JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 3)
The region represented by
{z = x + iy $$ \in $$ C : |z| – Re(z) $$ \le $$ 1} is also given by the
inequality : {z = x + iy $$ \in $$ C : |z| – Re(z) $$ \le $$ 1}
{z = x + iy $$ \in $$ C : |z| – Re(z) $$ \le $$ 1} is also given by the
inequality : {z = x + iy $$ \in $$ C : |z| – Re(z) $$ \le $$ 1}
y2 $$ \le $$ $$2\left( {x + {1 \over 2}} \right)$$
y2 $$ \le $$ $${x + {1 \over 2}}$$
y2 $$ \ge $$ 2(x + 1)
y2 $$ \ge $$ x + 1
Explanation
Given z = x + iy
|z| – Re(z) $$ \le $$ 1
$$ \Rightarrow $$ $$\sqrt {{x^2} + {y^2}} $$ - x $$ \le $$ 1
$$ \Rightarrow $$ $$\sqrt {{x^2} + {y^2}} $$ $$ \le $$ 1 + x
$$ \Rightarrow $$ x2 + y2 $$ \le $$ 1 + 2x + x2
$$ \Rightarrow $$ y2 $$ \le $$ 2x + 1
$$ \Rightarrow $$ y2 $$ \le $$ 2$$\left( {x + {1 \over 2}} \right)$$
|z| – Re(z) $$ \le $$ 1
$$ \Rightarrow $$ $$\sqrt {{x^2} + {y^2}} $$ - x $$ \le $$ 1
$$ \Rightarrow $$ $$\sqrt {{x^2} + {y^2}} $$ $$ \le $$ 1 + x
$$ \Rightarrow $$ x2 + y2 $$ \le $$ 1 + 2x + x2
$$ \Rightarrow $$ y2 $$ \le $$ 2x + 1
$$ \Rightarrow $$ y2 $$ \le $$ 2$$\left( {x + {1 \over 2}} \right)$$
Comments (0)
