a. For any non empty subsets A, B, C of universal set U, prove that: A (B C) = (A B) (A C) OR Define absolute value of a real number. Also prove that x y x + y , x, y R b. Sketch the graph of f(x) =…

a. Proof of $A-(B\cup C)=(A-B)\cap (A-C)$

Let $U$ be the universal set, and $A,B,C$ be non-empty subsets of $U$.

Step 1: Prove $A-(B\cup C)\subseteq (A-B)\cap (A-C)$

Let $x\in A-(B\cup C)$. This means:

• $x\in A$
• $x\notin B\cup C$ (i.e., $x\notin B$ and $x\notin C$)

Since $x\notin B$, $x\in A-B$. Since $x\notin C$, $x\in A-C$.

Thus, $x\in (A-B)\cap (A-C)$. Hence, $A-(B\cup C)\subseteq (A-B)\cap (A-C)$.

Step 2: Prove $(A-B)\cap (A-C)\subseteq A-(B\cup C)$

Let $x\in (A-B)\cap (A-C)$. This means:

• $x\in A-B$ (i.e., $x\in A$ and $x\notin B$)
• $x\in A-C$ (i.e., $x\in A$ and $x\notin C$)

Thus, $x\in A$ and $x\notin B\cup C$. Therefore, $x\in A-(B\cup C)$.

Hence, $(A-B)\cap (A-C)\subseteq A-(B\cup C)$.

Conclusion:

Since both subsets are contained in each other, we have: $$A-(B\cup C)=(A-B)\cap (A-C)$$

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b. Graph of $f(x)=(x-1{)}^3$

The function $f(x)=(x-1{)}^3$ is a cubic function shifted horizontally by 1 unit to the right.

Characteristics:

• Root: $f(x)=0$ when $x-1=0$, i.e., $x=1$.
• Y-intercept: When $x=0$, $f(0)=(0-1{)}^3=-1$.
• Behavior:
  • As $x\to -\infty$, $f(x)\to -\infty$.
  • As $x\to +\infty$, $f(x)\to +\infty$.
• Inflection point: The second derivative $f^{\prime \prime }(x)=6(x-1)$ is zero at $x=1$, indicating an inflection point at $(1,0)$.
• Monotonicity: The function is strictly increasing everywhere since its derivative $f^{\prime }(x)=3(x-1{)}^2\ge 0$ and is never zero except at $x=1$.

Graph:

The graph passes through the points $(-2,-27)$, $(0,-1)$, $(1,0)$, $(2,1)$, and $(3,8)$. The curve is smooth and continuous, crossing the x-axis at $x=1$. The inflection point is at $(1,0)$, where the concavity changes.