CBSE Class 11 Mathematics Relations and Functions Assignment Set 06

C - Functions

Question 1. Let \(A = \{1, 2, 3\}\), \(B = \{-1, 0, 1\}\). Which of the following relations from \(A\) to \(B\) are functions? Justify the answer.
(i) \(R = \{(1, -1), (2, 0), (3, 1)\}\)
(ii) \(R = \{(1, -1), (1, 0), (2, 0), (2, 1)\}\)
(iii) \(R = \{(1, 0), (2, -1), (3, 1)\}\)
(iv) \(R = \{(2, 0), (2, 1), (1, -1), (3, 1), (3, 0)\}\)
Answer:
(i) and (iii) are functions because every element of \(A\) has a unique image in \(B\).
(ii) and (iv) are not functions because some elements in \(A\) have more than one image (double images).
In simple words: A relation is a function only when each input from set \(A\) is paired with exactly one output in set \(B\).

Exam Tip: Check that every element in the domain appears exactly once as a first coordinate in the ordered pairs.

Question 2. Which of the following arrow diagrams does represent a function?
Answer:
(i) and (iii) are functions.
(ii) and (iv) are not functions.
In simple words: A diagram represents a function if every element in the first set has one and only one arrow pointing out to the second set.

Exam Tip: Ensure no element in the domain has multiple outgoing arrows, and every domain element has at least one outgoing arrow.

Question 3. Does the adjacent arrow diagram represent a function? If so, write its Domain, Co-domain, range
Answer:
Yes, it represents a function.
Domain = \(\{-1, 0, 1, 2, 3\}\)
Co-domain = \(\{0, 2, 4, 7\}\) (or the complete target set shown)
Range = \(\{0, 2, 7\}\)
In simple words: Domain is all input values, co-domain is all possible output values, and range is the set of actually matched output values.

Exam Tip: Range is a subset of the co-domain containing only those elements that have incoming arrows from the domain.

Question 4. If a function \(f\) is defined by \(f(x) = 2x - 1\), \(x \in \{1, 2, 0, -1\}\), Find its range and the linear function
Answer:
Substitute each element of the domain into \(f(x) = 2x - 1\):
For \(x = 1\): \(f(1) = 2(1) - 1 = 1\)
For \(x = 2\): \(f(2) = 2(2) - 1 = 3\)
For \(x = 0\): \(f(0) = 2(0) - 1 = -1\)
For \(x = -1\): \(f(-1) = 2(-1) - 1 = -3\)
Range = \(\{1, 3, -1, -3\}\)
Function as a set of ordered pairs: \(f = \{(1, 1), (2, 3), (0, -1), (-1, -3)\}\)
In simple words: Range is found by calculating the output value for every given input value.

Exam Tip: Clearly write out the substitution steps for each element in the given domain set.

Question 5. If a function \(f\) is defined by \(f(x) = 2x + 5\), Find the value of (i) \(f(0)\) (ii) \(f(\frac{1}{2})\) (iii) \(f(-1)\)
Answer:
(i) \(f(0) = 2(0) + 5 = 0 + 5 = 5\)
(ii) \(f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right) + 5 = 1 + 5 = 6\)
(iii) \(f(-1) = 2(-1) + 5 = -2 + 5 = 3\)
In simple words: Substitute the given number in place of \(x\) in the expression \(2x + 5\) and simplify.

Exam Tip: Be careful with signs when substituting negative numbers.

Question 6. The function \(t\) which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by \(t(c) = \frac{9c}{5} + 32\). Find \(c\) when \(t(c) = 212\)
Answer:
Given \(t(c) = \frac{9c}{5} + 32\)
Set \(t(c) = 212\):
\(\frac{9c}{5} + 32 = 212\)
\(\frac{9c}{5} = 212 - 32\)
\(\frac{9c}{5} = 180\)
\(9c = 180 \times 5\)
\(9c = 900\)
\(c = \frac{900}{9} = 100\)
In simple words: Replace \(t(c)\) with 212 and solve the linear equation for \(c\).

Question 7. Let \(f = \{(1, 1), (2, 3), (0, -1), (-1, -3)\}\) be a function from \(Z\) to \(Z\) defined by \(f(x) = ax + b\), for some integer \(a, b\). Find \(a, b\). \(Z\) is the set of integers.
Answer:
Given \(f(x) = ax + b\).
Using the ordered pair \((1, 1)\), we get \(f(1) = 1\):
\(a(1) + b = 1 \implies a + b = 1\) (Equation 1)
Using the ordered pair \((0, -1)\), we get \(f(0) = -1\):
\(a(0) + b = -1 \implies b = -1\)
Substitute \(b = -1\) into Equation 1:
\(a + (-1) = 1 \implies a = 1 + 1 = 2\)
Thus, \(a = 2\) and \(b = -1\).
In simple words: Pick two coordinate pairs from the function and solve the resulting simultaneous linear equations for \(a\) and \(b\).

Exam Tip: Choosing the pair with \(x = 0\) makes finding the value of \(b\) very direct and quick.

Question 8. Let \(f = \left\{\left(x, \frac{x^2}{1 + x^2}\right) : x \in R\right\}\) be a function from \(R\) into \(R\). Determine the range of \(f\)
Answer:
Let \(y = f(x) = \frac{x^2}{1 + x^2}\).
Since \(x^2 \geq 0\) for all \(x \in R\), the numerator is always non-negative, and the denominator \(1 + x^2 > x^2 \geq 0\), so \(y \geq 0\).
Also, rearrange the equation to express \(x^2\) in terms of \(y\):
\(y(1 + x^2) = x^2\)
\(y + yx^2 = x^2\)
\(y = x^2 - yx^2 = x^2(1 - y)\)
\(x^2 = \frac{y}{1 - y}\)
Since \(x^2 \geq 0\) for all real numbers \(x\), we must have:
\(\frac{y}{1 - y} \geq 0\)
This implies \(0 \leq y < 1\).
Range of \(f = \{y \in R : 0 \leq y < 1\}\) (or interval \([0, 1)\)).
In simple words: Express \(x^2\) in terms of \(y\) and apply the condition that the square of any real number is always greater than or equal to zero.

Exam Tip: Remember that division by zero is undefined, so the denominator \(1 - y\) cannot be zero, which gives the strict upper bound \(y < 1\).

Question 9. Find the domain and range of the following functions
(i) \(f(x) = -|x|\)
(ii) \(f(x) = \sqrt{9 - x^2}\)
(iii) \(f(x) = \sqrt{x - 1}\)
(iv) \(f(x) = |x - 1|\)
Answer:
(i) For \(f(x) = -|x|\):
Domain = \(R\) (all real numbers)
Range = \((-\infty, 0]\)
(ii) For \(f(x) = \sqrt{9 - x^2}\):
For domain, \(9 - x^2 \geq 0 \implies x^2 \leq 9 \implies -3 \leq x \leq 3\).
Domain = \([-3, 3]\)
Range = \([0, 3]\)
(iii) For \(f(x) = \sqrt{x - 1}\):
For domain, \(x - 1 \geq 0 \implies x \geq 1\).
Domain = \([1, \infty)\)
Range = \([0, \infty)\)
(iv) For \(f(x) = |x - 1|\):
Domain = \(R\)
Range = \([0, \infty)\)
In simple words: Domain is the set of all allowed inputs \(x\), and range is the set of all resulting outputs \(y\).

Exam Tip: For square root functions, ensure the expression inside the radical is greater than or equal to zero.

Question 10. If the real functions \(f\) and \(g\) are defined by \(f(x) = 2x - 1\) and \(g(x) = x + 1\), evaluate the following
(i) \((f + g)x\) (ii) \((f - g)x\) (iii) \(fg(x)\) (iv) \(\left(\frac{f}{g}\right)(x)\)
Answer:
Given \(f(x) = 2x - 1\) and \(g(x) = x + 1\) for \(x \in R\).
(i) \((f + g)x = f(x) + g(x) = (2x - 1) + (x + 1) = 3x, \, x \in R\)
(ii) \((f - g)x = f(x) - g(x) = (2x - 1) - (x + 1) = 2x - 1 - x - 1 = x - 2, \, x \in R\)
(iii) \(fg(x) = f(x) \cdot g(x) = (2x - 1)(x + 1) = 2x^2 + 2x - x - 1 = 2x^2 + x - 1, \, x \in R\)
(iv) \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{2x - 1}{x + 1}, \, x \neq -1\)
In simple words: Combine the two functions algebraically using addition, subtraction, multiplication, or division according to standard rules.

Exam Tip: Always state the domain restriction for division where the denominator cannot equal zero (\(x \neq -1\)).