Let $f: R \rightarrow (0, \infty)$ be a twice differentiable function such that $f(3) = 18$, $f'(3) = 0$, and $f''(3) = 4$. Then $\lim_{x \rightarrow 1} \left( \log_{e} \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-1)^{2}}} \right)$ is equal to:

Let $f: (-\infty, \infty) - \{0\} \rightarrow R$ be a differentiable function such that $f^{\prime}(1) = \lim_{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right)$. Then $\lim_{a \rightarrow \infty} \left[ \frac{a(a+1)}{2} \tan^{-1}\left(\frac{1}{a}\right) + a^2 - 2 \log_e a \right]$ is equal to

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 $f(x) = \begin{cases} \int_{0}^{x} |1-t| dt, & x > 1 \\ x - \frac{1}{2}, & x \leq 1 \end{cases}$. Then:

If $f(x) = 4x^3 - x^2 - 2x + 1$ and $g(x) = \begin{cases} \min \{f(t) : 0 \le t \le x\} & ; 0 \le x \le 1 \\ 3 - x & ; 1 < x \le 2 \end{cases}$,then the value of $g\left( \frac{1}{4} \right) + g\left( \frac{3}{4} \right) + g\left( \frac{5}{4} \right)$ is:

Let $f: R \rightarrow R$ be a function defined by $f(x) = \begin{cases} \max_{t \leq x} \{t^3 - 3t\} & x \leq 2 \\ x^2 + 2x - 6 & 2 < x < 3 \\ [x-3] + 9 & 3 \leq x \leq 5 \\ 2x + 1 & x > 5 \end{cases}$ where $[t]$ is the greatest integer less than or equal to $t$. Let $m$ be the number of points where $f$ is not differentiable and $I = \int_{-2}^{2} f(x) dx$. Then the ordered pair $(m, I)$ is equal to: