JEE 2022 — Mathematics PYQ
JEE | Mathematics | 2022Let α and β be the roots of the equation x2+(2i−1)=0. Then, the value of ∣α8+β8∣ is equal to:
Choose the correct answer:
- A.
50
(Correct Answer) - B.
250
- C.
1250
- D.
1500
50
Explanation
Solution
1. Identify the roots:
The equation is x2=1−2i.
Since α and β are roots of x2=c, we have α2=1−2i and β2=1−2i.
Actually, β=−α, so β2=(−α)2=α2.
2. Calculate α8 and β8:
Since α2=β2=1−2i:
α4=(α2)2=(1−2i)2=1+4i2−4i=1−4−4i=−3−4i
α8=(α4)2=(−3−4i)2=9+16i2+24i=9−16+24i=−7+24i
Similarly, β8=−7+24i.
3. Find the magnitude of the sum:
α8+β8=(−7+24i)+(−7+24i)=−14+48i
∣α8+β8∣=(−14)2+(48)2=196+2304=2500=50
Correct Option: (A)
Explanation
Solution
1. Identify the roots:
The equation is x2=1−2i.
Since α and β are roots of x2=c, we have α2=1−2i and β2=1−2i.
Actually, β=−α, so β2=(−α)2=α2.
2. Calculate α8 and β8:
Since α2=β2=1−2i:
α4=(α2)2=(1−2i)2=1+4i2−4i=1−4−4i=−3−4i
α8=(α4)2=(−3−4i)2=9+16i2+24i=9−16+24i=−7+24i
Similarly, β8=−7+24i.
3. Find the magnitude of the sum:
α8+β8=(−7+24i)+(−7+24i)=−14+48i
∣α8+β8∣=(−14)2+(48)2=196+2304=2500=50
Correct Option: (A)