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AP EAPCET 07-Jul-2022 Shift 1 Paper

Section: Mathematics

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Question : 1 of 160

Marks: +1, -0

If a function

f

f(x+1)+f(x-1)=\sqrt{2} f(x)

f(x+2)+f(x-2)=

$2 \cdot f(x)$
$f(x+1)-f(x-1)$
$4 \cdot f(x)$
$0$

Solution: 👈: Video Solution

Given,

f(x+1)+f(x-1)=\sqrt{2} f(x)\dots(i)

Replace

x

by

x+1

f(x+2)+f(x)=\sqrt{2} f(x+1)\dots(ii)

Replace

x

by

x-1

in Eq. (i),

f(x)+f(x-2)=\sqrt{2} f(x-1)\dots(iii)

On adding Eqs. (ii) and (iii),

f(x+2)+f(x-2)\;+2 f(x)

\;=\sqrt{2}[f(x+1)+f(x-1)]

\Rightarrow f(x+2)+f(x-2)\;=\sqrt{2}(\sqrt{2} f(x))-2 f(x)

\dots[\text{from Eq. (i)}]

\therefore f(x+2)+f(x-2)=0

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