Question

Let $f(x)$ be a function such that $\frac{f(x-1)+f(x+1)}{2 f(x)}=\sin 60^{\circ}$ and $f(7)=\frac{1}{2}$, then the value of $\displaystyle\sum_{r=0}^5 f (7+12 r )$ is equal to

• A. 3
• B. 6
• C. 7
• D. 15

Answer 1

Answer: (A)

$f ( x -1)+ f ( x +1)=\sqrt{3} f ( x )$....(1)
Replace $x$ by $(x-1)$ and $x$ by $(x+1)$, we get
$f ( x -2)+ f ( x )=\sqrt{3} f ( x -1) $....(2)
$f ( x )+ f ( x +2)=\sqrt{3} f ( x +1)$....(3)
Adding (2) and (3), we get
$ f ( x -2)+ f ( x +2)+2 f ( x )=\sqrt{3}( f ( x -1)+ f ( x +1))=3 f ( x )$
$\Rightarrow f ( x -2)+ f ( x +2)= f ( x ) $ ....(4)
$\text { Replace } x \text { by }( x +2) \text {, we get } $
$ f ( x )+ f ( x +4)= f ( x +2)$....(5)
Replace $x$ by $(x+2)$, we get
$f(x)+f(x+4)=f(x+2)$
Adding (4) and (5), we get
$f(x-2)+f(x+4)=0$
Replace $x$ by $(x+2)$
$f(x+6)+f(x)=0$....(6)
Replace $x$ by $(x+6)$, we get
$f ( x +12)+ f ( x +6)=0$....(7)
From (7) $-(6)$,
$f(x+12)-f(x)=0 \Rightarrow f(x+12)=f(x)$
$\therefore f ( x )$ is periodic with period 12 .
$\therefore \displaystyle\sum_{r=0}^5 f (7+12 r )= \displaystyle\sum_{ r =0}^5 f (7)=6 f (7)=3 $