CBSE Class 12 Mathematics Relations And Functions Class Test Set 12

Question. Let \( f: \{2, 3, 4, 5\} \rightarrow \{3, 4, 5, 9\} \) and \( g: \{3, 4, 5, 9\} \rightarrow \{7, 11, 15\} \) be functions defined by \( f(2) = 3 \), \( f(3) = 4 \), \( f(4) = 5 \), \( f(5) = 5 \), \( g(3) = g(4) = 7 \) and \( g(5) = g(9) = 11 \). Find \( gof \).
Answer:
We are given the functions \( f \) and \( g \). We need to find \( gof \).
The domain of \( gof \) is the domain of \( f \), which is \( \{2, 3, 4, 5\} \).
We calculate \( (gof)(x) = g(f(x)) \) for each element in the domain:
\( (gof)(2) = g(f(2)) = g(3) = 7 \)
\( (gof)(3) = g(f(3)) = g(4) = 7 \)
\( (gof)(4) = g(f(4)) = g(5) = 11 \)
\( (gof)(5) = g(f(5)) = g(5) = 11 \)
Therefore, the function \( gof \) is given by:
\( gof = \{(2, 7), (3, 7), (4, 11), (5, 11)\} \)

Question. If \( f: R \rightarrow R \) is given by \( f(x) = \left(3 - x^3\right)^{\frac{1}{3}} \). Find \( fof(x) \).
Answer:
We are given: \( f(x) = \left(3 - x^3\right)^{\frac{1}{3}} \)
We need to find \( (fof)(x) = f(f(x)) \).
Substitute \( f(x) \) into the definition of \( f \):
\( f(f(x)) = \left(3 - (f(x))^3\right)^{\frac{1}{3}} \)
\( = \left(3 - \left(\left(3 - x^3\right)^{\frac{1}{3}}\right)^3\right)^{\frac{1}{3}} \)
\( = \left(3 - \left(3 - x^3\right)\right)^{\frac{1}{3}} \)
\( = \left(3 - 3 + x^3\right)^{\frac{1}{3}} \)
\( = \left(x^3\right)^{\frac{1}{3}} \)
\( = x \)
Thus, \( fof(x) = x \).

Question. Let \( f: N \rightarrow Y \) be a function defined by \( f(x) = 4x+3 \) where \( Y = \{ y \in Y : y = 4x+3 \ \forall \ x \in N \} \). Show that \( f \) is invertible. Find the inverse.
Answer:
To show that \( f \) is invertible, we must prove that \( f \) is both one-one and onto.
1. One-one:
Let \( x_1, x_2 \in N \) such that \( f(x_1) = f(x_2) \).
\( 4x_1 + 3 = 4x_2 + 3 \)
\( 4x_1 = 4x_2 \)
\( x_1 = x_2 \)
Since \( f(x_1) = f(x_2) \implies x_1 = x_2 \), the function \( f \) is one-one.

2. Onto:
Let \( y \in Y \). By definition of \( Y \), \( y = 4x + 3 \) for some \( x \in N \).
Solving for \( x \) in terms of \( y \):
\( y = 4x + 3 \)
\( 4x = y - 3 \)
\( x = \frac{y-3}{4} \)
Since \( x \in N \), for every \( y \in Y \), there exists an element \( x = \frac{y-3}{4} \in N \) such that:
\( f(x) = f\left(\frac{y-3}{4}\right) = 4\left(\frac{y-3}{4}\right) + 3 = y \)
Thus, \( f \) is onto.

Since the function \( f \) is both one-one and onto, it is invertible.

Inverse of \( f \):
The inverse function \( f^{-1}: Y \rightarrow N \) is defined as:
\( f^{-1}(y) = \frac{y-3}{4} \)

Question. If \( f(x) = \frac{4x+3}{6x-4} \), \( x \neq \frac{2}{3} \) Show that \( fof(x)=x \) \( \forall \ x \neq \frac{2}{3} \). What is the inverse of \( f \)?
Answer:
We are given \( f(x) = \frac{4x+3}{6x-4} \) for \( x \neq \frac{2}{3} \).
To show \( fof(x) = x \):
\( fof(x) = f(f(x)) = f\left(\frac{4x+3}{6x-4}\right) \)
\( = \frac{4\left(\frac{4x+3}{6x-4}\right) + 3}{6\left(\frac{4x+3}{6x-4}\right) - 4} \)
Multiplying the numerator and denominator by \( (6x - 4) \):
\( = \frac{4(4x+3) + 3(6x-4)}{6(4x+3) - 4(6x-4)} \)
\( = \frac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16} \)
\( = \frac{34x}{34} \)
\( = x \)
Hence, \( fof(x) = x \) for all \( x \neq \frac{2}{3} \).

Inverse of \( f \):
Since \( f(f(x)) = x \) for all \( x \neq \frac{2}{3} \), we have \( f \circ f = I \) (the identity map).
This implies that the function \( f \) is its own inverse.
Therefore, \( f^{-1}(x) = f(x) = \frac{4x+3}{6x-4} \).