Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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

2

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

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

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

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

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

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

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

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

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

3

MCQ (Single Correct Answer)

If $${x_1},{x_2},{x_3}$$ and $${y_1},{y_2},{y_3}$$ are both in G.P. with the same common ratio, then the points $$\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$$ and $$\left( {{x_3},{y_3}} \right)$$

A

are vertices of a triangle

B

lie on a straight line

C

lie on an ellipse

D

lie on a circle

Taking co-ordinates as

$$\left( {{x \over r},{y \over r}} \right);\left( {x,y} \right)\,\,\& \,\,\left( {xr,yr} \right)$$

Then slope of line joining

$$\left( {{x \over r},{y \over r}} \right),\left( {x,y} \right) = {{y\left( {1 - {1 \over r}} \right)} \over {x\left( {1 - {1 \over r}} \right)}} = {y \over x}$$

and slope of line joining $$(x,y)$$ and $$(xr, yr)$$

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

$$\therefore$$ $${m_1} = {m_2}$$

$$ \Rightarrow $$ Points lie on the straight line.

$$\left( {{x \over r},{y \over r}} \right);\left( {x,y} \right)\,\,\& \,\,\left( {xr,yr} \right)$$

Then slope of line joining

$$\left( {{x \over r},{y \over r}} \right),\left( {x,y} \right) = {{y\left( {1 - {1 \over r}} \right)} \over {x\left( {1 - {1 \over r}} \right)}} = {y \over x}$$

and slope of line joining $$(x,y)$$ and $$(xr, yr)$$

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

$$\therefore$$ $${m_1} = {m_2}$$

$$ \Rightarrow $$ Points lie on the straight line.

4

MCQ (Single Correct Answer)

Locus of centroid of the triangle whose vertices are $$\left( {a\cos t,a\sin t} \right),\left( {b\sin t, - b\cos t} \right)$$ and $$\left( {1,0} \right),$$ where $$t$$ is a parameter, is

A

$${\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}$$

B

$${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}$$

C

$${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$

D

$${\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$

$$x = {{a\cos t + b\sin t + 1} \over 3}$$

$$ \Rightarrow a\cos t + b\sin t = 3x - 1$$

$$y = {{a\sin t - b\cos t} \over 3}$$

$$ \Rightarrow a\sin t - b\cos t = 3y$$

$$ \Rightarrow a\cos t + b\sin t = 3x - 1$$

$$y = {{a\sin t - b\cos t} \over 3}$$

$$ \Rightarrow a\sin t - b\cos t = 3y$$

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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

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Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations