First slide

Functions (XII)

Question

Let f(x) be a function such that $f (x - 1) +$ $f (x + 1) = \sqrt{2} f (x)$ for all $x \in R$ . If $f (3) = 5 then \sum_{r = 0}^{10} f (3 + 8 r)$ is equal to

Moderate

A

50
B

55
C

0
D

10

Solution

The given equation is
$f (x - 1) + f (x + 1) = \sqrt{2} f (x)$ …(i)
Replace x by x –1 in (i)
$f (x - 2) + f (x) = \sqrt{2} f (x - 1)$ …(ii)
Replace x by x +1 in (ii)
$f (x) + f (x + 2) = \sqrt{2} f (x + 1)$ …(iii)
Adding (ii) and (iii), we have
f(x – 2) + f(x + 2) + 2 f(x) = 2 ( f(x – 1) + f(x + 1))
$f (x - 2) + f (x + 2) + 2 f (x) = \sqrt{2} (f (x - 1) + f (x + 1)) = \sqrt{2} \sqrt{2} f (x) = 2 f (x)$
$\Rightarrow f (x - 2) = - f (x + 2)$
Replacing x by x + 2, we have
f(x) = –f(x + 4) = – ( f (x + 8)) = f(x + 8)
$\sum_{r = 0}^{10} f (3 + 8 r) = 11 \times f (3) = 11 \times 5 = 55$

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