Continuity Questions in English

(C) The signum function $\operatorname{sgn}(u)$ is discontinuous at $u = 0$.
For $f(x) = \operatorname{sgn}\left(3 \cos x - \frac{a}{3}\right)$ to be continuous for all $x$,the argument $3 \cos x - \frac{a}{3}$ must never be equal to $0$.
This implies $3 \cos x - \frac{a}{3} \neq 0$ for all $x$,which means $\cos x \neq \frac{a}{9}$.
Since the range of $\cos x$ is $[-1, 1]$,the equation $\cos x = \frac{a}{9}$ must have no solution.
This occurs if $\frac{a}{9} > 1$ or $\frac{a}{9} < -1$.
For $\frac{a}{9} > 1$,we have $a > 9$.
For $\frac{a}{9} < -1$,we have $a < -9$.
We are looking for the least positive integral value of $a$.
Since $a > 9$,the smallest integer $a$ is $10$.

(B) First,we check the continuity of $f(x)$ at $x = 0$. $\lim_{x \to 0} f(x) = \lim_{x \to 0} (3 - \sin(1/x))|x|$. Since $|\sin(1/x)| \le 1$,we have $2|x| \le (3 - \sin(1/x))|x| \le 4|x|$. By the Sandwich Theorem,$\lim_{x \to 0} f(x) = 0 = f(0)$. Thus,$f$ is continuous at $x = 0$.
Next,we examine the values of $f(x)$ near $x = 0$. For any $x \ne 0$,$|x| > 0$ and $(3 - \sin(1/x)) \ge 3 - 1 = 2 > 0$. Therefore,$f(x) = (3 - \sin(1/x))|x| > 0$ for all $x \ne 0$.
Since $f(0) = 0$ and $f(x) > 0$ for all $x \ne 0$ in the neighborhood of $0$,$f(x) \ge f(0)$ for all $x$ in the neighborhood. Hence,$f$ has a local minimum at $x = 0$.

(A) For $f(x)$ to be continuous at $x = 2,$ the condition $\lim_{x \to 2} f(x) = f(2)$ must hold.
First,we evaluate the limit $\lim_{x \to 2} ([x] + [-x])$.
We know that for any $x \notin \mathbb{Z},$ $[x] + [-x] = -1.$
Since the limit $x \to 2$ considers values of $x$ arbitrarily close to $2$ but not equal to $2,$ we have:
$\lim_{x \to 2} ([x] + [-x]) = -1.$
For continuity,we require $f(2) = \lim_{x \to 2} f(x).$
Given $f(2) = \lambda,$ we have $\lambda = -1.$

(C) To check continuity at $x = 1$:
$f(1) = |1 - 3| = |-2| = 2$.
Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (\frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}) = \frac{1}{4} - \frac{3}{2} + \frac{13}{4} = \frac{1 - 6 + 13}{4} = \frac{8}{4} = 2$.
Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} |x - 3| = |1 - 3| = 2$.
Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 2$,the function is continuous at $x = 1$.
To check continuity at $x = 3$:
The function is defined as $f(x) = |x - 3|$ in the neighborhood of $x = 3$.
Since $|x - 3|$ is a modulus function,it is continuous everywhere.
Thus,$f(x)$ is continuous at $x = 1$ and $x = 3$.

(A) For $g(f(x))$ to be continuous at $x = 1$,we must have $\lim_{x \to 1^-} g(f(x)) = \lim_{x \to 1^+} g(f(x)) = g(f(1))$.
First,evaluate the limits of $f(x)$ at $x = 1$:
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x - 1}{2} = 0$.
$\lim_{x \to 1^+} f(x) = 1/2$.
$f(1) = 1/2$.
Now,apply the condition for continuity of $g(f(x))$ at $x = 1$:
$g(\lim_{x \to 1^-} f(x)) = g(\lim_{x \to 1^+} f(x)) = g(f(1))$.
$g(0) = g(1/2) = g(1/2)$.
Calculate $g(0)$ and $g(1/2)$ using $g(x) = (2x + 1)(x - k) + 3$:
$g(0) = (2(0) + 1)(0 - k) + 3 = -k + 3$.
$g(1/2) = (2(1/2) + 1)(1/2 - k) + 3 = (1 + 1)(1/2 - k) + 3 = 2(1/2 - k) + 3 = 1 - 2k + 3 = 4 - 2k$.
Equating $g(0) = g(1/2)$:
$-k + 3 = 4 - 2k$.
$2k - k = 4 - 3$.
$k = 1$.

(A) We are given $f(x) = \lim_{n \to \infty} \frac{\sum_{k=1}^{n} [(kx)^2]}{n^3}$.
Using the property of the Greatest Integer Function,$[y] = y - \{y\}$,where $0 \le \{y\} < 1$,we have:
$f(x) = \lim_{n \to \infty} \frac{\sum_{k=1}^{n} ((kx)^2 - \{(kx)^2\})}{n^3}$.
$f(x) = \lim_{n \to \infty} \left( \frac{x^2 \sum_{k=1}^{n} k^2}{n^3} - \frac{\sum_{k=1}^{n} \{(kx)^2\}}{n^3} \right)$.
Since $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,the first term becomes $\lim_{n \to \infty} \frac{x^2 n(n+1)(2n+1)}{6n^3} = \frac{x^2}{6} \times 2 = \frac{x^2}{3}$.
The second term involves $\sum_{k=1}^{n} \{(kx)^2\}$. Since $0 \le \{(kx)^2\} < 1$,the sum is bounded by $n$. Thus,$\lim_{n \to \infty} \frac{\sum_{k=1}^{n} \{(kx)^2\}}{n^3} = 0$.
Therefore,$f(x) = \frac{x^2}{3}$.
Since $f(x) = \frac{x^2}{3}$ is a polynomial function,it is continuous for all $x \in R$.

(B) To check for continuity at $x = 0$,we evaluate $\lim_{x \to 0} f(x)$.
Given $f(x) = x \sin \left( \frac{1}{x} \right) \sin \left( \frac{1}{x \sin \left( \frac{1}{x} \right)} \right)$.
Since $|\sin \theta| \leq 1$,we have $|f(x)| = |x \sin \left( \frac{1}{x} \right) \sin \left( \frac{1}{x \sin \left( \frac{1}{x} \right)} \right)| \leq |x \sin \left( \frac{1}{x} \right)| \leq |x|$.
As $x \to 0$,$|x| \to 0$,so by the Squeeze Theorem,$\lim_{x \to 0} f(x) = 0 = f(0)$. Thus,$f(x)$ is continuous at $x = 0$.
To check for differentiability,we evaluate $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h \sin \left( \frac{1}{h} \right) \sin \left( \frac{1}{h \sin \left( \frac{1}{h} \right)} \right)}{h} = \lim_{h \to 0} \sin \left( \frac{1}{h} \right) \sin \left( \frac{1}{h \sin \left( \frac{1}{h} \right)} \right)$.
Let $g(h) = \sin \left( \frac{1}{h} \right) \sin \left( \frac{1}{h \sin \left( \frac{1}{h} \right)} \right)$. As $h \to 0$,$\sin \left( \frac{1}{h} \right)$ oscillates between $-1$ and $1$. The term $\frac{1}{h \sin \left( \frac{1}{h} \right)}$ also oscillates and takes arbitrarily large values,causing the second sine term to oscillate. Thus,the limit does not exist.
Therefore,$f(x)$ is continuous but not differentiable at $x = 0$.

(B) For the function $f(x) = \left[ \frac{(x - 2)^3}{a} \right] \sin(x - 2) + a \cos(x - 2)$ to be continuous on $[4, 6]$,the term inside the greatest integer function must be constant,or the jump discontinuity must be nullified by the multiplying term $\sin(x - 2)$.
In the interval $x \in [4, 6]$,the expression $(x - 2)^3$ ranges from $(4 - 2)^3 = 8$ to $(6 - 2)^3 = 64$.
If $0 < \frac{(x - 2)^3}{a} < 1$ for all $x \in [4, 6]$,then $\left[ \frac{(x - 2)^3}{a} \right] = 0$.
This occurs when $\frac{64}{a} \le 1$,which implies $a \ge 64$.
If $a > 64$,then $\left[ \frac{(x - 2)^3}{a} \right] = 0$ for all $x \in [4, 6]$,making $f(x) = a \cos(x - 2)$,which is continuous.
However,if $a = 64$,then at $x = 6$,$\left[ \frac{(6 - 2)^3}{64} \right] = [1] = 1$,and for $x < 6$,$\left[ \frac{(x - 2)^3}{64} \right] = 0$. The function would have a jump discontinuity at $x = 6$ unless the $\sin(x - 2)$ term is zero. Since $\sin(6 - 2) = \sin(4) \neq 0$,$a = 64$ is not a solution.
Thus,the condition for continuity is $a > 64$. Among the given options,$65$ is the only value satisfying $a > 64$.

(B) The function $f(x) = [2x^3 - 5]$ is discontinuous at points where the argument $2x^3 - 5$ is an integer.
We are looking for points $x \in (1, 2)$ such that $2x^3 - 5 = k$,where $k$ is an integer.
Since $x \in (1, 2)$,we have $1 < x < 2$.
Cubing the inequality,we get $1 < x^3 < 8$.
Multiplying by $2$,we get $2 < 2x^3 < 16$.
Subtracting $5$,we get $2 - 5 < 2x^3 - 5 < 16 - 5$,which simplifies to $-3 < 2x^3 - 5 < 11$.
Thus,the possible integer values for $k$ are $\{-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
There are $13$ such integer values.
For each integer $k$,there exists a unique $x = \sqrt[3]{\frac{k+5}{2}}$ in the interval $(1, 2)$.
Therefore,there are $13$ points of discontinuity in the interval $(1, 2)$.

(A) To check the continuity of $f(x)$ at $x = \frac{1}{2}$,we examine the limit as $x \to \frac{1}{2}$.
For $f(x)$ to be continuous at $x = a$,the limit $\lim_{x \to a} f(x)$ must exist and equal $f(a)$.
Here,$f(\frac{1}{2}) = \frac{1}{2}$ (since $\frac{1}{2}$ is rational).
For any sequence of rational numbers $r_n \to \frac{1}{2}$,$f(r_n) = r_n \to \frac{1}{2}$.
For any sequence of irrational numbers $i_n \to \frac{1}{2}$,$f(i_n) = 1 - i_n \to 1 - \frac{1}{2} = \frac{1}{2}$.
Since the limit exists and equals $f(\frac{1}{2})$,$f(x)$ is continuous at $x = \frac{1}{2}$.
Now,check differentiability: $f'(1/2) = \lim_{h \to 0} \frac{f(1/2 + h) - f(1/2)}{h}$.
If $1/2 + h$ is rational,$\frac{f(1/2+h) - f(1/2)}{h} = \frac{1/2+h - 1/2}{h} = 1$.
If $1/2 + h$ is irrational,$\frac{f(1/2+h) - f(1/2)}{h} = \frac{1 - (1/2+h) - 1/2}{h} = \frac{-h}{h} = -1$.
Since the limit does not exist,$f(x)$ is non-differentiable at $x = \frac{1}{2}$.

(D) We analyze the limit $f(x) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {4\sin^2 x} \right)}^{n}}}}{{{3^n} - {{\left( {4\cos^2 x} \right)}^{n}}}}$.
Case $1$: If $|4\sin^2 x| < 3$ and $|4\cos^2 x| < 3$,then $f(x) = 0$.
This occurs when $\sin^2 x < \frac{3}{4}$ and $\cos^2 x < \frac{3}{4}$,i.e.,$\frac{1}{4} < \sin^2 x < \frac{3}{4}$,which implies $x \in (m\pi + \frac{\pi}{6}, m\pi + \frac{\pi}{3})$.
Case $2$: If $|4\sin^2 x| > 3$ and $|4\cos^2 x| < 3$,then $f(x) = \infty$ (limit does not exist).
This occurs when $\sin^2 x > \frac{3}{4}$,i.e.,$x \in (m\pi + \frac{\pi}{3}, m\pi + \frac{2\pi}{3})$.
Case $3$: If $|4\cos^2 x| > 3$ and $|4\sin^2 x| < 3$,then $f(x) = \lim_{n \to \infty} \frac{(4\sin^2 x / 4\cos^2 x)^n}{(3/4\cos^2 x)^n - 1} = 0$.
At $x = m\pi \pm \frac{\pi}{6}$,the denominator approaches $0$ or the terms become undefined,leading to discontinuity. Thus,$f(x)$ is discontinuous at these points. Checking $f(0) = 0$ and $f(\pi/3) = 1$ confirms all statements are correct.

(A) For $f(x)$ to have a maximum at $x = -2$,the value of $f(-2)$ must be greater than or equal to the values of $f(x)$ in the neighborhood of $x = -2$.
First,let us analyze the function $g(x) = 2 - |x^2 + 5x + 6|$ for $x \neq -2$.
We know that $|x^2 + 5x + 6| \ge 0$,so $2 - |x^2 + 5x + 6| \le 2$.
The maximum value of $g(x)$ is $2$,which occurs when $x^2 + 5x + 6 = 0$,i.e.,$(x+2)(x+3) = 0$,so $x = -2$ or $x = -3$.
Since the function is defined as $f(x) = g(x)$ for $x \neq -2$ and $f(-2) = a^2 + 1$,for $f(x)$ to have a maximum at $x = -2$,we must have $f(-2) \ge \lim_{x \to -2} f(x)$.
Calculating the limit as $x \to -2$:
$\lim_{x \to -2} (2 - |x^2 + 5x + 6|) = 2 - |(-2)^2 + 5(-2) + 6| = 2 - |4 - 10 + 6| = 2 - 0 = 2$.
Thus,we require $a^2 + 1 \ge 2$.
$a^2 \ge 1$.
Taking the square root on both sides,we get $|a| \ge 1$.

(A) For the function $f(x)$ to be continuous at $x = 2$,the limit of $f(x)$ as $x \to 2$ must exist and be equal to $f(2)$.
Given $f(x) = [x] + [-x]$ for $x \neq 2$.
We know that for any real number $x$,$[x] + [-x] = \begin{cases} 0, & \text{if } x \in \mathbb{Z} \\ -1, & \text{if } x \notin \mathbb{Z} \end{cases}$.
Since $x = 2$ is an integer,we evaluate the limit as $x$ approaches $2$ from the left $(x \to 2^-)$ and from the right $(x \to 2^+)$.
For $x$ near $2$ but $x \neq 2$,$x$ is not an integer,so $[x] + [-x] = -1$.
Therefore,$\lim_{x \to 2} f(x) = -1$.
For continuity,$f(2) = \lim_{x \to 2} f(x)$,which implies $\lambda = -1$.

(C) Since $f(x)$ is continuous at $x = 2$,we have $\lim_{x \to 2} f(x) = f(2)$.
$\lim_{x \to 2} (x - 1)^{\frac{1}{2 - x}} = k$. This is a $1^{\infty}$ form.
Using the formula $\lim_{x \to a} [g(x)]^{h(x)} = e^{\lim_{x \to a} (g(x) - 1)h(x)}$,we get:
$k = e^{\lim_{x \to 2} (x - 1 - 1) \cdot \frac{1}{2 - x}}$
$k = e^{\lim_{x \to 2} \frac{x - 2}{2 - x}}$
$k = e^{\lim_{x \to 2} \frac{-(2 - x)}{2 - x}}$
$k = e^{-1}$.

(A) If the function is continuous at $x = 0$,then $\lim_{x \to 0} f(x)$ must exist and $f(0) = \lim_{x \to 0} f(x)$.
Now,$\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{1}{x} - \frac{k - 1}{e^{2x} - 1} \right)$.
$= \lim_{x \to 0} \left( \frac{e^{2x} - 1 - (k - 1)x}{x(e^{2x} - 1)} \right)$.
Using the expansion $e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots$,we get:
$= \lim_{x \to 0} \frac{(1 + 2x + 2x^2 + \dots) - 1 - kx + x}{x(2x + 2x^2 + \dots)}$.
$= \lim_{x \to 0} \frac{(3 - k)x + 2x^2 + \dots}{2x^2 + 2x^3 + \dots}$.
For the limit to exist,the coefficient of $x$ in the numerator must be zero,so $3 - k = 0$,which gives $k = 3$.
Substituting $k = 3$,the limit becomes $\lim_{x \to 0} \frac{2x^2 + \dots}{2x^2 + 2x^3 + \dots} = \frac{2}{2} = 1$.
Therefore,$f(0) = 1$,and the ordered pair is $(3, 1)$.

(C) For the function $f(x)$ to be continuous in $[0, \infty)$,it must be continuous at the transition points $x=1$ and $x=\sqrt{2}$.
$1$. Continuity at $x=1$:
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$
$\frac{2(1)^2}{a} = a \implies \frac{2}{a} = a \implies a^2 = 2 \implies a = \pm \sqrt{2}$.
$2$. Continuity at $x=\sqrt{2}$:
$\lim_{x \to \sqrt{2}^-} f(x) = \lim_{x \to \sqrt{2}^+} f(x) = f(\sqrt{2})$
$a = \frac{2b^2 - 4b}{(\sqrt{2})^3} = \frac{2b^2 - 4b}{2\sqrt{2}} = \frac{b^2 - 2b}{\sqrt{2}}$.
Case $1$: If $a = \sqrt{2}$,then $\sqrt{2} = \frac{b^2 - 2b}{\sqrt{2}} \implies b^2 - 2b = 2 \implies b^2 - 2b - 2 = 0$.
Using the quadratic formula $b = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} = \frac{2 \pm \sqrt{4+8}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$.
So,$(a, b) = (\sqrt{2}, 1 \pm \sqrt{3})$.
Comparing with the given options,the pair $(\sqrt{2}, 1 - \sqrt{3})$ matches option $C$.

(B) Given that $f(x)$ is continuous,$\lim_{x \to 0} f(g(x)) = f(\lim_{x \to 0} g(x))$.
First,evaluate the limit of the inner function: $\lim_{x \to 0} \frac{1 - \cos 3x}{x^2}$.
Using the identity $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$,we get $\lim_{x \to 0} \frac{2 \sin^2 \frac{3x}{2}}{x^2}$.
Multiply and divide by $\left( \frac{3}{2} \right)^2 = \frac{9}{4}$: $\lim_{x \to 0} 2 \cdot \frac{9}{4} \cdot \frac{\sin^2 \frac{3x}{2}}{\left( \frac{3x}{2} \right)^2} = \frac{9}{2} \cdot (1)^2 = \frac{9}{2}$.
Since $f(x)$ is continuous,$\lim_{x \to 0} f \left( \frac{1 - \cos 3x}{x^2} \right) = f \left( \lim_{x \to 0} \frac{1 - \cos 3x}{x^2} \right) = f \left( \frac{9}{2} \right)$.
Given $f \left( \frac{9}{2} \right) = \frac{2}{9}$,the final answer is $\frac{2}{9}$.

(B) For $f(x)$ at $x=0$:
$LHL = \lim_{h \to 0^-} (-h) \sin(-1/h) = \lim_{h \to 0^-} h \sin(1/h) = 0$.
$RHL = \lim_{h \to 0^+} h \sin(1/h) = 0$.
Since $f(0) = 0$,$LHL = RHL = f(0)$,so $f$ is continuous at $x=0$. Statement $I$ is true.
For $g(x) = x f(x) = \begin{cases} x^2 \sin(1/x), & x \ne 0 \\ 0, & x = 0 \end{cases}$.
To check differentiability at $x=0$,we find $g'(0) = \lim_{h \to 0} \frac{g(h) - g(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin(1/h) = 0$.
Since the limit exists and is finite,$g$ is differentiable at $x=0$. Statement $II$ is true.

(D) For the function $f(x)$ to be continuous at $x = \pi$,we must have $\lim_{x \to \pi} f(x) = f(\pi) = k$.
Let $x = \pi + h$,where $h \to 0$ as $x \to \pi$. Then $(\pi - x)^2 = (-h)^2 = h^2$.
$\lim_{h \to 0} \frac{\sqrt{2 + \cos(\pi + h)} - 1}{h^2} = k$.
Since $\cos(\pi + h) = -\cos h$,we have:
$k = \lim_{h \to 0} \frac{\sqrt{2 - \cos h} - 1}{h^2}$.
Rationalizing the numerator:
$k = \lim_{h \to 0} \frac{(\sqrt{2 - \cos h} - 1)(\sqrt{2 - \cos h} + 1)}{h^2(\sqrt{2 - \cos h} + 1)} = \lim_{h \to 0} \frac{2 - \cos h - 1}{h^2(\sqrt{2 - \cos h} + 1)}$.
$k = \lim_{h \to 0} \frac{1 - \cos h}{h^2(\sqrt{2 - \cos h} + 1)}$.
Using the identity $1 - \cos h = 2 \sin^2(h/2)$:
$k = \lim_{h \to 0} \frac{2 \sin^2(h/2)}{h^2(\sqrt{2 - \cos h} + 1)} = \lim_{h \to 0} \frac{2 \sin^2(h/2)}{4(h/2)^2(\sqrt{2 - \cos h} + 1)}$.
Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$k = \frac{2}{4} \times \frac{1}{\sqrt{2 - 1} + 1} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25$.

(A) Let $h(x) = f(x) - g(x) = x \log x - (2 - x) = x \log x + x - 2$.
We evaluate the function $h(x)$ at the endpoints of the interval $[1, 2]$:
$h(1) = 1 \cdot \log(1) + 1 - 2 = 0 + 1 - 2 = -1$.
$h(2) = 2 \cdot \log(2) + 2 - 2 = 2 \log(2) = \log(4)$.
Since $\log(4) > 0$ (as $4 > 1$),we have $h(1) < 0$ and $h(2) > 0$.
Since $h(x)$ is a continuous function on $[1, 2]$,by the Intermediate Value Theorem,there exists at least one $c \in (1, 2)$ such that $h(c) = 0$,which means $f(c) = g(c)$. Thus,Statement-$1$ is true.
For Statement-$2$,$f'(x) = \log x + x(1/x) = \log x + 1$. For $x \in [1, 2]$,$f'(x) \geq 1 > 0$,so $f(x)$ is increasing. $g'(x) = -1 < 0$,so $g(x)$ is decreasing. Since $f(1) < g(1)$ and $f(2) > g(2)$,the graphs must intersect in $(1, 2)$. Thus,Statement-$2$ is true and provides a correct explanation for Statement-$1$.

(A) Given $f(x) = [x] + |1 - x|$ for $x \in [-1, 3]$.
For $x \in [-1, 0)$,$[x] = -1$,so $f(x) = -1 + (1 - x) = -x$.
For $x \in [0, 1)$,$[x] = 0$,so $f(x) = 0 + (1 - x) = 1 - x$.
For $x \in [1, 2)$,$[x] = 1$,so $f(x) = 1 + (x - 1) = x$.
For $x \in [2, 3)$,$[x] = 2$,so $f(x) = 2 + (x - 1) = x + 1$.
At $x = 3$,$f(3) = [3] + |1 - 3| = 3 + 2 = 5$.
Checking continuity:
At $x=0$: $LHL = \lim_{x \to 0^-} (-x) = 0$,$RHL = \lim_{x \to 0^+} (1 - x) = 1$. Since $LHL \neq RHL$,$f$ is discontinuous at $x=0$.
At $x=1$: $LHL = \lim_{x \to 1^-} (1 - x) = 0$,$RHL = \lim_{x \to 1^+} (x) = 1$. Since $LHL \neq RHL$,$f$ is discontinuous at $x=1$.
At $x=2$: $LHL = \lim_{x \to 2^-} (x) = 2$,$RHL = \lim_{x \to 2^+} (x + 1) = 3$. Since $LHL \neq RHL$,$f$ is discontinuous at $x=2$.
At $x=3$: $LHL = \lim_{x \to 3^-} (x + 1) = 4$,$f(3) = 5$. Since $LHL \neq f(3)$,$f$ is discontinuous at $x=3$.
Thus,Statement $1$ is true. Statement $2$ provides an incorrect definition of $f(x)$,so Statement $2$ is false.

(D) Statement $1$ is true. This is the standard definition of continuity at a point $x_0$.
Statement $2$ is false. $A$ function $f$ is discontinuous at $x_0$ if it is not continuous at $x_0$. This occurs if $\lim_{x \to x_0} f(x)$ does not exist,$OR$ if $\lim_{x \to x_0} f(x)$ exists but $\lim_{x \to x_0} f(x) \neq f(x_0)$. Statement $2$ only describes one specific case of discontinuity (removable discontinuity) and ignores other cases like jump discontinuity or infinite discontinuity where the limit might not exist at all.

(D) For $f(x)$ to be continuous,it must be continuous at the transition points $x=1, x=3,$ and $x=5$.
At $x=1$:
$LHL = \lim_{x \to 1^-} f(x) = 5$
$RHL = \lim_{x \to 1^+} f(x) = a + b(1) = a + b$
For continuity,$a + b = 5$ $(i)$
At $x=3$:
$LHL = \lim_{x \to 3^-} f(x) = a + 3b$
$RHL = \lim_{x \to 3^+} f(x) = b + 5(3) = b + 15$
For continuity,$a + 3b = b + 15 \implies a + 2b = 15$ $(ii)$
At $x=5$:
$LHL = \lim_{x \to 5^-} f(x) = b + 5(5) = b + 25$
$RHL = \lim_{x \to 5^+} f(x) = 30$
For continuity,$b + 25 = 30 \implies b = 5$
Substituting $b=5$ into $(ii)$:
$a + 2(5) = 15 \implies a + 10 = 15 \implies a = 5$
Now check these values in $(i)$:
$a + b = 5 + 5 = 10 \neq 5$
Since the values $a=5$ and $b=5$ do not satisfy the condition at $x=1$,there are no values of $a$ and $b$ for which the function is continuous everywhere.

(A) We analyze the function $f(x)$ in different intervals:
For $x \in [-1, 0)$,$|x| = -x$ and $[x] = -1$,so $f(x) = -x - 1$.
For $x \in [0, 1)$,$|x| = x$ and $[x] = 0$,so $f(x) = x + 0 = x$.
For $x \in [1, 2)$,$|x| = x$,so $f(x) = x + x = 2x$.
For $x \in [2, 3]$,$|x| = x$,so $f(x) = x + x = 2x$.
Thus,$f(x) = \begin{cases} -x-1, & -1 \leq x < 0 \\ x, & 0 \leq x < 1 \\ 2x, & 1 \leq x \leq 3 \end{cases}$.
Checking continuity at $x=0$: $\lim_{x \to 0^-} f(x) = -1$ and $f(0) = 0$. Since $-1 \neq 0$,$f$ is discontinuous at $x=0$.
Checking continuity at $x=1$: $\lim_{x \to 1^-} f(x) = 1$ and $f(1) = 2(1) = 2$. Since $1 \neq 2$,$f$ is discontinuous at $x=1$.
Checking continuity at $x=2$: $\lim_{x \to 2^-} f(x) = 2(2) = 4$ and $f(2) = 2(2) = 4$. Since the limit equals the function value,$f$ is continuous at $x=2$.
Therefore,$f$ is discontinuous at only two points,$x=0$ and $x=1$.

(A) For the function $f(x)$ to be continuous at $x = 5$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function at $x = 5$ must be equal.
$LHL = \lim_{x \to 5^-} f(x) = a|\pi - 5| + 1 = a(5 - \pi) + 1$ (since $5 > \pi$,$|\pi - 5| = 5 - \pi$).
$RHL = \lim_{x \to 5^+} f(x) = b|\pi - 5| + 3 = b(5 - \pi) + 3$.
$f(5) = a|\pi - 5| + 1 = a(5 - \pi) + 1$.
Equating $LHL$ and $RHL$:
$a(5 - \pi) + 1 = b(5 - \pi) + 3$.
Rearranging the terms:
$a(5 - \pi) - b(5 - \pi) = 3 - 1$.
$(a - b)(5 - \pi) = 2$.
Therefore,$a - b = \frac{2}{5 - \pi}$.

(D) Given $f(x) = [x] - [\frac{x}{4}]$.
For the right-hand limit at $x = 4$:
$\lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} [x] - \lim_{x \to 4^+} [\frac{x}{4}] = 4 - 1 = 3$.
For the left-hand limit at $x = 4$:
$\lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} [x] - \lim_{x \to 4^-} [\frac{x}{4}] = 3 - 0 = 3$.
Also,$f(4) = [4] - [\frac{4}{4}] = 4 - 1 = 3$.
Since $\lim_{x \to 4^+} f(x) = \lim_{x \to 4^-} f(x) = f(4) = 3$,the function $f$ is continuous at $x = 4$.

(D) For the function $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
First,calculate the Right Hand Limit $(RHL)$:
$RHL = \lim_{x \to 0^+} \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1+x} - 1)}{x \cdot \sqrt{x}} = \lim_{x \to 0^+} \frac{\sqrt{1+x} - 1}{x}$.
Multiplying by the conjugate $\frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1}$,we get:
$RHL = \lim_{x \to 0^+} \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)} = \lim_{x \to 0^+} \frac{x}{x(\sqrt{1+x} + 1)} = \frac{1}{1+1} = \frac{1}{2}$.
Next,calculate the Left Hand Limit $(LHL)$:
$LHL = \lim_{x \to 0^-} \frac{\sin((p+1)x) + \sin x}{x} = \lim_{x \to 0^-} \left( \frac{\sin((p+1)x)}{x} + \frac{\sin x}{x} \right) = (p+1) + 1 = p+2$.
Since $f(0) = q$,for continuity,we require $LHL = RHL = f(0)$:
$p + 2 = 1/2 \implies p = 1/2 - 2 = -3/2$.
$q = 1/2$.
Thus,the ordered pair $(p, q) = (-3/2, 1/2)$.

(B) For the function $f(x)$ to be continuous at $x = 0$,we must have $k = \lim_{x \rightarrow 0} f(x)$.
$k = \lim_{x \rightarrow 0} \frac{1}{x} \log_{e}\left(\frac{1+3x}{1-2x}\right)$
Using the property of logarithms,$\log_{e}\left(\frac{a}{b}\right) = \log_{e}(a) - \log_{e}(b)$:
$k = \lim_{x \rightarrow 0} \frac{\log_{e}(1+3x) - \log_{e}(1-2x)}{x}$
$k = \lim_{x \rightarrow 0} \left( \frac{\log_{e}(1+3x)}{x} - \frac{\log_{e}(1-2x)}{x} \right)$
Using the standard limit $\lim_{u \rightarrow 0} \frac{\log_{e}(1+u)}{u} = 1$:
$k = \lim_{x \rightarrow 0} \left( 3 \cdot \frac{\log_{e}(1+3x)}{3x} - (-2) \cdot \frac{\log_{e}(1-2x)}{-2x} \right)$
$k = 3(1) - (-2)(1) = 3 + 2 = 5$.

(B) Given $A = \lim_{x \to 0} x[\frac{4}{x}]$.
Using the property $[y] = y - \{y\}$,we have $A = \lim_{x \to 0} x(\frac{4}{x} - \{\frac{4}{x}\}) = \lim_{x \to 0} (4 - x\{\frac{4}{x}\})$.
Since $0 \leq \{\frac{4}{x}\} < 1$,by the Squeeze Theorem,$\lim_{x \to 0} x\{\frac{4}{x}\} = 0$.
Thus,$A = 4$.
The function $f(x) = [x^2] \sin(\pi x)$ is discontinuous where $[x^2]$ is discontinuous,which occurs when $x^2$ is an integer,excluding points where $\sin(\pi x) = 0$.
For $x^2 = k$ (where $k \in \mathbb{Z}$),$f(x)$ is discontinuous unless $\sin(\pi \sqrt{k}) = 0$.
Checking the options for $A=4$: $\sqrt{A+1} = \sqrt{5}$. Since $x^2 = 5$ is an integer and $\sin(\pi \sqrt{5}) \neq 0$,the function is discontinuous at $x = \sqrt{5}$.

(D) For the function $f(x)$ to be continuous at $x = 0$, the left-hand limit, right-hand limit, and the value of the function at $x = 0$ must be equal.
$1$. Left-hand limit $(LHL)$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0} \left( \frac{\sin(a+2)x}{x} + \frac{\sin x}{x} \right) = (a+2) + 1 = a+3$.
$2$. Right-hand limit $(RHL)$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0} \frac{(x+3x^2)^{1/3} - x^{1/3}}{x^{4/3}} = \lim_{x \to 0} \frac{x^{1/3}((1+3x)^{1/3} - 1)}{x^{4/3}} = \lim_{x \to 0} \frac{(1+3x)^{1/3} - 1}{x}$.
Using the binomial expansion $(1+u)^n \approx 1 + nu$ for small $u$:
$\lim_{x \to 0} \frac{(1 + \frac{1}{3}(3x)) - 1}{x} = \lim_{x \to 0} \frac{1+x-1}{x} = 1$.
$3$. Value at $x=0$:
$f(0) = b$.
For continuity, $a+3 = b = 1$.
Thus, $a = -2$ and $b = 1$.
Therefore, $a+2b = -2 + 2(1) = 0$.

(A) First,note that the function is defined at the given point $x = 1$ and its value is $f(1) = 2(1) + 3 = 5$.
Next,find the limit of the function as $x$ approaches $1$:
$\lim_{x \to 1} f(x) = \lim_{x \to 1} (2x + 3) = 2(1) + 3 = 5$.
Since $\lim_{x \to 1} f(x) = 5 = f(1)$,the limit of the function at $x = 1$ is equal to the value of the function at $x = 1$.
Therefore,the function $f$ is continuous at $x = 1$.

(A) function $f(x)$ is continuous at a point $x = c$ if $\lim_{x \to c} f(x) = f(c)$.
Step $1$: Evaluate the function at $x = 0$.
$f(0) = 0^{2} = 0$.
Step $2$: Evaluate the limit of the function as $x$ approaches $0$.
$\lim_{x \to 0} f(x) = \lim_{x \to 0} x^{2} = 0^{2} = 0$.
Step $3$: Compare the limit and the function value.
Since $\lim_{x \to 0} f(x) = 0$ and $f(0) = 0$,we have $\lim_{x \to 0} f(x) = f(0)$.
Therefore,the function $f(x) = x^{2}$ is continuous at $x = 0$.

(A) By definition,the absolute value function is given by:
$f(x) = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \ge 0 \end{cases}$
Clearly,the function is defined at $x = 0$ and $f(0) = 0$.
The left-hand limit $(LHL)$ of $f$ at $x = 0$ is:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0$
The right-hand limit $(RHL)$ of $f$ at $x = 0$ is:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x) = 0$
Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0$,the function $f$ is continuous at $x = 0$.

(N/A) function $f(x)$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.
$1$. First,we find the value of the function at $x = 0$:
$f(0) = 1$.
$2$. Next,we find the limit of the function as $x$ approaches $0$:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^3 + 3) = 0^3 + 3 = 3$.
$3$. Comparing the limit and the function value:
Since $\lim_{x \to 0} f(x) = 3$ and $f(0) = 1$,we see that $\lim_{x \to 0} f(x) \neq f(0)$.
Therefore,the function $f$ is not continuous at $x = 0$.

(A) The function $f(x) = k$ is defined for all real numbers $x \in \mathbb{R}$.
By definition,for any real number $c$,the value of the function is $f(c) = k$.
To check for continuity at $x = c$,we evaluate the limit:
$\lim_{x \to c} f(x) = \lim_{x \to c} k = k$.
Since $\lim_{x \to c} f(x) = f(c) = k$ for any arbitrary real number $c$,the function $f(x) = k$ is continuous at every point in its domain,which is the set of all real numbers $\mathbb{R}$.

(N/A) The identity function $f(x) = x$ is defined for all real numbers $x \in \mathbb{R}$.
To check for continuity at any arbitrary real number $c$,we evaluate the limit of the function as $x$ approaches $c$:
$\lim_{x \to c} f(x) = \lim_{x \to c} x = c$.
Also,the value of the function at $x = c$ is $f(c) = c$.
Since $\lim_{x \to c} f(x) = f(c) = c$,the condition for continuity is satisfied for every real number $c$.
Therefore,the identity function $f(x) = x$ is continuous at every real number.

(A) We may rewrite $f$ as:
$f(x) = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \ge 0 \end{cases}$
For $x = 0$,the left-hand limit is $\lim_{x \to 0^-} (-x) = 0$ and the right-hand limit is $\lim_{x \to 0^+} (x) = 0$. Since $f(0) = 0$,the function is continuous at $x = 0$.
Let $c$ be a real number such that $c < 0$. Then $f(c) = -c$. Also,$\lim_{x \to c} f(x) = \lim_{x \to c} (-x) = -c$. Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous at all negative real numbers.
Now,let $c$ be a real number such that $c > 0$. Then $f(c) = c$. Also,$\lim_{x \to c} f(x) = \lim_{x \to c} (x) = c$. Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous at all positive real numbers.
Hence,$f(x) = |x|$ is continuous at all real points.

(N/A) The function $f(x) = x^{3} + x^{2} - 1$ is a polynomial function.
Polynomial functions are defined for all real numbers $c \in \mathbb{R}$.
The value of the function at $x = c$ is $f(c) = c^{3} + c^{2} - 1$.
Now,we evaluate the limit of the function as $x$ approaches $c$:
$\lim_{x \to c} f(x) = \lim_{x \to c} (x^{3} + x^{2} - 1) = c^{3} + c^{2} - 1$.
Since $\lim_{x \to c} f(x) = f(c)$ for any arbitrary real number $c$,the function $f$ is continuous at every point in its domain.
Therefore,$f$ is a continuous function on the set of all real numbers $\mathbb{R}$.

(N/A) Let $c$ be any non-zero real number in the domain of $f$.
We calculate the limit of the function as $x$ approaches $c$:
$\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} \frac{1}{x} = \frac{1}{c}$.
Now,we evaluate the function at $x = c$:
$f(c) = \frac{1}{c}$.
Since $\mathop {\lim }\limits_{x \to c} f(x) = f(c)$ for all $c \neq 0$,the function $f(x) = \frac{1}{x}$ is continuous at every point in its domain $(x \in \mathbb{R} \setminus \{0\})$.
Therefore,$f$ is a continuous function.

(N/A) The function $f$ is defined at all points of the real line.
Case $1$: If $c < 1$,then $f(c) = c + 2$. Therefore,$\lim_{x \to c} f(x) = \lim_{x \to c} (x + 2) = c + 2 = f(c)$.
Thus,$f$ is continuous at all real numbers less than $1$.
Case $2$: If $c > 1$,then $f(c) = c - 2$. Therefore,$\lim_{x \to c} f(x) = \lim_{x \to c} (x - 2) = c - 2 = f(c)$.
Thus,$f$ is continuous at all points $x > 1$.
Case $3$: If $c = 1$,then the left-hand limit of $f$ at $x = 1$ is
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 2) = 1 + 2 = 3$.
The right-hand limit of $f$ at $x = 1$ is
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - 2) = 1 - 2 = -1$.
Since the left-hand limit and right-hand limit of $f$ at $x = 1$ do not coincide,$f$ is not continuous at $x = 1$. Hence,$x = 1$ is the only point of discontinuity of $f$.

(C) The function $f$ is defined as:
$f(x) = \begin{cases} x + 2, & \text{if } x < 1 \\ 0, & \text{if } x = 1 \\ x - 2, & \text{if } x > 1 \end{cases}$
For any $x \neq 1$,the function is a polynomial,which is continuous. We only need to check the continuity at $x = 1$.
The left-hand limit $(LHL)$ at $x = 1$ is:
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 2) = 1 + 2 = 3$
The right-hand limit $(RHL)$ at $x = 1$ is:
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - 2) = 1 - 2 = -1$
The value of the function at $x = 1$ is $f(1) = 0$.
Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$,the limit does not exist at $x = 1$. Therefore,the function is discontinuous at $x = 1$.

(N/A) Observe that the function is defined at all real numbers except at $x = 0$. The domain of definition of this function is $D = D_1 \cup D_2$,where $D_1 = \{x \in \mathbb{R} : x < 0\}$ and $D_2 = \{x \in \mathbb{R} : x > 0\}$.
Case $1$: If $c \in D_1$,then $\lim_{x \to c} f(x) = \lim_{x \to c} (x + 2) = c + 2 = f(c)$. Hence,$f$ is continuous in $D_1$.
Case $2$: If $c \in D_2$,then $\lim_{x \to c} f(x) = \lim_{x \to c} (-x + 2) = -c + 2 = f(c)$. Hence,$f$ is continuous in $D_2$.
Since $f$ is continuous at all points in its domain,we conclude that $f$ is continuous on its domain. Note that the function is not defined at $x = 0$,so we do not discuss continuity at $x = 0$.

(N/A) Clearly,the function is defined at every real number. The graph of the function is shown in the figure. By inspection,it is prudent to partition the domain of definition of $f$ into three disjoint subsets of the real line.
Let $D_1 = \{ x \in \mathbb{R} : x < 0 \}$,$D_2 = \{ 0 \}$,and $D_3 = \{ x \in \mathbb{R} : x > 0 \}$.
Case $1$: At any point in $D_1$,we have $f(x) = x^2$,which is a polynomial function and is continuous everywhere in its domain.
Case $2$: At any point in $D_3$,we have $f(x) = x$,which is a polynomial function and is continuous everywhere in its domain.
Case $3$: Now we analyze the function at $x = 0$. The value of the function at $0$ is $f(0) = 0$.
The left-hand limit of $f$ at $0$ is:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^2 = 0^2 = 0$.
The right-hand limit of $f$ at $0$ is:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$.
Thus,$\lim_{x \to 0} f(x) = 0 = f(0)$,and hence $f$ is continuous at $x = 0$. Since $f$ is continuous at every point in its domain,$f$ is a continuous function.

$A$ function $p$ is defined as a polynomial function if it is of the form $p(x) = a_{0} + a_{1}x + \ldots + a_{n}x^{n}$,where $n$ is a natural number,$a_{n} \neq 0$,and $a_{i} \in \mathbb{R}$.
This function is defined for all real numbers $x \in \mathbb{R}$.
For any arbitrary real number $c$,the limit of the function as $x$ approaches $c$ is given by:
$\lim_{x \to c} p(x) = \lim_{x \to c} (a_{0} + a_{1}x + \ldots + a_{n}x^{n})$
Using the properties of limits,this becomes:
$\lim_{x \to c} p(x) = a_{0} + a_{1}c + \ldots + a_{n}c^{n} = p(c)$
Since $\lim_{x \to c} p(x) = p(c)$ for any real number $c$,the function $p(x)$ is continuous at every point in its domain.
Therefore,every polynomial function is continuous.

(N/A) First,observe that $f$ is defined for all real numbers. The graph of the function is shown in the figure. From the graph,it appears that $f$ is discontinuous at every integral point. Below,we verify if this is true.
Case $1$: Let $c$ be a real number which is not an integer. It is evident from the graph that for all real numbers close to $c$,the value of the function is equal to $[c]$; i.e.,$\lim_{x \to c} f(x) = \lim_{x \to c} [x] = [c]$. Also,$f(c) = [c]$,and hence the function is continuous at all real numbers that are not integers.
Case $2$: Let $c$ be an integer. Then we can find a sufficiently small real number $r > 0$ such that $[c - r] = c - 1$,whereas $[c + r] = c$.
This,in terms of limits,means that:
$\lim_{x \to c^-} f(x) = c - 1$ and $\lim_{x \to c^+} f(x) = c$.
Since these limits are not equal to each other for any integer $c$,the function is discontinuous at every integral point.

(N/A) rational function $f$ is defined as $f(x) = \frac{p(x)}{q(x)}$,where $p(x)$ and $q(x)$ are polynomial functions and $q(x) \neq 0$.
The domain of $f$ is the set of all real numbers $x$ such that $q(x) \neq 0$.
We know that every polynomial function is continuous everywhere on the set of real numbers $\mathbb{R}$.
According to the algebra of continuous functions,if $p(x)$ and $q(x)$ are continuous functions,then their quotient $\frac{p(x)}{q(x)}$ is also continuous at all points where the denominator $q(x) \neq 0$.
Since $p(x)$ and $q(x)$ are polynomials,they are continuous everywhere. Therefore,the rational function $f(x) = \frac{p(x)}{q(x)}$ is continuous at every point in its domain.

(N/A) To check the continuity of $f(x) = \sin x$,we need to verify if $\mathop {\lim }\limits_{x \to c} f(x) = f(c)$ for any real number $c$.
First,we use the fact that $\mathop {\lim }\limits_{x \to 0} \sin x = 0$.
Let $c$ be any arbitrary real number. We substitute $x = c + h$. As $x \to c$,it follows that $h \to 0$.
Now,we evaluate the limit:
$\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} \sin x$
$= \mathop {\lim }\limits_{h \to 0} \sin(c + h)$
$= \mathop {\lim }\limits_{h \to 0} [\sin c \cos h + \cos c \sin h]$
$= \sin c \cdot (\mathop {\lim }\limits_{h \to 0} \cos h) + \cos c \cdot (\mathop {\lim }\limits_{h \to 0} \sin h)$
$= \sin c \cdot (1) + \cos c \cdot (0)$
$= \sin c + 0 = \sin c$
Since $\mathop {\lim }\limits_{x \to c} f(x) = \sin c = f(c)$,the function $f(x) = \sin x$ is continuous for all real numbers $c$.

(N/A) The function $f(x) = \tan x = \frac{\sin x}{\cos x}$ is defined for all real numbers $x$ such that $\cos x \neq 0$,which means $x \neq (2n + 1) \frac{\pi}{2}$ for any integer $n$.
We know that both the sine function $g(x) = \sin x$ and the cosine function $h(x) = \cos x$ are continuous for all real numbers.
According to the algebra of continuous functions,if $g(x)$ and $h(x)$ are continuous functions,then their quotient $\frac{g(x)}{h(x)}$ is also continuous at all points where the denominator $h(x) \neq 0$.
Since $f(x) = \frac{\sin x}{\cos x}$ is the quotient of two continuous functions and is defined for all $x \in \mathbb{R} \setminus \{(2n + 1) \frac{\pi}{2} : n \in \mathbb{Z}\}$,it follows that $f(x) = \tan x$ is a continuous function on its entire domain.

(N/A) The function $f(x) = \sin(x^{2})$ is defined for all real numbers $x \in \mathbb{R}$.
We can express $f(x)$ as the composition of two functions $g(x)$ and $h(x)$,where $g(x) = \sin(x)$ and $h(x) = x^{2}$.
Then,$(g \circ h)(x) = g(h(x)) = g(x^{2}) = \sin(x^{2}) = f(x)$.
Since $g(x) = \sin(x)$ is a continuous function for all $x \in \mathbb{R}$ and $h(x) = x^{2}$ is a polynomial function which is continuous for all $x \in \mathbb{R}$,the composition of two continuous functions is also continuous.
Therefore,$f(x) = \sin(x^{2})$ is a continuous function for all $x \in \mathbb{R}$.