CBSE Class 12 Mathematics Relations And Functions Assignment Set 15

XII- Mathematics Assignment No-03 - Relations and Functions

Question 1. Define composite function, fog, gof for any given three sets A, B, C. Draw diagram, also
Answer: Let \(A\), \(B\), and \(C\) be three non-empty sets, and let \(f: A \to B\) and \(g: B \to C\) be two functions.
The composite function \(g \circ f\) is defined as a function from \(A\) to \(C\) given by \((g \circ f)(x) = g(f(x))\) for all \(x \in A\).
Similarly, if \(f: A \to B\) and \(g: B \to C\), the composite function \(f \circ g\) is not defined unless the codomain of \(g\) is a subset of the domain of \(f\). Here, \(g \circ f: A \to C\) and \(f \circ g: B \to B\) (if domain and codomain match appropriately).
In simple words: A composite function applies one function after another, feeding the output of the first function as the input into the second function.

Exam Tip: Always check the domain and codomain compatibility before defining composite functions. The range of the inner function must be a subset of the domain of the outer function.

Question 2. Let \(A = \{3, 4, 5, 6\}\), \(B = \{13, 14, 15, 16\}\), \(C = \{23, 24, 25\}\) given \(f: A \to B : f = \{(3,13), (4,14), (5,15), (6,16)\}\), \(g: B \to C : g = \{(13,23), (14,23), (15,24), (16,25)\}\).
Answer: To find \(g \circ f\), we evaluate \(g(f(x))\) for each element \(x \in A\):
For \(x = 3\): \((g \circ f)(3) = g(f(3)) = g(13) = 23\)
For \(x = 4\): \((g \circ f)(4) = g(f(4)) = g(14) = 23\)
For \(x = 5\): \((g \circ f)(5) = g(f(5)) = g(15) = 24\)
For \(x = 6\): \((g \circ f)(6) = g(f(6)) = g(16) = 25\)
Thus, \(g \circ f = \{(3, 23), (4, 23), (5, 24), (6, 25)\}\).
In simple words: Trace the path from set A through set B into set C using the given ordered pairs.

Exam Tip: Write out each composite step clearly to avoid mapping errors in discrete relations.

Question 3. Let \(f: R \to R\) and \(g: R \to R\) such that \(f(x) = x^2\) and \(g(x) = x + 1\). Show that \(f \circ g \neq g \circ f\).
Answer: First, find \((f \circ g)(x)\):
\((f \circ g)(x) = f(g(x)) = f(x + 1) = (x + 1)^2 = x^2 + 2x + 1\)
Next, find \((g \circ f)(x)\):
\((g \circ f)(x) = g(f(x)) = g(x^2) = x^2 + 1\)
Since \(x^2 + 2x + 1 \neq x^2 + 1\) for all \(x \neq 0\), we have \(f \circ g \neq g \circ f\).
In simple words: Function composition is not commutative in general, meaning the order in which you apply functions matters.

Exam Tip: Always evaluate both composite orders separately and compare their algebraic expressions.

Question 4. Let \(f: R \to R\) be defined by \(f(x) = 3x - 7\). Show that \(f\) is invertible. Also find its inverse function \(g = f^{-1}: R \to R\).
Answer: To show \(f\) is invertible, we prove it is one-one and onto.
1. One-one: Let \(f(x_1) = f(x_2)\).
\(3x_1 - 7 = 3x_2 - 7\)
\(3x_1 = 3x_2 \implies x_1 = x_2\)
Therefore, \(f\) is one-one.
2. Onto: Let \(y \in R\) be any element in the codomain. Then \(y = 3x - 7\), which gives \(x = \frac{y + 7}{3}\).
Since \(x \in R\) for every \(y \in R\), \(f\) is onto.
Since \(f\) is both one-one and onto, \(f\) is invertible.
To find the inverse, let \(y = f(x)\):
\(y = 3x - 7\)
\(x = \frac{y + 7}{3}\)
Thus, the inverse function is \(g(y) = f^{-1}(y) = \frac{y + 7}{3}\), or replacing \(y\) with \(x\): \(f^{-1}(x) = \frac{x + 7}{3}\).
In simple words: To find the inverse, set \(y = f(x)\) and solve for \(x\) in terms of \(y\).

Exam Tip: Proving invertibility requires establishing both injectivity (one-one) and surjectivity (onto), or alternatively showing the existence of an inverse function such that \(f \circ g = I\) and \(g \circ f = I\).

Question 5. Let \(f: A \to B\) and \(g: B \to C\). If \(f\) and \(g\) are onto functions, show that \(g \circ f\) is an onto function.
Answer: Let \(z\) be an arbitrary element of \(C\).
Since \(g: B \to C\) is an onto function, there exists an element \(y \in B\) such that \(g(y) = z\).
Also, since \(f: A \to B\) is an onto function, for this element \(y \in B\), there exists an element \(x \in A\) such that \(f(x) = y\).
Combining these two equations:
\((g \circ f)(x) = g(f(x)) = g(y) = z\)
Thus, for every element \(z \in C\), there exists an element \(x \in A\) such that \((g \circ f)(x) = z\).
Therefore, \(g \circ f\) is an onto function.
In simple words: If you can always trace backwards from the final set to the middle set, and from the middle set to the starting set, you can bridge the whole way from end to start.

Exam Tip: Use element-wise definitions of surjectivity to prove properties of composite functions smoothly.

Question 6. Let \(f: N \to R\) be a function defined by \(f(x) = 9x^2 + 12x + 11\), for some \(x \in N\). Show that \(f: N \to S\), where \(S\) is the range of \(f\), is invertible. Find the inverse of \(f\).
Answer: Let \(y\) be an arbitrary element of the range \(S\). Then \(y = f(x)\) for some \(x \in N\):
\(y = 9x^2 + 12x + 11\)
\(y = (3x + 2)^2 - 4 + 11\)
\(y = (3x + 2)^2 + 7\)
\((3x + 2)^2 = y - 7\)
\(3x + 2 = \sqrt{y - 7} \quad (\text{since } 3x + 2 > 0 \text{ for } x \in N)\)
\(3x = \sqrt{y - 7} - 2\)
\(x = \frac{\sqrt{y - 7} - 2}{3}\)
Since \(x\) is uniquely determined by \(y\), the function \(f\) is one-one. By definition, its codomain is its range \(S\), making it onto. Thus, \(f\) is invertible.
The inverse function is \(f^{-1}(y) = \frac{\sqrt{y - 7} - 2}{3}\), or \(f^{-1}(x) = \frac{\sqrt{x - 7} - 2}{3}\).
In simple words: Complete the square for the quadratic expression to easily solve for \(x\) in terms of \(y\).

Exam Tip: Completing the square is the standard technique for finding inverses of quadratic functions restricted to appropriate domains.

Question 7. If \(f(x) = \sqrt{x}\) (\(x \ge 0\)) and \(g(x) = x^2 - 1\) are two real functions, evaluate \(f \circ g\) and \(g \circ f\). Is \(f \circ g = g \circ f\)?
Answer: 1. Evaluation of \((f \circ g)(x)\):
\((f \circ g)(x) = f(g(x)) = f(x^2 - 1) = \sqrt{x^2 - 1}\)
2. Evaluation of \((g \circ f)(x)\):
\((g \circ f)(x) = g(f(x)) = g(\sqrt{x}) = (\sqrt{x})^2 - 1 = x - 1 \quad (\text{for } x \ge 0)\)
3. Comparison:
Clearly, \((f \circ g)(x) = \sqrt{x^2 - 1}\) and \((g \circ f)(x) = x - 1\).
Therefore, \(f \circ g \neq g \circ f\).
In simple words: Substituting the functions in different orders yields completely different algebraic formulas.

Exam Tip: Always pay attention to the domain restrictions when evaluating square root compositions.

Question 8. If \(f(x) = \frac{3x - 2}{2x - 3}\), show that \(f(f(x)) = x\) for \(x \in R - \{\frac{3}{2}\}\).
Answer: We need to evaluate \((f \circ f)(x) = f(f(x))\):
\(f(f(x)) = f\left(\frac{3x - 2}{2x - 3}\right)\)
Substitute \(\frac{3x - 2}{2x - 3}\) in place of \(x\) in the function definition:
\(f(f(x)) = \frac{3\left(\frac{3x - 2}{2x - 3}\right) - 2}{2\left(\frac{3x - 2}{2x - 3}\right) - 3}\)
Multiply numerator and denominator by \((2x - 3)\) to simplify:
\(f(f(x)) = \frac{3(3x - 2) - 2(2x - 3)}{2(3x - 2) - 3(2x - 3)}\)
Expand the numerator and denominator:
Numerator: \(9x - 6 - 4x + 6 = 5x\)
Denominator: \(6x - 4 - 6x + 9 = 5\)
\(f(f(x)) = \frac{5x}{5} = x\)
Hence proved.
In simple words: Applying a self-inverse function twice returns the original input value \(x\).

Exam Tip: This property implies that the function is its own inverse (\(f = f^{-1}\)).

Question 9. If \(f: R \to R\) such that \(f(x) = (3 - x^3)^{\frac{1}{3}}\), then show that \(f(f(x)) = x\).
Answer: We need to evaluate \((f \circ f)(x) = f(f(x))\):
\(f(f(x)) = f\left((3 - x^3)^{\frac{1}{3}}\right)\)
Substitute \((3 - x^3)^{\frac{1}{3}}\) into \(f(x)\):
\(f(f(x)) = \left(3 - \left((3 - x^3)^{\frac{1}{3}}\right)^3\right)^{\frac{1}{3}}\)
Simplify the cube term:
\(\left((3 - x^3)^{\frac{1}{3}}\right)^3 = 3 - x^3\)
Substitute this back into the expression:
\(f(f(x)) = \left(3 - (3 - x^3)\right)^{\frac{1}{3}}\)
\(f(f(x)) = \left(3 - 3 + x^3\right)^{\frac{1}{3}}\)
\(f(f(x)) = (x^3)^{\frac{1}{3}} = x\)
Hence proved.
In simple words: Cubing a cube root cancels out the exponents, leaving you back with the original variable.

Exam Tip: Be careful with exponent laws: \((a^m)^n = a^{mn}\).

Question 10. Consider \(f: \{1, 2, 3\} \to \{a, b, c\}\) given by \(f(1) = a\), \(f(2) = b\), \(f(3) = c\). Find \(f^{-1}\) and show that \((f^{-1})^{-1} = f\).
Answer: Given the function as a set of ordered pairs:
\(f = \{(1, a), (2, b), (3, c)\}\)
1. Find \(f^{-1}\):
By reversing the ordered pairs, we get:
\(f^{-1} = \{(a, 1), (b, 2), (c, 3)\}\) where \(f^{-1}: \{a, b, c\} \to \{1, 2, 3\}\).
2. Find \((f^{-1})^{-1}\):
By reversing the ordered pairs of \(f^{-1}\), we get:
\((f^{-1})^{-1} = \{(1, a), (2, b), (3, c)\} = f\).
Hence proved.
In simple words: Reversing the pairs twice brings you back to your starting set of pairs.

Exam Tip: For finite sets defined explicitly by mapping points, finding the inverse simply means swapping the coordinates of each ordered pair.