If f(x) satisfies the relation fleft( frac{5x | vedclass

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(B) The given equation is $f\left( \frac{5x - 3y}{2} \right) = \frac{5f(x) - 3f(y)}{2}$.This equation represents a linear functional equation of the f

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$\pi$

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(B) The given equation is $f\left( \frac{5x - 3y}{2} \right) = \frac{5f(x) - 3f(y)}{2}$.This equation represents a linear functional equation of the form $f(ax + (1-a)y) = af(x) + (1-a)f(y)$,which implies that $f(x)$ is a linear function of the form $f(x) = mx + c$.Given $f(0) = 1$,we have $c = 1$.Given $f'(0) = 2$,we have $m = 2$.Thus,$f(x) = 2x + 1$.We need to find the period of $\sin(f(x)) = \sin(2x + 1)$.The period of $\sin(kx + \phi)$ is given by $\frac{2\pi}{|k|}$.Here,$k = 2$,so the period is $\frac{2\pi}{2} = \pi$.

If $f(x)$ satisfies the relation $f\left( \frac{5x - 3y}{2} \right) = \frac{5f(x) - 3f(y)}{2}$ for all $x, y \in R$ and $f(0)=1, f'(0)=2$,then the period of $\sin(f(x))$ is:

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