Let $f$ be a function satisfying $f(xy) = \frac{f(x)}{y}$ for all positive real numbers $x$ and $y.$ If $f(30) = 20,$ then the value of $f(40)$ is-

If $f(x + y) = f(x) + f(y)$ for all $x, y \in R$ and $f(1) = 1$,then find the value of $\lim_{x \to 0} \frac{2^{f(\tan x)} - 2^{f(\sin x)}}{f(\tan x) - f(\sin x)}$.

Let $\sum\limits_{k = 1}^{10} {f(a + k)} = 16(2^{10} - 1),$ where the function $f$ satisfies $f(x + y) = f(x)f(y)$ for all natural numbers $x, y$ and $f(1) = 2.$ Then the natural number $a$ is

Let $f: R \rightarrow R$ be a twice differentiable function such that $(\sin x \cos y)(f(2x+2y) - f(2x-2y)) = (\cos x \sin y)(f(2x+2y) + f(2x-2y))$ for all $x, y \in R$. If $f'(0) = \frac{1}{2}$,then the value of $24f''\left(\frac{5\pi}{3}\right)$ is:

If $f(x)=x^2-2x+4$, then the set of values of $x$ satisfying $f(x-1)=f(x+1)$ is

$A$ function $f: R \rightarrow R$ satisfies the relation $f(x+y)=f(x) \cdot f(y), \forall x, y \in R$ and $f(x) \neq 0, \forall x \in R$. If $f$ is differentiable at $x=0$,$f^{\prime}(0)=4$,and $f(6)=3$,then $f^{\prime}(6)$ is equal to