Continuity Questions in English

(D) We analyze the continuity of each function on the interval $(0, \pi )$:
$1$. $f(x) = \cot x = \frac{\cos x}{\sin x}$. This function is discontinuous at $x = \pi /2$ is incorrect,it is discontinuous at $x = n\pi$. Within $(0, \pi )$,$\sin x$ is never $0$,so $f(x)$ is continuous.
$2$. $g(x) = \int_{0}^{x} t \sin \frac{1}{t} \, dt$. Since the integrand $t \sin(1/t)$ is bounded and continuous on $(0, \pi )$,the integral function $g(x)$ is continuous on $(0, \pi )$.
$3$. $h(x)$ is defined piecewise. At $x = 3\pi /4$,the left-hand limit is $1$. The right-hand limit is $2 \sin(\frac{2}{9} \cdot \frac{3\pi}{4}) = 2 \sin(\frac{\pi}{6}) = 2(1/2) = 1$. Since the limits match,$h(x)$ is continuous.
$4$. $l(x)$ is defined piecewise. At $x = \pi /2$,the left-hand limit is $\frac{\pi}{2} \sin(\frac{\pi}{2}) = \frac{\pi}{2}$. The right-hand limit is $\frac{\pi}{2} \sin(\frac{\pi}{2} + \pi) = \frac{\pi}{2} \sin(\frac{3\pi}{2}) = \frac{\pi}{2}(-1) = -\frac{\pi}{2}$. Since $\frac{\pi}{2} \neq -\frac{\pi}{2}$,$l(x)$ is discontinuous at $x = \pi /2$.

(C) The function is defined as $f(x) = \frac{3 \log_{\sin |x|} \cos x}{3 \log_{\sin |3x|} \cos (x/2)} = \frac{\ln \cos x}{\ln \sin |x|} \times \frac{\ln \sin |3x|}{\ln \cos (x/2)}$.
For $f(x)$ to be defined,the base of the logarithm must be positive and not equal to $1$,and the argument must be positive.
$1$. $\sin |x| > 0$ and $\sin |x| \neq 1 \implies |x| \in (0, \pi) \setminus \{\pi/2\}$. Given $|x| < \pi/3$,this is satisfied for all $x \in (-\pi/3, \pi/3) \setminus \{0\}$.
$2$. $\sin |3x| > 0$ and $\sin |3x| \neq 1 \implies |3x| \in (0, \pi) \setminus \{\pi/2\} \implies |x| \in (0, \pi/3) \setminus \{\pi/6\}$.
Thus,$f(x)$ is undefined at $x = \pm \pi/6$ and $x = 0$ (initially). Since $f(0)=4$,we check $\lim_{x \to 0} f(x)$.
Using $\ln \cos x \approx -x^2/2$,$\ln \sin |x| \approx \ln |x|$,$\ln \sin |3x| \approx \ln |3x|$,$\ln \cos (x/2) \approx -x^2/8$,we get $\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{-x^2/2}{\ln |x|} \cdot \frac{\ln |3x|}{-x^2/8} = 4 \lim_{x \to 0} \frac{\ln |3x|}{\ln |x|} = 4 \lim_{x \to 0} \frac{\ln 3 + \ln |x|}{\ln |x|} = 4(1) = 4$.
Since $\lim_{x \to 0} f(x) = f(0) = 4$,$f$ is continuous at $x = 0$.
The points of discontinuity are $x = \pi/6$ and $x = -\pi/6$. Total $2$ points.

(C) First,we find $f'(0)$ using the definition of the derivative:
$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^m \sin(1/h) - 0}{h} = \lim_{h \to 0} h^{m-1} \sin(1/h)$.
For this limit to exist and be equal to $0$,we must have $m-1 > 0$,i.e.,$m > 1$.
Now,for $x \neq 0$,$f'(x) = m x^{m-1} \sin(1/x) - x^{m-2} \cos(1/x)$.
For $f'(x)$ to be continuous at $x = 0$,we require $\lim_{x \to 0} f'(x) = f'(0) = 0$.
$\lim_{x \to 0} [m x^{m-1} \sin(1/x) - x^{m-2} \cos(1/x)] = 0$.
The first term $m x^{m-1} \sin(1/x)$ approaches $0$ if $m-1 > 0$,i.e.,$m > 1$.
The second term $x^{m-2} \cos(1/x)$ approaches $0$ if $m-2 > 0$,i.e.,$m > 2$.
Since $m \in N$,the smallest integer $m$ satisfying $m > 2$ is $m = 3$.

(A) The function $h(x)$ is known as the Thomae's function (or the popcorn function).
$1$. For any rational number $x = \frac{p}{q}$ (where $p, q$ are relatively prime),$h(x) = \frac{1}{q}$. As $x$ approaches a rational number,the values of $h(x)$ for nearby irrational numbers approach $0$. Since $\frac{1}{q} \neq 0$,the function is discontinuous at every rational point. Thus,statement $C$ is true.
$2$. For any irrational number $x$,$h(x) = 0$. For any $\epsilon > 0$,there are only finitely many rational numbers $\frac{p}{q}$ in any interval such that $\frac{1}{q} \geq \epsilon$. Thus,the limit of $h(x)$ as $x$ approaches any irrational number is $0$. Since $h(x) = 0$ at irrational points,the function is continuous at every irrational point. Thus,statement $B$ is true.
$3$. Since the function is continuous at irrational points and discontinuous at rational points,it is not discontinuous for all $x$ in $(0, \infty)$. Therefore,statement $A$ is false.
$4$. Since the function is discontinuous at every rational point,it is not differentiable (derivable) at any rational point. Thus,statement $D$ is true.
Conclusion: Statement $A$ does not hold good.

(D) To determine the continuity of the function $f(x)$,we check the limits at the transition points $x = 0$ and $x = 4$.
At $x = 0$:
Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \sqrt{-x} = 0$.
Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 0 = 0$.
Value of the function: $f(0) = 0$.
Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0$,the function is continuous at $x = 0$.
At $x = 4$:
Left-hand limit: $\lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} 0 = 0$.
Right-hand limit: $\lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} (x - 4) = 4 - 4 = 0$.
Value of the function: $f(4) = 0$.
Since $\lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = f(4) = 0$,the function is continuous at $x = 4$.
Since the function is continuous at all points in its domain,it is continuous on the entire real line.

(D) According to the algebra of continuous functions,if two functions $g(x)$ and $h(x)$ are continuous at a point $x = c$,then their quotient $\frac{g(x)}{h(x)}$ is also continuous at $x = c$,provided that the denominator $h(c) \neq 0$.
Since $g$ and $h$ are continuous on the open interval $(a, b)$,the function $f(x) = \frac{g(x)}{h(x)}$ is continuous at all points $x \in (a, b)$ where $h(x) \neq 0$.

(D) For $f(x)$ to be continuous at $x = 0$,$f(0) = \lim_{x \to 0} f(x)$.
$\lim_{x \to 0} \frac{x - e^x + \cos 2x}{x^2} = \lim_{x \to 0} \frac{x - (1 + x + \frac{x^2}{2!} + \dots) + (1 - \frac{(2x)^2}{2!} + \dots)}{x^2}$
$= \lim_{x \to 0} \frac{x - 1 - x - \frac{x^2}{2} + 1 - 2x^2}{x^2} = \lim_{x \to 0} \frac{-\frac{5}{2}x^2}{x^2} = -\frac{5}{2} = -2.5$.
Thus,$f(0) = -2.5$.
Now,$[f(0)] = [-2.5] = -3$.
And $\{f(0)\} = f(0) - [f(0)] = -2.5 - (-3) = 0.5$.
Therefore,$[f(0)] \cdot \{f(0)\} = (-3) \cdot (0.5) = -1.5$.

(C) To check for continuity at $x = 2$,we evaluate the limits from both sides.
For $x > 2$,$f(x) = \frac{2^x + 8 \cdot 2^{-x} - 6}{2^{-x/2} - 2 \cdot 2^{-x}}$. Let $u = 2^{-x/2}$,then $2^x = u^{-2}$ and $2^{-x} = u^2$. As $x \to 2^+$,$u \to 2^{-1} = 1/2$. The expression becomes $\frac{u^{-2} + 8u^2 - 6}{u - 2u^2} = \frac{1 + 8u^4 - 6u^2}{u^2(u - 2u^2)} = \frac{(2u^2 - 1)(4u^2 - 1)}{u^2(u(1 - 2u))} = \frac{-(2u^2 - 1)(2u - 1)(2u + 1)}{u^3(2u - 1)} = \frac{-(2u^2 - 1)(2u + 1)}{u^3}$. Substituting $u = 1/2$,we get $\frac{-(2(1/4) - 1)(1 + 1)}{1/8} = \frac{-( -1/2)(2)}{1/8} = 8$. Thus,$f(2^+) = 8$.
For $x < 2$,$f(x) = \frac{x^2 - 4}{x - \sqrt{3x - 2}}$. Rationalizing the denominator: $\frac{(x^2 - 4)(x + \sqrt{3x - 2})}{x^2 - (3x - 2)} = \frac{(x - 2)(x + 2)(x + \sqrt{3x - 2})}{x^2 - 3x + 2} = \frac{(x - 2)(x + 2)(x + \sqrt{3x - 2})}{(x - 2)(x - 1)} = \frac{(x + 2)(x + \sqrt{3x - 2})}{x - 1}$. Substituting $x = 2$,we get $\frac{(4)(2 + \sqrt{6 - 2})}{2 - 1} = 4(2 + 2) = 16$. Thus,$f(2^-) = 16$.
Since $f(2^+) \neq f(2^-)$,the function has a jump discontinuity at $x = 2$.

(A) First,we simplify the expression for $f(x)$ for $x \neq 0$:
If $x > 0$,then $|x| = x$,so $\frac{1}{|x|} + \frac{1}{x} = \frac{1}{x} + \frac{1}{x} = \frac{2}{x}$. Thus,$f(x) = (x + 1) e^{-2/x}$.
If $x < 0$,then $|x| = -x$,so $\frac{1}{|x|} + \frac{1}{x} = -\frac{1}{x} + \frac{1}{x} = 0$. Thus,$f(x) = (x + 1) e^0 = x + 1$.
Given $f(0) = 0$,the function is:
$f(x) = \begin{cases} (x + 1) e^{-2/x} & x > 0 \\ 0 & x = 0 \\ x + 1 & x < 0 \end{cases}$
Now,check continuity at $x = 0$:
Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x + 1) = 1$.
Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1) e^{-2/x} = (1) \cdot e^{-\infty} = 0$.
Since $\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$,the function is discontinuous at $x = 0$.
Therefore,option $(A)$ is false as the function is not continuous for all $x \in I$.

(B) To check the continuity at $x = 2$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 2$.
$1$. Left-hand limit $(LHL)$: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{x}{[x]}$. Since $1 \leqslant x < 2$,$[x] = 1$. Thus,$\lim_{x \to 2^-} \frac{x}{1} = 2$.
$2$. Right-hand limit $(RHL)$: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \sqrt{6-x} = \sqrt{6-2} = \sqrt{4} = 2$.
$3$. Value of the function: $f(2) = 1$.
Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 2$,but $f(2) = 1$,the limit exists but is not equal to the function value. This type of discontinuity,where the limit exists but the point is defined elsewhere,is called a removable discontinuity (specifically,an isolated point removable discontinuity).

(D) Explanation: Let $f(x)$ be a continuous function on the interval $[a, b]$ such that $f(x) \in \mathbb{Z}$ for all $x \in [a, b]$.
According to the Intermediate Value Theorem,if $f$ is continuous on $[a, b]$,then $f$ must take all values between $f(a)$ and $f(b)$.
Suppose $f$ is not constant. Then there exist $x_1, x_2 \in [a, b]$ such that $f(x_1) = n$ and $f(x_2) = m$ with $n \neq m$. Without loss of generality,assume $n < m$.
Since $f$ is continuous,for any value $y$ such that $n < y < m$,there must exist some $c \in (x_1, x_2)$ such that $f(c) = y$.
However,we are given that $f(x)$ is an integer for all $x \in [a, b]$.
If we choose $y = n + 0.5$,which is not an integer,this contradicts the condition that $f(x)$ must be an integer.
Therefore,the assumption that $f$ is not constant must be false.
Thus,$f(x)$ must be a constant function.

(A) Let $g(x) = \sin x - \sin^3 x = \sin x(1 - \sin^2 x) = \sin x \cos^2 x$.
For $x \in (0, \pi)$,$\sin x > 0$ and $\cos^2 x \geq 0$. Thus,$g(x) \geq 0$.
Specifically,for $x \neq \frac{\pi}{2}$,$g(x) > 0$.
Since $g(x) > 0$ for $x \in (0, \pi) \setminus \{\frac{\pi}{2}\}$,we have $|g(x)| = g(x)$.
Substituting this into the expression for $f(x)$:
$f(x) = \left[ \frac{2g(x) + g(x)}{2g(x) - g(x)} \right] = \left[ \frac{3g(x)}{g(x)} \right] = [3] = 3$.
Thus,$f(x) = 3$ for all $x \in (0, \pi) \setminus \{\frac{\pi}{2}\}$.
Given $f(\frac{\pi}{2}) = 3$,the function is $f(x) = 3$ for all $x \in (0, \pi)$.
$A$ constant function is continuous and differentiable everywhere in its domain.
Therefore,$f$ is continuous and differentiable at $x = \frac{\pi}{2}$.

(B) To determine the continuity and differentiability of $f(x)$ at $x = 1$,we first examine the domain of the function.
For $f(x)$ to be defined,all square root terms must have non-negative arguments:
$1$) $x^3 - 1 \geq 0 \implies x \geq 1$
$2$) $1 - x^4 \geq 0 \implies x^4 \leq 1 \implies -1 \leq x \leq 1$
$3$) $x - 1 \geq 0 \implies x \geq 1$
For the function to be defined,all conditions must be satisfied simultaneously. The intersection of $x \geq 1$ and $-1 \leq x \leq 1$ is only the point $x = 1$.
Since the domain of the function is the singleton set $\{1\}$,the function is only defined at $x = 1$.
Continuity and differentiability are defined for functions on an interval. Since $f(x)$ is not defined in any neighborhood of $x = 1$,it cannot be continuous or differentiable at $x = 1$ in the standard sense.
However,in the context of such problems,if we consider the domain constraints,the function is discontinuous at $x = 1$ because the limit does not exist.
Thus,the correct option is $B$.

(A) For $f(x)$ to be continuous at $x = 5$,the limit $\lim_{x \to 5} f(x)$ must exist and be equal to $f(5)$.
Given $f(x) = \frac{x^2 - bx + 25}{(x - 2)(x - 5)}$.
Since the denominator is zero at $x = 5$,the numerator must also be zero at $x = 5$ for the limit to exist.
Substituting $x = 5$ into the numerator: $5^2 - 5b + 25 = 0 \Rightarrow 25 - 5b + 25 = 0 \Rightarrow 50 = 5b \Rightarrow b = 10$.
Now,substitute $b = 10$ into the function: $f(x) = \frac{x^2 - 10x + 25}{(x - 2)(x - 5)} = \frac{(x - 5)^2}{(x - 2)(x - 5)}$.
For $x \neq 5$,we can simplify this to $f(x) = \frac{x - 5}{x - 2}$.
Now,calculate the limit: $\lim_{x \to 5} f(x) = \lim_{x \to 5} \frac{x - 5}{x - 2} = \frac{5 - 5}{5 - 2} = \frac{0}{3} = 0$.
Since $f$ is continuous at $x = 5$,$f(5) = \lim_{x \to 5} f(x) = 0$.

(D) Statement $I$ is false because differentiability on an open interval does not guarantee continuity at the endpoints.
Statement $II$ is false. Consider the function $f(x) = 1/x$ on the interval $(0, 1)$. This function is differentiable on $(0, 1)$ but is not bounded as $x \to 0^+$.
Statement $III$ is true. Since $f$ is differentiable on $(a, b)$,it is continuous on $(a, b)$. Therefore,$f$ is continuous on the closed sub-interval $[a_1, b_1]$. By the Intermediate Value Theorem,since $f(a_1) < 0 < f(b_1)$,there must exist at least one $c \in (a_1, b_1)$ such that $f(c) = 0$.

(C) To determine if $f$ can be continuous at $x = 2$,we examine the limit $\lim_{x \to 2} f(x)$.
Consider the left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{1}{x + 2^{\frac{1}{x - 2}}}$.
As $x \to 2^-$,$(x - 2) \to 0^-$,so $\frac{1}{x - 2} \to -\infty$.
Thus,$2^{\frac{1}{x - 2}} \to 2^{-\infty} = 0$.
Therefore,$\lim_{x \to 2^-} f(x) = \frac{1}{2 + 0} = 1/2$.
Now consider the right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{1}{x + 2^{\frac{1}{x - 2}}}$.
As $x \to 2^+$,$(x - 2) \to 0^+$,so $\frac{1}{x - 2} \to +\infty$.
Thus,$2^{\frac{1}{x - 2}} \to 2^{+\infty} = \infty$.
Therefore,$\lim_{x \to 2^+} f(x) = \frac{1}{2 + \infty} = 0$.
Since $\lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x)$,the limit does not exist at $x = 2$.
Consequently,$f$ cannot be continuous at $x = 2$ regardless of the value assigned to $f(2)$.

(A) For $f(x)$ to be continuous at $x = \frac{1}{2}$,we must have $\lim_{x \to \frac{1}{2}^-} f(x) = \lim_{x \to \frac{1}{2}^+} f(x) = f(\frac{1}{2}) = p$.
First,calculate the left-hand limit:
$\lim_{x \to \frac{1}{2}^-} \frac{1 + \cos 2\pi x}{1 - \sin \pi x} = \lim_{x \to \frac{1}{2}^-} \frac{2 \cos^2 \pi x}{1 - \sin \pi x} = \lim_{x \to \frac{1}{2}^-} \frac{2(1 - \sin^2 \pi x)}{1 - \sin \pi x} = \lim_{x \to \frac{1}{2}^-} 2(1 + \sin \pi x) = 2(1 + 1) = 4$.
Next,calculate the right-hand limit:
Let $t = \sqrt{2x - 1}$. As $x \to \frac{1}{2}^+$,$t \to 0^+$.
$\lim_{t \to 0^+} \frac{t}{\sqrt{4 + t} - 2} = \lim_{t \to 0^+} \frac{t(\sqrt{4 + t} + 2)}{(4 + t) - 4} = \lim_{t \to 0^+} (\sqrt{4 + t} + 2) = \sqrt{4} + 2 = 4$.
Since $\lim_{x \to \frac{1}{2}^-} f(x) = \lim_{x \to \frac{1}{2}^+} f(x) = 4$,the function is continuous at $x = \frac{1}{2}$ if $p = 4$.
Therefore,$f(x)$ is discontinuous at $x = \frac{1}{2}$ if $p \neq 4$,which means $p \in R - \{4\}$.

(A) For the function to be continuous at $x = -1$,the left-hand limit,right-hand limit,and the value of the function at $x = -1$ must be equal.
First,find the value of the function at $x = -1$:
$f(-1) = b([-1]^2 + [-1]) + 1 = b(1 - 1) + 1 = 1$.
Next,find the right-hand limit $(RHL)$ as $x \to -1^+$:
$\lim_{x \to -1^+} f(x) = \lim_{h \to 0} b([-1+h]^2 + [-1+h]) + 1 = b((-1)^2 + (-1)) + 1 = b(1-1) + 1 = 1$.
Now,find the left-hand limit $(LHL)$ as $x \to -1^-$:
$\lim_{x \to -1^-} f(x) = \lim_{h \to 0} \sin(\pi(-1-h+a)) = \sin(\pi(a-1))$.
For continuity,$LHL$ = $RHL$ = $f(-1)$,so $\sin(\pi(a-1)) = 1$.
This implies $\pi(a-1) = 2n\pi + \frac{\pi}{2}$ for any integer $n \in I$.
Dividing by $\pi$,we get $a - 1 = 2n + \frac{1}{2}$,which simplifies to $a = 2n + \frac{3}{2}$.
Since $b$ does not affect the limit at $x = -1$ (as the $RHL$ is independent of $b$),$b$ can be any real number $(b \in R)$.

(D) For $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x \to 0^+} f(x)$.
$f(0) = \lim_{x \to 0^+} \frac{\ln(e^{x^2} + 2\sqrt{x})}{\sqrt{x}}$
Using the standard limit $\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1$,we rewrite the expression as:
$f(0) = \lim_{x \to 0^+} \frac{\ln(1 + (e^{x^2} + 2\sqrt{x} - 1))}{e^{x^2} + 2\sqrt{x} - 1} \times \frac{e^{x^2} + 2\sqrt{x} - 1}{\sqrt{x}}$
As $x \to 0^+$,$e^{x^2} + 2\sqrt{x} - 1 \to 0$,so the first part of the limit is $1$.
$f(0) = 1 \times \lim_{x \to 0^+} \left( \frac{e^{x^2} - 1}{\sqrt{x}} + \frac{2\sqrt{x}}{\sqrt{x}} \right)$
$f(0) = \lim_{x \to 0^+} \left( \frac{e^{x^2} - 1}{x^2} \cdot x^{3/2} + 2 \right)$
Since $\lim_{x \to 0^+} \frac{e^{x^2} - 1}{x^2} = 1$ and $\lim_{x \to 0^+} x^{3/2} = 0$,the first term vanishes.
$f(0) = 0 + 2 = 2$.

(D) For $x \to 0^+$,$[x] = 0$ and $\{x\} = x$. Thus,$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\tan^2 x}{x^2 - 0^2} = \lim_{x \to 0^+} \left( \frac{\tan x}{x} \right)^2 = 1^2 = 1$.
For $x \to 0^-$,let $x = -h$ where $h > 0$. Then $[x] = -1$ and $\{x\} = x - [x] = -h - (-1) = 1 - h$.
$\lim_{x \to 0^-} f(x) = \lim_{h \to 0^+} \sqrt{(1-h) \cot(1-h)} = \sqrt{1 \cdot \cot(1)} = \sqrt{\cot 1}$.
Now,$\left( \lim_{x \to 0^-} f(x) \right)^2 = (\sqrt{\cot 1})^2 = \cot 1$.
Therefore,$\cot^{-1} \left( \lim_{x \to 0^-} f(x) \right)^2 = \cot^{-1}(\cot 1) = 1$.
Thus,both $(A)$ and $(C)$ are correct.

(D) function $f(x)$ has a removable discontinuity at $x = a$ if $\lim_{x \to a} f(x)$ exists but is not equal to $f(a)$ (or $f(a)$ is undefined).
For option $A$: $\lim_{x \to 0} \frac{1}{\ln |x|} = \frac{1}{-\infty} = 0$. Since the limit exists,it has a removable discontinuity.
For option $B$: $\lim_{x \to 0^+} \cos \left( \frac{\sin x}{x} \right) = \cos(1)$ and $\lim_{x \to 0^-} \cos \left( \frac{-\sin x}{x} \right) = \cos(-1) = \cos(1)$. Since the limit exists,it has a removable discontinuity.
For option $C$: $\lim_{x \to 0} x \sin \left( \frac{\pi}{x} \right)$. Since $|\sin(\pi/x)| \le 1$,by the Squeeze Theorem,$\lim_{x \to 0} x \sin \left( \frac{\pi}{x} \right) = 0$. Since the limit exists,it has a removable discontinuity.
Therefore,all the functions have a removable discontinuity at $x = 0$.

(D) The function is $f(x) = \cos \left[ \frac{\pi}{x} \right] \cos \left( \frac{\pi}{2} (x - 1) \right)$.
At $x = 0$,the function is undefined because the denominator in $\frac{\pi}{x}$ is zero.
At $x = 1$,$f(1) = \cos \left[ \frac{\pi}{1} \right] \cos \left( \frac{\pi}{2} (1 - 1) \right) = \cos(3) \cos(0) = \cos(3)$. The limit as $x \to 1$ exists and equals $f(1)$,so it is continuous at $x = 1$.
At $x = 2$,$f(2) = \cos \left[ \frac{\pi}{2} \right] \cos \left( \frac{\pi}{2} (2 - 1) \right) = \cos(1) \cos \left( \frac{\pi}{2} \right) = \cos(1) \times 0 = 0$. The limit as $x \to 2$ exists and equals $f(2)$,so it is continuous at $x = 2$.
Therefore,the function is continuous at both $x = 1$ and $x = 2$.

(D) Given the function $f(x) = x^3 + 7x - 1$.
Evaluating at the endpoints of the interval $[0, 1]$:
$f(0) = (0)^3 + 7(0) - 1 = -1$.
$f(1) = (1)^3 + 7(1) - 1 = 1 + 7 - 1 = 7$.
Since $f(x)$ is a polynomial,it is continuous on the interval $[0, 1]$.
Because $f(0) = -1$ and $f(1) = 7$,the function changes sign from negative to positive over the interval.
According to the Intermediate Value Theorem,if a continuous function $f$ takes on values of opposite signs at the endpoints of an interval,there must exist at least one value $c$ in that interval such that $f(c) = 0$.
Therefore,the theorem that describes this is the Intermediate Value Theorem.

(D) First,analyze the limit of $f(x)$ as $x \to \frac{1}{2}$.
$\lim_{x \to \frac{1}{2}} f(x) = \lim_{x \to \frac{1}{2}} \frac{(x - 1)(6x - 1)}{2x - 1}$.
As $x \to \frac{1}{2}$,the numerator $(x - 1)(6x - 1) \to (\frac{1}{2} - 1)(6(\frac{1}{2}) - 1) = (-\frac{1}{2})(2) = -1$.
The denominator $2x - 1 \to 0$.
Since the numerator approaches a non-zero constant $(-1)$ and the denominator approaches $0$,the limit $\lim_{x \to \frac{1}{2}} f(x)$ does not exist (it approaches $\pm \infty$).
Therefore,the function has a non-removable infinite discontinuity at $x = \frac{1}{2}$.

(C) To check continuity at $x = 0$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.
For $x > 0$,$[x] = 0$,so $\text{sgn}[x] = 0$. Thus,$f(x) = \frac{\ln(1+x^2)}{1-\cos x}$.
$RHL = \mathop {\lim }\limits_{h \to 0^+} \frac{\ln(1+h^2)}{1-\cos h} = \mathop {\lim }\limits_{h \to 0^+} \frac{\ln(1+h^2)}{h^2} \cdot \frac{h^2}{1-\cos h} = 1 \cdot 2 = 2$.
For $x < 0$,$[x] = -1$,so $\text{sgn}[x] = -1$. Thus,$f(x) = \frac{\ln(1-1+x^2)}{1-\cos x} = \frac{\ln(x^2)}{1-\cos x}$.
$LHL = \mathop {\lim }\limits_{h \to 0^+} \frac{\ln(h^2)}{1-\cos(h)} = \mathop {\lim }\limits_{h \to 0^+} \frac{\ln(h^2)}{h^2/2} = -\infty$.
Since the $LHL$ does not exist (it is $-\infty$),the limit $\mathop {\lim }\limits_{x \to 0} f(x)$ does not exist.
Therefore,$f(x)$ has a non-removable discontinuity at $x = 0$ for any value of $k$.

(D) Let $f(x) = [x] + \sqrt{\{x\}}$.
For any non-integer $x$,$[x]$ and $\{x\}$ are continuous,so $f(x)$ is continuous.
Now,check the continuity at an integer $n \in I$.
Left-hand limit: $\lim_{x \to n^-} f(x) = \lim_{x \to n^-} ([x] + \sqrt{\{x\}}) = (n-1) + \sqrt{1} = n-1+1 = n$.
Right-hand limit: $\lim_{x \to n^+} f(x) = \lim_{x \to n^+} ([x] + \sqrt{\{x\}}) = n + \sqrt{0} = n$.
Value of the function: $f(n) = [n] + \sqrt{\{n\}} = n + 0 = n$.
Since $\lim_{x \to n^-} f(x) = \lim_{x \to n^+} f(x) = f(n) = n$,the function is continuous at all integers.
Therefore,$f(x)$ is continuous $\forall x \in R$.

(B) To find the limit of $f(x)$ as $x \to 0$,we use the Taylor series expansions:
$e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots = 1 + 2x + 2x^2 + \frac{4x^3}{3} + \dots$
$(1 + 4x)^{1/2} = 1 + \frac{1}{2}(4x) + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(4x)^2 + \dots = 1 + 2x - 2x^2 + \dots$
$\ln(1 - x^2) = -x^2 - \frac{(-x^2)^2}{2} - \dots = -x^2 - \frac{x^4}{2} - \dots$
Substituting these into the expression for $f(x)$:
$f(x) = \frac{(1 + 2x + 2x^2 + \dots) - (1 + 2x - 2x^2 + \dots)}{-x^2 - \frac{x^4}{2} - \dots}$
$f(x) = \frac{4x^2 + \dots}{-x^2 - \dots}$
Taking the limit as $x \to 0$:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{4x^2}{-x^2} = -4$
Since the limit exists and is finite,$f$ has a removable discontinuity at $x = 0$ and $f(0) = -4$.

(B) Let us analyze the function $f(x)$ for different intervals.
For $x \notin I$,$f(x) = sgn([x])$.
If $x \in (0, 1)$,$[x] = 0$,so $f(x) = sgn(0) = 0$.
If $x \in (1, 2)$,$[x] = 1$,so $f(x) = sgn(1) = 1$.
If $x \in (-1, 0)$,$[x] = -1$,so $f(x) = sgn(-1) = -1$.
For $x \in I$,$f(x) = [sgn(x)]$.
If $x = 0$,$f(0) = [sgn(0)] = [0] = 0$.
If $x > 0$ and $x \in I$,$f(x) = [sgn(x)] = [1] = 1$.
If $x < 0$ and $x \in I$,$f(x) = [sgn(x)] = [-1] = -1$.
Combining these,the function is discontinuous at every integer $n \in I$ because the left-hand limit and right-hand limit do not match the function value at integers. Specifically,for any integer $n$,$\lim_{x \to n^-} f(x) = sgn(n-1)$ and $\lim_{x \to n^+} f(x) = sgn(n)$. These are not equal to $f(n) = [sgn(n)]$. Thus,the function is discontinuous at every integer.

(B) The function $f(x) = x^2[\sin^{-1}x]$ is discontinuous where the greatest integer function $[\sin^{-1}x]$ is discontinuous.
For the function $g(x) = [\sin^{-1}x]$,the points of discontinuity occur when $\sin^{-1}x$ is an integer.
The domain of $\sin^{-1}x$ is $[-1, 1]$. The possible integer values for $\sin^{-1}x$ are $\{-1, 0, 1\}$.
$1$. $\sin^{-1}x = -1 \implies x = \sin(-1) = -\sin 1$.
$2$. $\sin^{-1}x = 0 \implies x = \sin(0) = 0$.
$3$. $\sin^{-1}x = 1 \implies x = \sin(1) = \sin 1$.
However,we are given that $\alpha, \beta \in R - \{0\}$. Thus,we exclude $x = 0$.
The points of discontinuity in the given domain are $\alpha = \sin 1$ and $\beta = -\sin 1$.
Therefore,the value of $\alpha + \beta = \sin 1 + (-\sin 1) = 0$.

(C) Given $f(x) = \begin{cases} x\{x\}^2, & x \notin I \\ x, & x \in I \end{cases}$ for $x \in [0,3]$.
For $x \in (0,1)$,${x} = x$,so $f(x) = x(x)^2 = x^3$. At $x=0, f(0)=0$. At $x=1, f(1)=1$.
For $x \in (1,2)$,${x} = x-1$,so $f(x) = x(x-1)^2$. At $x=2, f(2)=2$.
For $x \in (2,3)$,${x} = x-2$,so $f(x) = x(x-2)^2$. At $x=3, f(3)=3$.
Thus,$f(x) = \begin{cases} 0, & x=0 \\ x^3, & 0 < x < 1 \\ 1, & x=1 \\ x(x-1)^2, & 1 < x < 2 \\ 2, & x=2 \\ x(x-2)^2, & 2 < x < 3 \\ 3, & x=3 \end{cases}$.
Checking continuity: $\lim_{x \to 1^-} f(x) = 1^3 = 1$,$f(1)=1$,$\lim_{x \to 1^+} f(x) = 1(1-1)^2 = 0$. Since $1 \neq 0$,$f(x)$ is discontinuous at $x=1$.
Similarly,$\lim_{x \to 2^-} f(x) = 2(2-1)^2 = 2$,$f(2)=2$,$\lim_{x \to 2^+} f(x) = 2(2-2)^2 = 0$. Since $2 \neq 0$,$f(x)$ is discontinuous at $x=2$.
Thus,$f(x)$ is discontinuous at $x=1$ and $x=2$. There are two points of discontinuity.
Option $C$ is correct because the function is non-differentiable at $x=1, 2$ (discontinuity) and also at $x=0, 3$ (endpoints) and potentially other points,but the number of points of discontinuity is $2$.

(A) $1$. For $x \in (0, 1)$,$[x] = 0 \Rightarrow f(x) = x\sqrt{1 - 0} = x$. Since $f'(x) = 1 > 0$,$f(x)$ is increasing in $(0, 1)$.
$2$. At $x = 1$,$f(1) = 1\sqrt{1 - 1^2} = 0$. For $x \in (0, 1)$,$f(x) = x$. As $x \to 1^-$,$f(x) \to 1$. Since $f(1) = 0$ and $f(x) > 0$ for $x$ near $1$ from the left,$x = 1$ is not a point of local maxima.
$3$. The domain is determined by $1 - [x]^2 \ge 0 \Rightarrow [x]^2 \le 1 \Rightarrow -1 \le [x] \le 1$. This implies $x \in [-1, 2)$.
$4$. The function is defined as:
$f(x) = \begin{cases} 0, & -1 \le x < 0 \\ x, & 0 \le x < 1 \\ 0, & 1 \le x < 2 \end{cases}$
Since $f(x) \ge 0$ for all $x$ in the domain,it is not a negative function.
$5$. Since $f(x)$ is discontinuous at $x = 1$,Rolle's theorem is not applicable on $[0, 1]$.

(B) For $f(x)$ to be continuous at $x = 0$,$\text{LHL} = \text{RHL} = f(0)$.
$\text{LHL} = \lim_{x \to 0^-} \frac{a + 3\cos x}{x^2}$.
For the limit to exist,the numerator must approach $0$ as $x \to 0$,so $a + 3(1) = 0 \Rightarrow a = -3$.
Using the Taylor expansion $\cos x \approx 1 - \frac{x^2}{2}$,we get $\text{LHL} = \lim_{x \to 0^-} \frac{-3 + 3(1 - \frac{x^2}{2})}{x^2} = \lim_{x \to 0^-} \frac{-3/2 x^2}{x^2} = -\frac{3}{2}$.
$\text{RHL} = \lim_{x \to 0^+} b \tan \left( \frac{\pi}{[x + 3]} \right) = b \tan \left( \frac{\pi}{3} \right) = b\sqrt{3}$.
Since $\text{LHL} = \text{RHL}$,we have $b\sqrt{3} = -\frac{3}{2} \Rightarrow b = -\frac{\sqrt{3}}{2}$.
Thus,$a = -3$ and $b = -\frac{\sqrt{3}}{2}$.

(D) function $f(x)$ is discontinuous where it is undefined.
$1$. The denominator $x(x-3)(x-5)$ is zero at $x = 0, 3, 5$. Thus,the function is undefined at these points.
$2$. The term $\tan x$ is undefined when $x = (2n+1)\frac{\pi}{2}$ for $n \in Z$.
$3$. The term $\tan^{-1}\left(\frac{1}{x-1}\right)$ is defined for all $x \neq 1$. At $x = 1$,the limit $\lim_{x \to 1^+} \tan^{-1}\left(\frac{1}{x-1}\right) = \frac{\pi}{2}$ and $\lim_{x \to 1^-} \tan^{-1}\left(\frac{1}{x-1}\right) = -\frac{\pi}{2}$. Since the left-hand limit and right-hand limit are not equal,the function is discontinuous at $x = 1$.
Combining all these,the set of points of discontinuity is $\{0, 1, 3, 5\} \cup \{(2n+1)\frac{\pi}{2}, n \in Z\}$.

(B) The function is defined as $f(x) = \frac{1}{1 - e^{-\frac{x+1}{x-2}}}$.
First,the function is undefined when the denominator is zero,i.e.,$x - 2 = 0$,which gives $x = 2$.
Second,the function is discontinuous when the denominator of the entire expression is zero,i.e.,$1 - e^{-\frac{x+1}{x-2}} = 0$.
This implies $e^{-\frac{x+1}{x-2}} = 1$,which means $-\frac{x+1}{x-2} = 0$.
This occurs when the numerator is zero,i.e.,$x + 1 = 0$,which gives $x = -1$.
Thus,the points of discontinuity are $x = -1$ and $x = 2$.
Therefore,the number of points of discontinuity is $2$.

(A) Given that $f(x)$ is a continuous,periodic even function with period $T = 4$.
Since $f(0) = 1$ and $f(2) = -1$,by the Intermediate Value Theorem,there must be at least one root in the interval $(0, 2)$.
Because $f$ is an even function,$f(-x) = f(x)$,so there must be at least one root in the interval $(-2, 0)$.
Thus,there are at least $2$ roots in one period interval $(-2, 2)$.
The interval $[-10, 10]$ has a length of $20$,which covers $20/4 = 5$ periods.
Therefore,the minimum number of roots in $[-10, 10]$ is $2 \times 5 = 10$.

(C) Given $f(x) = \sin(\{2^x + [2^x] + [3^{-x}]\})$.
Since $[2^x]$ and $[3^{-x}]$ are integers,we use the property $\{x + n\} = \{x\}$ for any integer $n$.
Thus,$f(x) = \sin(\{2^x + [2^x] + [3^{-x}]\}) = \sin(\{2^x + [3^{-x}]\}) = \sin(\{2^x\})$.
The function $f(x) = \sin(\{2^x\})$ is discontinuous where $2^x$ is an integer,because the fractional part function $\{u\}$ is discontinuous at all integers $u$.
For $x \in [0, 4]$,$2^x$ takes values in the interval $[2^0, 2^4] = [1, 16]$.
The integers in the interval $[1, 16]$ are $1, 2, 3, \dots, 16$.
However,$\{2^x\}$ is continuous at $x$ where $2^x$ is an integer if the limit from the left and right are equal.
At $x=0$,$2^x=1$,$\{2^x\}$ jumps from $0$ to $0$ (as $x \to 0^+$,$2^x \to 1^+$,$\{2^x\} \to 0$). Thus,it is continuous at $x=0$.
For $x > 0$,$2^x = k$ (where $k \in \{2, 3, \dots, 16\}$),the function $\{2^x\}$ has a jump discontinuity.
There are $15$ such values of $k$ $(k=2, 3, \dots, 16)$.
Therefore,the number of points of discontinuity is $15$.

(C) The function $f(x) = \operatorname{sgn}(g(x))$ is discontinuous where $g(x) = 0$.
Let $g(x) = (x^2 - kx + 6)(\sin x - 1/2)$.
The points of discontinuity occur where $x^2 - kx + 6 = 0$ or $\sin x = 1/2$.
For $\sin x = 1/2$ in $(0, 6)$,the solutions are $x = \pi/6 \approx 0.52$ and $x = 5\pi/6 \approx 2.62$.
These provide $2$ points of discontinuity.
For $f(x)$ to have exactly $4$ points of discontinuity,the quadratic $x^2 - kx + 6 = 0$ must have $2$ distinct roots in $(0, 6)$ that are not equal to $\pi/6$ or $5\pi/6$.
For the quadratic $x^2 - kx + 6 = 0$ to have $2$ distinct roots in $(0, 6)$:
$1) D = k^2 - 24 > 0 \Rightarrow k > \sqrt{24} \approx 4.89$.
$2) 0 < \text{vertex} < 6 \Rightarrow 0 < k/2 < 6 \Rightarrow 0 < k < 12$.
$3) f(0) = 6 > 0$ (always true).
$4) f(6) = 36 - 6k + 6 > 0 \Rightarrow 42 > 6k \Rightarrow k < 7$.
Combining these,$k \in (4.89, 7)$.
The integral values for $k$ are $5$ and $6$.
Since we need the maximum integral value,$k = 6$.

(B) Let $g(x) = \frac{x^2 + 1}{x^2[|x|] + 1}$.
Case $1$: If $|x| < 1$,then $[|x|] = 0$. Thus,$g(x) = \frac{x^2 + 1}{0 + 1} = x^2 + 1$. Since $|x| < 1$,$0 \le x^2 < 1$,so $1 \le g(x) < 2$. Therefore,$[g(x)] = 1$.
Case $2$: If $|x| \ge 1$,then $[|x|] \ge 1$. Let $n = [|x|]$,where $n \in \{1, 2, 3, \dots\}$. Then $g(x) = \frac{x^2 + 1}{nx^2 + 1}$.
As $x^2 \to \infty$,$g(x) \to \frac{1}{n}$. Since $n \ge 1$,$0 < \frac{1}{n} \le 1$. Specifically,for $x^2 \ge 1$,$g(x) = \frac{x^2 + 1}{nx^2 + 1} \le 1$. Also,$g(x) > 0$. Thus,$[g(x)] = 0$ for $|x| \ge 1$ except when $g(x) = 1$,which happens at $x^2 = 1$ (i.e.,$x = \pm 1$).
At $x = \pm 1$,$g(1) = \frac{1+1}{1(1)+1} = 1$,so $[g(1)] = 1$.
Thus,$f(x) = 1$ for $|x| < 1$ and $f(x) = 0$ for $|x| > 1$,with $f(1) = f(-1) = 1$.
The function jumps from $1$ to $0$ at $x = 1$ and $x = -1$. Hence,it is discontinuous at two points.

(B) Given $f(x) = | | |x + [x]| - 3[x] | - 5[x] |$ on $[-2, 2]$.
We analyze the function in intervals based on $[x]$:
For $x \in [-2, -1)$,$[x] = -2$. Then $f(x) = | | |x - 2| - 3(-2) | - 5(-2) | = | | |x - 2| + 6 | + 10 | = | |2 - x + 6| + 10 | = |8 - x + 10| = |18 - x| = 18 - x$.
For $x \in [-1, 0)$,$[x] = -1$. Then $f(x) = | | |x - 1| - 3(-1) | - 5(-1) | = | | |x - 1| + 3 | + 5 | = | |1 - x + 3| + 5 | = |4 - x + 5| = |9 - x| = 9 - x$.
For $x \in [0, 1)$,$[x] = 0$. Then $f(x) = | | |x + 0| - 3(0) | - 5(0) | = | |x| | = x$.
For $x \in [1, 2)$,$[x] = 1$. Then $f(x) = | | |x + 1| - 3(1) | - 5(1) | = | |x + 1 - 3| - 5 | = | |x - 2| - 5 | = |(2 - x) - 5| = |-x - 3| = x + 3$.
For $x = 2$,$[x] = 2$. Then $f(2) = | | |2 + 2| - 3(2) | - 5(2) | = | |4 - 6| - 10 | = |-2 - 10| = 12$.
Checking continuity at boundaries:
At $x = -1$: $LHL = 18 - (-1) = 19$,$RHL = 9 - (-1) = 10$. Discontinuous.
At $x = 0$: $LHL = 9 - 0 = 9$,$RHL = 0$. Discontinuous.
At $x = 1$: $LHL = 1$,$RHL = 1 + 3 = 4$. Discontinuous.
At $x = 2$: $LHL = 2 + 3 = 5$,$f(2) = 12$. Discontinuous.
Thus,there are $4$ points of discontinuity: $x = -1, 0, 1, 2$.

(D) function has an irremovable discontinuity at $x = a$ if the limit $\lim_{x \to a} f(x)$ does not exist.
$(1)$ For $f(x) = \frac{1}{\ln |x|}$,$\lim_{x \to 0} \frac{1}{\ln |x|} = 0$. Since the limit exists,it has a removable discontinuity.
$(2)$ For $f(x) = \cos \left( \frac{|\sin x|}{x} \right)$:
$\lim_{x \to 0^+} \cos \left( \frac{\sin x}{x} \right) = \cos(1)$
$\lim_{x \to 0^-} \cos \left( \frac{-\sin x}{x} \right) = \cos(-1) = \cos(1)$.
Since the limit exists,it has a removable discontinuity.
$(3)$ For $f(x) = x \sin \frac{\pi}{x}$,by the Squeeze Theorem,$\lim_{x \to 0} x \sin \frac{\pi}{x} = 0$. Since the limit exists,it has a removable discontinuity.
$(4)$ For $f(x) = \frac{1}{1 + 2^{\cot x}}$:
$\lim_{x \to 0^+} \frac{1}{1 + 2^{\cot x}} = \frac{1}{1 + 2^{\infty}} = 0$
$\lim_{x \to 0^-} \frac{1}{1 + 2^{\cot x}} = \frac{1}{1 + 2^{-\infty}} = \frac{1}{1 + 0} = 1$.
Since the left-hand limit $(1)$ is not equal to the right-hand limit $(0)$,the limit does not exist. Thus,it has an irremovable (jump) discontinuity at $x = 0$.

(D) For $f(x)$ to have the greatest value at $x = 1$,we must satisfy $f(x) \le f(1)$ for all $x$ in the domain.
First,calculate $f(1)$ using the first part of the function: $f(1) = 1^3 - 1^2 + 10(1) - 5 = 1 - 1 + 10 - 5 = 5$.
For $x \le 1$,$f(x) = x^3 - x^2 + 10x - 5$. Let $g(x) = x^3 - x^2 + 10x - 5$. Then $g'(x) = 3x^2 - 2x + 10$. The discriminant of $g'(x)$ is $D = (-2)^2 - 4(3)(10) = 4 - 120 = -116 < 0$. Since the leading coefficient is positive,$g'(x) > 0$ for all $x$,meaning $f(x)$ is strictly increasing for $x \le 1$. Thus,$f(x) \le f(1)$ is satisfied for all $x \le 1$.
For $x > 1$,we require $f(x) \le f(1)$,which means $-2x + \log_2(b^2 - 2) \le 5$. As $x \to 1^+$,the limit is $-2(1) + \log_2(b^2 - 2) = -2 + \log_2(b^2 - 2)$. For the function to be continuous or have a maximum at $x=1$,we need $-2 + \log_2(b^2 - 2) \le 5$.
Also,the argument of the logarithm must be positive: $b^2 - 2 > 0 \implies b^2 > 2 \implies b \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)$.
Solving $-2 + \log_2(b^2 - 2) \le 5$:
$\log_2(b^2 - 2) \le 7$
$b^2 - 2 \le 2^7$
$b^2 - 2 \le 128$
$b^2 \le 130$.
Combining $b^2 > 2$ and $b^2 \le 130$,we get $2 < b^2 \le 130$,which implies $b \in [-\sqrt{130}, -\sqrt{2}) \cup (\sqrt{2}, \sqrt{130}]$.
Therefore,the correct option is $D$.

(C) Let $g(x) = x|x^3|$. Since $g(x) = x^4$ for $x \ge 0$ and $g(x) = -x^4$ for $x < 0$,$g(x)$ is $3$ times differentiable at $x=0$ and $g''(0) = 0$. Thus,$g''(x)$ is continuous at $x=0$.
Now consider $h(x) = x^p \sin \left( \frac{1}{x} \right)$ for $x \neq 0$ and $h(0) = 0$.
For $h'(0)$ to exist,$\lim_{h \to 0} \frac{h^p \sin(1/h)}{h} = 0$,which requires $p > 1$. Then $h'(0) = 0$.
For $x \neq 0$,$h'(x) = p x^{p-1} \sin \left( \frac{1}{x} \right) - x^{p-2} \cos \left( \frac{1}{x} \right)$.
For $h''(0)$ to exist,$\lim_{h \to 0} \frac{h'(h) - h'(0)}{h} = \lim_{h \to 0} \left( p h^{p-2} \sin \left( \frac{1}{h} \right) - h^{p-3} \cos \left( \frac{1}{h} \right) \right) = 0$,which requires $p > 3$. Then $h''(0) = 0$.
For $x \neq 0$,$h''(x) = p(p-1) x^{p-2} \sin \left( \frac{1}{x} \right) - p x^{p-3} \cos \left( \frac{1}{x} \right) - (p-2) x^{p-3} \cos \left( \frac{1}{x} \right) - x^{p-4} \sin \left( \frac{1}{x} \right) = p(p-1) x^{p-2} \sin \left( \frac{1}{x} \right) - 2(p-1) x^{p-3} \cos \left( \frac{1}{x} \right) - x^{p-4} \sin \left( \frac{1}{x} \right)$.
For $h''(x)$ to be continuous at $x=0$,we need $\lim_{x \to 0} h''(x) = h''(0) = 0$. This requires the power of $x$ in all terms to be positive,specifically $p-4 > 0$,so $p > 4$.

(C) Define a new function $g(x) = f(x) - x$ for $x \in [0, 1]$.
Since $f : [0, 1] \to [0, 1]$,we have $0 \leq f(x) \leq 1$ for all $x \in [0, 1]$.
At $x = 0$,$g(0) = f(0) - 0 = f(0) \geq 0$.
At $x = 1$,$g(1) = f(1) - 1 \leq 0$ (since $f(1) \leq 1$).
If $g(0) = 0$ or $g(1) = 0$,then $x = 0$ or $x = 1$ is a solution to $f(x) = x$.
If $g(0) > 0$ and $g(1) < 0$,then by the Intermediate Value Theorem,since $g(x)$ is continuous,there must exist at least one $c \in (0, 1)$ such that $g(c) = 0$,which implies $f(c) = c$.
Therefore,the equation $f(x) = x$ must have at least one solution in $[0, 1]$.

(D) Given the function $f(x) = \begin{cases} -x^3 + 1, & x \leq 1 \\ |x - 1| + \lambda, & x > 1 \end{cases}$.
At $x = 1$,the value of the function is $f(1) = -(1)^3 + 1 = 0$.
For $x > 1$,$f(x) = |x - 1| + \lambda = (x - 1) + \lambda$.
As $x \to 1^+$,$f(x) \to \lambda$.
For $x = 1$ to be a point of minima,we must have $f(1) \leq f(x)$ for all $x$ in a neighborhood of $1$.
Since $f(1) = 0$,we require $0 \leq \lambda$ for the function to be continuous or to maintain the minimum at $x=1$ relative to the right-hand side limit.
Specifically,if $\lambda > 0$,then for $x$ slightly greater than $1$,$f(x) = (x-1) + \lambda > 0 = f(1)$.
Thus,$f(x)$ has a point of minima at $x = 1$ for all $\lambda > 0$.

(D) To find $\lim_{x \to 0} f(f(x))$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$ at $x = 0$.
For the $LHL$ $(x \to 0^-)$:
As $x \to 0^-$,$f(x) = 2 - 3x$. Since $x$ is slightly less than $0$,$f(x)$ is slightly greater than $2$.
Let $u = f(x)$. As $x \to 0^-$,$u \to 2^+$.
Since $u > 0$,we use the definition $f(u) = 3 + u$.
Thus,$\lim_{x \to 0^-} f(f(x)) = \lim_{u \to 2^+} (3 + u) = 3 + 2 = 5$.
For the $RHL$ $(x \to 0^+)$:
As $x \to 0^+$,$f(x) = 3 + x$. Since $x$ is slightly greater than $0$,$f(x)$ is slightly greater than $3$.
Let $u = f(x)$. As $x \to 0^+$,$u \to 3^+$.
Since $u > 0$,we use the definition $f(u) = 3 + u$.
Thus,$\lim_{x \to 0^+} f(f(x)) = \lim_{u \to 3^+} (3 + u) = 3 + 3 = 6$.
Since the $LHL$ $(5)$ is not equal to the $RHL$ $(6)$,the limit does not exist.

(B) For $f(x)$ to be continuous at $x = 3$,the left-hand limit must equal the right-hand limit: $\lim_{x \to 3^-} (x^2 - a) = \lim_{x \to 3^+} (b\sqrt{x - 2} + a) \implies 9 - a = b(1) + a \implies 2a + b = 9$.
For $f(x)$ to be continuous at $x = 6$,the left-hand limit must equal the right-hand limit: $\lim_{x \to 6^-} (b\sqrt{x - 2} + a) = \lim_{x \to 6^+} (2x + b) \implies b\sqrt{4} + a = 12 + b \implies 2b + a = 12 + b \implies a + b = 12$.
Solving the system of equations: $2a + b = 9$ and $a + b = 12$. Subtracting the second from the first gives $a = -3$. Substituting $a = -3$ into $a + b = 12$ gives $b = 15$.
Now,$f(1) = 1^2 - (-3) = 4$ and $f(3) = b\sqrt{3 - 2} + a = 15(1) - 3 = 12$.
Finally,$\frac{f(1) - f(3)}{4} = \frac{4 - 12}{4} = \frac{-8}{4} = -2$.

(A) Given,$f(x) = \max(\sin x, \sin^{-1}(\cos x))$.
We know that $\sin^{-1}(\cos x) = \sin^{-1}(\sin(\frac{\pi}{2} - x))$.
For $x \in [-\pi, \pi]$,this simplifies to:
$\sin^{-1}(\cos x) = \begin{cases} \frac{\pi}{2} - x, & 0 \le x \le \pi \\ \frac{\pi}{2} + x, & -\pi \le x < 0 \end{cases}$.
Both $\sin x$ and $\sin^{-1}(\cos x)$ are continuous functions on their respective domains. The maximum of two continuous functions is also a continuous function.
By plotting the graphs of $y = \sin x$ and $y = \sin^{-1}(\cos x)$,we observe that the two curves intersect at certain points,but there are no breaks or jumps in the resulting function $f(x) = \max(\sin x, \sin^{-1}(\cos x))$.
Therefore,$f(x)$ is continuous everywhere.

(A) For $f(x)$ to be continuous at $x = 0$,we must have $k = \lim_{x \to 0} f(x)$.
$k = \lim_{x \to 0} \frac{(1 + \tan x)^{\frac{1}{x}} - e}{x} = \lim_{x \to 0} \frac{e^{\frac{1}{x} \ln(1 + \tan x)} - e}{x}$.
Factoring out $e$,we get $k = e \lim_{x \to 0} \frac{e^{\frac{1}{x} \ln(1 + \tan x) - 1} - 1}{x}$.
Using the limit $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$,we multiply and divide by $(\frac{1}{x} \ln(1 + \tan x) - 1)$:
$k = e \lim_{x \to 0} \left( \frac{e^{\frac{1}{x} \ln(1 + \tan x) - 1} - 1}{\frac{1}{x} \ln(1 + \tan x) - 1} \right) \cdot \left( \frac{\ln(1 + \tan x) - x}{x^2} \right)$.
Using the expansion $\ln(1 + t) = t - \frac{t^2}{2} + \dots$ and $\tan x = x + \frac{x^3}{3} + \dots$,we have $\ln(1 + \tan x) = (x + \frac{x^3}{3}) - \frac{(x + \dots)^2}{2} = x - \frac{x^2}{2} + \dots$.
Thus,$\lim_{x \to 0} \frac{\ln(1 + \tan x) - x}{x^2} = \lim_{x \to 0} \frac{x - \frac{x^2}{2} - x}{x^2} = -\frac{1}{2}$.
Therefore,$k = e \cdot 1 \cdot (-\frac{1}{2}) = -\frac{e}{2}$.

(D) First,factorize the expression inside the absolute value: $x^3 - 2x^2 - x + 2 = x^2(x - 2) - 1(x - 2) = (x^2 - 1)(x - 2) = (x - 1)(x + 1)(x - 2)$.
Thus,$f(x) = [x]|(x - 1)(x + 1)(x - 2)|$.
The function $g(x) = |(x - 1)(x + 1)(x - 2)|$ is a continuous polynomial function.
The function $h(x) = [x]$ is discontinuous at all integers $x \in \mathbb{Z}$.
In the interval $[-\frac{3}{2}, \frac{9}{2}]$,the integers are $\{-1, 0, 1, 2, 3, 4\}$.
Let $f(x) = [x] \cdot g(x)$. The product is discontinuous at $x = k$ if $g(k) \neq 0$ and $[x]$ is discontinuous at $x = k$.
Check the values of $g(k)$ for $k \in \{-1, 0, 1, 2, 3, 4\}$:
$g(-1) = |(-1-1)(-1+1)(-1-2)| = 0$.
$g(0) = |(0-1)(0+1)(0-2)| = |(-1)(1)(-2)| = 2 \neq 0$.
$g(1) = |(1-1)(1+1)(1-2)| = 0$.
$g(2) = |(2-1)(2+1)(2-2)| = 0$.
$g(3) = |(3-1)(3+1)(3-2)| = |(2)(4)(1)| = 8 \neq 0$.
$g(4) = |(4-1)(4+1)(4-2)| = |(3)(5)(2)| = 30 \neq 0$.
Since $g(k) = 0$ at $x = -1, 1, 2$,the function $f(x)$ is continuous at these points because $\lim_{x \to k} [x]g(x) = g(k) = 0$.
Thus,$f(x)$ is discontinuous only at $x = 0, 3, 4$.
The number of points of discontinuity is $3$.

(A) The function is defined as $f(x) = \operatorname{sgn}(x^2 - 3x + 2)$ for $x \in \mathbb{Q}$ and $f(x) = 0$ for $x \notin \mathbb{Q}$.
We know that $\operatorname{sgn}(x^2 - 3x + 2) = \operatorname{sgn}((x-1)(x-2))$.
The signum function takes values $\{-1, 0, 1\}$.
For $f(x)$ to be continuous at a point $x = a$,the limit $\lim_{x \to a} f(x)$ must exist and equal $f(a)$.
Since $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$,for the limit to exist,the function must take the same value for both rational and irrational sequences approaching $a$.
Thus,we require $\operatorname{sgn}(a^2 - 3a + 2) = 0$.
This occurs when $a^2 - 3a + 2 = 0$,which gives $a = 1$ or $a = 2$.
At $x = 1$,$f(1) = \operatorname{sgn}(0) = 0$. For any sequence $x_n \to 1$,$f(x_n)$ will approach $0$ because the values of $\operatorname{sgn}(x^2 - 3x + 2)$ are $0$ at $x=1$ and $x=2$,and the function is $0$ for irrational numbers.
Specifically,near $x=1$,the function values are $\{-1, 0, 1\}$ for rationals and $0$ for irrationals. The limit exists only if the function value is $0$ for all sequences,which happens at the roots.
Thus,$f(x)$ is continuous at $x=1$ and $x=2$.
There are $2$ such points.