Continuity Questions in English

(D) To find $\lim_{x \to 1} f(x)$,we evaluate the left-hand limit and the right-hand limit at $x = 1$.
Left-hand limit $(LHL)$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2) = (1)^2 = 1$.
Right-hand limit $(RHL)$: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x) = 1$.
Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = 1$,the limit exists and is equal to $1$.

(D) To determine if the limit exists at $x = 1$,we calculate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$.
$LHL = \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} 3(1 - h) = 3(1 - 0) = 3$.
$RHL = \lim_{x \to 1^+} f(x) = \lim_{h \to 0} f(1 + h) = \lim_{h \to 0} [5 - 3(1 + h)] = 5 - 3(1 + 0) = 5 - 3 = 2$.
Since $LHL \neq RHL$ (i.e.,$3 \neq 2$),the limit $\lim_{x \to 1} f(x)$ does not exist.

(D) Given $f(x) = |x - 2|$.
First,we find the value of the function at $x = 2$: $f(2) = |2 - 2| = 0$.
Next,we calculate the left-hand limit $(LHL)$ as $x \to 2^-$:
$\mathop {\lim }\limits_{x \to 2^-} f(x) = \mathop {\lim }\limits_{h \to 0} f(2 - h) = \mathop {\lim }\limits_{h \to 0} |2 - h - 2| = \mathop {\lim }\limits_{h \to 0} |-h| = 0$.
Now,we calculate the right-hand limit $(RHL)$ as $x \to 2^+$:
$\mathop {\lim }\limits_{x \to 2^+} f(x) = \mathop {\lim }\limits_{h \to 0} f(2 + h) = \mathop {\lim }\limits_{h \to 0} |2 + h - 2| = \mathop {\lim }\limits_{h \to 0} |h| = 0$.
Since $\mathop {\lim }\limits_{x \to 2^-} f(x) = \mathop {\lim }\limits_{x \to 2^+} f(x) = f(2) = 0$,the function $f(x)$ is continuous at $x = 2$.

(B) For the function $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the limit of $f(x)$ as $x \to \frac{\pi}{2}$ must equal $f\left(\frac{\pi}{2}\right)$.
Given $f\left(\frac{\pi}{2}\right) = 3$.
We evaluate the limit: $\lim_{x \to \frac{\pi}{2}} \frac{k\cos x}{\pi - 2x}$.
Let $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}$,$h \to 0$.
Substituting this into the limit: $\lim_{h \to 0} \frac{k\cos(\frac{\pi}{2} + h)}{\pi - 2(\frac{\pi}{2} + h)} = \lim_{h \to 0} \frac{k(-\sin h)}{\pi - \pi - 2h} = \lim_{h \to 0} \frac{-k\sin h}{-2h} = \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h}$.
Since $\lim_{h \to 0} \frac{\sin h}{h} = 1$,the limit is $\frac{k}{2}$.
Equating the limit to the function value: $\frac{k}{2} = 3 \implies k = 6$.

(B) For the function $f(x)$ to be continuous at $x = 0$,the value $f(0)$ must be equal to $\lim_{x \to 0} f(x)$.
Given $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$.
Using the standard limit $\lim_{x \to 0} \frac{\log(1 + kx)}{x} = k$,we can rewrite the limit as:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{\log(1 + ax)}{x} - \frac{\log(1 - bx)}{x} \right)$
$= \lim_{x \to 0} \left( a \cdot \frac{\log(1 + ax)}{ax} - (-b) \cdot \frac{\log(1 - bx)}{-bx} \right)$
$= a(1) + b(1) = a + b$.
Therefore,$f(0) = a + b$.

(A) For $f(x)$ to be continuous at $x = 2$,we must have $\lim_{x \to 2} f(x) = f(2) = k$.
First,we evaluate the limit: $\lim_{x \to 2} \frac{x^3 + x^2 - 16x + 20}{(x - 2)^2}$.
Since substituting $x = 2$ gives the indeterminate form $\frac{0}{0}$,we factor the numerator.
By synthetic division or polynomial division,$x^3 + x^2 - 16x + 20 = (x - 2)^2(x + 5)$.
Thus,$\lim_{x \to 2} \frac{(x - 2)^2(x + 5)}{(x - 2)^2} = \lim_{x \to 2} (x + 5) = 2 + 5 = 7$.
Therefore,$k = 7$.

(A) For a 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.
$1$. Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 + k) = 0^2 + k = k$.
$2$. Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x^2 - k) = -0^2 - k = -k$.
$3$. Value of the function: $f(0) = 0^2 + k = k$.
For continuity,$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0)$.
Therefore,$k = -k$.
$2k = 0 \implies k = 0$.

(B) For a function $f(x)$ to be continuous at $x = 0$,the limit of the function as $x$ approaches $0$ must equal the value of the function at $x = 0$,i.e.,$\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$.
Given $f(x) = (1 + x)^{1/x}$.
We know that the standard limit $\mathop {\lim }\limits_{x \to 0} (1 + x)^{1/x} = e$.
Therefore,for the function to be continuous at $x = 0$,we must have $f(0) = e$.

(D) To check the continuity at $x = 1/2$,we evaluate the left-hand limit,right-hand limit,and the function value.
$1$. Left-hand limit: $\lim_{x \to 1/2^-} f(x) = \lim_{x \to 1/2^-} x = 1/2$.
$2$. Right-hand limit: $\lim_{x \to 1/2^+} f(x) = \lim_{x \to 1/2^+} (1 - x) = 1 - 1/2 = 1/2$.
$3$. Function value: $f(1/2) = 1$.
Since $\lim_{x \to 1/2^-} f(x) = \lim_{x \to 1/2^+} f(x) = 1/2$,the limit $\lim_{x \to 1/2} f(x)$ exists and is equal to $1/2$.
However,$\lim_{x \to 1/2} f(x) = 1/2 \neq f(1/2) = 1$.
Therefore,$f(x)$ is discontinuous at $x = 1/2$.

(B) Given $f(x) = \begin{cases} \frac{x^2}{a} - a, & x < a \\ 0, & x = a \\ a - \frac{x^2}{a}, & x > a \end{cases}$.
First,we find the value of the function at $x = a$: $f(a) = 0$.
Next,we calculate the left-hand limit $(LHL)$ at $x = a$:
$\lim_{x \to a^-} f(x) = \lim_{h \to 0} f(a - h) = \lim_{h \to 0} \left( \frac{(a - h)^2}{a} - a \right) = \frac{a^2}{a} - a = a - a = 0$.
Then,we calculate the right-hand limit $(RHL)$ at $x = a$:
$\lim_{x \to a^+} f(x) = \lim_{h \to 0} f(a + h) = \lim_{h \to 0} \left( a - \frac{(a + h)^2}{a} \right) = a - \frac{a^2}{a} = a - a = 0$.
Since $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) = 0$,the function $f(x)$ is continuous at $x = a$.

(C) Given $f(x) = \begin{cases} e^{1/x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$.
First,we evaluate the left-hand limit at $x = 0$:
$\lim_{x \to 0^-} f(x) = \lim_{h \to 0} e^{-1/h} = e^{-\infty} = 0$.
Next,we evaluate the right-hand limit at $x = 0$:
$\lim_{x \to 0^+} f(x) = \lim_{h \to 0} e^{1/h} = e^{\infty} = \infty$.
Since the right-hand limit does not exist (it is infinite),the function $f(x)$ is not continuous at $x = 0$.
Therefore,the correct option is $(c)$.

(C) To check the continuity of $f(x)$ at $x = 1$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 1$.
Given $f(1) = 2$.
For $x \ne 1$,$f(x) = \frac{x^2 - 4x + 3}{x^2 - 1} = \frac{(x - 1)(x - 3)}{(x - 1)(x + 1)} = \frac{x - 3}{x + 1}$.
Now,calculate the limit as $x \to 1$:
$\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x - 3}{x + 1} = \frac{1 - 3}{1 + 1} = \frac{-2}{2} = -1$.
Since $\lim_{x \to 1^+} f(x) = -1$ and $\lim_{x \to 1^-} f(x) = -1$,the limit exists and is equal to $-1$.
However,$f(1) = 2$.
Since $\lim_{x \to 1} f(x) \ne f(1)$,the function $f(x)$ is discontinuous at $x = 1$.

(A) rational function $f(x) = \frac{P(x)}{Q(x)}$ is discontinuous at points where the denominator $Q(x) = 0$.
Given $f(x) = \frac{x + 1}{x^2 + x - 12}$.
To find the points of discontinuity,we set the denominator equal to zero:
$x^2 + x - 12 = 0$
Factoring the quadratic equation:
$x^2 + 4x - 3x - 12 = 0$
$x(x + 4) - 3(x + 4) = 0$
$(x - 3)(x + 4) = 0$
Thus,$x = 3$ and $x = -4$.
Therefore,the function is discontinuous at $x = 3$ and $x = -4$.

(C) To check the continuity at $x = 0$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 0$.
$1$. Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (\frac{\sin x}{x} + \cos x) = 1 + 1 = 2$.
$2$. Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (\frac{\sin x}{x} + \cos x) = 1 + 1 = 2$.
$3$. Value of the function: $f(0) = 2$.
Since $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) = 2$,the function $f(x)$ is continuous at $x = 0$.

(C) To check the continuity of $f(x)$ at $x = 0$,we evaluate the limit as $x \to 0$.
For $x \neq 0$,$f(x) = x^2 \sin \frac{1}{x}$.
We know that $-1 \le \sin \frac{1}{x} \le 1$ for all $x \neq 0$.
Multiplying by $x^2$,we get $-x^2 \le x^2 \sin \frac{1}{x} \le x^2$.
As $x \to 0$,both $-x^2 \to 0$ and $x^2 \to 0$.
By the Squeeze Theorem,$\lim_{x \to 0} f(x) = 0$.
Since $f(0) = 0$,we have $\lim_{x \to 0} f(x) = f(0)$.
Therefore,$f(x)$ is continuous at $x = 0$.

(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 $(LHL)$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} k(2x - x^2) = k(2(0) - (0)^2) = 0$.
$2$. Right-hand limit $(RHL)$: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \cos x = \cos(0) = 1$.
$3$. Value of the function: $f(0) = \cos(0) = 1$.
Since the $LHL$ $(0)$ is not equal to the $RHL$ $(1)$,there is no value of $k$ that can make the function continuous at $x = 0$.

(C) To check the continuity at $x = 0$,we evaluate the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the function value $f(0)$.
$1$. Function value: $f(0) = 0$.
$2$. Left-hand limit $(LHL)$: $\lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) = \lim_{h \to 0} \frac{-h}{e^{-1/h} + 1}$.
As $h \to 0^+$,$1/h \to \infty$,so $e^{-1/h} \to 0$.
Thus,$\lim_{h \to 0} \frac{-h}{0 + 1} = 0$.
$3$. Right-hand limit $(RHL)$: $\lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) = \lim_{h \to 0} \frac{h}{e^{1/h} + 1}$.
As $h \to 0^+$,$e^{1/h} \to \infty$,so $\frac{h}{e^{1/h} + 1} \to 0$.
Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0$,the function $f(x)$ is continuous at $x = 0$.

(B) To check the continuity at $x = 0$,we calculate the limit of $f(x)$ as $x \to 0$.
Given $f(x) = (1 + 2x)^{1/x}$ for $x \ne 0$.
We know the standard limit $\lim_{u \to 0} (1 + u)^{1/u} = e$.
Here,let $u = 2x$. As $x \to 0$,$u \to 0$.
Then $f(x) = (1 + 2x)^{1/x} = [(1 + 2x)^{1/(2x)}]^2$.
Taking the limit as $x \to 0$:
$\lim_{x \to 0} f(x) = \lim_{u \to 0} [(1 + u)^{1/u}]^2 = e^2$.
Since $\lim_{x \to 0} f(x) = e^2$ and $f(0) = e^2$,the function is continuous at $x = 0$.
Thus,$\lim_{x \to 0^+} f(x) = e^2$ and $\lim_{x \to 0^-} f(x) = e^2$.
Comparing with the options,option $(b)$ is correct.

(D) To determine the behavior of the function at $x = 0$,we calculate the left-hand limit and the right-hand limit.
Right-hand limit $(RHL)$:
$\mathop {\lim }\limits_{x \to 0^+} f(x) = \mathop {\lim }\limits_{h \to 0} 2^{1/(0+h)} = \mathop {\lim }\limits_{h \to 0} 2^{1/h} = \infty$.
Left-hand limit $(LHL)$:
$\mathop {\lim }\limits_{x \to 0^-} f(x) = \mathop {\lim }\limits_{h \to 0} 2^{1/(0-h)} = \mathop {\lim }\limits_{h \to 0} 2^{-1/h} = \mathop {\lim }\limits_{h \to 0} \frac{1}{2^{1/h}} = \frac{1}{\infty} = 0$.
Since the $RHL$ is $\infty$ and the $LHL$ is $0$,the limit $\mathop {\lim }\limits_{x \to 0} f(x)$ does not exist.
Therefore,none of the given options $(A)$,$(B)$,or $(C)$ are correct.

(C) To check the continuity of $f(x)$ at $x = 0$,we evaluate the limit as $x \to 0$.
Given $f(x) = \frac{\sin(x^2)}{x}$ for $x \ne 0$.
We can rewrite this as $f(x) = x \cdot \frac{\sin(x^2)}{x^2}$.
Now,$\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( x \cdot \frac{\sin(x^2)}{x^2} \right)$.
Since $\lim_{x \to 0} \frac{\sin(x^2)}{x^2} = 1$ and $\lim_{x \to 0} x = 0$,we have $\lim_{x \to 0} f(x) = 0 \cdot 1 = 0$.
Since $\lim_{x \to 0} f(x) = f(0) = 0$,the function $f(x)$ is continuous at $x = 0$.

(C) To check the continuity of $f(x)$ at $x = 0$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 0$.
$1$. Left-hand limit $(LHL)$: $\mathop {\lim }\limits_{x \to 0^-} f(x) = \mathop {\lim }\limits_{x \to 0^-} (x - 1) = 0 - 1 = -1$.
$2$. Right-hand limit $(RHL)$: $\mathop {\lim }\limits_{x \to 0^+} f(x) = \mathop {\lim }\limits_{x \to 0^+} (x^2) = 0^2 = 0$.
$3$. Value of the function at $x = 0$: $f(0) = \frac{1}{4}$.
Since $\mathop {\lim }\limits_{x \to 0^-} f(x) \neq \mathop {\lim }\limits_{x \to 0^+} f(x)$,the limit $\mathop {\lim }\limits_{x \to 0} f(x)$ does not exist.
Therefore,$f(x)$ is discontinuous at $x = 0$.

(A) The function $f(x) = \log x$ (where the base is typically $e$ or $10$) is defined only for $x > 0$.
In its domain $(0, \infty)$,the logarithmic function is a continuous function.
Therefore,the graph of $f(x) = \log x$ is continuous for all $x$ in its domain.
Thus,statement $(a)$ is correct.

(B) For a function $f(x)$ to be continuous at $x = a$,the condition $\lim_{x \to a} f(x) = f(a)$ must be satisfied.
Here,$a = 1$,so we need $\lim_{x \to 1} f(x) = f(1)$.
Given $f(1) = k$.
Now,calculate the limit: $\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x^2 - 1}{x - 1}$.
Since this is a $0/0$ form,we factor the numerator: $\lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1} (x + 1)$.
Substituting $x = 1$,we get $1 + 1 = 2$.
Therefore,for continuity,$k = 2$.

(B) The function $f(x) = \frac{x}{[x]}$ is defined only when $[x] \neq 0$.
$(i)$ For $0 \le x < 1$,$[x] = 0$,so $f(x)$ is undefined in the interval $(0, 1)$. Therefore,the function is discontinuous for all $x \in (0, 1)$.
$(ii)$ At any integer $n \in \mathbb{Z} \setminus \{0, 1\}$,the left-hand limit is $\lim_{x \to n^-} \frac{x}{[x]} = \frac{n}{n-1}$ and the right-hand limit is $\lim_{x \to n^+} \frac{x}{[x]} = \frac{n}{n} = 1$. Since $\frac{n}{n-1} \neq 1$ for $n \neq 0$,the function is discontinuous at all integers.
$(iii)$ At $x = 1$,$\lim_{x \to 1^-} f(x)$ is undefined (as $[x]=0$ for $x \in [0, 1)$),and $\lim_{x \to 1^+} f(x) = 1$. Thus,it is discontinuous at $x = 1$.
Hence,the function is discontinuous at all integers and in the interval $(0, 1)$.

(B) To check the continuity of $f(x)$ at $x = 0$,we evaluate the limit:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin^2(ax)}{x^2} = \lim_{x \to 0} \left( \frac{\sin(ax)}{ax} \right)^2 \cdot a^2 = (1)^2 \cdot a^2 = a^2$.
The value of the function at $x = 0$ is $f(0) = 1$.
For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \to 0} f(x) = f(0)$,which implies $a^2 = 1$,or $a = \pm 1$.
If $a \neq \pm 1$,then $a^2 \neq 1$,so $\lim_{x \to 0} f(x) \neq f(0)$.
Thus,$f(x)$ is discontinuous at $x = 0$ when $a \neq \pm 1$.

(B) To check continuity at $x = 0$:
$\lim_{x \to 0^-} f(x) = -(0)^2 = 0$
$\lim_{x \to 0^+} f(x) = 5(0) - 4 = -4$
Since $\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$,$f(x)$ is discontinuous at $x = 0$.
To check continuity at $x = 1$:
$\lim_{x \to 1^-} f(x) = 5(1) - 4 = 1$
$\lim_{x \to 1^+} f(x) = 4(1)^2 - 3(1) = 1$
$f(1) = 5(1) - 4 = 1$
Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$,$f(x)$ is continuous at $x = 1$.
To check continuity at $x = 2$:
$\lim_{x \to 2^-} f(x) = 4(2)^2 - 3(2) = 16 - 6 = 10$
$\lim_{x \to 2^+} f(x) = 3(2) + 4 = 10$
$f(2) = 3(2) + 4 = 10$
Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$,$f(x)$ is continuous at $x = 2$.
Thus,option $B$ is correct.

(C) To check the continuity of $f(x)$ at $x = 0$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 0$.
$1$. Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \sin^{-1}|x| = \sin^{-1}(0) = 0$.
$2$. Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \sin^{-1}|x| = \sin^{-1}(0) = 0$.
$3$. 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 $f(x)$ is continuous at $x = 0$.

(B) For a function $f(x)$ to be continuous at $x = 0$,the limit of the function as $x$ approaches $0$ must be equal to the value of the function at $x = 0$.
That is,$\lim_{x \to 0} f(x) = f(0)$.
Given $f(x) = \frac{\sin 2x}{5x}$ for $x \ne 0$,we calculate the limit:
$\lim_{x \to 0} \frac{\sin 2x}{5x} = \lim_{x \to 0} \left( \frac{\sin 2x}{2x} \times \frac{2x}{5x} \right)$
$= \lim_{x \to 0} \left( \frac{\sin 2x}{2x} \right) \times \frac{2}{5}$
Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,we have $1 \times \frac{2}{5} = \frac{2}{5}$.
Since $f(0) = k$,we get $k = \frac{2}{5}$.

(C) To check the continuity at $x = 1$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$.
$1$. Left-hand limit $(LHL)$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (1 + x^2) = 1 + (1)^2 = 2$.
$2$. Right-hand limit $(RHL)$: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (1 - x) = 1 - 1 = 0$.
Since $\lim_{x \to 1^-} f(x) \ne \lim_{x \to 1^+} f(x)$,the limit does not exist at $x = 1$.
Therefore,$f(x)$ is discontinuous at $x = 1$.

(D) Given $f(x) = \begin{cases} \frac{x^2 - 1}{x + 1}, & x \neq -1 \\ -2, & x = -1 \end{cases}$.
For $x \neq -1$,$f(x) = \frac{(x - 1)(x + 1)}{x + 1} = x - 1$.
Now,calculate the left-hand limit: $\lim_{x \to (-1)^-} f(x) = \lim_{x \to -1} (x - 1) = -1 - 1 = -2$.
Calculate the right-hand limit: $\lim_{x \to (-1)^+} f(x) = \lim_{x \to -1} (x - 1) = -1 - 1 = -2$.
Since $\lim_{x \to (-1)^-} f(x) = \lim_{x \to (-1)^+} f(x) = f(-1) = -2$,the function is continuous at $x = -1$.
Therefore,all the given statements are correct.

(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: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (\frac{5}{2} - x) = \frac{5}{2} - 2 = \frac{1}{2}$.
$2$. Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x - \frac{3}{2}) = 2 - \frac{3}{2} = \frac{1}{2}$.
$3$. Value of the function: $f(2) = 1$.
Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = \frac{1}{2}$,the limit $\lim_{x \to 2} f(x)$ exists and is equal to $\frac{1}{2}$.
However,$\lim_{x \to 2} f(x) \neq f(2)$ because $\frac{1}{2} \neq 1$.
Therefore,$f(x)$ is discontinuous at $x = 2$.

(B) The function $f(x) = |x - b|$ is a modulus function.
For any real number $b$,the function $f(x) = |x - b|$ is continuous for all $x \in \mathbb{R}$.
Specifically,at $x = b$,we have $\lim_{x \to b} f(x) = \lim_{x \to b} |x - b| = 0$.
Also,$f(b) = |b - b| = 0$.
Since $\lim_{x \to b} f(x) = f(b)$,the function is continuous at $x = b$.

(B) To check the continuity of $f(x)$ at $x = a$,we evaluate the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the function value $f(a)$.
$1$. Left-hand limit $(LHL)$: $\lim_{x \to a^-} f(x) = \lim_{x \to a^-} \frac{|x - a|}{x - a}$. Since $x < a$,$|x - a| = -(x - a)$,so $\lim_{x \to a^-} \frac{-(x - a)}{x - a} = -1$.
$2$. Right-hand limit $(RHL)$: $\lim_{x \to a^+} f(x) = \lim_{x \to a^+} \frac{|x - a|}{x - a}$. Since $x > a$,$|x - a| = (x - a)$,so $\lim_{x \to a^+} \frac{x - a}{x - a} = 1$.
$3$. Function value: $f(a) = 1$.
Since $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$,the limit $\lim_{x \to a} f(x)$ does not exist. Therefore,$f(x)$ is discontinuous at $x = a$.

(C) To check the continuity of $f(x)$ at $x = 1$,we evaluate the limit and the function value.
First,calculate the limit as $x$ approaches $1$:
$\lim_{x \to 1} f(x) = \lim_{x \to 1} x^2 = (1)^2 = 1$.
Next,identify the value of the function at $x = 1$:
$f(1) = 2$.
Since $\lim_{x \to 1} f(x) \neq f(1)$ (because $1 \neq 2$),the function $f(x)$ is discontinuous at $x = 1$.

(A) To check the continuity of $f(x)$ 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: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (1 + x) = 1 + 2 = 3$.
$2$. Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (5 - x) = 5 - 2 = 3$.
$3$. Value of the function: $f(2) = 1 + 2 = 3$.
Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 3$,the function $f(x)$ is continuous at $x = 2$.

(C) To check the continuity at $x = \frac{3\pi}{4}$,we evaluate the left-hand limit,right-hand limit,and the value of the function at that point.
First,the value of the function at $x = \frac{3\pi}{4}$ is $f\left(\frac{3\pi}{4}\right) = 1$.
Next,the left-hand limit is $\lim_{x \to \frac{3\pi}{4}^-} f(x) = 1$.
Then,the right-hand limit is $\lim_{x \to \frac{3\pi}{4}^+} f(x) = \lim_{h \to 0} 2\sin \left(\frac{2}{9} \left(\frac{3\pi}{4} + h\right)\right) = 2\sin \left(\frac{2}{9} \cdot \frac{3\pi}{4}\right) = 2\sin \left(\frac{\pi}{6}\right) = 2 \cdot \frac{1}{2} = 1$.
Since $\lim_{x \to \frac{3\pi}{4}^-} f(x) = \lim_{x \to \frac{3\pi}{4}^+} f(x) = f\left(\frac{3\pi}{4}\right) = 1$,the function $f(x)$ is continuous at $x = \frac{3\pi}{4}$.

(A) To check the continuity at $x = \frac{\pi}{2}$,we evaluate the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the function value.
$1$. Left-hand limit $(LHL)$: $\lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}} (x \sin x) = \frac{\pi}{2} \sin(\frac{\pi}{2}) = \frac{\pi}{2} \times 1 = \frac{\pi}{2}$.
$2$. Right-hand limit $(RHL)$: $\lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{\pi}{2} \sin(\pi + x) = \frac{\pi}{2} \sin(\pi + \frac{\pi}{2}) = \frac{\pi}{2} \sin(\frac{3\pi}{2}) = \frac{\pi}{2} \times (-1) = -\frac{\pi}{2}$.
$3$. Function value: $f(\frac{\pi}{2}) = \frac{\pi}{2} \sin(\frac{\pi}{2}) = \frac{\pi}{2}$.
Since $\lim_{x \to \frac{\pi}{2}^-} f(x) \neq \lim_{x \to \frac{\pi}{2}^+} f(x)$,the function $f(x)$ is discontinuous at $x = \frac{\pi}{2}$.

(A) 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 $(LHL)$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1 - \cos 4x}{x^2} = \lim_{x \to 0^-} \frac{2 \sin^2(2x)}{x^2} = \lim_{x \to 0^-} 2 \times \left( \frac{\sin 2x}{2x} \right)^2 \times 4 = 2 \times 1^2 \times 4 = 8$.
$2$. Right-hand limit $(RHL)$: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}$.
Rationalizing the denominator: $\lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{16 + \sqrt{x} - 16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{\sqrt{x}} = \lim_{x \to 0^+} (\sqrt{16 + \sqrt{x}} + 4) = \sqrt{16} + 4 = 4 + 4 = 8$.
$3$. Since the function is continuous at $x = 0$,$f(0) = \text{LHL} = \text{RHL}$.
Therefore,$a = 8$.

(D) For a function $f(x)$ 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.
$1$. Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (ax^2 - b) = a(1)^2 - b = a - b$.
$2$. Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + 1) = 1 + 1 = 2$.
$3$. Value at $x = 1$: $f(1) = 2$.
For continuity,$a - b = 2$.
Checking the options:
- For $A$: $a = 2, b = 0 \Rightarrow 2 - 0 = 2$ (Correct).
- For $B$: $a = 1, b = -1 \Rightarrow 1 - (-1) = 2$ (Correct).
- For $C$: $a = 4, b = 2 \Rightarrow 4 - 2 = 2$ (Correct).
Since all options satisfy the condition $a - b = 2$,the correct answer is $D$.

(B) To check the continuity of $f(x)$ at $x = 0$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 0$.
$1$. Left-hand limit $(LHL)$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x - (-x)}{x} = \lim_{x \to 0^-} \frac{2x}{x} = 2$.
$2$. Right-hand limit $(RHL)$: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x - x}{x} = \lim_{x \to 0^+} \frac{0}{x} = 0$.
$3$. Value of the function: $f(0) = 2$.
Since the left-hand limit $(2)$ is not equal to the right-hand limit $(0)$,the limit $\lim_{x \to 0} f(x)$ does not exist.
Therefore,$f(x)$ is discontinuous at $x = 0$.

(B) To check the continuity of $f(x)$ at $x = 2$,we evaluate the limit of $f(x)$ as $x \to 2$:
$\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x^4 - 16}{x - 2}$
Using the identity $a^2 - b^2 = (a - b)(a + b)$,we have $x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$.
Thus,$\lim_{x \to 2} \frac{(x - 2)(x + 2)(x^2 + 4)}{x - 2} = \lim_{x \to 2} (x + 2)(x^2 + 4) = (2 + 2)(2^2 + 4) = 4 \times 8 = 32$.
Since $\lim_{x \to 2} f(x) = 32$ and $f(2) = 16$,we see that $\lim_{x \to 2} f(x) \neq f(2)$.
Therefore,$f(x)$ is discontinuous at $x = 2$.

(B) To check the continuity of $f(x)$ at $x = 1$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$.
$LHL$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2) = (1)^2 = 1$.
$RHL$: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + 5) = 1 + 5 = 6$.
Since the left-hand limit $(1)$ is not equal to the right-hand limit $(6)$,the limit $\lim_{x \to 1} f(x)$ does not exist.
Therefore,$f(x)$ is discontinuous at $x = 1$.

(B) For the function $f(x)$ to be continuous at $x = -5$,the limit of $f(x)$ as $x \to -5$ must equal $f(-5)$.
Given $f(-5) = a$.
Now,calculate the limit:
$\lim_{x \to -5} f(x) = \lim_{x \to -5} \frac{x^2 + 3x - 10}{x^2 + 2x - 15}$
Factorizing the numerator and denominator:
$x^2 + 3x - 10 = (x + 5)(x - 2)$
$x^2 + 2x - 15 = (x + 5)(x - 3)$
So,$\lim_{x \to -5} \frac{(x + 5)(x - 2)}{(x + 5)(x - 3)} = \lim_{x \to -5} \frac{x - 2}{x - 3}$
Substituting $x = -5$:
$\frac{-5 - 2}{-5 - 3} = \frac{-7}{-8} = \frac{7}{8}$
Since the function is continuous,$a = \frac{7}{8}$.

(D) For a function $f(x)$ to be continuous at $x = 3$,the following condition must be satisfied:
$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$
Given $f(3) = 4$.
Calculating the left-hand limit:
$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x + \lambda) = 3 + \lambda$
Calculating the right-hand limit:
$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (3x - 5) = 3(3) - 5 = 9 - 5 = 4$
Equating the limits to $f(3)$:
$3 + \lambda = 4$
$\lambda = 4 - 3$
$\lambda = 1$

(D) For a function $f(x)$ to be continuous at $x = 0$,the limit $\lim_{x \to 0} f(x)$ must exist and be equal to $f(0)$.
Here,$f(0) = k$.
We examine the limit $\lim_{x \to 0} \sin \frac{1}{x}$.
As $x \to 0$,the argument $\frac{1}{x}$ approaches $\infty$ or $-\infty$.
The function $\sin \frac{1}{x}$ oscillates infinitely between $-1$ and $1$ as $x$ approaches $0$.
Since the limit does not approach a unique finite value,the limit $\lim_{x \to 0} \sin \frac{1}{x}$ does not exist.
Therefore,there is no value of $k$ that can make the function continuous at $x = 0$.

(D) For $f(x)$ to be continuous at $x = 4$,the left-hand limit,right-hand limit,and the value of the function at $x = 4$ must be equal.
Left-hand limit $(LHL)$: $\lim_{x \to 4^-} f(x) = \lim_{h \to 0} f(4 - h) = \lim_{h \to 0} (\frac{4 - h - 4}{|4 - h - 4|} + a) = \lim_{h \to 0} (\frac{-h}{|-h|} + a) = -1 + a$.
Right-hand limit $(RHL)$: $\lim_{x \to 4^+} f(x) = \lim_{h \to 0} f(4 + h) = \lim_{h \to 0} (\frac{4 + h - 4}{|4 + h - 4|} + b) = \lim_{h \to 0} (\frac{h}{|h|} + b) = 1 + b$.
Value of the function: $f(4) = a + b$.
For continuity: $a - 1 = a + b = b + 1$.
From $a - 1 = a + b$,we get $b = -1$.
From $a + b = b + 1$,we get $a = 1$.
Thus,$a = 1$ and $b = -1$.

(C) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
Given $f(x) = \frac{(27 - 2x)^{1/3} - 3}{9 - 3(243 + 5x)^{1/5}}$.
As $x \to 0$,the expression takes the indeterminate form $\frac{0}{0}$.
Using $L$'$H$ôpital's Rule,we differentiate the numerator and denominator with respect to $x$:
Numerator derivative: $\frac{d}{dx} [(27 - 2x)^{1/3} - 3] = \frac{1}{3}(27 - 2x)^{-2/3} \cdot (-2) = -\frac{2}{3}(27 - 2x)^{-2/3}$.
Denominator derivative: $\frac{d}{dx} [9 - 3(243 + 5x)^{1/5}] = -3 \cdot \frac{1}{5}(243 + 5x)^{-4/5} \cdot 5 = -3(243 + 5x)^{-4/5}$.
Now,taking the limit as $x \to 0$:
$\lim_{x \to 0} f(x) = \frac{-\frac{2}{3}(27)^{-2/3}}{-3(243)^{-4/5}} = \frac{\frac{2}{3}(3^3)^{-2/3}}{3(3^5)^{-4/5}} = \frac{\frac{2}{3}(3^{-2})}{3(3^{-4})} = \frac{\frac{2}{3} \cdot \frac{1}{9}}{3 \cdot \frac{1}{81}} = \frac{2/27}{3/81} = \frac{2/27}{1/27} = 2$.
Thus,$f(0) = 2$.

(A) For the function $f(x)$ to be continuous at $x = 0$,the limit of the function as $x \to 0$ must be equal to $f(0)$.
Given $f(0) = k$,we need to evaluate $\lim_{x \to 0} (\cos x)^{1/x}$.
Let $L = \lim_{x \to 0} (\cos x)^{1/x}$.
Taking the natural logarithm on both sides,we get $\ln L = \lim_{x \to 0} \frac{1}{x} \ln(\cos x)$.
Using $L$'$H$ôpital's rule for the indeterminate form $\frac{0}{0}$:
$\ln L = \lim_{x \to 0} \frac{\frac{d}{dx}(\ln(\cos x))}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{\frac{1}{\cos x} \cdot (-\sin x)}{1} = \lim_{x \to 0} (-\tan x) = 0$.
Since $\ln L = 0$,we have $L = e^0 = 1$.
Therefore,$k = 1$.

(A) To check the continuity of the function $f(x)$ 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: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x - 1) = 2 - 1 = 1$.
$2$. Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x - 3) = 2(2) - 3 = 4 - 3 = 1$.
$3$. Value of the function at $x = 2$: $f(2) = 2(2) - 3 = 1$.
Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 1$,the function is continuous at $x = 2$.
For $x < 2$,$f(x) = x - 1$,which is a polynomial function and thus continuous.
For $x > 2$,$f(x) = 2x - 3$,which is also a polynomial function and thus continuous.
Therefore,the function $f(x)$ is continuous for all real values of $x$.

(C) Given that the function $f(x)$ is continuous in the interval $(-\infty, 6)$,it must be continuous at the points $x = 1$ and $x = 3$.
For continuity at $x = 1$:
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$
$1 + \sin(\frac{\pi}{2}) = a(1) + b$
$1 + 1 = a + b \implies a + b = 2$ ..... $(i)$
For continuity at $x = 3$:
$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$
$a(3) + b = 6 \tan(\frac{3\pi}{12})$
$3a + b = 6 \tan(\frac{\pi}{4})$
$3a + b = 6(1) \implies 3a + b = 6$ ..... $(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(3a + b) - (a + b) = 6 - 2$
$2a = 4 \implies a = 2$
Substituting $a = 2$ into equation $(i)$:
$2 + b = 2 \implies b = 0$
Thus,the values are $a = 2$ and $b = 0$.