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1
JEE Advanced 2013 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Match the chemical conversions in List I with the appropriate reagents in List II and select the correct answer using the code given below the lists :

2
JEE Advanced 2013 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-1
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 =
3
JEE Advanced 2013 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
Let $\omega=\frac{\sqrt{3}+i}{2}$ and $P=\left\{\omega^n: n=1,2,3, \ldots\right\}$. Further
$\mathrm{H}_1=\left\{z \in \mathrm{C}: \operatorname{Re} z<\frac{1}{2}\right\}$ and
$\mathrm{H}_2=\left\{z \in \mathrm{C}: \operatorname{Re} z<\frac{-1}{2}\right\}$, where C is the
set of all complex numbers. If $z_1 \in \mathrm{P} \cap \mathrm{H}_1, z_2 \in$ $\mathrm{P} \cap \mathrm{H}_2$ and O
represents the origin, then $\angle z_1 \mathrm{O} z_2=$
4
JEE Advanced 2013 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
Two lines $${L_1}:x = 5,{y \over {3 - \alpha }} = {z \over { - 2}}$$ and $${L_2}:x = \alpha ,{y \over { - 1}} = {z \over {2 - \alpha }}$$ are coplanar. Then $$\alpha $$ can take value(s)
Paper Analysis
Total Questions
Chemistry 20
Mathematics 20
Physics 20
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