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If f Be A Function Such That f(x+1) + f(x-1) = Root2 * f(x). Find f(x+4)/f(x)

Sujoy Das
07/02/2018 0 0


Method 1:

Best approach: Let f(x-1) = 1 and f(x) = 1

So the sequence becomes : 1, 1, (√2) - 1, 1 - (√2), -1, -1, 1 - (√2), (√2) - 1, 1, 1, ...

i.e. f(x) is a periodic function with period 8 and after every four terms, terms start repeating with negative sign as shown above.

Method 2:

Let f(x)=a and f(x-1)=b then solve in terms of a and b we will get f(x+4)=-a so ans is -1

f(x+2) + f(x) = rt2 . f(x+1)...... i

f(x+3) + f(x+1) = rt2 . f(x+2) ............ii

f(x+4) + f(x+2) = rt2 . f(x+3)

Or f(x+4) + f(x+2) = 2f(x+2) - rt2 . f(x+1)....... from ii

So f(x+4) = f(x+2) - rt2. f(x+1)

So f(x+4) = - f(x)..from i

So answer = -1

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