Solution

Using perceptron learning rule: Δw = η (t − y) x, where y = 1 if y_in > 0 else 0.

S1 (0,0,t=0): y_in=0 ⇒ y=0 ⇒ no change ⇒ (w1,w2)=(0,0)

S2 (0,1,t=1): y_in=0 ⇒ y=0 ⇒ Δw=0.1(1−0)(0,1) ⇒ (w1,w2)=(0,0.1)

S3 (1,0,t=1): y_in=0 ⇒ y=0 ⇒ Δw=0.1(1−0)(1,0) ⇒ (w1,w2)=(0.1,0.1)

S4 (1,1,t=1): y_in=0.1+0.1=0.2>0 ⇒ y=1 ⇒ no change

Final Weights: w1=0.1, w2=0.1

Answer: (1)

Solution

Solution

Explanation:

• A → IV → Intersection → $\min(\mu_A(x), \mu_B(x))$
• B → III → Bounded Sum → $\min(1, \mu_A(x) + \mu_B(x))$
• C → II → Bounded Difference → $\max(0, \mu_A(x) - \mu_B(x))$
• D → I → Algebraic Sum → $\mu_A(x) + \mu_B(x) - \mu_A(x)\mu_B(x)$

Solution

Solution

Solution

Solution

Solution

For a fuzzy set $A$, the $\lambda$-cut is defined as: $$A_\lambda = \{x \mid \mu_A(x) \ge \lambda\}$$

• Union Property: $(A \cup B)_\lambda = A_\lambda \cup B_\lambda$ ✅
• Intersection Property: $(A \cap B)_\lambda = A_\lambda \cap B_\lambda$ ✅
• Monotonicity: If $\lambda \le \beta$, then $A_\beta \subseteq A_\lambda$ ✅
• Complement: $\mu_{\overline{A}}(x) = 1 - \mu_A(x)$ ⇒ $(\overline{A})_\lambda = \{x \mid 1 - \mu_A(x) \ge \lambda\} = \{x \mid \mu_A(x) \le 1 - \lambda\}$ ❌

Hence, $(\overline{A})_\lambda$ is not equal to $A_\lambda$, even when $\lambda = 1$. Therefore, the statement “except when $\lambda = 1$” is incorrect.

✅ Final Answer: Option 3 is not valid.

Solution

Solution