Search any question & find its solution
Question:
Answered & Verified by Expert
If a function $f$ satisfies $f(x+1)+f(x-1)=\sqrt{2} f(x)$, then $f(x+2)+f(x-2)=$
Options:
Solution:
2839 Upvotes
Verified Answer
The correct answer is:
0
Given,
$f(x+1)+f(x-1)=\sqrt{2} f(x)$ ...(i)
Replace $x$ by $x+1$
$f(x+2)+f(x)=\sqrt{2} f(x+1)$ ...(ii)
Replace $x$ by $x-1$ in Eq. (i),
$f(x)+f(x-2)=\sqrt{2} f(x-1)$ ...(iii)
On adding Eqs. (ii) and (iii),
$\begin{aligned} f(x+2)+f(x-2) & +2 f(x) \\ = & \sqrt{2}[f(x+1)+f(x-1)]\end{aligned}$
$\therefore f(x+2)+f(x-2)=0$
$f(x+1)+f(x-1)=\sqrt{2} f(x)$ ...(i)
Replace $x$ by $x+1$
$f(x+2)+f(x)=\sqrt{2} f(x+1)$ ...(ii)
Replace $x$ by $x-1$ in Eq. (i),
$f(x)+f(x-2)=\sqrt{2} f(x-1)$ ...(iii)
On adding Eqs. (ii) and (iii),
$\begin{aligned} f(x+2)+f(x-2) & +2 f(x) \\ = & \sqrt{2}[f(x+1)+f(x-1)]\end{aligned}$
$\therefore f(x+2)+f(x-2)=0$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.