x2−42kx+2e4lnk−1=0 or, x2−42kx+2k4−1=0 α+β=42k and α⋅β=2k4−1 Squaring both sides, we get (α+β)2=(42k)2⇒α2+β2+2αβ=32k266+2αβ=32k266+2(2k4−1)=32k266+4k4−2=32k2⇒4k4−32k2+64=0 or, k4−8k2+16=0⇒(k2)2−8k2+16=0⇒(k2−4)(k2−4)=0⇒k2=4,k2=4⇒k=±2 Now, α3+β3=(α+β)(α2+β2−αβ) ∴α3+β3=(42k)[66−(2k4−1)] Putting k=−2,(k=+2 cannot be taken because it does not satisfy the above equation) ∴α3+β3=(42(−2))[66−2(−2)4−1] α3+β3=(−82)(66−32+1)= (−82)(35) ∴α3+β3=−2802
If α and β are roots of the equation, x^2-4 √2 k x+2 e^4 ln k-1=0 for some k, and α^2+β^2=66, then α^3+β^3 is equal to:
JEE Main 2014 — Mathematics Algebra
If α and β are roots of the equation, x2−42kx+2e4lnk−1=0 for some k, and α2+β2=66, then α3+β3 is equal to:
Held on 11 Apr 2014 · Verified 6 Jul 2026.
Options
- A
2482
- B
2802
- C
−322
- D
−2802
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