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A function $f :R \rightarrow R$ stisfies

$\\\sin x \cos y ( f ( 2x + 2y )) - f ( 2x - 2y ) = \ \cos x \sin y ( f ( 2x + 2y )+f ( 2x - 2 y ))$

$I\! \! f \: f ' (0) = 1/2 , then :$
Option: 1
$f''(x ) = f ( x ) = 0$

Option: 2
$4 f''(x ) + f ( x ) = 0$

Option: 3
$f''(x ) + f ( x ) = 0$

Option: 4
$4f''(x ) - f ( x ) = 0$

Answers (1)

best_answer

Trigonometric functions -

$\frac{d}{dx}(sinx)=cosx$

$\frac{d}{dx}(secx)=secx\:tanx$

$\frac{d}{dx}(cosec\:f(x))=-cosec\:\left \{ f(x) \right \}\:cot\:\left \{ f(x) \right \}\;f'(x)$

-

$\frac{f ( 2x+2y )}{f ( 2x -2y )} = \frac{\sin ( x+ y )}{\sin ( x-y )}$

$\frac{f \left ( \alpha \right )}{\sin ( \alpha /2 )}= \frac{ f ( \beta)}{\sin \beta /2 }$ = k

$f(x)=k\sin x/2$

$f^{{}'}(x) = k/2$ $\cos x/2$ $f^{''}$ $(x) = -k/4 \sin x/2$

$4 f^{''}(x) + f(x) = 0$

Posted by

Kuldeep Maurya

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