CBSE Class 12 Mathematics Relations And Functions Assignment Set 17

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 \(f: A \to B\) and \(g: B \to C\) be two functions.
The composite function of \(f\) and \(g\), denoted by \(gof\), is defined as the function \(gof: A \to C\) given by \((gof)(x) = g(f(x))\) for all \(x \in A\).
Similarly, if \(f: A \to B\) and \(g: B \to C\), the composition \(fog\) is generally defined when \(f: B \to C\) and \(g: A \to B\), mapping from \(B\) to \(A\). More generally, for \(f: X \to Y\) and \(g: Y \to Z\), \(gof: X \to Z\) is defined.
In simple words: A composite function chains two functions together so the output of the first function becomes the input of the second function.

Exam Tip: Always check the domains and codomains before defining composition; 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 \(f: A \to B : f = \{(3, 13), (4, 14), (5, 15), (6, 16)\}\) and \(g: B \to C : g = \{(13, 23), (14, 23), (15, 24), (16, 25)\}\). Find \(gof\).
Answer: To find \((gof)(x) = g(f(x))\) for each element \(x \in A\):
For \(x = 3\), \(f(3) = 13\) and \(g(13) = 23\), so \((gof)(3) = 23\)
For \(x = 4\), \(f(4) = 14\) and \(g(14) = 23\), so \((gof)(4) = 23\)
For \(x = 5\), \(f(5) = 15\) and \(g(15) = 24\), so \((gof)(5) = 24\)
For \(x = 6\), \(f(6) = 16\) and \(g(16) = 25\), so \((gof)(6) = 25\)
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 for each individual element.

Exam Tip: Write out the step-by-step intermediate mappings clearly to avoid missing any ordered pairs.

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 \(fog \neq gof\).
Answer: First, compute \((fog)(x)\):
\((fog)(x) = f(g(x)) = f(x + 1) = (x + 1)^2 = x^2 + 2x + 1\)
Next, compute \((gof)(x)\):
\((gof)(x) = g(f(x)) = g(x^2) = x^2 + 1\)
Comparing the two results, \(x^2 + 2x + 1 \neq x^2 + 1\) for \(x \neq 0\).
Therefore, \(fog \neq gof\).
In simple words: Changing the order in which functions are applied generally changes the final outcome.

Exam Tip: To show two functions are not equal, it is sufficient to evaluate them for a specific value of \(x\) (e.g., \(x = 1\)) and show the outputs differ.

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 (injective) and onto (surjective).
1. One-one: Let \(f(x_1) = f(x_2)\)
\(3x_1 - 7 = 3x_2 - 7\)
\(3x_1 = 3x_2 \implies x_1 = x_2\). Thus, \(f\) is one-one.
2. Onto: Let \(y \in R\) (codomain). Then \(y = 3x - 7 \implies 3x = y + 7 \implies x = \frac{y + 7}{3} \in R\).
For every \(y \in R\), there exists \(x = \frac{y + 7}{3} \in R\) such that \(f(x) = 3\left(\frac{y + 7}{3}\right) - 7 = y\). Thus, \(f\) is onto.
Since \(f\) is both one-one and onto, it is invertible.
To find the inverse function \(g = f^{-1}\), set \(y = f(x)\) and solve for \(x\):
\(g(y) = \frac{y + 7}{3}\), so \(f^{-1}(x) = \frac{x + 7}{3}\).
In simple words: A function has an inverse if every output comes from a unique input and every possible output is reached.

Exam Tip: Always verify invertibility by proving bijective nature before explicitly deriving the inverse formula.

Question 5. If \(f: A \to B\) and \(g: B \to C\) and \(f\) and \(g\) are onto functions, show that \(gof\) is an onto function.
Answer: Let \(z\) be an arbitrary element of set \(C\).
Since \(g: B \to C\) is an onto function, for \(z \in C\), there exists an element \(y \in B\) such that \(g(y) = z\).
Also, since \(f: A \to B\) 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, \(gof: A \to C\) 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 surjectivity systematically from the outer codomain back to the inner domain.

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 \(y \in S\) (where \(S\) is the range of \(f\)). Then, \(y = 9x^2 + 12x + 11\) for some \(x \in N\).
Completing the square for the quadratic expression:
\(y = (3x)^2 + 2(3x)(2) + 4 + 7\)
\(y = (3x + 2)^2 + 7\)
\((3x + 2)^2 = y - 7\)
\(3x + 2 = \sqrt{y - 7}\) (since \(3x + 2 > 0\) for \(x \in N\))
\(3x = \sqrt{y - 7} - 2\)
\(x = \frac{\sqrt{y - 7} - 2}{3}\)
This shows that for every \(y \in S\), there is a unique \(x \in N\). Thus, \(f\) is one-one and onto (by definition of codomain \(S\)), hence 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: Express the input variable \(x\) in terms of the output variable \(y\) and verify uniqueness.

Exam Tip: Completing the square is the standard technique for inverting quadratic functions restricted to natural or positive domains.

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: 1. Evaluate \((fog)(x)\):
\((fog)(x) = f(g(x)) = f(x^2 - 1) = \sqrt{x^2 - 1}\)
2. Evaluate \((gof)(x)\):
\((gof)(x) = g(f(x)) = g(\sqrt{x}) = (\sqrt{x})^2 - 1 = x - 1\) (for \(x \geq 0\))
3. Comparing \(fog\) and \(gof\):
Clearly, \(\sqrt{x^2 - 1} \neq x - 1\). Therefore, \(fog \neq gof\).
In simple words: Substitute one function into the other in both directions and compare the resulting algebraic expressions.

Exam Tip: Pay close attention to domain restrictions such as \(x \geq 0\) and \(x^2 - 1 \geq 0\) when dealing with square root functions.

Question 8. If \(f(x) = \frac{3x - 2}{2x - 3}\), show \(f(f(x)) = x\) for \(x \in R - \{\frac{3}{2}\}\).
Answer: Compute \((fof)(x) = f(f(x))\):
\(f(f(x)) = \frac{3\left(\frac{3x - 2}{2x - 3}\right) - 2}{2\left(\frac{3x - 2}{2x - 3}\right) - 3}\)
Multiply numerator and denominator by \((2x - 3)\) to simplify:
Numerator = \(3(3x - 2) - 2(2x - 3) = 9x - 6 - 4x + 6 = 5x\)
Denominator = \(2(3x - 2) - 3(2x - 3) = 6x - 4 - 6x + 9 = 5\)
\(f(f(x)) = \frac{5x}{5} = x\).
In simple words: Applying this specific fractional function twice brings you right back to your original input value \(x\).

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

Question 9. If \(f: R \to R\) such that \(f(x) = (3 - x^3)^{\frac{1}{3}}\), then show \(fof(x) = x\).
Answer: Compute \((fof)(x) = f(f(x))\):
\((fof)(x) = f\left((3 - x^3)^{\frac{1}{3}}\right)\)
Substitute \(f(x)\) into the function definition:
\((fof)(x) = \left(3 - \left((3 - x^3)^{\frac{1}{3}}\right)^3\right)^{\frac{1}{3}}\)
Simplify the cube term:
\(\left((3 - x^3)^{\frac{1}{3}}\right)^3 = 3 - x^3\)
Substitute back into the expression:
\((fof)(x) = \left(3 - (3 - x^3)\right)^{\frac{1}{3}} = (3 - 3 + x^3)^{\frac{1}{3}} = (x^3)^{\frac{1}{3}} = x\).
In simple words: The cube root and the cube power cancel each other out when composed.

Exam Tip: Be careful with signs when distributing negative signs inside brackets during simplification.

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 \(f = \{(1, a), (2, b), (3, c)\}\).
1. Find \(f^{-1}\):
By reversing the ordered pairs, \(f^{-1} = \{(a, 1), (b, 2), (c, 3)\}\) where \(f^{-1}: \{a, b, c\} \to \{1, 2, 3\}\).
2. Find \((f^{-1})^{-1}\):
Reversing the ordered pairs of \(f^{-1}\) again:
\((f^{-1})^{-1} = \{(1, a), (2, b), (3, c)\} = f\).
In simple words: Inverting a function twice returns the original function mapping.

Exam Tip: For finite set mappings, finding inverses simply requires swapping the coordinates of each ordered pair.