Limits, Continuity and Differentiability

For every pair of continuous functions $f,g:[0,1]\to\mathbb R$ such that
$$\max\{f(x):x\in[0,1]\}=\max\{g(x):x\in[0,1]\},$$
the correct statement(s) is(are):

Let $f:[a,b]\to[1,\infty)$ be a continuous function and let $g:\mathbb R\to\mathbb R$ be defined as
$$g(x)=\begin{cases}0&\text{if }x<a,\\\displaystyle\int_a^xf(t)\,dt&\text{if }a\le x\le b,\\\displaystyle\int_a^bf(t)\,dt&\text{if }x>b.\end{cases}$$
Then

Let $f:\mathbb R\to\mathbb R$ and $g:\mathbb R\to\mathbb R$ be respectively given by $f(x)=|x|+1$ and $g(x)=x^2+1$. Define $h:\mathbb R\to\mathbb R$ by
$$h(x)=\begin{cases}\max\{f(x),g(x)\}&\text{if }x\le0,\\\min\{f(x),g(x)\}&\text{if }x>0.\end{cases}$$
The number of points at which $h(x)$ is not differentiable is

The largest value of the nonnegative integer $a$ for which
$$\lim_{x\to1}\left\{\frac{-ax+\sin(x-1)+a}{x+\sin(x-1)-1}\right\}^{\frac{1-x}{1-\sqrt x}}=\frac14$$
is

Let $f_1:\mathbb R\to\mathbb R$, $f_2:[0,\infty)\to\mathbb R$, $f_3:\mathbb R\to\mathbb R$ and $f_4:\mathbb R\to[0,\infty)$ be defined by
$$f_1(x)=\begin{cases}|x| & \text{if } x<0,\\ e^x & \text{if } x\ge0;\end{cases}\qquad f_2(x)=x^2;$$
$$f_3(x)=\begin{cases}\sin x & \text{if } x<0,\\ x & \text{if } x\ge0\end{cases}$$
and
$$f_4(x)=\begin{cases}f_2(f_1(x)) & \text{if } x<0,\\ f_2(f_1(x))-1 & \text{if } x\ge0.\end{cases}$$

$\displaystyle\lim_{x\to0}\frac{\sin\left(\pi\cos^2x\right)}{x^2}$ is equal to :

If $f(x)$ is continuous and $f\left(\frac92\right)=\frac29$, then $\displaystyle\lim_{x\to0}f\left(\frac{1-\cos3x}{x^2}\right)$ is equal to :

If
$$\lim_{x\to2}\frac{\tan(x-2)\left\{x^2+(k-2)x-2k\right\}}{x^2-4x+4}=5,$$
then $k$ is equal to :

Let $f(x)=x|x|$, $g(x)=\sin x$ and $h(x)=(g\circ f)(x)$. Then

Let $f,g:\mathbb R\to\mathbb R$ be two functions defined by
$$f(x)=\begin{cases}x\sin\left(\dfrac1x\right),&x\ne0\\0,&x=0\end{cases},\ \text{and } g(x)=xf(x)$$

Statement I : $f$ is a continuous function at $x=0$.

Statement II : $g$ is a differentiable function at $x=0$.

If the function
$$f(x)=\begin{cases}\dfrac{\sqrt{2+\cos x}-1}{(\pi-x)^2},&x\ne\pi\\k,&x=\pi\end{cases}$$
is continuous at $x=\pi$, then k equals :

Let $f:\mathbb R\to\mathbb R$ be a function such that $|f(x)|\le x^2$, for all $x\in\mathbb R$. Then, at $x=0$, $f$ is :

For $a\in\mathbb R$ (the set of all real numbers), $a\ne-1$,
$$\lim_{n\to\infty}\frac{1^a+2^a+\cdots+n^a}{(n+1)^{a-1}\left[(na+1)+(na+2)+\cdots+(na+n)\right]}=\frac1{60}.$$
Then $a=$

Let $f(x)=\begin{cases}x^2\left|\cos\dfrac\pi x\right|,&x\ne0\\0,&x=0\end{cases}$, $x\in\mathbb R$, then $f$ is

If $\displaystyle\lim_{x\to\infty}\left(\frac{x^2+x+1}{x+1}-ax-b\right)=4$, then

Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation
$$\left(\sqrt[3]{1+a}-1\right)x^2+\left(\sqrt{1+a}-1\right)x+\left(\sqrt[6]{1+a}-1\right)=0$$
where $a>-1$. Then $\displaystyle\lim_{a\to0^+}\alpha(a)$ and $\displaystyle\lim_{a\to0^+}\beta(a)$ are

For every integer $n$, let $a_n$ and $b_n$ be real numbers. Let function $f:\mathbb R\to\mathbb R$ be given by
$$f(x)=\begin{cases}a_n+\sin\pi x, & \text{for } x\in[2n,2n+1]\\ b_n+\cos\pi x, & \text{for } x\in(2n-1,2n)\end{cases},\quad\text{for all integers } n.$$
If $f$ is continuous, then which of the following hold(s) for all $n$?

Let $f:\mathbb R\to\mathbb R$ be a function such that
$$f(x+y)=f(x)+f(y),\quad\forall x,y\in\mathbb R.$$
If $f(x)$ is differentiable at $x=0$, then

If
$$\lim_{x\to0}\left[1+x\ln(1+b^2)\right]^{\frac1x}=2b\sin^2\theta,\quad b>0\ \text{and}\ \theta\in(-\pi,\pi],$$
then the value of $\theta$ is

If
$$f(x)=\begin{cases}-x-\dfrac\pi2, & x\le-\dfrac\pi2\\[4pt]-\cos x, & -\dfrac\pi2<x\le0\\[4pt]x-1, & 0<x\le1\\[2pt]\ln x, & x>1,\end{cases}$$
then