) and its domain
Answer:
Given: \( f(x) = 3x + 5 \) and \( g(x) = 6x - 1 \)

(i) \( (f + g)(x) = f(x) + g(x) \)
\( = (3x + 5) + (6x - 1) \)
\( = 9x + 4 \)

(ii) \( (f - g)(2) = f(2) - g(2) \)
\( = [3(2) + 5] - [6(2) - 1] \)
\( = [6 + 5] - [12 - 1] \)
\( = 11 - 11 \)
\( = 0 \)

(iii) \( (fg)(3) = f(3) \cdot g(3) \)
\( = [3(3) + 5] \cdot [6(3) - 1] \)
\( = (14)(17) \)
\( = 238 \)

(iv) \( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{3x + 5}{6x - 1} \)
For the function to be defined, the denominator cannot be zero:
\( 6x - 1 \neq 0 \)
\( \implies x \neq \frac{1}{6} \)
Therefore, Domain = \( R - \left\{\frac{1}{6}\right\} \)
In simple words: We perform basic arithmetic operations on functions by combining their algebraic expressions, and for division, we must exclude any value that makes the denominator zero.

🎯 Exam Tip: When finding the domain of a rational function, always set the denominator not equal to zero to find the excluded values from the set of real numbers.

Question 9. If \( f(x) = 2x^2 + 3 \), \( g(x) = 5x - 2 \), then find
(i) fog
(ii) gof
(iii) fof
(iv) gog
Answer:
Given: \( f(x) = 2x^2 + 3 \) and \( g(x) = 5x - 2 \). These functions map real numbers to other real numbers through algebraic operations.

(i) \( (fog)(x) = f(g(x)) \)
\( = f(5x - 2) \)
\( = 2(5x - 2)^2 + 3 \)
\( = 2(25x^2 - 20x + 4) + 3 \)
\( = 50x^2 - 40x + 8 + 3 \)
\( = 50x^2 - 40x + 11 \)

(ii) \( (gof)(x) = g(f(x)) \)
\( = g(2x^2 + 3) \)
\( = 5(2x^2 + 3) - 2 \)
\( = 10x^2 + 15 - 2 \)
\( = 10x^2 + 13 \)

(iii) \( (fof)(x) = f(f(x)) \)
\( = f(2x^2 + 3) \)
\( = 2(2x^2 + 3)^2 + 3 \)
\( = 2(4x^4 + 12x^2 + 9) + 3 \)
\( = 8x^4 + 24x^2 + 18 + 3 \)
\( = 8x^4 + 24x^2 + 21 \)

(iv) \( (gog)(x) = g(g(x)) \)
\( = g(5x - 2) \)
\( = 5(5x - 2) - 2 \)
\( = 25x - 10 - 2 \)
\( = 25x - 12 \)
In simple words: Composite functions mean putting one function inside another. To find fog, we take the entire formula of g and plug it into the x of function f, then simplify the math.

🎯 Exam Tip: When finding composite functions, always substitute the inner function carefully into every occurrence of the variable in the outer function and simplify step-by-step to avoid algebraic errors.