NTA Abhyas JEE Main2020MathematicsDeterminantsPractice
If α 2 n α 2 n + 2 α 2 n + 4 β 2 n β 2 n + 2 β 2 n + 4 γ 2 n γ 2 n + 2 γ 2 n + 4 = 1 β 2 - 1 α 2 1 γ 2 - 1 β 2 1 α 2 - 1 γ 2 where α 2 ,   β 2 and γ 2 are all distinct , then the value of n is equal to
Options
- A4
- B- 4
- C3
- D- 2
Correct answer
D. - 2
Step-by-step solution
Given, α 2 n β 2 n γ 2 n 1 α 2 α 4 1 β 2 β 4 1 γ 2 γ 4 = α 2 - β 2 β 2 - γ 2 ( γ 2 - α 2 ) α 4 β 4 γ 4 ⇒ α 2 n β 2 n γ 2 n [ α 2 - β 2 β 2 - γ 2 γ 2 - α 2 ] = α 2 - β 2 β 2 - γ 2 ( γ 2 - α 2 ) α β γ 4 On comparison, we get, n = - 2