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1

### JEE Main 2019 (Online) 12th January Morning Slot

Let S = {1, 2, 3, … , 100}. The number of non-empty subsets A of S such that the product of elements in A is even is :
A
250 – 1
B
250 (250 $$-$$ 1)
C
2100 $$-$$ 1
D
250 + 1

## Explanation

S = {1,2,3, . . . .100}

= Total non empty subsets-subsets with product of element is odd

= 2100 $$-$$ 1 $$-$$ 1[(250 $$-$$ 1)]

= 2100 $$-$$ 250

= 250 (250 $$-$$ 1)
2

### JEE Main 2019 (Online) 11th January Evening Slot

The number of functions f from {1, 2, 3, ...., 20} onto {1, 2, 3, ...., 20} such that f(k) is a multiple of 3, whenever k is a multiple of 4, is :
A
65 $$\times$$ (15)!
B
56 $$\times$$ 15
C
(15)! $$\times$$ 6!
D
5! $$\times$$ 6!

## Explanation

f(k) = 3m (3, 6, 9, 12, 15, 18)

for k = 4, 8, 12, 16, 20

6.5.4.3.2 ways

For rest numbers 15! ways

Total ways = 6! (15!)
3

### JEE Main 2019 (Online) 10th January Morning Slot

In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is
A
42
B
102
C
1
D
38

## Explanation

Let n(A) = number of students opted mathematic = 70,

n(B) = number of studens opted Physics = 46,

n(C) = number of students opted Chemistry = 28,

n(A $$\cap$$ B) = 23,

n(B $$\cap$$ C) = 9,

n(A $$\cap$$ C) = 14,

n(A $$\cap$$ B $$\cap$$ C) = 4,

Now n(A $$\cup$$ B $$\cup$$ C)

= n(A) + n(B) + n(C) $$-$$ n(A $$\cap$$ B) $$-$$ n(B $$\cap$$ C)

$$-$$ n(A $$\cap$$ C) + n(A $$\cap$$ B $$\cap$$ C)

= 70 + 46 + 28 $$-$$ 23 $$-$$ 9 $$-$$ 14 + 4 = 102

So number of students not opted for any course

Total $$-$$ n(A $$\cup$$ B $$\cup$$ C)

= 140 $$-$$ 102 = 38
4

### JEE Main 2018 (Offline)

Two sets A and B are as under :

A = {($$a$$, b) $$\in$$ R $$\times$$ R : |$$a$$ - 5| < 1 and |b - 5| < 1};

B = {($$a$$, b) $$\in$$ R $$\times$$ R : 4($$a$$ - 6)2 + 9(b - 5)2 $$\le$$ 36 };

Then
A
neither A $$\subset$$ B nor B $$\subset$$ A
B
B $$\subset$$ A
C
A $$\subset$$ B
D
A $$\cap$$ B = $$\phi$$ ( an empty set )

## Explanation

Given,

$$4{\left( {a - 6} \right)^2} + 9{\left( {b - 5} \right)^2} \le 36$$

Let $$a - 6 = x$$ and $$b - 5 = y$$

$$\therefore\,\,\,\,$$ $$4{x^2} + 9{y^2} \le 36$$

$$\Rightarrow \,\,\,\,{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$

This is a equation of ellipse.

This ellipse will look like this,

According to set A,

$$\left| {a - 5} \right| < 1$$

as $$a - 6 = x$$ then $$a - 5 = x + 1$$

$$\therefore\,\,\,$$ $$\left| {x + 1} \right| < 1$$

$$\Rightarrow \,\,\, - 1 < x + 1 < 1$$

$$\Rightarrow \,\,\, - 2 < x < 0$$

$$\left| {b - 5} \right| < 1$$

as $$b - 5 = y$$

$$\therefore\,\,\,$$ $$\left| y \right| < 1$$

$$\Rightarrow \,\,\,\, - 1 < y < 1$$

This will look like this,

By combining both those graphs it will look like this,

To check entire set A is inside of B or not put any paint of set A on the equation $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$ and if this inequality satisfies, then it is inside B.

By looking at the graph you can surely say (0, 1) or (0, $$-$$1) inside the graph. But to check point ($$-2$$, 1) or ($$-$$2, $$-$$1) is inside of the ellipse or not, put on the $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1.$$

Putting ($$-$$2, 1) on the inequality,

LHS $$= {{\left( { - 2} \right){}^2} \over 9} + {{{1^2}} \over 4}$$

$$= {4 \over 9} + {1 \over 4}$$

$$= {{25} \over {36}} < 1$$

$$\therefore\,\,\,$$ Inequality holds.

So, $$\left( { - 2,1} \right)$$ is inside the ellipse.

Similarly by checking we can see $$\left( { - 2, - 1} \right)$$ is also inside the ellipse.

Hence, we can say entire set A is inside of the set B.

$$\therefore\,\,\,\,$$ A $$\subset$$ B

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