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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
If z is a complex number such that $$\,\left| z \right| \ge 2\,$$, then the minimum value of $$\,\,\left| {z + {1 \over 2}} \right|$$ :
A
is strictly greater that $${{5 \over 2}}$$
B
is strictly greater that $${{3 \over 2}}$$ but less than $${{5 \over 2}}$$
C
is equal to $${{5 \over 2}}$$
D
lie in the interval (1, 2)

Explanation

We know minimum value of

$$\,\,\,\left| {{Z_1} + {Z_2}} \right|\,\,\,$$ is $$\,\,\,\left| {\left| {{Z_1}} \right| - \left| {{Z_2}} \right|} \right|$$

Thus minimum value of

$$\,\,\,\left| {Z + {1 \over 2}} \right|\,\,\,$$ is $$\,\,\,\left| {\left| Z \right| - {1 \over 2}} \right|$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \le \left| {Z + {1 \over 2}} \right| \le \left| Z \right| + {1 \over 2}$$

Since, $$\,\,\,\left| Z \right| \ge 2$$

$$\therefore$$ $$\,\,\,2 - {1 \over 2} < \left| {Z + {1 \over 2}} \right| < 2 + {1 \over 2}$$

$$ \Rightarrow {3 \over 2} < \left| {Z + {1 \over 2}} \right| < {5 \over 2}$$
2

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
If z is a complex number of unit modulus and argument $$\theta $$, then arg $$\left( {{{1 + z} \over {1 + \overline z }}} \right)$$ equals :
A
$$ - \theta \,\,$$
B
$${\pi \over 2} - \theta \,$$
C
$$\theta \,$$
D
$$\,\pi - \theta \,\,$$

Explanation

Given $$\,\,\,\,\left| z \right| = 1,\,\,\arg \,z = \theta $$

As we know, $$\,\,\,\,\overrightarrow z = {1 \over z}$$

$$\therefore$$ $$\,\,\,\,\arg \left( {{{1 + z} \over {1 + \overrightarrow z }}} \right) = \arg \left( {{{1 + z} \over {1 + {1 \over z}}}} \right)$$

$$ = \arg \left( z \right) = \theta .$$
3

AIEEE 2012

MCQ (Single Correct Answer)
If $$z \ne 1$$ and $$\,{{{z^2}} \over {z - 1}}\,$$ is real, then the point represented by the complex number z lies :
A
either on the real axis or a circle passing through the origin.
B
on a circle with centre at the origin
C
either on real axis or on a circle not passing through the origin.
D
on the imaginary axis.

Explanation

Let $$z = x + iy$$

$$\therefore$$ $$\,\,\,\,{z^2} = {x^2} - {y^2} + 2ixy$$

Now $${{{z^2}} \over {z - 1}}$$ is real

$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{z^2}} \over {z - 1}}} \right) = 0$$

$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{x^2} - {y^2} + 2ixy} \over {\left( {x - 1} \right) + iy}}} \right) = 0$$

$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left[ {\left( {{x^2} - {y^2} + 2ixy} \right)\left. {\left( {x - 1} \right) - iy} \right)} \right] = 0$$

$$ \Rightarrow 2xy\left( {x - 1} \right) - y\left( {{x^2} - {y^2}} \right) = 0$$

$$ \Rightarrow y\left( {{x^2} + {y^2} - 2x} \right) = 0$$

$$ \Rightarrow y = 0;\,{x^2} + {y^2} - 2x = 0$$

$$\therefore$$ $$\,\,\,\,$$ $$z$$ lies either on real axis or on a circle through origin.
4

AIEEE 2011

MCQ (Single Correct Answer)
If $$\omega ( \ne 1)$$ is a cube root of unity, and $${(1 + \omega )^7} = A + B\omega \,$$. Then $$(A,B)$$ equals
A
(1 ,1)
B
(1, 0)
C
(- 1 ,1)
D
(0 ,1)

Explanation

$${\left( {1 + \omega } \right)^7} = A + B\omega ;\,\,\,\,{\left( { - {\omega ^2}} \right)^7} = A + B\omega $$

$$ - {\omega ^2} = A + B\omega ;\,\,\,\,\,\,\,\,\,\,1 + \omega = A + B\omega $$

$$ \Rightarrow A = 1,B = 1.$$

Questions Asked from Complex Numbers

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
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JEE Main 2017 (Offline) (1)
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JEE Main 2016 (Offline) (1)
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JEE Main 2014 (Offline) (1)
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