Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

Let $$PS$$ be the median of the triangle with vertices $$P(2, 2)$$, $$Q(6, -1)$$ and $$R(7, 3)$$. The equation of the line passing through $$(1, -1)$$ band parallel to PS is:

A

$$4x + 7y + 3 = 0$$

B

$$2x - 9y - 11 = 0$$

C

$$4x - 7y - 11 = 0$$

D

$$2x + 9y + 7 = 0$$

Let $$P,Q,R,$$ be the vertices of $$\Delta PQR$$

Since $$PS$$ is the median, $$S$$ is mid-point of $$QR$$

So, $$S = \left( {{{7 + 6} \over 2},{{3 - 1} \over 2}} \right) = \left( {{{13} \over 2},1} \right)$$

Now, slope of $$PS$$ $$ = {{2 - 1} \over {2 - {{13} \over 2}}} = - {2 \over 9}$$

Since, required line is parallel to $$PS$$ therefore slope of required line $$=$$ slope of $$PS$$

Now, equation of line passing through $$(1, -1)$$ and having slope $$ - {2 \over 9}$$ is

$$y - \left( { - 1} \right) = - {2 \over 9}\left( {x - 1} \right)$$

$$9y + 9 = - 2x + 2$$

$$ \Rightarrow 2x + 9y + 7 = 0$$

Since $$PS$$ is the median, $$S$$ is mid-point of $$QR$$

So, $$S = \left( {{{7 + 6} \over 2},{{3 - 1} \over 2}} \right) = \left( {{{13} \over 2},1} \right)$$

Now, slope of $$PS$$ $$ = {{2 - 1} \over {2 - {{13} \over 2}}} = - {2 \over 9}$$

Since, required line is parallel to $$PS$$ therefore slope of required line $$=$$ slope of $$PS$$

Now, equation of line passing through $$(1, -1)$$ and having slope $$ - {2 \over 9}$$ is

$$y - \left( { - 1} \right) = - {2 \over 9}\left( {x - 1} \right)$$

$$9y + 9 = - 2x + 2$$

$$ \Rightarrow 2x + 9y + 7 = 0$$

2

MCQ (Single Correct Answer)

The $$x$$-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as $$(0, 1) (1, 1)$$ and $$(1, 0)$$ is :

A

$$2 + \sqrt 2 $$

B

$$2 - \sqrt 2 $$

C

$$1 + \sqrt 2 $$

D

$$1 - \sqrt 2 $$

From the figure, we have

$$a = 2,b = 2\sqrt 2 ,c = 2$$

$${x_1} = 0,\,{x^2} = 0,\,{x_3} = 2$$

Now, $$x$$-co-ordinate of incenter is given as

$${{a{x_1} + b{x_2} + c{x_3}} \over {a + b + c}}$$

$$ \Rightarrow x$$-coordinate of incentre

$$ = {{2 \times 0 + 2\sqrt 2 .0 + 2.2} \over {2 + 2 + 2\sqrt 2 }}$$

$$=$$ $${2 \over {2 + \sqrt 2 }} = 2 - \sqrt 2 $$

$$a = 2,b = 2\sqrt 2 ,c = 2$$

$${x_1} = 0,\,{x^2} = 0,\,{x_3} = 2$$

Now, $$x$$-co-ordinate of incenter is given as

$${{a{x_1} + b{x_2} + c{x_3}} \over {a + b + c}}$$

$$ \Rightarrow x$$-coordinate of incentre

$$ = {{2 \times 0 + 2\sqrt 2 .0 + 2.2} \over {2 + 2 + 2\sqrt 2 }}$$

$$=$$ $${2 \over {2 + \sqrt 2 }} = 2 - \sqrt 2 $$

3

MCQ (Single Correct Answer)

A ray of light along $$x + \sqrt 3 y = \sqrt 3 $$ gets reflected upon reaching $$X$$-axis, the equation of the reflected ray is

A

$$y = x + \sqrt 3 $$

B

$$\sqrt 3 y = x - \sqrt 3 $$

C

$$y = \sqrt 3 x - \sqrt 3 $$

D

$$\sqrt 3 y = x - 1$$

$$x + \sqrt 3 y = \sqrt 3 $$ or $$y = - {1 \over {\sqrt 3 }}x + 1$$

Let $$\theta$$ be the angle which the line makes with the positive x-axis.

$$\therefore$$ $$\tan \theta = - {1 \over {\sqrt 3 }} = \tan \left( {\pi - {\pi \over 6}} \right)$$ or $$\theta = \pi - {\pi \over 6}$$

$$\therefore$$ $$\angle ABC = {\pi \over 6}$$; $$\therefore$$ $$\angle DBE = {\pi \over 6}$$

$$\therefore$$ the equation of the line BD is,

$$y = \tan {\pi \over 6}x + c$$ or $$y = {x \over {\sqrt 3 }} + c$$ ..... (1)

The line $$x + \sqrt 3 y = \sqrt 3 $$ intersects the x-axis at $$B(\sqrt 3 ,0)$$ and, the line (1) passes through $$B(\sqrt 3 ,0)$$.

$$\therefore$$ $$0 = {{\sqrt 3 } \over {\sqrt 3 }} + c$$ or, c = $$-$$1

Hence, the equation of the reflected ray is,

$$y = {x \over {\sqrt 3 }} - 1$$ or $$y\sqrt 3 = x - \sqrt 3 $$

4

MCQ (Single Correct Answer)

If the line $$2x + y = k$$ passes through the point which divides the line segment joining the points $$(1, 1)$$ and $$(2, 4)$$ in the ratio $$3 : 2$$, then $$k$$ equals :

A

$${{29 \over 5}}$$

B

$$5$$

C

$$6$$

D

$${{11 \over 5}}$$

The point which divides the line segment joining the points (1, 1) and (2, 4) in the ratio 3 : 2 is

$$ = \left( {{{3 \times 2 + 2 \times 1} \over {3 + 2}},{{3 \times 4 + 2 \times 1} \over {3 + 2}}} \right)$$

$$ = \left( {{{6 + 2} \over 5},{{12 + 2} \over 5}} \right) = \left( {{8 \over 5},{{14} \over 5}} \right)$$

Since the line 2x + y = k passes through this point,

$$\therefore$$ $$2 \times {8 \over 5} + {{14} \over 5} = k$$ or $${{30} \over 5} = k$$ or, k = 6

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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

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