The function $f(x) = \begin{cases} |x - 3| & x \geqslant 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4} & x < 1 \end{cases}$ is :

Let $f$ and $g$ be real-valued functions defined on the interval $(-1, 1)$ such that $g^{\prime \prime}(x)$ is continuous,$g(0) \neq 0$,$g^{\prime}(0) = 0$,$g^{\prime \prime}(0) \neq 0$,and $f(x) = g(x) \sin x$.
$STATEMENT-1$: $\lim_{x \rightarrow 0} [g(x) \cot x - g(0) \operatorname{cosec} x] = f^{\prime \prime}(0)$.
$STATEMENT-2$: $f^{\prime}(0) = g(0)$.

Let $f: R \to R$ be a twice differentiable function such that the quadratic equation $f(x)m^{2}-2f^{\prime}(x)m+f^{\prime\prime}(x)=0$ in $m$ has two equal roots for every $x \in R$. If $f(0)=1$,$f^{\prime}(0)=2$ and $(\alpha, \beta)$ is the largest interval in which the function $g(x) = f(\log_{e}x-x)$ is increasing,then $\alpha+\beta$ is equal to:

$\frac{1}{e^{3x}}(e^x + e^{5x}) = a_0 + a_1x + a_2x^2 + \ldots$
$\Rightarrow 2a_1 + 2^3a_3 + 2^5a_5 + \ldots$ is equal to

Let $f : R \to R$ be a differentiable function satisfying $f''(3) + f'(2) = 0$. Then $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + f\left( {3 + x} \right) - f\left( 3 \right)}}{{1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)^{\frac{1}{x}}}$ is equal to

If $y=|\cos x-\sin x|+|\tan x-\cot x|$,then $\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{3}}+\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{6}}=$