Let f(x) be a function such that f(x – 1) + f(x + 1) = 
f(x) ∀ x ∈ R. If f(5) = 100
then 
= ..............
Text Solution
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Ans. 5000
Sol. f(x – 1) + f(x + 1) = 
f(x)
x ⇒ x + 1 f(x) + f(x + 2) = 
f(x + 1) ...(1)
x ⇒ x – 1 f(x– 2) + f(x) = 
f(x – 1) ...(2)
From (1) + (2)
f(x + 2) + f(x – 2) + 2f(x)
= 
{f(x + 1) + f(x – 1)} ...(3)
f(x + 2) + f(x – 2) = f(x)
x ⇒ x + 2 f(x + 4) + f(x) = f(x + 2) ...(4)
(3) + (4) f(x – 2) + f(x + 4) = 0 ...(5)
x ⇒ x + 6 f(x + 4) + f(x + 10) = 0 ..(6)
(5) – (6) f(x – 2) = f (x + 10)
f(x) = f(x + 12) period is 12
Now 
⇒ f(5) + f(5 + 12.1)
+ f (5 + 12.2) +..........+ f(5 + 49.12)
= 
= 50 × f(5) = 50 × 100 = 5000
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