110 JEE MAIN CHAPTERWISE EXPLORER
28. (*) : sin –1x = 2 sin– 1 a s-a s-b s-c
<<
Þ DD D
p p éQ sin-1 x = 2sin -1 a ù
Þ - 2 £ 2 sin– 1a £ 2 êú Þ a > b > c
êand - p £ sin -1 x £ p ú
pp ë2 2 û p
Þ - £ sin –1 a £ 30. (a) : Using tan –1q + cot– 1q = = x
44 2
p
Þ sin çèæ - p ö £ a £ sin çèæ p ö \ sin x = sin = 1
4 ÷ø 4 ÷ø 2
Þ - 1 £ a £ 1 31. (b) : tan x + sec x = 2 cos x
22
1 + sin x = 2 cos 2 x sin 150° = ½ sin 30° = ½
1 + sin x = 2(1 – sin 2 x)
1 Þ 2 sin2 x + sin x – 1 = 0
Þ |a| £ No choice is matched.
Þ (2 sin x – 1) (1 + sin x) = 0
2
29. (a) : As r 1 > r 2 > r3 1 sin 270° = – 1
Þ D > D > D Þ sin x = , sin x = –1
s - a s - b s - c
2
so there are three solution like x = 30°, 150°, 270°
Mathematical Logic 111
CHAPTER MATHEMATICAL LOGIC
18
1. Consider : (a) There is a rational number x Î S such that x £ 0.
Statement1 : (p Ù ~ q) Ù (~ p Ù q) is a fallacy. (b) There is no rational number x Î S such that x £ 0.
Statement2 : (p ® q) « (~ q ® ~ p) is a tautology. (c) Every rational number x Î S satisfies x £ 0.
(a) Statement1 is true, Statement2 is true, Statement2 is (d) x Î S and x £ 0 Þ x is not rational. (2010)
not a correct explanation for Statement1. 5. Statement1 : ~ (p « ~q) is equivalent to p « q.
(b) Statement1 is true, Statement2 is false.
(c) Statement1 is false, Statement2 is true. Statement2 : ~ (p « ~q) is a tautology.
(d) Statement1 is true, Statement2 is true, Statement2 is a (a) Statement1 is true, Statement2 is true; Statement2 is
correct explanation for Statement1. (2013) not a correct explanation for Statement1
2. The negation of the statement “If I become a teacher, then I (b) Statement 1 is true, Statement2 is false
will open a school”, is (c) Statement1 is false, Statement2 is true
(a) Neither I will become a teacher nor I will open a school. (d) Statement1 is true, Statement2 is true; Statement2 is
(b) I will not become a teacher or I will open a school. correct explanation for Statement1 (2009)
(c) I will become a teacher and I will not open a school. 6. The statement p ® (q ® p) is equivalent to
(d) Either I will not become a teacher or I will not open a (a) p ® ( p « q) (b) p ® ( p ® q)
school. (2012) (c) p ® ( p Ú q) (d) p ® ( p Ù q)
3. Consider the following statements (2008)
P : Suman is brilliant 7. Let p be the statement “x is an irrational number”, q be the
statement “y is a transcendental number”, and r be the statement
Q : Suman is rich “x is a rational number iff y is a transcendental number”.
R : Suman is hones t
The negation of the statement “Suman is brilliant and dishonest Statement1 : r is equivalent to either q or p.
Statement2 : r is equivalent to ~ ( p « ~ q).
if and only if Suman is rich” can be expressed as (a) Statement1 is true, Statement2 is false
(b) Statemen1 is false, Statement2 is true
(a) ~ Q « ~ P Ù R (b) ~ (P Ù ~ R) « Q (c) Statement1 is true, Statement2 is true; Statement2 is a
(c) ~ P Ù (Q « ~ R) (d) ~ (Q « (P Ù ~ R)) correct explanation for Statement1
(d) Statement1 is true, Statement2 is true; Statement2 is
(2011)
not a correct explanation for Statement1 (2008)
4. Let S be a nonempty subset of R. Consider the following
statement:
P : There is a rational number x Î S such that x > 0.
Which of the following statements is the negation of the
statement P ?
Answer Key
1. (a) 2. (c) 3. (d) 4. (c) 5. (b) 6. (c)
7. (a)
112 JEE MAIN CHAPTERWISE EXPLORER
1. (a) : 1 st solution : Let's prepare the truth table for the statements. 3. (d) : The statement can be written as P Ù ~R Û Q
p q ~p ~q p Ù ~q ~p Ù q (p Ù ~q) Ù (~p Ù q) Thus the negation is ~ (Q « P Ù ~ R)
T T F F F F F 4. (c) : The given statement is
F P : at least one rational x Î S such that x > 0.
T F F T T F F The negation would be : There is no rational number
F x Î S such that x > 0
F T T F F T which is equivalent to all rational numbers x Î S satisfy
x £ 0.
F F T T F F
Then Statement1 is fallacy.
p q ~ p ~ q p ® q ~ q ® p (p ® q) ® (~ q ® p) 5. (b) : Let’s prepare the truth table
T T F F T T T p q ~q p « q p « ~q ~(p « ~q)
T T F T F T
T F F T F F T T F T F T F
F T T F T T T F T F F T F
F F T T T T T F F T T F T
Then Statement2 is tautology. As the column for ~(p « ~q) and (p « q) is the same, we conclude
that ~(p « ~q) is equivalent to (p « q).
2 nd solution : ~ (~ p Ú q) Ù ~ (~ q Ú p) ~(p « ~q) is NOT a tautology because it’s statement value is not
º ~ ((~ p Ú q) Ú (~ q Ú p)) º ~ ((p ® q) Ú (q ® p)) º ~ T always true.
Thus Statement1 is true because its negation is false. 6. (c) : Let’s simplify the statement
((p ® q) ® (~ q ® ~ p) Ù ((~ q ® ~ p) ® (p ® q)) p ® (q ® p) = ~ p Ú (q ® p) = ~ p Ú (~ q Ú p)
= ((~ p Ú q) ® (q Ú ~ p) Ù ((q Ú ~ p) ® (~ p Ú q)) = – p Ú p Ú ~ q = p ® (p Ú q)
º T Ù T º T. Then Statement2 is true. 7. (a) : The given statement r º ~ p « q
The Statement1 is r 1 º (p Ù ~ q) Ú (~ p Ù q)
2. (c) : The given statement is The Statement2 is
‘‘If I become a teacher, then I will open a school’’ r2 º ~ (p « ~ q) = (p Ù q) Ú (~ q Ù ~ p)
Negation of the given statement is we can establish that r = r 1
‘‘ I will become a teacher and I will not open a school’’ Thus Statement1 is true but Statement2 is false.
(. . . ~ (p ® q) = p Ù ~ q)