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1

### JEE Advanced 2015 Paper 2 Offline

Numerical
For any integer k, let $${a_k} = \cos \left( {{{k\pi } \over 7}} \right) + i\,\,\sin \left( {{{k\pi } \over 7}} \right)$$, where $$i = \sqrt { - 1} \,$$. The value of the expression $${{\sum\limits_{k = 1}^{12} {\left| {{\alpha _{k + 1}} - {a_k}} \right|} } \over {\sum\limits_{k = 1}^3 {\left| {{\alpha _{4k - 1}} - {\alpha _{4k - 2}}} \right|} }}$$ is

## Explanation

Given, $${a_k} = \cos \left( {{{k\pi } \over 7}} \right) + i\sin \left( {{{k\pi } \over 7}} \right) = {e^{{{k\pi } \over 7}i}}$$

We have to find $${{\sum\limits_{k = 1}^{12} {\left| {{a_{k + 1}} - {a_k}} \right|} } \over {\sum\limits_{k = 1}^3 {\left| {{a_{4k - 1}} - {a_{4k - 2}}} \right|} }}$$

$${a_{k + 1}} = \cos \left( {{{k + 1} \over 7}} \right)\pi + i\sin \left( {{{k + 1} \over 7}} \right)\pi = {e^{i\left( {{{k + 1} \over 7}} \right)\pi }}$$

$$\therefore$$ $${a_{k + 1}} - {a_k} = {e^{\left( {{{k + 1} \over 7}} \right)\pi i}} - {e^{{{k\pi } \over 7}i}}$$

$$= {e^{{{k\pi } \over 7}i}}\,.\,\,{e^{{\pi \over 7}i}} - {e^{{{k\pi } \over 7}i}}$$

$$= {e^{{{k\pi } \over 7}i}}\left( {{e^{{\pi \over 7}i}} - 1} \right)$$

$$\therefore$$ $$\left| {{a_{k + 1}} - {a_k}} \right| = \left| {{e^{{{k\pi } \over 7}i}}\left( {{e^{{\pi \over 7}i}} - 1} \right)} \right|$$

$$= \left| {{e^{{{k\pi } \over 7}i}}} \right|\left| {{e^{{\pi \over 7}i}} - 1} \right|$$

$$= \left| {{e^{{\pi \over 7}i}} - 1} \right|$$

If $$z = {e^{i\theta }} = \cos \theta + i\sin \theta$$

then $$\left| z \right| = \sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } = 1$$

that is why $$\left| {{e^{{{k\pi } \over 7}i}}} \right| = 1$$

Now, $${a_{4k - 1}} = {e^{\left( {{{4k - 1} \over 7}} \right)\pi i}}$$

$${a_{4k - 2}} = {e^{\left( {{{4k - 2} \over 7}} \right)\pi i}}$$

$${a_{4k - 1}} - {a_{4k - 2}} = {e^{\left( {{{4k - 1} \over 7}} \right)\pi i}} - {e^{\left( {{{4k - 2} \over 7}} \right)\pi i}}$$

$$= {e^{{{4k\pi } \over 7}i}}\,.\,{e^{ - {\pi \over 7}i}} - {e^{{{4k\pi } \over 7}}}\,.\,{e^{ - {{2\pi } \over 7}i}}$$

$$= {e^{{{4k\pi } \over 7}i}}\left( {{e^{ - {\pi \over 7}i}} - {e^{ - {{2\pi } \over 7}i}}} \right)$$

$$\therefore$$ $$\left| {{a_{4k - 1}} - {a_{4k - 2}}} \right| = \left| {{e^{{{4k\pi } \over 7}i}}} \right|\left| {{e^{ - {\pi \over 7}i}} - {e^{ - {{2\pi } \over 7}i}}} \right|$$

$$= \left| {{e^{ - {\pi \over 7}i}} - {e^{ - {{2\pi } \over 7}i}}} \right|$$

Now, $${{\sum\limits_{k = 1}^{12} {\left| {{a_{k + 1}} - {a_k}} \right|} } \over {\sum\limits_{k = 1}^3 {\left| {{a_{4k - 1}} - {a_{4k - 2}}} \right|} }}$$

$$= {{12\left| {{e^{{\pi \over 7}i}} - 1} \right|} \over {3\left| {{e^{ - {\pi \over 7}i}} - {e^{ - {{2\pi } \over 7}i}}} \right|}}$$

$$= 4\,.\,{{\left| {{e^{{\pi \over 7}i}} - 1} \right|} \over {\left| {\left( {{e^{ - {\pi \over 7}i}}1 - {e^{ - {\pi \over 7}i}}} \right)} \right|}}$$

$$= 4\,.\,{{\left| {{e^{{\pi \over 7}i}} - 1} \right|} \over {\left| {1 - {e^{ - {\pi \over 7}i}}} \right|}}$$

$$= 4\,.\,{{\left| {{e^{{\pi \over 7}i}}\left( {1 - {e^{ - {\pi \over 7}i}}} \right)} \right|} \over {\left| {1 - {e^{ - {\pi \over 7}i}}} \right|}}$$

$$= 4\,.\,\left| {{e^{{\pi \over 7}i}}} \right|$$

= 4 as $$\left| {{e^{{\pi \over 7}i}}} \right| = 1$$

2

### IIT-JEE 2011 Paper 2 Offline

Numerical
Let $$\omega = {e^{{{i\pi } \over 3}}}$$, and a, b, c, x, y, z be non-zero complex numbers such that
$$a + b + c = x$$
$$a + b\omega + c{\omega ^2} = y$$
$$a + b{\omega ^2} + c\omega = z$$

Then the value of $${{{{\left| x \right|}^2} + {{\left| y \right|}^2} + {{\left| z \right|}^2}} \over {{{\left| a \right|}^2} + {{\left| b \right|}^2} + {{\left| c \right|}^2}}}$$ is

3

### IIT-JEE 2011 Paper 1 Offline

Numerical
If z is any complex number satisfying $$\,\left| {z - 3 - 2i} \right| \le 2$$, then the minimum value of $$\left| {2z - 6 + 5i} \right|$$ is

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