CBSE Class 12 Mathematics Relations And Functions Class Test Set 10

Question 1. Let \(f : \{2, 3, 4, 5\} \to \{3, 4, 5, 9\}\) and \(g : \{3, 4, 5, 9\} \to \{7, 11, 15\}\) 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: To find the composite function \(gof\), we need to evaluate \(g(f(x))\) for each element \(x\) in the domain of \(f\), which is \(\{2, 3, 4, 5\}\).
For \(x = 2\):
\((gof)(2) = g(f(2)) = g(3) = 7\)
For \(x = 3\):
\((gof)(3) = g(f(3)) = g(4) = 7\)
For \(x = 4\):
\((gof)(4) = g(f(4)) = g(5) = 11\)
For \(x = 5\):
\((gof)(5) = g(f(5)) = g(5) = 11\)
Thus, \(gof = \{(2, 7), (3, 7), (4, 11), (5, 11)\}\).
In simple words: We substitute the output of the function \(f\) into the function \(g\) for each input value.

Exam Tip: Always check that the range of the inner function is a subset of the domain of the outer function before finding the composite function.

Question 2. If \(f : R \to R\) is given by \(f(x) = (3 - x^3)^{\frac{1}{3}}\). Find \(fof(x)\).
Answer: We are given the function \(f(x) = (3 - x^3)^{\frac{1}{3}}\).
To find \(fof(x)\), we evaluate \(f(f(x))\):
\(fof(x) = f(f(x)) = f\left((3 - x^3)^{\frac{1}{3}}\right)\)
Substitute \(f(x)\) into the function definition in place of \(x\):
\(fof(x) = \left(3 - \left((3 - x^3)^{\frac{1}{3}}\right)^3\right)^{\frac{1}{3}}\)
Simplify the exponent inside the expression:
\(fof(x) = \left(3 - (3 - x^3)\right)^{\frac{1}{3}}\)
\(fof(x) = \left(3 - 3 + x^3\right)^{\frac{1}{3}}\)
\(fof(x) = (x^3)^{\frac{1}{3}} = x\)
In simple words: Applying the function twice brings us back to the original value \(x\).

Exam Tip: Be careful with cube roots and powers; cubing a cube root term completely eliminates the fractional exponent.

Question 3. Let \(f : N \to Y\) be a function defined by \(f(x) = 4x + 3\) where \(Y = \{y \in Y : y = 4x + 3 \text{ for } x \in N\}\). Show that \(f\) is invertible. Find the inverse.
Answer: To show that \(f\) is invertible, we must prove that \(f\) is bijective (one-one and onto).
1. One-one (Injective):
Let \(x_1, x_2 \in N\) such that \(f(x_1) = f(x_2)\).
\(4x_1 + 3 = 4x_2 + 3\)
\(4x_1 = 4x_2\)
\(x_1 = x_2\)
Therefore, \(f\) is one-one.
2. Onto (Surjective):
The co-domain of \(f\) is given as \(Y = \{y \in Y : y = 4x + 3 \text{ for } x \in N\}\), which is identical to the range of \(f\). Thus, every element in the co-domain has a pre-image in \(N\). Therefore, \(f\) is onto.
Since \(f\) is both one-one and onto, \(f\) is invertible.
3. Finding the Inverse:
Let \(y = f(x)\), so \(y = 4x + 3\).
Express \(x\) in terms of \(y\):
\(y - 3 = 4x\)
\(x = \frac{y - 3}{4}\)
Thus, the inverse function is \(f^{-1}(y) = \frac{y - 3}{4}\) or replacing \(y\) with \(x\), \(f^{-1}(x) = \frac{x - 3}{4}\).
In simple words: A function is invertible if every output comes from one unique input. We find the inverse by solving the equation backward for \(x\).

Exam Tip: Always explicitly state both one-one and onto conditions when proving invertibility unless using the theorem that every bijective function is invertible.

Question 4. If \(f(x) = \frac{4x + 3}{6x - 4}\), \(x \neq \frac{2}{3}\). Show that \(fof(x) = x\) for all \(x \neq \frac{2}{3}\). What is the inverse of \(f\)?
Answer: 1. Show that \(fof(x) = x\):
\(fof(x) = f(f(x)) = f\left(\frac{4x + 3}{6x - 4}\right)\)
Substitute the expression into \(f\) in place of \(x\):
\(fof(x) = \frac{4\left(\frac{4x + 3}{6x - 4}\right) + 3}{6\left(\frac{4x + 3}{6x - 4}\right) - 4}\)
Multiply numerator and denominator by \((6x - 4)\) to simplify:
Numerator = \(4(4x + 3) + 3(6x - 4) = 16x + 12 + 18x - 12 = 34x\)
Denominator = \(6(4x + 3) - 4(6x - 4) = 24x + 18 - 24x + 16 = 34\)
\(fof(x) = \frac{34x}{34} = x\)
Hence, \(fof(x) = x\).
2. Finding the inverse of \(f\):
Since \(fof(x) = x\), it implies that \(f\) is its own inverse (involutory function).
Therefore, \(f^{-1}(x) = f(x) = \frac{4x + 3}{6x - 4}\).
In simple words: When a composite function \(fof(x)\) simplifies to \(x\), it proves that the function is its own inverse.

Exam Tip: If \(fof(x) = x\) is proven, you can immediately conclude that \(f^{-1}(x) = f(x)\) without repeating the full algebraic rearrangement steps.