1

### JEE Main 2018 (Online) 15th April Evening Slot

Consider the following two statements :

Statement p :
The value of sin 120o can be derived by taking $\theta = {240^o}$ in the equation
2sin${\theta \over 2} = \sqrt {1 + \sin \theta } - \sqrt {1 - \sin \theta }$

Statement q :
The angles A, B, C and D of any quadrilateral ABCD satisfy the equation
cos$\left( {{1 \over 2}\left( {A + C} \right)} \right) + \cos \left( {{1 \over 2}\left( {B + D} \right)} \right) = 0$

Then the truth values of p and q are respectively :
A
F, T
B
T, F
C
T, T
D
F, F

## Explanation

Statement p :
sin 120o = cos 30o = ${{\sqrt 3 } \over 2}$ $\Rightarrow$ 2 sin 120o = $\sqrt 3$

So, $\sqrt {1 + \sin {{240}^o}} - \sqrt {1 - \sin {{240}^o}}$

$= \sqrt {{{1 - \sqrt 3 } \over 2}} - \sqrt {{{1 + \sqrt 3 } \over 2}} \ne \sqrt 3$

Statement q :
So,   A + B + C + D = 2$\pi$

$\Rightarrow$ ${{A + C} \over 2} + {{B + D} \over 2} = \pi$

$\Rightarrow$  cos$\left( {{{A + C} \over 2}} \right) + \cos \left( {{{B + D} \over 2}} \right)$

= cos $\left( {{{A + C} \over 2}} \right)$ $-$ cos$\left( {{{A + C} \over 2}} \right) = 0$

Therefore, statement p is false and statement q is true.
2

### JEE Main 2018 (Online) 16th April Morning Slot

If p $\to$ ($\sim$ p$\vee$ $\sim$ q) is false, then the truth values of p and q are respectively :
A
F, F
B
T, F
C
F, T
D
T, T

## Explanation

p q ~p ~q ~p $\vee$ ~q p $\to$ (~p $\vee$ ~q)
T T F F F F
T F F T T T
F T T F T T
F F T T T T

From the truth table,

p $\to$ (~p $\vee$ ~q) is false only when p and q both are true.
3

### JEE Main 2019 (Online) 9th January Morning Slot

If the Boolean expression
(p $\oplus$ q) $\wedge$ (~ p $\odot$ q) is equivalent
to p $\wedge$ q, where $\oplus , \odot \in \left\{ { \wedge , \vee } \right\}$, then the
ordered pair $\left( { \oplus , \odot } \right)$ is :
A
$\left( { \vee , \wedge } \right)$
B
$\left( { \vee , \vee } \right)$
C
$\left( { \wedge , \vee } \right)$
D
$\left( { \wedge , \wedge } \right)$

## Explanation Given that, (p $\oplus$ q) $\wedge$ (~ p $\odot$ q) $\equiv$ p $\wedge$ q.

From the truth table you can see this equivalence only hold when ordered pair $\left( { \oplus , \odot } \right)$ is = $\left( { \wedge , \vee } \right)$
4

### JEE Main 2019 (Online) 9th January Evening Slot

The logical statement

[ $\sim$ ( $\sim$ p $\vee$ q) $\vee$ (p $\wedge$ r)] $\wedge$ ($\sim$ q $\wedge$ r) is equivalent to :
A
( $\sim$ p $\wedge$ $\sim$ q) $\wedge$ r
B
$\sim$ p $\vee$ r
C
(p $\wedge$ r) $\wedge$ $\sim$ q
D
(p $\wedge$ $\sim$ q) $\vee$ r

## Explanation 