1

### JEE Main 2017 (Online) 9th April Morning Slot

Contrapositive of the statement

‘If two numbers are not equal, then their squares are not equal’, is :
A
If the squares of two numbers are equal, then the numbers are equal.
B
If the squares of two numbers are equal, then the numbers are not equal.
C
If the squares of two numbers are not equal, then the numbers are not equal.
D
If the squares of two numbers are not equal, then the numbers are equal.

## Explanation

Let,

p : two numbers are not equal

q : squares of two numbers are not equal

Contrapositive of p $\to$ q is $\sim$q $\to$ $\sim$p.

$\therefore$ $\sim$q $\to$ $\sim$p means "If the squares of two numbers are equal, then the numbers are equal".
2

### JEE Main 2018 (Offline)

The Boolean expression

$\sim \left( {p \vee q} \right) \vee \left( { \sim p \wedge q} \right)$ is equvalent to
A
${ \sim q}$
B
${ \sim p}$
C
p
D
q

## Explanation From the table you can see ~p and
~(p $\vee$ q) $\vee$ (~p $\wedge$ q) are equivalent.
3

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

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

## Explanation From the truth table you can see (p $\wedge$ ~q) $\wedge$ (p $\wedge$ r) $\to$ ~p $\vee$ q is false only when values of (p, q, r) is (T, F, T).
4

### 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.