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Fractional Part Function

If a non-zero...

Question

If a non-zero function $f$ satisfies the relation $f (x + y) + f (x - y) = 2 f (x) . f (y)$ for all $x, y$ in $R$ and $f (0) \neq 0$ ; then $f (10) - f (- 10) =$

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Solution

The correct option is B $0$
Given, $f (x + y) + f (x - y) = 2 f (x) \cdot f (y)$

Put $y = 0, x = x$
$f (x) + f (x) = 2 f (x) f (0)$
$2 f (x) = 2 f (x) f (0)$
$f (x) = 0$ that is not possible
$f (0) = 1$

Take $x = 0$ and $y = 10$

$f (10) + f (- 10) = 2 f (10) f (0)$

$f (10) + f (- 10) = 2 f (10)$

$f (- 10) = f (10)$

$∴ f (10) - f (- 10) = 0$

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If a non-zero function f satisfies the relation f(x+y)+f(x−y)=2f(x).f(y) for all x,y in R and f(0)≠0; then f(10)−f(−10)=

The correct option is B 0Given, f(x+y)+f(x−y)=2f(x)⋅f(y)Put y=0,x=xf(x)+f(x)=2f(x)f(0)2f(x)=2f(x)f(0)f(x)=0 that is not possiblef(0)=1Take x=0 and y=10f(10)+f(−10)=2f(10)f(0)f(10)+f(−10)=2f(10)f(−10)=f(10)∴f(10)−f(−10)=0

If a non-zero function $f$ satisfies the relation $f (x + y) + f (x - y) = 2 f (x) . f (y)$ for all $x, y$ in $R$ and $f (0) \neq 0$ ; then $f (10) - f (- 10) =$