- Option : A
- Explanation :

No. of host = 1500

No. of host bits = [log_{2}1500] = 11 bits

∴ Total possible hosts = 2^{11}= 2^{3}× 2^{8}

n is the netmask bits,

Range of addresses is = 2^{32 – n}

2^{11}= 2^{32-n}

Available IP address ⇒ 202.61.0.0/17

So the IP address follow the pattern

0.0 to 7.255

8.0 to 15.255

16.0 to 23.255

:

64.0 to ...

:

104.0 to ...

∴ The possible IP addresses are 202.61.64.0/21 & 202.61.104.0/21

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- Option : C
- Explanation :

L_{1}L(m) = ϕ ⇒ emptiness problem of TM.

TM is undecidable under emptiness.

L_{2}= where a TM visits a particular state in finite steps in decidable, as we can do this with UTM.

L_{3}= L(m) is non-Recursive,

Clearly from the diagram

L(A) ⇒ non recursive language accepted by TM

L(B) ⇒ non-recursive language not accepted by TM.

∴ it is a non-trivial property, hence undecidable.

L_{4}= Undecidable problem using rice-theorem.

Hence, L_{1}, L_{3}, and L_{4}are undecidable.

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- Option : B
- Explanation :

In non-persistent HTTP, each packet takes 2 RTT (Round trip Time): one for TCP connection, one or HTTP Text (Image file As, it is given text and 5 images that totals 6 objects.)

So, it takes 12 RTT in total. But,

12 RTT includes 6 HTTP connections + 6TCP connections.

So, the minimum number of TCP connections required is 6.

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- Option : C
- Explanation :

According to Lagrange’s theorem, state that for any finite group G, the order (number of element) of every subgroup t_{1}of G divides the order of G. therefore, possible subgroup of group of 35 elements.

{1, 5, 7, 35}

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