1a. Given X=2,3,4,5, Y=4,5,6,7,8, Z=10. 1b. In a class of 36 students, 19 read Biology, 16 read Chemistry and 7 read Physics. If 4 read Biology and Chemistry, 5 read Biology and Physics, 6 read Chemistry and Physics, and no student reads all three subjects. Find how many students read(i) both Biology and Chemistry (ii) Y Z (iii) Z Y.

Step 1: Solve Question 1a.
Given the sets:
$X = \{2, 3, 4, 5\}$
$Y = \{4, 5, 6, 7, 8\}$
$Z = \{7, 8, 9, 10\}$

a) (i) Find $X \cap Y$:
The intersection of $X$ and $Y$ consists of elements common to both sets.

$$X \cap Y = \{2, 3, 4, 5\} \cap \{4, 5, 6, 7, 8\}$$ $$X \cap Y = \boxed{\{4, 5\}}$$

a) (ii) Find $Y \cap Z$:
The intersection of $Y$ and $Z$ consists of elements common to both sets.

$$Y \cap Z = \{4, 5, 6, 7, 8\} \cap \{7, 8, 9, 10\}$$ $$Y \cap Z = \boxed{\{7, 8\}}$$

a) (iii) Find $Z \setminus Y$:
The set difference $Z \setminus Y$ consists of elements that are in $Z$ but not in $Y$.

$$Z \setminus Y = \{7, 8, 9, 10\} \setminus \{4, 5, 6, 7, 8\}$$ $$Z \setminus Y = \boxed{\{9, 10\}}$$

Step 2: Solve Question 1b.
Given:
Total students $N(U) = 36$
Students reading Biology $N(B) = 19$
Students reading Chemistry $N(C) = 16$
Students not reading both subjects $N((B \cup C)') = 5$

First, find the number of students who read at least one subject:

$$N(B \cup C) = N(U) - N((B \cup C)')$$ $$N(B \cup C) = 36 - 5 = 31$$

The formula for the union of two sets is $N(B \cup C) = N(B) + N(C) - N(B \cap C)$.

b) (i) Find how many students read both Biology and Chemistry:

$$31 = 19 + 16 - N(B \cap C)$$ $$31 = 35 - N(B \cap C)$$ $$N(B \cap C) = 35 - 31$$ $$N(B \cap C) = \boxed{4}$$

b) (ii) Find how many students read only Biology:
Students who read only Biology are those in Biology but not in the intersection.

$$N(\text{only B}) = N(B) - N(B \cap C)$$ $$N(\text{only B}) = 19 - 4$$ $$N(\text{only B}) = \boxed{15}$$

b) (iii) Find how many students read only Chemistry:
Students who read only Chemistry are those in Chemistry but not in the intersection.

$$N(\text{only C}) = N(C) - N(B \cap C)$$ $$N(\text{only C}) = 16 - 4$$ $$N(\text{only C}) = \boxed{12}$$

Venn Diagram Illustration:

• Draw two overlapping circles. Label one "Biology" and the other "Chemistry".
• The intersection (both B and C) contains 4 students.
• The part of the Biology circle outside the intersection (only B) contains 15 students.
• The part of the Chemistry circle outside the intersection (only C) contains 12 students.
• The area outside both circles (neither B nor C) contains 5 students.
• Total students: $15 + 4 + 12 + 5 = 36$.

Step 3: Solve Question 2 (Proof by Induction).
Prove that $1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$.

Base Case (n=1):
LHS: $1^2 = 1$
RHS: $\frac{1(1+1)(2 \cdot 1+1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = \frac{6}{6} = 1$
Since LHS = RHS, the statement is true for $n=1$.

Inductive Hypothesis:
Assume the statement is true for some positive integer $k$:

$$1^2 + 2^2 + 3^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6}$$

Inductive Step (n=k+1):
We need to prove that the statement is true for $n=k+1$:

$$1^2 + 2^2 + 3^2 + \dots + k^2 + (k+1)^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$ $$1^2 + 2^2 + 3^2 + \dots + k^2 + (k+1)^2 = \frac{(k+1)(k+2)(2k+3)}{6}$$

Start with the LHS:

$$\text{LHS} = (1^2 + 2^2 + \dots + k^2) + (k+1)^2$$

Using the inductive hypothesis:

$$\text{LHS} = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$$

Factor out $(k+1)$:

$$\text{LHS} = (k+1) \left[ \frac{k(2k+1)}{6} + (k+1) \right]$$ $$$$