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Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$ and $$\left( {{p^2} + 1} \right){}^2x + \left( {{p^2} + 1} \right)y + 2q$$ $$=0$$ are perpendicular to a common line for :

A

exactly one values of $$p$$

B

exactly two values of $$p$$

C

more than two values of $$p$$

D

no value of $$p$$

If the lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$

and $${\left( {{p^2} + 1} \right)^2}x + \left( {{p^2} + 1} \right)y + 2q = 0$$

are perpendicular to a common line then these lines -

must be parallel to each other,

$$\therefore$$ $${m_1} = {m_2} \Rightarrow - {{p\left( {{p^2} + 1} \right)} \over { - 1}} = - {{{{\left( {{p^2} + 1} \right)}^2}} \over {{p^2} + 1}}$$

$$ \Rightarrow \left( {{p^2} + 1} \right)\left( {p + 1} \right) = 0$$

$$ \Rightarrow p = - 1$$

$$\therefore$$ $$p$$ can have exactly one value.

and $${\left( {{p^2} + 1} \right)^2}x + \left( {{p^2} + 1} \right)y + 2q = 0$$

are perpendicular to a common line then these lines -

must be parallel to each other,

$$\therefore$$ $${m_1} = {m_2} \Rightarrow - {{p\left( {{p^2} + 1} \right)} \over { - 1}} = - {{{{\left( {{p^2} + 1} \right)}^2}} \over {{p^2} + 1}}$$

$$ \Rightarrow \left( {{p^2} + 1} \right)\left( {p + 1} \right) = 0$$

$$ \Rightarrow p = - 1$$

$$\therefore$$ $$p$$ can have exactly one value.

2

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 $$D{}_{\min } = {3 \over {4\sqrt 2 }} = {{3\sqrt 2 } \over 8}$$

3

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

4

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

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

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

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

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Complex Numbers

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