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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

If $$|z - 25i| \le 15$$, then Maximum arg(z) $$-$$ Minimum arg(z) is equal to

(arg z is the principal value of argument of z)

A
$$2{\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$
B
$$2{\cos ^{ - 1}}\left( {{4 \over 5}} \right)$$
C
$${\pi \over 2} + {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$
D
$${\sin ^{ - 1}}\left( {{3 \over 5}} \right) - {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$

যদি $$|z - 25i| \le 15$$ হয়, তবে সর্বোচ্চ arg(z) $$-$$ সর্বনিম্ন arg(z) হবে

(arg z, z-এর আরগুমেন্টের মুখ্যমান বুঝাবে)

A
$$2{\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$
B
$$2{\cos ^{ - 1}}\left( {{4 \over 5}} \right)$$
C
$${\pi \over 2} + {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$
D
$${\sin ^{ - 1}}\left( {{3 \over 5}} \right) - {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$
2

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
Let C denote the set of all complex numbers. Define A = {(z, w) | z, w$$\in$$C and |z| = |w|}, B = {z, w} | z, w$$\in$$C and z2 = w2}. Then
A
A = B
B
A $$ \subset $$ B
C
B $$\subset$$ A
D
A $$\cap$$ B = $$\varphi $$

Explanation

We have, z2 = w2

$$\Rightarrow$$ z2 $$-$$ w2 = 0

$$\Rightarrow$$ (z $$-$$ w) (z + w) = 0

$$\Rightarrow$$ z = w, z = $$-$$w

$$\Rightarrow$$ | z | = | w |

$$\therefore$$ B $$\subset$$ A
মনে কর C সকল জটিল রাশির সেট। A = {(z, w) | z, w$$\in$$C এবং |z| = |w|}, B = {z, w} | z, w$$\in$$C এবং z2 = w2}। সেক্ষেত্রে
A
A = B
B
A $$ \subset $$ B
C
B $$\subset$$ A
D
A $$\cap$$ B = $$\varphi $$

Explanation

আমাদের আছে, z2 = w2

$$\Rightarrow$$ z2 $$-$$ w2 = 0

$$\Rightarrow$$ (z $$-$$ w) (z + w) = 0

$$\Rightarrow$$ z = w, z = $$-$$w

$$\Rightarrow$$ | z | = | w |

$$\therefore$$ B $$\subset$$ A
3

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
If |z| = 1 and z $$\ne$$ $$\pm$$ 1, then all the points representing $${z \over {1 - {z^2}}}$$ lie on
A
a line not passing through the origin
B
the line y = x
C
the X-axis
D
the Y-axis

Explanation

Let z = ei$$\theta$$

Also, let $$w = {{{e^{i\theta }}} \over {1 - {e^{2i\theta }}}} = {1 \over {{e^{ - i\theta }} - {e^{i\theta }}}}$$

$$ = {1 \over { - 2\,.\,i\sin \theta }} = {i \over {2\sin \theta }}$$

$$\Rightarrow$$ w is purely imaginary

$$\Rightarrow$$ Locus of point is Y-axis.
প্রদত্ত |z| = 1 1এবং z $$\ne$$ $$\pm$$ 1, সেক্ষেত্রে $${z \over {1 - {z^2}}}$$ এর প্রতিনিধিত্বকারী সকল বিন্দু
A
মূলবিন্দুগামী নয় এমন সরলরেখার উপর অবস্থিত থাকবে
B
y = x সরলরেখার উপর অবস্থিত
C
x-অক্ষের উপর অবস্থিত
D
y-অক্ষের উপর অবস্থিত

Explanation

ধরা যাক z = ei$$\theta$$

এছাড়াও, ধরা যাক $$w = {{{e^{i\theta }}} \over {1 - {e^{2i\theta }}}} = {1 \over {{e^{ - i\theta }} - {e^{i\theta }}}}$$

$$ = {1 \over { - 2\,.\,i\sin \theta }} = {i \over {2\sin \theta }}$$

$$\Rightarrow$$ w সম্পূর্ণ কাল্পনিক

$$\Rightarrow$$ বিন্দুর অবস্থান হল Y-অক্ষ।
4

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
The equation $$z\bar z + (2 - 3i)z + (2 + 3i)\bar z + 4 = 0$$ represents a circle of radius
A
2 unit
B
3 unit
C
4 unit
D
6 unit

Explanation

Given,

$$z\bar z + (2 - 3i)z + (2 + 3i)\bar z + 4 = 0$$

This is the form of

$$z\bar z + \bar az + a\bar z + b$$

Here, a = 2 + 3i, b = 4

$$ \therefore $$ Radius of circle = $$\sqrt {{{\left| a \right|}^2} - b} $$

= $$\sqrt {(4 + 9) - 4} $$

= $$\sqrt {13 - 4} = 3$$

$$z\overline z + (2 - 3i)z + (2 + 3i)\overline z + 4 = 0$$ সমীকরণটি যে বৃত্ত সূচিত করে, তার ব্যাসার্ধ হল -

A
2 একক
B
3 একক
C
4 একক
D
6 একক

Explanation

$$z\overline z + (2 - 3i)z + (2 + 3i)\overline z + 4 = 0$$

বা, $${x^2} + {y^2} + (2 - 3i + 2 - 3i)x + iy(2 - 3i - 2 - 3i) + 4 = 0$$

বা, $${x^2} + {y^2} + 4x + 6y + 4 = 0$$

ব্যাসার্ধ $$ = \sqrt {4 + 9 - 4} = 3$$ একক

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