JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023Let S={z∈C−{i,2i}:z2−3iz−2z2+8iz−15∈R}. If α−1113i∈S,a∈R−{0}, then 242α2 is equal to ________.
Choose the correct answer:
- A.
1680
(Correct Answer) - B.
1580
- C.
1480
- D.
1380
1680
Explanation
-
Simplify Expression: z2−3iz−2z2+8iz−15=(z−i)(z−2i)(z+3i)(z+5i)=1+z2−3iz−211iz−13.
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Condition for Real: Since 1∈R, we need z2−3iz−211iz−13∈R.
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Substitution: Given z=α−1113i⟹11iz−13=iα.
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Expression becomes 1+z2−3iz−2iα∈R.
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This implies (z2−3iz−2) must be purely imaginary.
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-
Locus Calculation: Let z=x+iy.
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Re(z2−3iz−2)=x2−y2+3y−2=0.
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x2=y2−3y+2=(y−1)(y−2).
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Final Value: Putting x=α and y=11−13:
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α2=(11−13−1)(11−13−2)=(11−24)(11−35)=12124×35.
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242α2=242⋅12124×35=2⋅840=1680.
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Answer: 1680
Explanation
-
Simplify Expression: z2−3iz−2z2+8iz−15=(z−i)(z−2i)(z+3i)(z+5i)=1+z2−3iz−211iz−13.
-
Condition for Real: Since 1∈R, we need z2−3iz−211iz−13∈R.
-
Substitution: Given z=α−1113i⟹11iz−13=iα.
-
Expression becomes 1+z2−3iz−2iα∈R.
-
This implies (z2−3iz−2) must be purely imaginary.
-
-
Locus Calculation: Let z=x+iy.
-
Re(z2−3iz−2)=x2−y2+3y−2=0.
-
x2=y2−3y+2=(y−1)(y−2).
-
-
Final Value: Putting x=α and y=11−13:
-
α2=(11−13−1)(11−13−2)=(11−24)(11−35)=12124×35.
-
242α2=242⋅12124×35=2⋅840=1680.
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Answer: 1680