JEE MAIN - Mathematics (2021 - 27th July Evening Shift - No. 4)
Let C be the set of all complex numbers. Let
S1 = {z$$\in$$C : |z $$-$$ 2| $$\le$$ 1} and
S2 = {z$$\in$$C : z(1 + i) + $$\overline z $$(1 $$-$$ i) $$\ge$$ 4}.
Then, the maximum value of $${\left| {z - {5 \over 2}} \right|^2}$$ for z$$\in$$S1 $$\cap$$ S2 is equal to :
S1 = {z$$\in$$C : |z $$-$$ 2| $$\le$$ 1} and
S2 = {z$$\in$$C : z(1 + i) + $$\overline z $$(1 $$-$$ i) $$\ge$$ 4}.
Then, the maximum value of $${\left| {z - {5 \over 2}} \right|^2}$$ for z$$\in$$S1 $$\cap$$ S2 is equal to :
$${{3 + 2\sqrt 2 } \over 4}$$
$${{5 + 2\sqrt 2 } \over 2}$$
$${{3 + 2\sqrt 2 } \over 2}$$
$${{5 + 2\sqrt 2 } \over 4}$$
Explanation
|t $$-$$ 2| $$\le$$ 1
Put t = x + iy
_27th_July_Evening_Shift_en_4_2.png)
(x $$-$$ 2)2 + y2 $$\le$$ 1
Also, t(1 + i) + $$\overline t$$(1 $$-$$ i) $$\ge$$ 4
x $$-$$ y $$\ge$$ 2
Let point on circle be A(2 + cos$$\theta$$, sin$$\theta$$)
$$\theta \in \left[ { - {{3\pi } \over 4},{\pi \over 4}} \right]$$
$${(AP)^2} = {\left( {2 + \cos \theta - {5 \over 2}} \right)^2} + \sin \theta $$
$$ = {\cos ^2}\theta - \cos \theta + {1 \over 4} + {\sin ^2}\theta $$
$$ = {5 \over 4} - \cos \theta $$
For (AP)2 maximum $$\theta = - {{3\pi } \over 4}$$
$${(AP)^2} = {5 \over 4} + {1 \over {\sqrt 2 }} = {{5\sqrt 2 + 4} \over {4\sqrt 2 }}$$
Put t = x + iy
(x $$-$$ 2)2 + y2 $$\le$$ 1
Also, t(1 + i) + $$\overline t$$(1 $$-$$ i) $$\ge$$ 4
x $$-$$ y $$\ge$$ 2
Let point on circle be A(2 + cos$$\theta$$, sin$$\theta$$)
$$\theta \in \left[ { - {{3\pi } \over 4},{\pi \over 4}} \right]$$
$${(AP)^2} = {\left( {2 + \cos \theta - {5 \over 2}} \right)^2} + \sin \theta $$
$$ = {\cos ^2}\theta - \cos \theta + {1 \over 4} + {\sin ^2}\theta $$
$$ = {5 \over 4} - \cos \theta $$
For (AP)2 maximum $$\theta = - {{3\pi } \over 4}$$
$${(AP)^2} = {5 \over 4} + {1 \over {\sqrt 2 }} = {{5\sqrt 2 + 4} \over {4\sqrt 2 }}$$
Comments (0)
