Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If the sum of the slopes of the lines given by $${x^2} - 2cxy - 7{y^2} = 0$$ is four times their product $$c$$ has the value

A

$$-2$$

B

$$-1$$

C

$$2$$

D

$$1$$

Let the lines be $$y = {m_1}x$$ and $$y = {m_2}x$$ then

$${m_1} + {m_2} = - {{2c} \over 7}$$ and $${m_1}{m_2} = - {1 \over 7}$$

Given $${m_1} + {m_2} = 4m{}_1{m_2}$$

$$ \Rightarrow {{2c} \over 7} = - {4 \over 7} \Rightarrow c = 2$$

$${m_1} + {m_2} = - {{2c} \over 7}$$ and $${m_1}{m_2} = - {1 \over 7}$$

Given $${m_1} + {m_2} = 4m{}_1{m_2}$$

$$ \Rightarrow {{2c} \over 7} = - {4 \over 7} \Rightarrow c = 2$$

2

MCQ (Single Correct Answer)

Let $$A\left( {2, - 3} \right)$$ and $$B\left( {-2, 1} \right)$$ be vertices of a triangle $$ABC$$. If the centroid of this triangle moves on the line $$2x + 3y = 1$$, then the locus of the vertex $$C$$ is the line

A

$$3x - 2y = 3$$

B

$$2x - 3y = 7$$

C

$$3x + 2y = 5$$

D

$$2x + 3y = 9$$

Let the vertex $$C$$ be $$(h,k),$$ then the

centroid of $$\Delta ABC$$ is $$\left( {{{2 + (- 2) + h} \over 3},{{ - 3 + 1 + k} \over 3}} \right)$$

or $$\left( {{h \over 3},{{ - 2 + k} \over 3}} \right).$$ It lies on $$2x+3y=1$$

$$ \Rightarrow {{2h} \over 3} - 2 + k = 1$$

$$ \Rightarrow 2h + 3k = 9$$

$$ \therefore $$ Locus of $$C$$ is $$2x+3y=9$$

centroid of $$\Delta ABC$$ is $$\left( {{{2 + (- 2) + h} \over 3},{{ - 3 + 1 + k} \over 3}} \right)$$

or $$\left( {{h \over 3},{{ - 2 + k} \over 3}} \right).$$ It lies on $$2x+3y=1$$

$$ \Rightarrow {{2h} \over 3} - 2 + k = 1$$

$$ \Rightarrow 2h + 3k = 9$$

$$ \therefore $$ Locus of $$C$$ is $$2x+3y=9$$

3

MCQ (Single Correct Answer)

The equation of the straight line passing through the point $$(4, 3)$$ and making intercepts on the co-ordinate axes whose sum is $$-1$$ is

A

$${x \over 2} - {y \over 3} = 1$$ and $${x \over -2} +{y \over 1} = 1$$

B

$${x \over 2} - {y \over 3} = -1$$ and $${x \over -2} +{y \over 1} = -1$$

C

$${x \over 2} + {y \over 3} = 1$$ and $${x \over 2} +{y \over 1} = 1$$

D

$${x \over 2} + {y \over 3} = -1$$ and $${x \over -2} +{y \over 1} = -1$$

Let the required line be $${x \over a} + {y \over b} = 1.......\left( 1 \right)$$

then $$a+b=-1$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.........\left( 2 \right)$$

$$(1)$$ passes through $$(4,3), $$ $$ \Rightarrow {4 \over a} + {3 \over b} = 1$$

$$ \Rightarrow 4b + 3a = ab\,\,...............\left( 3 \right)$$

Eliminating $$b$$ from $$(2)$$ and $$(3),$$ we get

$${a^2} - 4 = 0 \Rightarrow a = \pm 2 \Rightarrow b = - 3$$

$$1$$

$$\therefore$$ Equation of straight lines are

$${x \over 2} + {y \over { - 3}} = 1$$

or $${x \over { - 2}} + {y \over 1} = 1$$

then $$a+b=-1$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.........\left( 2 \right)$$

$$(1)$$ passes through $$(4,3), $$ $$ \Rightarrow {4 \over a} + {3 \over b} = 1$$

$$ \Rightarrow 4b + 3a = ab\,\,...............\left( 3 \right)$$

Eliminating $$b$$ from $$(2)$$ and $$(3),$$ we get

$${a^2} - 4 = 0 \Rightarrow a = \pm 2 \Rightarrow b = - 3$$

$$1$$

$$\therefore$$ Equation of straight lines are

$${x \over 2} + {y \over { - 3}} = 1$$

or $${x \over { - 2}} + {y \over 1} = 1$$

4

MCQ (Single Correct Answer)

If the equation of the locus of a point equidistant from the point $$\left( {{a_{1,}}{b_1}} \right)$$ and $$\left( {{a_{2,}}{b_2}} \right)$$ is

$$\left( {{a_1} - {b_2}} \right)x + \left( {{a_1} - {b_2}} \right)y + c = 0$$ , then the value of $$'c'$$ is

$$\left( {{a_1} - {b_2}} \right)x + \left( {{a_1} - {b_2}} \right)y + c = 0$$ , then the value of $$'c'$$ is

A

$$\sqrt {{a_1}^2 + {b_1}^2 - {a_2}^2 - {b_2}^2} $$

B

$${1 \over 2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right)$$

C

$${{a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2}$$

D

$${1 \over 2}\left( {{a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2} \right)$$.

$${\left( {x - {a_1}} \right)^2} + \left( {y - {b_1}} \right){}^2 = \left( {x - {a_2}} \right){}^2 + {\left( {y - {b_2}} \right)^2}$$

$$\left( {a{}_1 - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y$$

$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ + {1 \over 2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right) = 0$$

$$c = {1 \over 2}\left( {{a_2}^2 + {b_2}^2 - a{{{}_1}^2} - {b_1}^2} \right)$$

$$\left( {a{}_1 - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y$$

$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ + {1 \over 2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right) = 0$$

$$c = {1 \over 2}\left( {{a_2}^2 + {b_2}^2 - a{{{}_1}^2} - {b_1}^2} \right)$$

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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