Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

For any two sets A and B, A $$-$$ (A $$-$$ B) equals

A

B

B

A $$-$$ B

C

A $$\cap$$ B

D

A^{c} $$\cap$$ B^{c}

A $$-$$ (A $$-$$ B)

= A $$\cap$$ (A $$\cap$$ B^{c})^{c} = A $$\cap$$ (A^{c} $$\cup$$ B)

= $$\phi$$ $$\cup$$ (A $$\cap$$ B) = A $$\cap$$ B

2

MCQ (Single Correct Answer)

Let A = {1, 2, 3} and B = {2, 3, 4}, then which of the following relations is a function from A to B?

A

{(1, 2), (2, 3), (3, 4), (2, 2)}

B

{(1, 2), (2, 3), (1, 3)}

C

{(1, 2), (2, 3), (3, 3)}

D

{(1, 2), (2, 3), (3, 4)}

If a relation is function from set A to set B, then first element of ordered pairs in relation should not be repeated and second element of ordered pairs should be in set B. So Relation {(1, 3), (2, 3), (3, 3)} is a function from A = {1, 2, 3} to B = {2, 3, 4}.

3

MCQ (Single Correct Answer)

A function f : A $$\to$$ B, where A = {x/$$-$$1 $$\le$$ x $$\le$$ 1} and B = {y/1 $$\le$$ y $$\le$$ 2} is defined by the rule y = f(x) = 1 + x^{2}. Which of the following statement is then true?

A

f is injective but not surjective

B

f is surjective but not injective

C

f is both injective and surjective

D

f is neither injective nor surjective

$$\because$$ $$-$$ 1 $$\le$$ x $$\le$$ 1 $$\Rightarrow$$ 0 $$\le$$ x^{2} $$\le$$ 1

$$\Rightarrow$$ 1 $$\le$$ x^{2} + 1 $$\le$$ 2 (adding 1)

$$\Rightarrow$$ 1 $$\le$$ f(x) $$\le$$ 2 $$\Rightarrow$$ R_{f} = [1, 2] = Co-domain B

$$\therefore$$ f is surjective.

Let $$x = - {1 \over 2},{1 \over 2} \in A$$ $$\because$$ $$ - {1 \over 2} \ne {1 \over 2} \Rightarrow {\left( { - {1 \over 2}} \right)^2} = {\left( {{1 \over 2}} \right)^2}$$

$$ \Rightarrow {1 \over 4} = {1 \over 4} \Rightarrow 1 + {1 \over 4} = 1 + {1 \over 4} \Rightarrow f\left( { - {1 \over 2}} \right) = f\left( {{1 \over 2}} \right)$$

$$\therefore$$ f is not one-one

($$\because$$ $${x_1} \ne {x_2} \Rightarrow f({x_1}) = f({x_2})$$ then f is not one-one)

$$\therefore$$ f is surjective but not injective.

4

MCQ (Single Correct Answer)

The mapping f : N $$\to$$ N given by f(n) = 1 + n^{2}, n $$\in$$ N where N is the set of natural numbers, is

A

One to one and onto

B

Onto but not one-to-one

C

One-to-one but not onto

D

Neither one-to-one nor onto

Range of mapping = {f(1), f(2), f(3), ....}

= {2, 5, 10, ....} $$\ne$$ N (co-domain)

$$\therefore$$ mapping is not onto

If n = n_{1}, n_{2} $$\in$$ N such that n_{1} $$\ne$$ n_{2}

$$ \Rightarrow n_1^2 \ne n_2^2$$ ($$\because$$ n_{1} and n_{2} are +ve)

$$ \Rightarrow 1 + n_1^2 \ne 1 + n_2^2$$

$$ \Rightarrow f({n_1}) \ne f({n_2})$$

$$\therefore$$ Mapping is one-one

$$\therefore$$ Mapping is one-one and not onto.

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

WB JEE 2022 (3)

WB JEE 2021 (2)

WB JEE 2020 (3)

WB JEE 2019 (3)

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Logarithms

Sequence and Series

Quadratic Equations

Complex Numbers

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Matrices and Determinants

Vector Algebra

Three Dimensional Geometry

Probability

Statistics

Sets and Relations

Functions

Definite Integration

Application of Integration

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Differential Equations

Straight Lines and Pair of Straight Lines

Circle

Parabola

Ellipse and Hyperbola