JEE MAIN - Mathematics (2021 - 24th February Morning Shift - No. 16)
If the least and the largest real values of a, for which the
equation z + $$\alpha $$|z – 1| + 2i = 0 (z $$ \in $$ C and i = $$\sqrt { - 1} $$) has a solution, are p and q respectively; then 4(p2 + q2) is equal to __________.
equation z + $$\alpha $$|z – 1| + 2i = 0 (z $$ \in $$ C and i = $$\sqrt { - 1} $$) has a solution, are p and q respectively; then 4(p2 + q2) is equal to __________.
Answer
10
Explanation
$$x + iy + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} + 2i = 0$$
$$ \therefore $$ y + 2 = 0 and $$x + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} = 0$$
y = $$-$$2 & $${x^2} = {\alpha ^2}({x^2} - 2x + 1 + 4)$$
$${\alpha ^2} = {{{x^2}} \over {{x^2} - 2x + 5}} \Rightarrow {x^2}({\alpha ^2} - 1) - 2x{\alpha ^2} + 5{\alpha ^2} = 0$$
$$x \in R \Rightarrow D \ge 0$$
$$4{\alpha ^4} - 4({\alpha ^2} - 1)5{\alpha ^2} \ge 0$$
$${\alpha ^2}[4{\alpha ^2} - 2{\alpha ^2} + 20] \ge 0$$
$${\alpha ^2}[ - 16{\alpha ^2} + 20] \ge 0$$
$${\alpha ^2}\left[ {{\alpha ^2} - {5 \over 4}} \right] \le 0$$
$$0 \le {\alpha ^2} \le {5 \over 4}$$
$$ \therefore $$ $${\alpha ^2} \in \left[ {0,{5 \over 4}} \right]$$
$$ \therefore $$ $$\alpha \in \left[ { - {{\sqrt 5 } \over 2},{{\sqrt 5 } \over 2}} \right]$$
then $$4[{(q)^2} + {(p)^2}] = 4\left[ {{5 \over 4} + {5 \over 4}} \right] = 10$$
$$ \therefore $$ y + 2 = 0 and $$x + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} = 0$$
y = $$-$$2 & $${x^2} = {\alpha ^2}({x^2} - 2x + 1 + 4)$$
$${\alpha ^2} = {{{x^2}} \over {{x^2} - 2x + 5}} \Rightarrow {x^2}({\alpha ^2} - 1) - 2x{\alpha ^2} + 5{\alpha ^2} = 0$$
$$x \in R \Rightarrow D \ge 0$$
$$4{\alpha ^4} - 4({\alpha ^2} - 1)5{\alpha ^2} \ge 0$$
$${\alpha ^2}[4{\alpha ^2} - 2{\alpha ^2} + 20] \ge 0$$
$${\alpha ^2}[ - 16{\alpha ^2} + 20] \ge 0$$
$${\alpha ^2}\left[ {{\alpha ^2} - {5 \over 4}} \right] \le 0$$
$$0 \le {\alpha ^2} \le {5 \over 4}$$
$$ \therefore $$ $${\alpha ^2} \in \left[ {0,{5 \over 4}} \right]$$
$$ \therefore $$ $$\alpha \in \left[ { - {{\sqrt 5 } \over 2},{{\sqrt 5 } \over 2}} \right]$$
then $$4[{(q)^2} + {(p)^2}] = 4\left[ {{5 \over 4} + {5 \over 4}} \right] = 10$$
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