Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The shortest distance between the line $$y - x = 1$$ and the curve $$x = {y^2}$$ is :

A

$${{2\sqrt 3 } \over 8}$$

B

$${{3\sqrt 2 } \over 5}$$

C

$${{\sqrt 3 } \over 4}$$

D

$${{3\sqrt 2 } \over 8}$$

Let $$\left( {{a^2},a} \right)$$ be the point of shortest distance on $$x = {y^2}$$

Then distance between $$\left( {{a^2},a} \right)$$ and line $$x - y + 1 = 0$$

is given by

$$\,\,\,\,\,\,\,\,D = {{{a^2} - a + 1} \over {\sqrt 2 }} = {1 \over {\sqrt 2 }}\left[ {{{\left( {a - {1 \over 2}} \right)}^2} + {3 \over 4}} \right]$$

It is min when $$a = {1 \over 2}$$ and $$D{}_{\min } = {3 \over {4\sqrt 2 }} = {{3\sqrt 2 } \over 8}$$

Then distance between $$\left( {{a^2},a} \right)$$ and line $$x - y + 1 = 0$$

is given by

$$\,\,\,\,\,\,\,\,D = {{{a^2} - a + 1} \over {\sqrt 2 }} = {1 \over {\sqrt 2 }}\left[ {{{\left( {a - {1 \over 2}} \right)}^2} + {3 \over 4}} \right]$$

It is min when $$a = {1 \over 2}$$ and $$D{}_{\min } = {3 \over {4\sqrt 2 }} = {{3\sqrt 2 } \over 8}$$

2

MCQ (Single Correct Answer)

The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept -4. Then a possible value of k is

A

1

B

2

C

-2

D

-4

Slope of $$PQ = {{3 - 4} \over {k - 1}} = {{ - 1} \over {k - 1}}$$

$$\therefore$$ Slope of perpendicular bisector of

$$PQ = \left( {k - 1} \right)$$

Also mid point of

$$PQ\left( {{{k + 1} \over 2},{7 \over 2}} \right).$$

Equation of perpendicular bisector is

$$y - {7 \over 2} = \left( {k - 1} \right)\left( {x - {{k + 1} \over 2}} \right)$$

$$ \Rightarrow 2y - 7 = 2\left( {k - 1} \right)x - \left( {{k^2} - 1} \right)$$

$$ \Rightarrow 2\left( {k - 1} \right)x - 2y + \left( {8 - {k^2}} \right) = 0$$

$$\therefore$$ $$y$$-intercept $$ = {{8 - {k^2}} \over { - 2}} = - 4$$

$$ \Rightarrow $$ $$8 - {k^2} = - 8$$ or $${k^2} = 16 \Rightarrow k = \pm 4$$

$$\therefore$$ Slope of perpendicular bisector of

$$PQ = \left( {k - 1} \right)$$

Also mid point of

$$PQ\left( {{{k + 1} \over 2},{7 \over 2}} \right).$$

Equation of perpendicular bisector is

$$y - {7 \over 2} = \left( {k - 1} \right)\left( {x - {{k + 1} \over 2}} \right)$$

$$ \Rightarrow 2y - 7 = 2\left( {k - 1} \right)x - \left( {{k^2} - 1} \right)$$

$$ \Rightarrow 2\left( {k - 1} \right)x - 2y + \left( {8 - {k^2}} \right) = 0$$

$$\therefore$$ $$y$$-intercept $$ = {{8 - {k^2}} \over { - 2}} = - 4$$

$$ \Rightarrow $$ $$8 - {k^2} = - 8$$ or $${k^2} = 16 \Rightarrow k = \pm 4$$

3

MCQ (Single Correct Answer)

If one of the lines of $$m{y^2} + \left( {1 - {m^2}} \right)xy - m{x^2} = 0$$ is a bisector of angle between the lines $$xy = 0,$$ then $$m$$ is

A

$$1$$

B

$$2$$

C

$$-1/2$$

D

$$-2$$

Equation of bisectors of lines, $$xy=0$$ are $$y = \pm x$$

$$\therefore$$ Put $$y = \pm \,x$$ in the given equation

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

$$\therefore$$ $$m{x^2} + \left( {1 - {m^2}} \right){x^2} - m{x^2} = 0$$

$$ \Rightarrow 1 - {m^2} = 0 \Rightarrow m = \pm 1$$

$$\therefore$$ Put $$y = \pm \,x$$ in the given equation

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

$$\therefore$$ $$m{x^2} + \left( {1 - {m^2}} \right){x^2} - m{x^2} = 0$$

$$ \Rightarrow 1 - {m^2} = 0 \Rightarrow m = \pm 1$$

4

MCQ (Single Correct Answer)

Let $$P = \left( { - 1,0} \right),\,Q = \left( {0,0} \right)$$ and $$R = \left( {3,3\sqrt 3 } \right)$$ be three point. The equation of the bisector of the angle $$PQR$$ is

A

$${{\sqrt 3 } \over 2}x + y = 0$$

B

$$x + \sqrt {3y} = 0$$

C

$$\sqrt 3 x + y = 0$$

D

$$x + {{\sqrt 3 } \over 2}y = 0$$

$$\left( {0,0} \right),\,\left( {3,3\sqrt 3 } \right)$$ respectively

Slope of $$QR$$ $$ = {{{y_2} - {y_1}} \over {{x_2} - {x_1}}} = {{3\sqrt 3 } \over 3}$$

$$ \Rightarrow \tan \theta = \sqrt 3 \Rightarrow \theta = {\pi \over 3}$$

$$ \Rightarrow \angle RQX = {\pi \over 3}$$

$$\therefore$$ $$\angle RQP = \pi - {\pi \over 3} = {{2\pi } \over 3}$$

Let $$QM$$ bisects the $$\angle PQR,$$

$$\therefore$$ Slope of the line $$QM=tan$$ $${{2\pi } \over 3} = - \sqrt 3 $$

$$\therefore$$ Equation of line $$QM$$ is $$\left( {y - 0} \right) = - \sqrt 3 \left( {x - 0} \right)$$

$$ \Rightarrow y = - \sqrt 3 \,x \Rightarrow \sqrt 3 x + y = 0$$

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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