Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

The differential equation of the family of circles with fixed radius $$5$$ units and centre on the line $$y = 2$$ is

A

$$\left( {x - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}$$

B

$$\left( {y - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}$$

C

$${\left( {y - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}$$

D

$${\left( {x - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}$$

Let the center of the circle be $$(h, 2)$$

$$\therefore$$ Equation of circle is

$${\left( {x - h} \right)^2} + \left( {y - 2} \right){}^2 = 25\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Differentiating with respect to $$x,$$ we get

$$2\left( {x - h} \right) + 2\left( {y - 2} \right){{dy} \over {dx}} = 0$$

$$ \Rightarrow x - h = - \left( {y - 2} \right){{dy} \over {dx}}$$

Substituting in equation $$(1)$$ we get

$${\left( {y - 2} \right)^2}{\left( {{{dy} \over {dx}}} \right)^2} + {\left( {y - 2} \right)^2} = 25$$

$$ \Rightarrow {\left( {y - 2} \right)^2}{\left( {y'} \right)^2} = 25 - {\left( {y - 2} \right)^2}$$

$$\therefore$$ Equation of circle is

$${\left( {x - h} \right)^2} + \left( {y - 2} \right){}^2 = 25\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Differentiating with respect to $$x,$$ we get

$$2\left( {x - h} \right) + 2\left( {y - 2} \right){{dy} \over {dx}} = 0$$

$$ \Rightarrow x - h = - \left( {y - 2} \right){{dy} \over {dx}}$$

Substituting in equation $$(1)$$ we get

$${\left( {y - 2} \right)^2}{\left( {{{dy} \over {dx}}} \right)^2} + {\left( {y - 2} \right)^2} = 25$$

$$ \Rightarrow {\left( {y - 2} \right)^2}{\left( {y'} \right)^2} = 25 - {\left( {y - 2} \right)^2}$$

2

MCQ (Single Correct Answer)

The point diametrically opposite to the point $$P(1, 0)$$ on the circle $${x^2} + {y^2} + 2x + 4y - 3 = 0$$ is

A

$$(3, -4)$$

B

$$(-3, 4)$$

C

$$(-3, -4)$$

D

$$(3, 4)$$

The given circle is $${x^2} + {y^2} + 2x + 4y - 3 = 0$$

Center $$(-1,-2)$$

Let $$Q$$ $$\left( {\alpha ,\beta } \right)$$ be the point diametrically opposite to the point $$P(1,0),$$

then $${{1 + \alpha } \over 2} = - 1$$ and $${{0 + \beta } \over 2} = - 2$$

$$ \Rightarrow \alpha = - 3,\beta = - 4,$$ So, $$Q$$ is $$\left( { - 3, - 4} \right)$$

Center $$(-1,-2)$$

Let $$Q$$ $$\left( {\alpha ,\beta } \right)$$ be the point diametrically opposite to the point $$P(1,0),$$

then $${{1 + \alpha } \over 2} = - 1$$ and $${{0 + \beta } \over 2} = - 2$$

$$ \Rightarrow \alpha = - 3,\beta = - 4,$$ So, $$Q$$ is $$\left( { - 3, - 4} \right)$$

3

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 $$\left( { - 1, + 1} \right)$$

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

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

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

$$D \ge 0$$

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

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

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

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

4

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

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

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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