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1

MCQ (Single Correct Answer)

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

2^{50} – 1

B

2^{50} (2^{50} $$-$$ 1)

C

2^{100} $$-$$ 1

D

2^{50} + 1

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

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

= 2^{100} $$-$$ 1 $$-$$ 1[(2^{50} $$-$$ 1)]

= 2^{100} $$-$$ 2^{50}

= 2^{50} (2^{50} $$-$$ 1)

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

= 2

= 2

= 2

2

MCQ (Single Correct Answer)

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

6^{5} $$ \times $$ (15)!

B

5^{6} $$ \times $$ 15

C

(15)! $$ \times $$ 6!

D

5! $$ \times $$ 6!

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!)

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

6.5.4.3.2 ways

For rest numbers 15! ways

Total ways = 6! (15!)

3

MCQ (Single Correct Answer)

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

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

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

MCQ (Single Correct Answer)

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

B = {($$a$$, b) $$ \in $$ R $$ \times $$ R : 4($$a$$ - 6)

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 )

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

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

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

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

JEE Main 2022 (Online) 29th July Morning Shift (1)

JEE Main 2022 (Online) 28th July Evening Shift (1)

JEE Main 2022 (Online) 28th July Morning Shift (1)

JEE Main 2022 (Online) 27th July Morning Shift (1)

JEE Main 2022 (Online) 29th June Morning Shift (1)

JEE Main 2022 (Online) 28th June Evening Shift (1)

JEE Main 2022 (Online) 25th June Evening Shift (1)

JEE Main 2019 (Online) 12th April Evening Slot (1)

JEE Main 2019 (Online) 9th April Evening Slot (1)

JEE Main 2019 (Online) 12th January Evening Slot (1)

JEE Main 2019 (Online) 12th January Morning Slot (1)

JEE Main 2019 (Online) 11th January Evening Slot (1)

JEE Main 2019 (Online) 10th January Morning Slot (1)

JEE Main 2018 (Offline) (1)

JEE Main 2015 (Offline) (1)

AIEEE 2012 (1)

Complex Numbers

Quadratic Equation and Inequalities

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Sequences and Series

Matrices and Determinants

Vector Algebra and 3D Geometry

Probability

Statistics

Mathematical Reasoning

Sets and Relations

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations