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1.Relation and Function
normal
If $f(x)$ satisfies the relation $f\left( {\frac{{5x - 3y}}{2}} \right) = \frac{{5f(x) - 3f(y)}}{2}\forall x,y\, \in \,R$ and $f(0)=1, f'(0)=2$ then the period of $sin(f(x))$ is
A
$2\pi $
B
$\pi $
C
$3\pi $
D
$4\pi $
Solution
Give equation can be written as
$f\left(\frac{5 x-3 y}{2}\right)=\frac{5 f(x)-3 f(y)}{5-3}$
Which satisfies section formula for abscissa on $LHS$ and ordinate on $RHS$
hence $f(x)$ must be linear function
Let $\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}$
$\therefore f(0)=b=1$
and $f^{\prime}(0)=a=2$
$\Rightarrow f(x)=2 x+1$
$\therefore$ period of $\sin (2 x+1)$ is $\pi$
Standard 12
Mathematics
