Given data,
Sequence F​ ₀₀​ defined as
F​ ₀₀​ (0) = 1,
F​ ₀₀​ (1) = 1,
F​ ₀₀​ (n) = ((10 ∗ F​ ₀₀​ (n – 1) + 100)/ F​ ₀₀​ (n – 2)) for n ≥ 2
Let n=2
F​ ₀₀​ (2) = (10 * F​ ₀₀​ (1) + 100) / F​ ₀₀​ (2 – 2)
= (10 * 1 + 100) / 1
= (10 + 100) / 1
= 110
Let n=3
F​ ₀₀​ (3) = (10 * F​ ₀₀​ (2) + 100) / F​ ₀₀​ (3 – 2)
= (10 * 110 + 100) / 1
= (1100 + 100) / 1
= 1200
Similarly, n=4
F​ ₀₀​ (4) = (10 * F​ ₀₀​ (3) + 1₀₀) / F​ ₀₀​ (4 – 2)
= (12100) / 110
= 110
F​ ₀₀​ (5) = (10 * F​ ₀₀​ (4) + 100) / F​ ₀₀​ (5 – 2)
= (10*110 + 100) / 1200
= 1
The sequence will be (1, 110, 1200,110, 1).

Absurd→ Clearly impossible being contrary to some evident truth.
Ambiguous→ Capable of more than one interpretation or meaning.
Axiom→ An assertion that is accepted and used without a proof.
Conjecture→ An opinion preferably based on some experience or wisdom

Step-1: Given data,
f(x) = x​ ³​ – 4x, g(x) = 1/(x^{​ 2}​ + 1) and h(x) = x​ ⁴
hog(x)=h(1/(x​ ²​ + 1))
=h(1/(x​ ²​ )+1)​ ⁴
= 1/(x​ ²​ +1)​ ⁴
= (x​ ²​ +1)​ ^-4
hogof(x)= hog(x​ ³​ -4x)
= h(1/(x​ ³​ -4x)^{​ 2}​ +1)
= h(1/(x​ ³​ -4x)^{​ 2}​ +1)​ 4
= h((x​ ³​ -4x)​ ²​ +1)^{​ -4}
So, option D id is correct answer.

Method-1:
0 and 100 → Counting sequentially:
0,6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96 Total=17
–6 and 34 → Counting sequentially: -6,0,6,12,18,24,30
Total=7
Method-2: 0 and 100 → Maximum number is 100. Divide ⌊100/6⌋ = 16+1 =17
Here, +1 because of 0.
–6 and 34 → Maximum number is 34. Divide ⌊34/6⌋ = 5+1+1 =7
Here, +1 because of 0 and +1 for -6

With the help of dirac’s theorem, we can prove above three statements.

Option-A: Let P be TRUE and R be false, then the conclusion PVR will be TRUE. Now if we make Q as false then premises (P→Q)∧(R→ S)will be false because P→ Q is false. Hence this option is not correct.
Option-B: Let P be TRUE and S be false then the conclusion PVS is TRUE. Now, if we make R as TRUE then the premises (P→ Q)∧(R→ S) will be false because (R→ S) will be false. Hence this option is not correct.
Option-C: Let Q be false, R be TRUE then conclusion QVR will be TRUE. Now if we make S as FALSE then Premises (P→ Q)∧(R→ S) will be FALSE because (R→ S) will be false. Hence, this option is not correct.
Option-D: Let Q be TRUE and S be FALSE then conclusion QVS will be TRUE. Now if we make R as TRUE then premises (P→ Q)∧(R→ S) will be FALSE because (R→ S) will be false.
Therefore None of the given options are correct.
Note: As per UGC NET key, given option D as correct answer.

→ ECL is the fastest of all logic families. High speed in ECL is possible because transistors are used in difference amplifier configuration, in which they are never driven into Saturation.
→ Emitter-coupled logic (ECL) is a high-speed integrated circuit bipolar transistor logic family. ECL uses an overdriven BJT differential amplifier with single-ended input and limited emitter current to avoid the saturated (fully on) region of operation and its slow turn-off behavior.
→ As the current is steered between two legs of an emitter-coupled pair, ECL is sometimes called current-steering logic (CSL), current-mode logic (CML) or current-switch emitter-follower (CSEF) logic.[

In a JK flip-flop, Q_n=Q(bar) iff J=K=1.
State sequence of down counter is as follows:

Step-1: Convert octal number into binary number
(2357)​ ₈​ = (010 011 101 111)_{​ 2}
Step-2: Divide 4 bits from LSB then will get hexadecimal number
0100 1110 1111
2 E F
(2EF)​ ₁₆​ = (2357)_{​ 8}

→ Header files contain definitions of Functions and Variables, which is imported or used into any C++ program by using the pre-processor #include statement. Header file have an extension ".h" which contains C++ function declaration and macro definition.
→ Header file is a library file, we can not passed to a function in C++.
→ We can pass constant,Structure and Array

→ Function overloading allows you to use the same name for different functions, to perform, either same or different functions in the same class.
→ Compiler sets up a separate function for every definition of function.
→ Two ways to use overloaded function is
1. By changing number of Arguments.
2. By having different types of argument.