CBSE Class 12 Mathematics Relations And Functions Class Test Set 18

Question 1. If \(f: R \to R\) is defined by \(f(x) = x^2 - 3x + 2\). Find \(f(f(x))\).
Answer: Given function is \(f(x) = x^2 - 3x + 2\).
We need to find \(f(f(x))\), which means substituting \(f(x)\) into itself:
\(f(f(x)) = (f(x))^2 - 3(f(x)) + 2\)
Substitute \(f(x) = x^2 - 3x + 2\) into the expression:
\(f(f(x)) = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2\)
Now, expand the square term:
\((x^2 - 3x + 2)^2 = (x^2)^2 + (-3x)^2 + (2)^2 + 2(x^2)(-3x) + 2(-3x)(2) + 2(x^2)(2)\)
\(= x^4 + 9x^2 + 4 - 6x^3 - 12x + 4x^2\)
\(= x^4 - 6x^3 + 13x^2 - 12x + 4\)
Next, expand the linear term:
\(-3(x^2 - 3x + 2) = -3x^2 + 9x - 6\)
Combine all the terms together:
\(f(f(x)) = (x^4 - 6x^3 + 13x^2 - 12x + 4) + (-3x^2 + 9x - 6) + 2\)
Group the like powers of \(x\):
\(= x^4 - 6x^3 + (13x^2 - 3x^2) + (-12x + 9x) + (4 - 6 + 2)\)
\(= x^4 - 6x^3 + 10x^2 - 3x\)
In simple words: To find a function of a function, replace every \(x\) in the original function formula with the formula of the function itself and simplify.

Exam Tip: Be very careful with algebraic expansions and sign conventions when opening brackets for polynomials.

Question 2. Show that the function \(f: R \to R\) given by \(f(x) = x^3\) is injective. Is it surjective also?
Answer: Injectivity (One-to-One):
Let \(x_1, x_2 \in R\) such that \(f(x_1) = f(x_2)\).
Then, \(x_1^3 = x_2^3\)
Taking cube roots on both sides gives \(x_1 = x_2\).
Since \(f(x_1) = f(x_2) \implies x_1 = x_2\), the function \(f\) is injective (one-to-one).
Surjectivity (Onto):
Let \(y\) be an arbitrary element in the codomain \(R\). For \(f\) to be surjective, there must exist an \(x \in R\) such that \(f(x) = y\).
\(x^3 = y \implies x = y^{1/3}\)
Since \(R\) is the set of real numbers, for every real number \(y\), its cube root \(y^{1/3}\) is also a unique real number in the domain \(R\).
Thus, for every \(y \in R\), there exists \(x = y^{1/3} \in R\) such that \(f(x) = (y^{1/3})^3 = y\).
Therefore, \(f\) is surjective.
In simple words: A function is injective if different inputs give different outputs, and surjective if every possible output is reached. The cube function satisfies both properties on real numbers.

Exam Tip: Always state the general definitions clearly: \(f(x_1) = f(x_2) \implies x_1 = x_2\) for injective, and range equals codomain for surjective.

Question 3. Let \(*\) be the binary operation on \(N\) defined by \(a * b = \text{H.C.F of } a \text{ and } b\). Is \(*\) commutative? Is \(*\) associative?
Answer: Commutativity:
For any \(a, b \in N\):
\(a * b = \text{H.C.F of } a \text{ and } b\)
\(b * a = \text{H.C.F of } b \text{ and } a\)
Since the Highest Common Factor of two numbers is independent of the order of the numbers, \(\text{H.C.F}(a, b) = \text{H.C.F}(b, a)\).
Therefore, \(a * b = b * a\). Hence, \(*\) is commutative.
Associativity:
For any \(a, b, c \in N\):
\((a * b) * c = (\text{H.C.F}(a, b)) * c = \text{H.C.F}(\text{H.C.F}(a, b), c) = \text{H.C.F}(a, b, c)\)
Similarly, \(a * (b * c) = a * (\text{H.C.F}(b, c)) = \text{H.C.F}(a, \text{H.C.F}(b, c)) = \text{H.C.F}(a, b, c)\)
Since \((a * b) * c = a * (b * c) = \text{H.C.F}(a, b, c)\), the operation \(*\) is associative.
In simple words: The H.C.F. of numbers does not depend on the order in which you take them or group them together, so the operation is both commutative and associative.

Exam Tip: Use the property that H.C.F. of a set of numbers is unique regardless of grouping to prove associativity quickly.

Question 4. Let \(L\) be the set of all lines in \(xy\)-plane and \(R\) be the relation in \(L\) defined by \(R = \{(L_1, L_2) : L_1 \perp L_2 \text{ for } L_1, L_2 \in L\}\). Show \(R\) is an equivalence relation. Find the set of all lines related to the line \(y = 2x + 5\).
Answer: Note: The printed source defines the relation as perpendicularity (\(\perp\)), but perpendicularity is symmetric and irreflexive, making it a similarity/tolerance relation rather than an equivalence relation. However, interpreting the standard textbook problem of this type where equivalence relations apply to parallelism: let us examine the standard parallel lines relation \(R = \{(L_1, L_2) : L_1 \parallel L_2\}\).
1. Reflexivity:
Every line is parallel to itself, i.e., \(L_1 \parallel L_1\) for every \(L_1 \in L\).
Therefore, \((L_1, L_1) \in R\). Thus, \(R\) is reflexive.
2. Symmetry:
Let \((L_1, L_2) \in R\).
\(\implies L_1 \parallel L_2\)
\(\implies L_2 \parallel L_1\)
\(\implies (L_2, L_1) \in R\). Thus, \(R\) is symmetric.
3. Transitivity:
Let \((L_1, L_2) \in R\) and \((L_2, L_3) \in R\).
\(\implies L_1 \parallel L_2\) and \(L_2 \parallel L_3\)
\(\implies L_1 \parallel L_3\)
\(\implies (L_1, L_3) \in R\). Thus, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
Set of all lines related to \(y = 2x + 5\):
The given line is \(y = 2x + 5\), which has a slope \(m = 2\).
The set of all lines related to this line consists of all lines parallel to it.
All parallel lines in a plane share the exact same slope but have different y-intercepts \(c\).
Therefore, the set of all lines related to \(y = 2x + 5\) is given by:
\(\{y = 2x + c : c \in R\}\).
In simple words: Lines being parallel satisfies all conditions of an equivalence relation (reflexive, symmetric, transitive). Parallel lines always have the same slope.

Exam Tip: For equivalence relations on lines, remember that parallelism forms equivalence classes of lines with identical slopes, whereas perpendicularity does not.