Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

If $$\left( {a,{a^2}} \right)$$ falls inside the angle made by the lines $$y = {x \over 2},$$ $$x > 0$$ and $$y = 3x,$$ $$x > 0,$$ then a belong to

A

$$\left( {0,{1 \over 2}} \right)$$

B

$$\left( {3,\infty } \right)$$

C

$$\left( {{1 \over 2},3} \right)$$

D

$$\left( {-3,-{1 \over 2}} \right)$$

Clearly for point $$P,$$

$${a^2} - 3a < 0$$**and** $${a^2} - {a \over 2} > 0 \Rightarrow {1 \over 2} < a < 3$$

$${a^2} - 3a < 0$$

2

MCQ (Single Correct Answer)

A straight line through the point $$A (3, 4)$$ is such that its intercept between the axes is bisected at $$A$$. Its equation is

A

$$x + y = 7$$

B

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

C

$$4x + 3y = 24$$

D

$$3x + 4y = 25$$

As is the mid point of $$PQ,$$ therefore

$${{a + 0} \over 2} = 3,{{0 + b} \over 2} = 4 \Rightarrow a = 6,b = 8$$

$$\therefore$$ Equation of line is $${x \over 6} + {y \over 8} = 1$$

or $$4x + 3y = 24$$

3

MCQ (Single Correct Answer)

If a vertex of a triangle is $$(1, 1)$$ and the mid points of two sides through this vertex are $$(-1, 2)$$ and $$(3, 2)$$ then the centroid of the triangle is

A

$$\left( { - 1,{7 \over 3}} \right)$$

B

$$\left( {{{ - 1} \over 3},{7 \over 3}} \right)$$

C

$$\left( { 1,{7 \over 3}} \right)$$

D

$$\left( {{{ 1} \over 3},{7 \over 3}} \right)$$

Vertex of triangle is $$\left( {1,\,1} \right)$$ and midpoint of sides through -

this vertex is $$\left( { - 1,\,2} \right)$$ and $$\left( {3,2} \right)$$

$$ \Rightarrow $$ vertex $$B$$ and $$C$$ come out to be $$\left( { - 3,3} \right)$$ and $$\left( {5,3} \right)$$

$$\therefore$$ centroid is $${{1 - 3 + 5} \over 3},{{1 + 3 + 5} \over 3} \Rightarrow \left( {1,{7 \over 3}} \right)$$

this vertex is $$\left( { - 1,\,2} \right)$$ and $$\left( {3,2} \right)$$

$$ \Rightarrow $$ vertex $$B$$ and $$C$$ come out to be $$\left( { - 3,3} \right)$$ and $$\left( {5,3} \right)$$

$$\therefore$$ centroid is $${{1 - 3 + 5} \over 3},{{1 + 3 + 5} \over 3} \Rightarrow \left( {1,{7 \over 3}} \right)$$

4

MCQ (Single Correct Answer)

The line parallel to the $$x$$ - axis and passing through the intersection of the lines $$ax + 2by + 3b = 0$$ and $$bx - 2ay - 3a = 0,$$ where $$(a, b)$$ $$ \ne $$ $$(0, 0)$$ is

A

below the $$x$$ - axis at a distance of $${3 \over 2}$$ from it

B

below the $$x$$ - axis at a distance of $${2 \over 3}$$ from it

C

above the $$x$$ - axis at a distance of $${3 \over 2}$$ from it

D

above the $$x$$ - axis at a distance of $${2 \over 3}$$ from it

The line passing through the intersection of lines

$$ax + 2by = 3b = 0$$

and $$bx - 2ay - 3a = 0$$ is

$$ax + 2by + 3b + \lambda \left( {bx - 2ay - 3a} \right) = 0$$

$$ \Rightarrow \left( {a + b\lambda } \right)x + \left( {2b - 2a\lambda } \right)y + 3b - 3\lambda a = 0$$

As this line is parallel to $$x$$-axis.

$$\therefore$$ $$a + b\lambda = 0 \Rightarrow \lambda = - a/b$$

$$ \Rightarrow ax + 2by + 3b - {a \over b}\left( {bx - 2ay - 3a} \right) = 0$$

$$ \Rightarrow ax + 2by + 3b - ax + {{2{a^2}} \over b}y + {{3{a^2}} \over b} = 0$$

$$y\left( {2b + {{2{a^2}} \over b}} \right) + 3b + {{3{a^2}} \over b} = 0$$

$$y\left( {{{2{b^2} + 2{a^2}} \over b}} \right) = - \left( {{{3{b^2} + 3{a^2}} \over b}} \right)$$

$$y = {{ - 3\left( {{a^2} + {b^2}} \right)} \over {2\left( {{b^2} + {a^2}} \right)}} = {{ - 3} \over 2}$$

So it is $$3/2$$ units below $$x$$-axis.

$$ax + 2by = 3b = 0$$

and $$bx - 2ay - 3a = 0$$ is

$$ax + 2by + 3b + \lambda \left( {bx - 2ay - 3a} \right) = 0$$

$$ \Rightarrow \left( {a + b\lambda } \right)x + \left( {2b - 2a\lambda } \right)y + 3b - 3\lambda a = 0$$

As this line is parallel to $$x$$-axis.

$$\therefore$$ $$a + b\lambda = 0 \Rightarrow \lambda = - a/b$$

$$ \Rightarrow ax + 2by + 3b - {a \over b}\left( {bx - 2ay - 3a} \right) = 0$$

$$ \Rightarrow ax + 2by + 3b - ax + {{2{a^2}} \over b}y + {{3{a^2}} \over b} = 0$$

$$y\left( {2b + {{2{a^2}} \over b}} \right) + 3b + {{3{a^2}} \over b} = 0$$

$$y\left( {{{2{b^2} + 2{a^2}} \over b}} \right) = - \left( {{{3{b^2} + 3{a^2}} \over b}} \right)$$

$$y = {{ - 3\left( {{a^2} + {b^2}} \right)} \over {2\left( {{b^2} + {a^2}} \right)}} = {{ - 3} \over 2}$$

So it is $$3/2$$ units below $$x$$-axis.

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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