1

### 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.

2

### 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
3

### IIT-JEE 1998

MCQ (More than One Correct Answer)
The value of the sum $$\,\,\sum\limits_{n = 1}^{13} {({i^n}} + {i^{n + 1}})$$ , where i = $$\sqrt { - 1}$$, equals
A
i
B
i - 1
C
- i
D
0
4

### IIT-JEE 1998

MCQ (More than One Correct Answer)
If $${\omega}$$ is an imaginary cube root of unity, then $${(1\, + \omega \, - {\omega ^2})^7}$$ equals
A
$$128\omega$$
B
$$- 128\omega$$
C
$$128{\omega ^2}$$
D
$$- 128{\omega ^2}$$

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