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1

### JEE Advanced 2014 Paper 2 Offline

Let $${z_k}$$ = $$\cos \left( {{{2k\pi } \over {10}}} \right) + i\,\,\sin \left( {{{2k\pi } \over {10}}} \right);\,k = 1,2....,9$$

List-I

P. For each $${z_k}$$ = there exits as $${z_j}$$ such that $${z_k}$$.$${z_j}$$ = 1
Q. There exists a $$k \in \left\{ {1,2,....,9} \right\}$$ such that $${z_1}.z = {z_k}$$ has no solution z in the set of complex numbers
R. $${{\left| {1 - {z_1}} \right|\,\left| {1 - {z_2}} \right|\,....\left| {1 - {z_9}} \right|} \over {10}}$$ equals
S. $$1 - \sum\limits_{k = 1}^9 {\cos \left( {{{2k\pi } \over {10}}} \right)}$$ equals

List-II

1. True
2. False
3. 1
4. 2
A
P = 1, Q = 2, R = 4, S = 3
B
P = 2, Q = 1, R = 3, S = 4
C
P = 1, Q = 2, R = 3, S = 4
D
P =2, Q = 1, R = 4, S = 3
2

### JEE Advanced 2013 Paper 2 Offline

Let $$S = {S_1} \cap {S_2} \cap {S_3}$$, where $${S_1} = \left\{ {z \in C:\left| z \right| < 4} \right\},{S_2} = \left\{ {z \in C:{\mathop{\rm Im}\nolimits} \left[ {{{z - 1 + \sqrt 3 i} \over {1 - \sqrt 3 i}}} \right] > 0} \right\}$$ and $${S_3} = \left\{ {z \in C:{\mathop{\rm Re}\nolimits} z > 0} \right\}\,$$.

$$\,\mathop {\min }\limits_{z \in S} \left| {1 - 3i - z} \right| =$$

A
$${{2 - \sqrt 3 } \over 2}$$
B
$${{2 + \sqrt 3 } \over 2}$$
C
$${{3 - \sqrt 3 } \over 2}$$
D
$${{3 + \sqrt 3 } \over 2}$$
3

### JEE Advanced 2013 Paper 2 Offline

Let $$S = {S_1} \cap {S_2} \cap {S_3}$$, where $${S_1} = \left\{ {z \in C:\left| z \right| < 4} \right\},{S_2} = \left\{ {z \in C:{\mathop{\rm Im}\nolimits} \left[ {{{z - 1 + \sqrt 3 i} \over {1 - \sqrt 3 i}}} \right] > 0} \right\}$$ and $${S_3} = \left\{ {z \in C:{\mathop{\rm Re}\nolimits} z > 0} \right\}\,$$.

Area of S =

A
$${{10\pi } \over 3}$$
B
$${{20\pi } \over 3}$$
C
$${{16\pi } \over 3}$$
D
$${{32\pi } \over 3}$$
4

### JEE Advanced 2013 Paper 1 Offline

Let complex numbers $$\alpha \,and\,{1 \over {\overline \alpha }}\,$$ lie on circles $${\left( {x - {x_0}} \right)^2} + \,\,{\left( {y - {y_0}} \right)^2} = {r^2}$$ and $$\,{\left( {x - {x_0}} \right)^2} + \,\,{\left( {y - {y_0}} \right)^2} = 4{r^2}$$ respextively. If $${z_0} = {x_0} + i{y_0}$$ satisfies the equation $$2{\left| {{z_0}} \right|^2}\, = {r^2} + 2,\,then\,\left| a \right| =$$
A
$${1 \over {\sqrt 2 }}$$
B
$${1 \over 2}\,$$
C
$${1 \over {\sqrt 7 }}$$
D
$${1 \over 3}$$

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