Question

Let $f(x)$ be a function such that $\frac{f(x-1)+f(x+1)}{2f(x)}=\sin 60^{\circ }$ and $f(7)=\frac{1}{2}$, then the value of $\sum _{r=0}^{5}f(7+12r)$ 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
$f(x-2)+f(x+2)+2f(x)=\sqrt{3}(f(x-1)+f(x+1))=3f(x)$
$\Rightarrow f(x-2)+f(x+2)=f(x)$ ....(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 .
$\therefore \sum _{r=0}^{5}f(7+12r)=\sum _{r=0}^{5}f(7)=6f(7)=3$