Question

If p,q,r,s are statements, where: $p:A^{2}-B^{2}=(A-B)(A+B)$ for matrices $AB\ne BA$; $q:5\le5$; $r:{^{8}C_{1}+{}^{8}C_{2}+...+^{8}C_{8}=256}$; $s$: Max value of ${}^{8}C_{r}$ is 70. The statement with truth value true is:

• A. $(p\wedge\sim r)\vee(\sim q\wedge\sim s)$
• B. $(p\vee\sim q)\leftrightarrow(\sim r\rightarrow s)$
• C. $(p\leftrightarrow q)\wedge(\sim p\vee\sim q)$
• D. $(s\vee\sim p)\leftrightarrow(\sim p\wedge\sim r)$

Hint:

Carefully evaluate individual statement values before checking logical combinations.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
First, determine the individual truth values of statements $p, q, r$, and $s$. Then check the compound statements in options.

Step 3: Detailed Explanation:
- $p$: For matrices, $(A-B)(A+B) = A^2 + AB - BA - B^2$. This equals $A^2 - B^2$ only if $AB = BA$. Given $AB \neq BA$, so $p$ is False (F).
- $q$: $5 \le 5$ is True (T) because $5 = 5$.
- $r$: $\sum_{i=0}^8 {}^8C_i = 2^8 = 256$. The given sum is from $i=1$ to $8$.
So sum $= 256 - {}^8C_0 = 256 - 1 = 255 \neq 256$. $r$ is False (F).
- $s$: Max value of ${}^nC_r$ is at middle. For $n=8$, max is ${}^8C_4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$. $s$ is True (T).
Values: $p=\text{F}, q=\text{T}, r=\text{F}, s=\text{T}$.
Checking Option D:
$(s \vee \sim p) \leftrightarrow (\sim p \wedge \sim r)$ $(\text{T} \vee \text{T}) \leftrightarrow (\text{T} \wedge \text{T}) \equiv \text{T} \leftrightarrow \text{T} \equiv \text{T}$.

Step 4: Final Answer:
Option D has the truth value True.