Question

Question

Answer 1

\(\large \color{black}{\begin{align} & \{a,b\}\in \mathbb{Z} ,\ \ \ a,b >2,\ a>b\ \hspace{.33em}\\~\\
& \normalsize \text{which of the following will be always true ?} \hspace{.33em}\\~\\
& a.)\ \dfrac{a+b}{2}>\dfrac{a-b}{2} \hspace{.33em}\\~\\
& b.)\ \dfrac{a+b}{2}\geq \dfrac{a-b}{2} \hspace{.33em}\\~\\
& c.)\ a^2-b^2<a^3-b^3\hspace{.33em}\\~\\
& d.)\ \normalsize \text{Both a and c} \hspace{.33em}\\~\\
& e.)\ \normalsize \text{ a, b and c} \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

B is confusing, but I thinks it's true
The answer is E since we are dealing with positive numbers and bigger than one

Answer 3

but in the book it is given as option d.)

Answer 4

Not too good at calc but here it goes...

What I did was, was I basically picked random numbers that are greater than 2. (I did multiple trials).

Trail 1: a = 8 while b = 6
A. True
B. True
C. True
Trial 2: a = 12 while b = 4
A. True
B. True
C. True

So I would have to agree with @Ahmad-nedal that the answer is E. However, I am not in calc therefore I don't serve much of a purpose. Perhaps it is d because of the {a,b} E Z.

Answer 5

for a and b note that \(a+b\) is strictly greater than \(a-b\)

Answer 6

I understand the reason, some books deal with ineaqalities from different aspects.
Although the equality stated in part B will not be achieved, but the argument still holds for any a,b in z that satsifies the constraints given. So I insist on my opinion

Answer 7

for c you can divide both sides of inequality by \(a-b\) which is a positive number and get\[a+b<a^2+ab+b^2\]which is true clearly

Answer 8

is the answer option e.) ?

Answer 9

Yes

Answer 10

it's d

Answer 11

ok thnx

Answer 12

It is given that a and b are integers greater than 2 and that a is always greater than b.
You can choose a number.
Let as say a=4 and b=3
For a)= \[\frac{ 4+3 }{ 2 }>\frac{ 4-3 }{ 2 }=\frac{ 7 }{ 2 }>\frac{ 1 }{ 2 }\]
Similarly at c) \[4^{2}-3^{2}<4^{3}-3^{3}=7<37\]
Therefore both a and c are true.