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1

MCQ (Single Correct Answer)

The line parallel to the $$x$$ - axis and passing through the intersection of the lines $$ax + 2by + 3b = 0$$ and $$bx - 2ay - 3a = 0,$$ where $$(a, b)$$ $$ \ne $$ $$(0, 0)$$ is

A

below the $$x$$ - axis at a distance of $${3 \over 2}$$ from it

B

below the $$x$$ - axis at a distance of $${2 \over 3}$$ from it

C

above the $$x$$ - axis at a distance of $${3 \over 2}$$ from it

D

above the $$x$$ - axis at a distance of $${2 \over 3}$$ from it

The line passing through the intersection of lines

$$ax + 2by = 3b = 0$$

and $$bx - 2ay - 3a = 0$$ is

$$ax + 2by + 3b + \lambda \left( {bx - 2ay - 3a} \right) = 0$$

$$ \Rightarrow \left( {a + b\lambda } \right)x + \left( {2b - 2a\lambda } \right)y + 3b - 3\lambda a = 0$$

As this line is parallel to $$x$$-axis.

$$\therefore$$ $$a + b\lambda = 0 \Rightarrow \lambda = - a/b$$

$$ \Rightarrow ax + 2by + 3b - {a \over b}\left( {bx - 2ay - 3a} \right) = 0$$

$$ \Rightarrow ax + 2by + 3b - ax + {{2{a^2}} \over b}y + {{3{a^2}} \over b} = 0$$

$$y\left( {2b + {{2{a^2}} \over b}} \right) + 3b + {{3{a^2}} \over b} = 0$$

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

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

So it is $$3/2$$ units below $$x$$-axis.

$$ax + 2by = 3b = 0$$

and $$bx - 2ay - 3a = 0$$ is

$$ax + 2by + 3b + \lambda \left( {bx - 2ay - 3a} \right) = 0$$

$$ \Rightarrow \left( {a + b\lambda } \right)x + \left( {2b - 2a\lambda } \right)y + 3b - 3\lambda a = 0$$

As this line is parallel to $$x$$-axis.

$$\therefore$$ $$a + b\lambda = 0 \Rightarrow \lambda = - a/b$$

$$ \Rightarrow ax + 2by + 3b - {a \over b}\left( {bx - 2ay - 3a} \right) = 0$$

$$ \Rightarrow ax + 2by + 3b - ax + {{2{a^2}} \over b}y + {{3{a^2}} \over b} = 0$$

$$y\left( {2b + {{2{a^2}} \over b}} \right) + 3b + {{3{a^2}} \over b} = 0$$

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

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

So it is $$3/2$$ units below $$x$$-axis.

2

MCQ (Single Correct Answer)

If one of the lines given by $$6{x^2} - xy + 4c{y^2} = 0$$ is $$3x + 4y = 0,$$ then $$c$$ equals

A

$$-3$$

B

$$-1$$

C

$$3$$

D

$$1$$

$$3x+4y=0$$ is one of the lines of the pair

$$6{x^2} - xy + 4c{y^2} = 0,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$

Put $$\,\,\,\,\,y = - {3 \over 4}x,$$

we get $$6{x^2} + {3 \over 4}{x^2} + 4c{\left( { - {3 \over 4}x} \right)^2} = 0$$

$$ \Rightarrow 6 + {3 \over 4} + {{9c} \over 4} = 0 \Rightarrow c = - 3$$

$$6{x^2} - xy + 4c{y^2} = 0,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$

Put $$\,\,\,\,\,y = - {3 \over 4}x,$$

we get $$6{x^2} + {3 \over 4}{x^2} + 4c{\left( { - {3 \over 4}x} \right)^2} = 0$$

$$ \Rightarrow 6 + {3 \over 4} + {{9c} \over 4} = 0 \Rightarrow c = - 3$$

3

MCQ (Single Correct Answer)

If the sum of the slopes of the lines given by $${x^2} - 2cxy - 7{y^2} = 0$$ is four times their product $$c$$ has the value

A

$$-2$$

B

$$-1$$

C

$$2$$

D

$$1$$

Let the lines be $$y = {m_1}x$$ and $$y = {m_2}x$$ then

$${m_1} + {m_2} = - {{2c} \over 7}$$ and $${m_1}{m_2} = - {1 \over 7}$$

Given $${m_1} + {m_2} = 4m{}_1{m_2}$$

$$ \Rightarrow {{2c} \over 7} = - {4 \over 7} \Rightarrow c = 2$$

$${m_1} + {m_2} = - {{2c} \over 7}$$ and $${m_1}{m_2} = - {1 \over 7}$$

Given $${m_1} + {m_2} = 4m{}_1{m_2}$$

$$ \Rightarrow {{2c} \over 7} = - {4 \over 7} \Rightarrow c = 2$$

4

MCQ (Single Correct Answer)

Let $$A\left( {2, - 3} \right)$$ and $$B\left( {-2, 1} \right)$$ be vertices of a triangle $$ABC$$. If the centroid of this triangle moves on the line $$2x + 3y = 1$$, then the locus of the vertex $$C$$ is the line

A

$$3x - 2y = 3$$

B

$$2x - 3y = 7$$

C

$$3x + 2y = 5$$

D

$$2x + 3y = 9$$

Let the vertex $$C$$ be $$(h,k),$$ then the

centroid of $$\Delta ABC$$ is $$\left( {{{2 + (- 2) + h} \over 3},{{ - 3 + 1 + k} \over 3}} \right)$$

or $$\left( {{h \over 3},{{ - 2 + k} \over 3}} \right).$$ It lies on $$2x+3y=1$$

$$ \Rightarrow {{2h} \over 3} - 2 + k = 1$$

$$ \Rightarrow 2h + 3k = 9$$

$$ \therefore $$ Locus of $$C$$ is $$2x+3y=9$$

centroid of $$\Delta ABC$$ is $$\left( {{{2 + (- 2) + h} \over 3},{{ - 3 + 1 + k} \over 3}} \right)$$

or $$\left( {{h \over 3},{{ - 2 + k} \over 3}} \right).$$ It lies on $$2x+3y=1$$

$$ \Rightarrow {{2h} \over 3} - 2 + k = 1$$

$$ \Rightarrow 2h + 3k = 9$$

$$ \therefore $$ Locus of $$C$$ is $$2x+3y=9$$

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Complex Numbers

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