Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The point ($$-$$4, 5) is the vertex of a square and one of its diagonals is 7x $$-$$ y + 8 = 0. The equation of the other diagonal is

A

7x $$-$$ y + 23 = 0

B

7y + x = 30

C

7y + x = 31

D

x $$-$$ 7y = 30

Point ($$-$$4, 5) does not lie on the diagonal 7x $$-$$ y + 8 = 0, so point will lie on the other diagonal, also diagonals are perpendicular

$$\therefore$$ Slope of other diagonal = $$-$$1/7

$$\therefore$$ Equation of other diagonal is

$$y - 5 = - {1 \over 7}(x + 4) \Rightarrow 7y + x = 31$$

2

MCQ (Single Correct Answer)

The number of points on the line x + y = 4 which are unit distance apart from the line 2x + 2y = 5 is

A

0

B

1

C

2

D

infinity

Since x + y = 4 and 2x + 2y = 5 are parallel.

Take (4, 0) on the line x + y = 4

Distance of (4, 0) from the line 2x + 2y $$-$$ 5 = 0 is

$${{|2\,.\,4 + 2\,.\,0 - 5|} \over {\sqrt {{2^2} + {2^2}} }} = {3 \over {2\sqrt 2 }} = {{3\sqrt 2 } \over 4} > 1$$ (unit distance)

$$\because$$ Both lines are parallel and at a distance greater than unity

$$\therefore$$ There is no point on the line x + y = 4.

3

MCQ (Single Correct Answer)

If C is the reflection of A(2, 4) in x-axis and B is the reflection of C in y-axis, then $$|AB|$$ is

A

20

B

2$$\sqrt5$$

C

4$$\sqrt5$$

D

4

Reflection of point (h, k) about x-axis is (h, $$-$$k)

Reflection of point (h, k) about y-axis is ($$-$$h, k)

$$\therefore$$ Reflection point C of A(2, 4) in x-axis is (2, $$-$$4) and reflection point B of C(2, $$-$$4) in y-axis is ($$-$$2, $$-$$4)

$$\therefore$$ $$|AB| = \sqrt {{{(2 + 2)}^2} + {{(4 + 4)}^2}} = 4\sqrt 5 $$

4

MCQ (Single Correct Answer)

A line through the point A(2, 0) which makes an angle of 30$$^\circ$$ with the positive direction of x-axis is rotated about A in clockwise direction through an angle 15$$^\circ$$. Then the equation of the straight line in the new position is

A

$$(2 - \sqrt 3 )x + y - 4 + 2\sqrt 3 = 0$$

B

$$(2 - \sqrt 3 )x - y - 4 + 2\sqrt 3 = 0$$

C

$$(2 - \sqrt 3 )x - y + 4 + 2\sqrt 3 = 0$$

D

$$(2 - \sqrt 3 )x + y + 4 + 2\sqrt 3 = 0$$

Equation of line which is passing through (x_{1}, y_{1}) and angle of inclination $$\theta$$ with +ve x axis in anti-clockwise is

$$y - {y_1} = \tan \theta (x - {x_1})$$

$$\therefore$$ $$(y - 0) = \tan 15^\circ (x - 2)$$

$$[\tan 15^\circ = \tan (45^\circ - 30^\circ ) = {{\tan 45^\circ - \tan 30^\circ } \over {1 + \tan 45^\circ \tan 30^\circ }} = {{\sqrt 3 - 1} \over {\sqrt 3 + 1}} = 2 - \sqrt 3 ]$$

$$ \Rightarrow (2 - \sqrt 3 )x - y - 4 + 2\sqrt 3 = 0$$

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

WB JEE 2022 (2)

WB JEE 2021 (4)

WB JEE 2020 (4)

WB JEE 2019 (4)

WB JEE 2018 (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