Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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

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

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

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

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

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

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

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

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

2

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.

3

MCQ (Single Correct Answer)

If the pair of straight lines $${x^2} - 2pxy - {y^2} = 0$$ and $${x^2} - 2qxy - {y^2} = 0$$ be such that each pair bisects the angle between the other pair, then

A

$$pq = -1$$

B

$$p = q$$

C

$$p = -q$$

D

$$pq = 1$$.

Equation of bisectors of second pair of straight lines is,

$$q{x^2} + 2xy - q{y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

It must be identical to the first pair

$${x^2} - 2\,pxy - {y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(... 2 \right)$$

from $$(1)$$ and $$(2)$$ $${q \over 1} = {2 \over { - 2p}} = {{ - q} \over { - 1}}$$

$$ \Rightarrow pq = - 1.$$

$$q{x^2} + 2xy - q{y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

It must be identical to the first pair

$${x^2} - 2\,pxy - {y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(... 2 \right)$$

from $$(1)$$ and $$(2)$$ $${q \over 1} = {2 \over { - 2p}} = {{ - q} \over { - 1}}$$

$$ \Rightarrow pq = - 1.$$

4

MCQ (Single Correct Answer)

A square of side a lies above the $$x$$-axis and has one vertex at the origin. The side passing through the origin makes an angle $$\alpha \left( {0 < \alpha < {\pi \over 4}} \right)$$ with the positive direction of x-axis. The equation of its diagonal not passing through the origin is

A

$$y\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a$$

B

$$y\left( {\cos \alpha - \sin \alpha } \right) - x\left( {\sin \alpha - \cos \alpha } \right) = a$$

C

$$y\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\sin \alpha - \cos \alpha } \right) = a$$

D

$$y\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\sin \alpha + \cos \alpha } \right) = a$$

Co-ordinate of $$A = \left( {a\,\cos \,\alpha ,\,\,a\,\sin \,\alpha } \right)$$

Equation of $$OB,$$

$$y = \tan \left( {{\pi \over 4} + \alpha } \right)x$$

$$CA{ \bot ^r}$$ to $$OB$$

$$\therefore$$ slope of $$CA=-$$ $$\cot \left( {{\pi \over 4} + \alpha } \right)$$

Equation of $$CA$$

$$y - a\sin \alpha = - cot\left( {{\pi \over 4} + \alpha } \right)\left( {x - a\,\cos \,\alpha } \right)$$

$$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {\tan \left( {{\pi \over 4} + \alpha } \right)} \right) = \left( {a\,\cos \,\alpha - x} \right)$$

$$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {{{\tan {\pi \over 4} + \tan \alpha } \over {1 - \tan {\pi \over 4}\tan \alpha }}} \right)\left( {a\,\cos \,\alpha - x} \right)$$

$$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {1 + \tan \alpha } \right) = \left( {a\cos \alpha - x} \right)\left( {1 - \tan \alpha } \right)$$

$$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {\cos \alpha + \sin \alpha } \right) = \left( {a\cos \alpha - x} \right)\left( {\cos \alpha - \sin \alpha } \right)$$

$$ \Rightarrow y\left( {\cos + \sin \alpha } \right) - a\sin \alpha \cos \alpha - a{\sin ^2}\alpha $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = a{\cos ^2}\alpha - a\cos \alpha \sin \alpha - x\left( {\cos \alpha - \sin \alpha } \right)$$

$$ \Rightarrow y\left( {\cos \alpha + sin\alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a$$

$$y\left( {\sin \alpha + \cos \alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a.$$

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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