# CSE - Questions and solutions set-13

Computer science and engineering(CSE) - Question & solution set-13
Question:-
Option (A)
0.093
Option (B)
0.090
Option (C)
0.078
Option (D)
None of these
Correct Option:
A
Question Solution:

Root lies between the 0 and 1

f'(x) = 4x - 5

Let x0 = 0.45

f(0.45) = 2(0.45)2 - 5(0.45) + 0.45 = - 1.395

f(0.45) = 4(0.45) - 5 = -3.2

Now x1 = = = 0.45 - 0.435= 0.015

f(0.015) = 2(0.015)2 - 5(0.015) + 0.45 = 0.375

f'(0.015) = 4(0.015) - 5 = -4.94

x2 = = 0.015 + 0.075 = 0.090

f(0.090) = 2(0.090)2 - 5(0.090) + 0.45 = 0.016

f'(0.090) = 4(0.090) - 5 = -4.64

x3 = = 0.090 + 0.003 = 0.093

f(0.093) = 2(0.093)2 - 5(0.093) + 0.45 = 0.002

f'(0.093) = 4(0.093) - 5 = -4.628

x4 = = 0.093 + 0.0003= 0.0933

For the given expression root is 0.093

Question:-
Option (A)
Tautology
Option (B)
Option (C)
Option (D)
Not decidable
Correct Option:
A
Question Solution:

If we have to prove that it is a contradiction then [(p -> r) ^ (q -> r)] should be true and [(p ∨ q) -> r] should be false. Then only the given preposition becomes false. In all other cases it is true according to the definition of "->" operator. Let us take [(p -> r) ^ (q -> r)] as preposition (1) and [(p ∨ q) -> r] as preposition (2).

Preposition (2) is false only if

(p ∨ q) - T ...(3)

r - F...(4)

Preposition (1) is true only (p -> r) and (q -> r) both are true. But r is false according to the equation (4). So p and q both should be false to make (p -> r) and (q -> r). True which doesn't satisfy equation (3)? So the given preposition is a tautology.

Question:-
Option (A)
120 elements
Option (B)
60 elements
Option (C)
5 elements
Option (D)
None of these
Correct Option:
B
Question Solution:

The alternating group of degree n contains (n! /2) elements.

Here the value of n=5, so there are 5!/2 elements, means 60 elements.

Question:-
Option (A)
The Poset P = {a, b, c, d, e} with hasse diagram is a lattice
Option (B)
For the lattice {x, a, b, y} is a sub lattice
Option (C)
The lattice is is distributive
Option (D)
None of the above
Correct Option:
B
Question Solution:

(a) is not true.

In case of (c)

a ∨ (c ∨ d) = a ^ I = a

(a ^ c) ∨ (a ^ d) = 0 ∨ 0 = 0

Hence not distributive.

1

a

d

0

c