Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Consider a family of circles which are passing through the point $$(-1, 1)$$ and are tangent to $$x$$-axis. If $$(h, k)$$ are the coordinate of the centre of the circles, then the set of values of $$k$$ is given by the interval

A

$$ - {1 \over 2} \le k \le {1 \over 2}$$

B

$$k \le {1 \over 2}$$

C

$$0 \le k \le {1 \over 2}$$

D

$$k \ge {1 \over 2}$$

Equation of circle whose center is $$\left( {h,k} \right)$$

i.e $${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {k^2}$$

(radius of circle $$=k$$ because circle is tangent to $$x$$-axis)

Equation of circle passing through $$\left( { - 1, + 1} \right)$$

$$\therefore$$ $${\left( { - 1, - h} \right)^2} + {\left( {1 - k} \right)^2} = {k^2}$$

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

$$ \Rightarrow {h^2} + 2h - 2k + 2 = 0$$

$$D \ge 0$$

$$\therefore$$ $${\left( 2 \right)^2} - 4 \times 1.\left( { - 2k + 2} \right) \ge 0$$

$$ \Rightarrow 4 - 4\left( { - 2k + 2} \right) \ge 0$$

$$ \Rightarrow 1 + 2k - 2 \ge 0$$

$$ \Rightarrow k \ge {1 \over 2}$$

i.e $${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {k^2}$$

(radius of circle $$=k$$ because circle is tangent to $$x$$-axis)

Equation of circle passing through $$\left( { - 1, + 1} \right)$$

$$\therefore$$ $${\left( { - 1, - h} \right)^2} + {\left( {1 - k} \right)^2} = {k^2}$$

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

$$ \Rightarrow {h^2} + 2h - 2k + 2 = 0$$

$$D \ge 0$$

$$\therefore$$ $${\left( 2 \right)^2} - 4 \times 1.\left( { - 2k + 2} \right) \ge 0$$

$$ \Rightarrow 4 - 4\left( { - 2k + 2} \right) \ge 0$$

$$ \Rightarrow 1 + 2k - 2 \ge 0$$

$$ \Rightarrow k \ge {1 \over 2}$$

2

MCQ (Single Correct Answer)

Let $$C$$ be the circle with centre $$(0, 0)$$ and radius $$3$$ units. The equation of the locus of the mid points of the chords of the circle $$C$$ that subtend an angle of $${{2\pi } \over 3}$$ at its center is

A

$${x^2} + {y^2} = {3 \over 2}$$

B

$${x^2} + {y^2} = 1$$

C

$${x^2} + {y^2} = {{27} \over 4}$$

D

$${x^2} + {y^2} = {{9} \over 4}$$

Let $$M\left( {h,k} \right)$$ be the mid point of chord $$AB$$ where

$$\angle AOB = {{2\pi } \over 3}$$

$$\therefore$$ $$\angle AOM = {\pi \over 3}.$$ Also $$OM=$$ $$3\cos {\pi \over 3} = {3 \over 2}$$

$$ \Rightarrow \sqrt {{h^2} + k{}^2} = {3 \over 2}$$

$$ \Rightarrow {h^2} + {k^2} = {9 \over 4}$$

$$\therefore$$ Locus of $$\left( {h,k} \right)$$ is

$${x^2} + {y^2} = {9 \over 4}$$

$$\angle AOB = {{2\pi } \over 3}$$

$$\therefore$$ $$\angle AOM = {\pi \over 3}.$$ Also $$OM=$$ $$3\cos {\pi \over 3} = {3 \over 2}$$

$$ \Rightarrow \sqrt {{h^2} + k{}^2} = {3 \over 2}$$

$$ \Rightarrow {h^2} + {k^2} = {9 \over 4}$$

$$\therefore$$ Locus of $$\left( {h,k} \right)$$ is

$${x^2} + {y^2} = {9 \over 4}$$

3

MCQ (Single Correct Answer)

If the lines $$3x - 4y - 7 = 0$$ and $$2x - 3y - 5 = 0$$ are two diameters of a circle of area $$49\pi $$ square units, the equation of the circle is

A

$$\,{x^2} + {y^2} + 2x\, - 2y - 47 = 0\,$$

B

$$\,{x^2} + {y^2} + 2x\, - 2y - 62 = 0\,$$

C

$${x^2} + {y^2} - 2x\, + 2y - 62 = 0$$

D

$${x^2} + {y^2} - 2x\, + 2y - 47 = 0$$

Point of intersection of $$3x - 4y - 7 = 0$$ and

$$2x - 3y - 5 = 0$$ is $$\left( {1, - 1} \right)$$ which is the center of the

circle and radius $$=7$$

$$\therefore$$ Equation is $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 49$$

$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0$$

$$2x - 3y - 5 = 0$$ is $$\left( {1, - 1} \right)$$ which is the center of the

circle and radius $$=7$$

$$\therefore$$ Equation is $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 49$$

$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0$$

4

MCQ (Single Correct Answer)

If the pair of lines $$a{x^2} + 2\left( {a + b} \right)xy + b{y^2} = 0$$ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then

A

$$3{a^2} - 10ab + 3{b^2} = 0$$

B

$$3{a^2} - 2ab + 3{b^2} = 0$$

C

$$3{a^2} + 10ab + 3{b^2} = 0$$

D

$$3{a^2} + 2ab + 3{b^2} = 0$$

As per question area of one sector $$=3$$ area of another sector

$$ \Rightarrow $$ at center by one sector $$ = 3 \times $$ angle at center by another sector

Let one angle be $$\theta $$ then other $$=30$$

Clearly $$\theta + 3\theta = 180 \Rightarrow \theta = {45^ \circ }$$

$$\therefore$$ Angle between the diameters represented by combined equation

$$a{x^2} + 2\left( {a + b\,\,\,xy} \right) + b{y^2} = 0$$ is $${45^ \circ }$$

$$\therefore$$ Using $$tan$$ $$\theta $$ $$ = {{2\sqrt {{h^2} - ab} } \over {a + b}}$$

we get $$\tan \,{45^ \circ } = {{2\sqrt {{{\left( {a + b} \right)}^2} - ab} } \over {a + b}}$$

$$ \Rightarrow 1 = {{2\sqrt {{a^2} + {b^2} + ab} } \over {a + b}}$$

$$ \Rightarrow {\left( {a + b} \right)^2} = 4\left( {{a^2} + {b^2} + ab} \right)$$

$$ \Rightarrow {a^2} + {b^2} + 2ab = 4{a^2} + 4{b^2} + 4ab$$

$$ \Rightarrow 3{a^2} + 3{b^2} + 2ab = 0$$

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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