CBSE Class 12 Mathematics Relations And Functions Assignment Set 16

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. Let \(f: A \rightarrow B\) and \(g: B \rightarrow C\) be two functions.
The composite function of \(f\) and \(g\), denoted by \(gof\), is defined as the function \(gof: A \rightarrow C\) given by \(gof(x) = g(f(x))\) for all \(x \in A\).
Similarly, if \(f: A \rightarrow B\) and \(g: B \rightarrow C\), the composite function \(fog\) is defined only when the range of \(g\) is a subset of the domain of \(f\). (In standard textbook problems where domains match appropriately, \(fog: B \rightarrow B\) or similar is defined as \(fog(x) = f(g(x))\)).
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 range compatibility before defining composite functions like \(fog\) and \(gof\).

Question 2. Let \(A = \{3, 4, 5, 6\}\), \(B = \{13, 14, 15, 16\}\), \(C = \{23, 24, 25\}\) given \(f: A \rightarrow B : f = \{(3, 13), (4, 14), (5, 15), (6, 16)\}\) \(g: B \rightarrow C : g = \{(13, 23), (14, 23), (15, 24), (16, 25)\}\)
Answer: To find \(gof\), we evaluate \(g(f(x))\) for each element \(x \in A\):
For \(x = 3\), \(g(f(3)) = g(13) = 23\)
For \(x = 4\), \(g(f(4)) = g(14) = 23\)
For \(x = 5\), \(g(f(5)) = g(15) = 24\)
For \(x = 6\), \(g(f(6)) = g(16) = 25\)
Therefore, \(gof = \{(3, 23), (4, 23), (5, 24), (6, 25)\}\).
In simple words: Follow the mapping from set A through set B to set C step by step.

Exam Tip: Write down intermediate values clearly to avoid mapping errors in set-based composition problems.

Question 3. Let \(f: R \rightarrow R\) and \(g: R \rightarrow R\) such that \(f(x) = x^2\) and \(g(x) = x + 1\). Show that \(fog \neq gof\)
Answer: First, let us evaluate \(fog(x)\):
\(fog(x) = f(g(x)) = f(x + 1) = (x + 1)^2 = x^2 + 2x + 1\)
Next, let us evaluate \(gof(x)\):
\(gof(x) = g(f(x)) = g(x^2) = x^2 + 1\)
Comparing both expressions:
\(fog(x) = x^2 + 2x + 1\) and \(gof(x) = x^2 + 1\)
Since \(x^2 + 2x + 1 \neq x^2 + 1\) for all \(x\) (specifically for \(x \neq 0\)), we have \(fog \neq gof\).
In simple words: Function composition is generally not commutative, meaning the order in which you apply functions matters.

Exam Tip: To prove two functions are not equal, show their values differ for at least one value in their domain (or show algebraically that their expressions are distinct).

Question 4. Let \(f: R \rightarrow R\) be defined by \(f(x) = 3x - 7\). Show that \(f\) is invertible. Also find its inverse function \(g = f^{-1} : R \rightarrow R\).
Answer: Step 1: Check for injectivity (one-to-one):
Let \(f(x_1) = f(x_2)\)
\(3x_1 - 7 = 3x_2 - 7\)
\(3x_1 = 3x_2 \implies x_1 = x_2\)
Thus, \(f\) is one-to-one.
Step 2: Check for surjectivity (onto):
Let \(y \in R\) (co-domain). Then \(y = 3x - 7\)
\(3x = y + 7 \implies x = \frac{y + 7}{3} \in R\)
For every \(y \in R\), there exists \(x = \frac{y + 7}{3} \in R\) such that \(f(x) = y\). Thus, \(f\) is onto.
Since \(f\) is both one-to-one and onto, \(f\) is invertible.
Step 3: Find the inverse function \(g = f^{-1}\):
\(g(y) = \frac{y + 7}{3}\), so replacing \(y\) with \(x\), we get \(f^{-1}(x) = \frac{x + 7}{3}\).
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 one-to-one and onto properties, or showing the existence of a function \(g\) such that \(gof = I_A\) and \(fog = I_B\).

Question 5. If \(f: A \rightarrow B\) and \(g: B \rightarrow C\) and \(f\) and \(g\) are onto functions, show that \(gof\) is an onto function.
Answer: Let \(z\) be an arbitrary element of \(C\).
Since \(g: B \rightarrow C\) is an onto function, for every \(z \in C\), there exists an element \(y \in B\) such that \(g(y) = z\).
Also, since \(f: A \rightarrow B\) is an onto function, for this \(y \in B\), there exists an element \(x \in A\) such that \(f(x) = y\).
Combining these relations:
\(gof(x) = g(f(x)) = g(y) = z\)
Thus, for every \(z \in C\), there exists an element \(x \in A\) such that \(gof(x) = z\).
Therefore, the composite function \(gof: A \rightarrow C\) 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 trace all the way from the final set back to the starting set.

Exam Tip: Use the definition of onto functions systematically by taking an arbitrary element in the target set and working backwards.

Question 6. Let \(f: N \rightarrow R\) be a function defined by \(f(x) = 9x^2 + 12x + 11\), for some \(x \in N\). Show that \(f: N \rightarrow S\), where \(S\) is the range of \(f\) is invertible. Find the inverse of \(f\).
Answer: Let \(y\) be an arbitrary element of range \(S\). Then \(y = f(x)\) for some \(x \in N\):
\(y = 9x^2 + 12x + 11\)
\(y = (3x)^2 + 2(3x)(2) + 4 + 7\)
\(y = (3x + 2)^2 + 7\)
\((3x + 2)^2 = y - 7\)
\(3x + 2 = \sqrt{y - 7}\) (since \(3x + 2 > 0\) for \(x \in N\))
\(3x = \sqrt{y - 7} - 2\)
\(x = \frac{\sqrt{y - 7} - 2}{3}\)
Since this expression uniquely determines \(x\) for each \(y \in S\), the function \(f\) is one-to-one (injective) and onto by definition of range \(S\). Thus, \(f\) is invertible.
The inverse function is given by \(f^{-1}(y) = \frac{\sqrt{y - 7} - 2}{3}\), or replacing \(y\) with \(x\):
\(f^{-1}(x) = \frac{\sqrt{x - 7} - 2}{3}\).
In simple words: Complete the square for the quadratic expression to express \(x\) explicitly in terms of \(y\).

Exam Tip: When restricting the codomain to the range \(S\), surjectivity is automatic; you only need to prove injectivity or directly construct the unique inverse mapping.

Question 7. If \(f(x) = \sqrt{x}\) (\(x \geq 0\)) and \(g(x) = x^2 - 1\) are two real functions. Evaluate \(fog\) and \(gof\). Is \(fog = gof\)?
Answer: Finding \(fog(x)\):
\(fog(x) = f(g(x)) = f(x^2 - 1) = \sqrt{x^2 - 1}\)
Finding \(gof(x)\):
\(gof(x) = g(f(x)) = g(\sqrt{x}) = (\sqrt{x})^2 - 1 = x - 1\)
Comparing both composite functions:
\(fog(x) = \sqrt{x^2 - 1}\) and \(gof(x) = x - 1\)
Clearly, \(fog(x) \neq gof(x)\).
In simple words: Substituting functions in different orders yields completely different algebraic expressions.

Exam Tip: Pay close attention to domain restrictions such as \(x \geq 0\) when evaluating square root functions.

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 \(f(f(x))\):
\(f(f(x)) = f\left(\frac{3x - 2}{2x - 3}\right)\)
Substitute \(\frac{3x - 2}{2x - 3}\) in place of \(x\) in the function \(f(x)\):
\(f(f(x)) = \frac{3\left(\frac{3x - 2}{2x - 3}\right) - 2}{2\left(\frac{3x - 2}{2x - 3}\right) - 3}\)
Multiply numerator and denominator by \((2x - 3)\) to simplify:
Numerator = \(3(3x - 2) - 2(2x - 3) = 9x - 6 - 4x + 6 = 5x\)
Denominator = \(2(3x - 2) - 3(2x - 3) = 6x - 4 - 6x + 9 = 5\)
\(f(f(x)) = \frac{5x}{5} = x\)
Hence proved.
In simple words: Applying a self-inverse function twice returns the original input value \(x\).

Exam Tip: Clearing complex fractions by multiplying numerator and denominator by the common denominator is a fast and reliable technique.

Question 9. If \(f: R \rightarrow R\) such that \(f(x) = (3 - x^3)^{\frac{1}{3}}\), then show that \(f(f(x)) = x\).
Answer: We need to evaluate \(f(f(x))\):
\(f(f(x)) = f\left((3 - x^3)^{\frac{1}{3}}\right)\)
Substitute \((3 - x^3)^{\frac{1}{3}}\) in place of \(x\) in \(f(x)\):
\(f(f(x)) = \left(3 - \left((3 - x^3)^{\frac{1}{3}}\right)^3\right)^{\frac{1}{3}}\)
Simplify the cube: \(\left((3 - x^3)^{\frac{1}{3}}\right)^3 = 3 - x^3\)
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: Cube roots and cubes cancel each other out when nested.

Exam Tip: Apply exponent laws carefully: \((a^m)^n = a^{mn}\).

Question 10. Consider \(f: \{1, 2, 3\} \rightarrow \{a, b, c\}\) given by \(f(1) = a\), \(f(2) = b\), \(f(3) = c\). Find \(f^{-1}\) and show that \((f^{-1})^{-1} = f\).
Answer: Given \(f = \{(1, a), (2, b), (3, c)\}\).
Since \(f\) is one-to-one and onto, its inverse \(f^{-1} : \{a, b, c\} \rightarrow \{1, 2, 3\}\) exists and is given by:
\(f^{-1} = \{(a, 1), (b, 2), (c, 3)\}\).
Now, to find \((f^{-1})^{-1}\), we invert the pairs of \(f^{-1}\):
\((f^{-1})^{-1} = \{(1, a), (2, b), (3, c)\} = f\).
Hence proved.
In simple words: Inverting a function twice brings you right back to the original function.

Exam Tip: For finite set mappings, reversing ordered pairs \((x, y)\) to \((y, x)\) directly gives the inverse function.