If $f(x) = \frac{2x + 1}{3x - 2}$,then $(fof)(2)$ is equal to

If $f(x) = \sin^2 x + \sin^2(x + \frac{\pi}{3}) + \cos x \cos(x + \frac{\pi}{3})$ and $g(\frac{5}{4}) = 1$,then $(g \circ f)(x) = $

Consider the function $f: R \rightarrow R$ defined by $f(x)=\frac{2x}{\sqrt{1+9x^2}}$. If the composition of $f$,$\underbrace{(f \circ f \circ \ldots \circ f)}_{10 \text{ times }}(x) = \frac{2^{10}x}{\sqrt{1+9\alpha x^2}}$,then the value of $\sqrt{3\alpha+1}$ is equal to:

Show that if $f: A \rightarrow B$ and $g: B \rightarrow C$ are one-one,then $g \circ f: A \rightarrow C$ is also one-one.

Let $f: \{1,3,4\} \rightarrow \{1,2,5\}$ and $g: \{1,2,5\} \rightarrow \{1,3\}$ be given by $f = \{(1,2), (3,5), (4,1)\}$ and $g = \{(1,3), (2,3), (5,1)\}$. Write down $g \circ f$.

Let $f(x) = \sin \left(\frac{\pi}{6} \sin \left(\frac{\pi}{2} \sin x\right)\right)$ for all $x \in R$ and $g(x) = \frac{\pi}{2} \sin x$ for all $x \in R$. Let $(f \circ g)(x)$ denote $f(g(x))$ and $(g \circ f)(x)$ denote $g(f(x))$. Then which of the following is (are) true?
$(A)$ Range of $f$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$(B)$ Range of $f \circ g$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$(C)$ $\lim _{x \rightarrow 0} \frac{f(x)}{g(x)} = \frac{\pi}{6}$
$(D)$ There is an $x \in R$ such that $(g \circ f)(x) = 1$