If `alpha,beta!=0`
, and `f(n)""=alpha^n+beta^n`
and `{:|(3, 1+f(1),1+f(2)), (1+f(1),1+f(2),1+f(3)), (1+f(2),1+f(3),1+f(4))|:}=K(1-alpha)^2(1-beta)^2(alpha-beta)^2`
, then K is
equal to
(1) `alphabeta`
(2) `1/(alphabeta)`
(3) `1`
(4) `-1`
A
`alphabeta`
B
`(1)/(alphabeta)`
C
`1`
D
`-1`
Text Solution
AI Generated Solution
To solve the given problem step by step, we will analyze the determinant and use the properties of determinants to find the value of \( K \).
### Step 1: Define the function \( f(n) \)
Given that \( f(n) = \alpha^n + \beta^n \), we can compute the first few values:
- \( f(1) = \alpha + \beta \)
- \( f(2) = \alpha^2 + \beta^2 \)
- \( f(3) = \alpha^3 + \beta^3 \)
- \( f(4) = \alpha^4 + \beta^4 \)
...
Topper's Solved these Questions
COORDINATE GEOMETRY
JEE MAINS PREVIOUS YEAR ENGLISH|Exercise All Questions|2 Videos
DIFFERENTIAL EQUATIONS
JEE MAINS PREVIOUS YEAR ENGLISH|Exercise All Questions|10 Videos
Similar Questions
Explore conceptually related problems
If alpha,beta!=0 , and f(n)""=alpha^n+beta^n and |3 1+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)|=K(1-alpha)^2(1-beta)^2(alpha-beta)^2 , then K is equal to (1) alphabeta (2) 1/(alphabeta) (3) 1 (4) -1
f(1)=1, n ge 1 f(n+1)=2f(n)+1 then f(n)=
If cos(alpha-beta)=3sin(alpha+beta),t h e n 1/(1-3sin2alpha)+1/(1-3sin2beta)= (a) 1/2 (b) (-1)/2 (c) 1/4 (d) (-1)/4
Let f(theta)=cottheta/(1+cottheta) and alpha+beta=(5pi)/4 then the value f(alpha)f (beta) is
If cos(alpha-beta)=3sin(alpha+beta),t h e n1/(1-3sin2alpha)+1/(1-3sin2beta)= 1/2 (b) (-1)/2 (c) 1/4 (d) (-1)/4
Let f(n)=2cosn xAAn in N , then f(1)f(n+1)-f(n) is equal to f(n+3) (b) f(n+2) f(n+1)f(2) (d) f(n+2)f(2)
If f(x)=(cotx)/(1+cotx)andalpha+beta=(5pi)/(4) , then find f(alpha).f(beta) .
If f(n)=|(n!, (n+1)!, (n+2)!),((n+1)!, (n+2)!, (n+3)!), ((n+2)!, (n+3)!, (n+4)!)| Then the value of 1/1020[(f(100))/(f(99))] is
If f(alpha,beta)=|(cos alpha,-sin alpha,1),(sin alpha,cos alpha,1),(cos(alpha+beta),-sin(alpha+beta),1)|, then
