Discussion

Explanation:

g(x + 1) + g(x – 1) = g(x)

∴ g(x + 1) = g(x) – g(x – 1)

Now, let g(x − 1) = a and g(x) = b

∴ g(x + 1) = b – a

∴ g(x + 2) = b – a – b = –a

∴ g(x + 3) = −a – b + a = –b

∴ g(x + 4) = −b + a = a – b

∴ g(x + 5) = a – b + b = a = g(x – 1)

∴ g(x + 6) = a – a + b = b = g(x)

and so on.

Thus we observe that the values of g(x + 6) and g(x) are always equal.

Hence, option (d).

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