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1.

In a binary tree, the number of internal nodes of degree 1 is 5, and the number of internal nodes of degree 2 is 10. The number of leaf nodes in the binary tree is(A) 10(B) 11(C) 12(D) 15

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2.

Which regular expression best describes the language accepted by the non-deterministic automaton below?(A) (a + b)* a(a + b)b(B) (abb)*(C) (a + b)* a(a + b)* b(a + b)*(D) (a + b)*

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3.

In the context-free grammar below, S is the start symbol, a and b are terminals, and ϵ denotes the empty stringS → aSa | bSb | a | b | ϵWhich of the following strings is NOT generated by the grammar?(A) aaaa(B) baba(C) abba(D) babaaabab

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4.

In the automaton below, s is the start state and t is the only final state.Consider the strings u = abbaba, v = bab, and w = aabb. Which of the following statements is true?(A) The automaton accepts u and v but not w(B) The automaton accepts each of u, v, and w(C) The automaton rejects each of u, v, and w(D) The automaton accepts u but rejects v and w

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5.

Given a boolean function f (x1, x2, …, xn), which of the following equations is NOT true(A) f (x1, x2, …, xn) = x1’f(x1, x2, …, xn) + x1f(x1, x2, …, xn)(B) f (x1, x2, …, xn) = x2f(x1, x2, …, xn) + x2’f(x1, x2, …,xn)(C) f (x1, x2, …, xn) = xn’f(x1, x2, …, 0) + xnf(x1, x2, …,1)(D) f (x1, x2, …, xn) = f(0, x2, …, xn) + f(1, x2, .., xn)

Answer» Answer: (D)
Explanation: Option A: f (x1, x2, …, xn) = x1’f(x1, x2, …, xn) + x1f(x1, x2, …, xn)
Case 1: taking x1=0
RHS = 1.f(x1, x2, …, xn) + 0.f(x1, x2, …, xn)
RHS =f(x1, x2, …, xn).

Case 2: taking x1=1
RHS = 0.f(x1, x2, …, xn) + 1.f(x1, x2, …, xn)
RHS =f(x1, x2, …, xn).
In both cases RHS=LHS, so, (A) is true

Option B: f (x1, x2, …, xn) = x2f(x1, x2, …, xn) + x2’f(x1, x2, …, xn)
Case 1: taking x2=0
RHS= 0.f(x1, x2, …, xn) + 1.f(x1, x2…,xn)
RHS =f(x1, x2, …, xn).

Case 2: taking x2=1
RHS = 1.f(x1, x2, …, xn) + 0.f(x1, x2, …, xn)
RHS =f(x1, x2, …, xn).

In both cases RHS=LHS, so, (B) is true.

Option C: f (x1, x2, …, xn) = xn’f(x1, x2, …, 0) + xnf(x1, x2, …,1)
Case 1: taking xn=0
RHS= 1.f(x1, x2, …, 0) + 0.f(x1, x2, …, 1)
RHS =f(x1, x2, …, 0)

Case 2: taking xn=1
RHS = 0.f(x1, x2, …, 0) + 1.f(x1, x2, …, 1)
RHS =f(x1, x2, …, 1)In both cases RHS=LHS, so, (C) is true.

Option D: f (x1, x2, …, xn) = f(0, x2, …, xn) + f(1, x2, .., xn)
Here, no way to equate LHS and RHS so ‘NOT true’. NO term depends on value of ‘x1’.

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6.

(A) I and II only(B) II and III only(C) I and III only(D) None of these

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