1

JEE Advanced 2016 Paper 2 Offline

MCQ (More than One Correct Answer)
Let $$a,\,b \in R\,and\,{a^{2\,}} + {b^2} \ne 0$$. Suppose
$$S = \left\{ {Z \in C:Z = {1 \over {a + ibt}}, + \in R,t \ne 0} \right\}$$, where $$i = \sqrt { - 1} $$. Ifz = x + iy and z $$ \in $$ S, then (x, y) lies on
A
the circle with radius $${{1 \over {2a}}}$$and centre $$\left\{ {{1 \over {2a}},\,0} \right\}\,for\,a > 0\,,b \ne \,0$$
B
the circle with radius $$-{{1 \over {2a}}}$$and centre $$\left\{ -{{1 \over {2a}},\,0} \right\}\,for\,a < 0\,,b \ne \,0$$
C
the x-axis for $$a \ne \,\,0,\,b \ne \,0$$
D
the y-axis for $$a = \,\,0,\,b \ne \,0$$

Explanation

Here, $$x + iy = {1 \over {a + ibt}} \times {{a - ibt} \over {a - ibt}}$$

$$\therefore$$ $$x + iy = {{a - ibt} \over {{a^2} + {b^2}{t^2}}}$$

Let a $$\ne$$ 0, b $$\ne$$ 0

$$\therefore$$ $$x = {a \over {{a^2} + {b^2}{t^2}}}$$ and $$y = {{ - bt} \over {{a^2} + {b^2}{t^2}}}$$

$$ \Rightarrow {y \over x} = {{ - bt} \over a} \Rightarrow t = {{ay} \over {bx}}$$

On putting $$x = {a \over {{a^2} + {b^2}{t^2}}}$$, we get

$$x\left( {{a^2} + {b^2}\,.\,{{{a^2}{y^2}} \over {{b^2}{x^2}}}} \right) = a$$

$$ \Rightarrow {a^2}({x^2} + {y^2}) = ax$$

or, $${x^2} + {y^2} - {x \over a} = 0$$ ...... (i)

or, $${\left( {x - {1 \over {2a}}} \right)^2} + {y^2} = {1 \over {4{a^2}}}$$

$$\therefore$$ Option (a) is correct.

For a $$\ne$$ 0 and b = 0,

$$x + iy = {1 \over a} \Rightarrow x = {1 \over a},\,y = 0$$

$$\Rightarrow$$ z lies on X-axis.

$$\therefore$$ Option (c) is correct.

For a = 0 and b $$\ne$$ 0,

$$x + iy = {1 \over {ibt}}$$

$$ \Rightarrow x = 0,\,y = - {1 \over {bt}}$$

$$\Rightarrow$$ z lies on Y-axis.

$$\therefore$$ Option (d) is correct.

2

JEE Advanced 2013 Paper 2 Offline

MCQ (More than One Correct Answer)
Let $$w = {{\sqrt 3 + i} \over 2}$$ and P = { $${w^n}$$ : n = 1, 2, 3, ...}. Further and , where is the set of all complex numbers. If $${z_1} \in P \cap {H_1},\,{z_2} \in \,P \cap {H_2}$$ and 0 represents the origin, then $$\angle \,{z_1}\,o{z_2} = $$
A
$${\pi \over 2}$$
B
$${\pi \over 6}\,$$
C
$${{2\pi } \over 3}$$
D
$${{5\pi } \over 6}$$
3

IIT-JEE 2010 Paper 1 Offline

MCQ (More than One Correct Answer)
Let $${{z_1}}$$ and $${{z_2}}$$ be two distinct complex number and let z =( 1 - t)$${{z_1}}$$ + t$${{z_2}}$$ for some real number t with 0 < t < 1. IfArg (w) denote the principal argument of a non-zero complex number w, then
A
$$\left| {z - {z_1}} \right| + \left| {z - {z_2}} \right| = \left| {{z_1} - {z_2}} \right|$$
B
Arg $$(z - {z_1})$$ = Arg$$(z - {z_2})$$
C
$$\left| {\matrix{ {z - {z_1}} & {\overline z - {{\overline z }_1}} \cr {{z_2} - {z_1}} & {{{\overline z }_2} - {{\overline z }_1}} \cr } } \right|$$ = 0
D
Arg $$(z - {z_1})$$ = Arg$$({z_2} - {z_1})$$

Explanation

Given, $$z = {{(1 - t){z_1} + t\,{z_2}} \over {(1 - t) + t}}$$

Clearly, z divides z1 and z2 in the ratio of t : (1 $$-$$ t), 0 < t < 1

$$\Rightarrow$$ AP + BP = AB

i.e., $$|z - {z_1}| + |z - {z_2}| = |{z_1} - {z_2}|$$

$$\Rightarrow$$ Option (a) is true.

and $$\arg (z - {z_1}) = \arg ({z_2} - z) = \arg ({z_2} - {z_1})$$

$$\Rightarrow$$ (b) is false and (d) is true.

Also, $$\arg (z - {z_1}) = \arg ({z_2} - {z_1})$$

$$ \Rightarrow \arg \left( {{{z - {z_1}} \over {{z_2} - {z_1}}}} \right) = 0$$

$$\therefore$$ $${{z - {z_1}} \over {{z_2} - {z_1}}}$$ is purely real.

$$ \Rightarrow {{z - {z_1}} \over {{z_2} - {z_1}}} = {{\overline z - {{\overline z }_1}} \over {{{\overline z }_2} - {{\overline z }_1}}}$$

or, $$\left| {\matrix{ {z - {z_1}} & {\overline z - {{\overline z }_1}} \cr {{z_2} - {z_1}} & {{{\overline z }_2} - {{\overline z }_1}} \cr } } \right| = 0$$

$$\therefore$$ Option (c) is correct.

Hence, (a, c, d) is the correct option.

4

IIT-JEE 1998

MCQ (More than One Correct Answer)
If $$\,\left| {\matrix{ {6i} & { - 3i} & 1 \cr 4 & {3i} & { - 1} \cr {20} & 3 & i \cr } } \right| = x + iy$$ , then
A
x = 3, y = 2
B
x = 1, y = 3
C
x = 0, y = 3
D
x = 0, y = 0

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