Continuity Questions in English

(N/A) Let $g(x) = 1 - x + |x|$ and $h(x) = |x|$ for all real $x$.
Then the composite function $(h \circ g)(x) = h(g(x)) = h(1 - x + |x|) = |1 - x + |x|| = f(x)$.
Since $h(x) = |x|$ is a continuous function for all real $x$,and $g(x) = 1 - x + |x|$ is the sum of a polynomial function $(1 - x)$ and the modulus function $(|x|)$,both of which are continuous,$g(x)$ is also continuous.
Since $f(x)$ is the composition of two continuous functions $h$ and $g$,$f(x)$ is a continuous function for all real $x$.

The given function is $f(x)=5x-3$.
At $x=0$,$f(0)=5(0)-3=-3$.
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (5x-3) = 5(0)-3 = -3$.
Since $\lim_{x \to 0} f(x) = f(0)$,the function $f$ is continuous at $x=0$.
At $x=-3$,$f(-3)=5(-3)-3=-15-3=-18$.
$\lim_{x \to -3} f(x) = \lim_{x \to -3} (5x-3) = 5(-3)-3 = -18$.
Since $\lim_{x \to -3} f(x) = f(-3)$,the function $f$ is continuous at $x=-3$.
At $x=5$,$f(5)=5(5)-3=25-3=22$.
$\lim_{x \to 5} f(x) = \lim_{x \to 5} (5x-3) = 5(5)-3 = 22$.
Since $\lim_{x \to 5} f(x) = f(5)$,the function $f$ is continuous at $x=5$.

(A) The given function is $f(x) = 2x^{2} - 1$.
First,we find the value of the function at $x = 3$:
$f(3) = 2(3)^{2} - 1 = 2(9) - 1 = 18 - 1 = 17$.
Next,we calculate the limit of the function as $x$ approaches $3$:
$\lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^{2} - 1) = 2(3)^{2} - 1 = 18 - 1 = 17$.
Since $\lim_{x \to 3} f(x) = f(3) = 17$,the function $f(x)$ is continuous at $x = 3$.

(N/A) The given function is $f(x) = x - 5$.
It is evident that $f$ is defined at every real number $k$ and its value at $k$ is $f(k) = k - 5$.
We calculate the limit of the function as $x$ approaches $k$:
$\lim_{x \to k} f(x) = \lim_{x \to k} (x - 5) = k - 5$.
Since $\lim_{x \to k} f(x) = f(k)$,the function $f$ is continuous at every real number $k$.
Therefore,$f(x) = x - 5$ is a continuous function.

(N/A) The given function is $f(x) = \frac{1}{x-5}$,where $x \neq 5$.
For any real number $k$ such that $k \neq 5$,we evaluate the limit:
$\lim_{x \to k} f(x) = \lim_{x \to k} \frac{1}{x-5} = \frac{1}{k-5}$.
Also,the value of the function at $x = k$ is $f(k) = \frac{1}{k-5}$.
Since $\lim_{x \to k} f(x) = f(k)$ for all $k \in \mathbb{R} \setminus \{5\}$,the function $f$ is continuous at every point in its domain.
Therefore,$f$ is a continuous function.

(N/A) The given function is $f(x) = \frac{x^{2} - 25}{x + 5}$ for $x \neq -5$.
For any real number $c \neq -5$,we have:
$\lim_{x \to c} f(x) = \lim_{x \to c} \frac{x^{2} - 25}{x + 5} = \lim_{x \to c} \frac{(x + 5)(x - 5)}{x + 5} = \lim_{x \to c} (x - 5) = c - 5$.
Also,$f(c) = \frac{c^{2} - 25}{c + 5} = \frac{(c + 5)(c - 5)}{c + 5} = c - 5$.
Since $\lim_{x \to c} f(x) = f(c)$ for all $c \neq -5$,the function $f$ is continuous at every point in its domain.

The given function is $f(x) = |x - 5| = \begin{cases} 5 - x, & \text{if } x < 5 \\ x - 5, & \text{if } x \ge 5 \end{cases}$.
This function $f$ is defined for all real numbers.
Let $c$ be any real number. Then $c < 5$,$c = 5$,or $c > 5$.
Case $I$: $c < 5$.
Then $f(c) = 5 - c$.
$\lim_{x \to c} f(x) = \lim_{x \to c} (5 - x) = 5 - c$.
Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous for all $c < 5$.
Case $II$: $c = 5$.
Then $f(5) = 5 - 5 = 0$.
Left-hand limit: $\lim_{x \to 5^-} f(x) = \lim_{x \to 5} (5 - x) = 5 - 5 = 0$.
Right-hand limit: $\lim_{x \to 5^+} f(x) = \lim_{x \to 5} (x - 5) = 5 - 5 = 0$.
Since $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5)$,$f$ is continuous at $x = 5$.
Case $III$: $c > 5$.
Then $f(c) = c - 5$.
$\lim_{x \to c} f(x) = \lim_{x \to c} (x - 5) = c - 5$.
Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous for all $c > 5$.
Conclusion: Since $f$ is continuous at all real numbers,it is a continuous function.

(N/A) The given function is $f(x) = x^{n}$.
It is evident that $f$ is defined at all positive integers $n$,and its value at $x = n$ is $f(n) = n^{n}$.
Now,we evaluate the limit of the function as $x$ approaches $n$:
$\lim_{x \to n} f(x) = \lim_{x \to n} (x^{n}) = n^{n}$.
Since $\lim_{x \to n} f(x) = f(n) = n^{n}$,the condition for continuity is satisfied.
Therefore,the function $f(x) = x^{n}$ is continuous at $x = n$,where $n$ is a positive integer.

(N/A) The given function is $f(x) = \begin{cases} x, & \text{if } x \le 1 \\ 5, & \text{if } x > 1 \end{cases}$.
At $x=0$:
The function is defined at $x=0$ and $f(0) = 0$.
The limit is $\lim_{x \to 0} f(x) = \lim_{x \to 0} x = 0$.
Since $\lim_{x \to 0} f(x) = f(0)$,the function is continuous at $x=0$.
At $x=1$:
The function is defined at $x=1$ and $f(1) = 1$.
The left-hand limit is $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$.
The right-hand limit is $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5 = 5$.
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 not continuous at $x=1$.
At $x=2$:
The function is defined at $x=2$ and $f(2) = 5$.
The limit is $\lim_{x \to 2} f(x) = \lim_{x \to 2} 5 = 5$.
Since $\lim_{x \to 2} f(x) = f(2)$,the function is continuous at $x=2$.

(C) The given function is $f(x) = \begin{cases} 2x + 3, & \text{if } x \le 2 \\ 2x - 3, & \text{if } x > 2 \end{cases}$.
It is evident that the function $f$ is defined for all real numbers. Let $c$ be any real number. We consider three cases:
Case $I$: $c < 2$. Here $f(x) = 2x + 3$. $\lim_{x \to c} f(x) = 2c + 3 = f(c)$. Thus,$f$ is continuous for all $x < 2$.
Case $II$: $c > 2$. Here $f(x) = 2x - 3$. $\lim_{x \to c} f(x) = 2c - 3 = f(c)$. Thus,$f$ is continuous for all $x > 2$.
Case $III$: $c = 2$. We check the limits at $x = 2$.
Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x + 3) = 2(2) + 3 = 7$.
Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x - 3) = 2(2) - 3 = 1$.
Since the left-hand limit $(7)$ is not equal to the right-hand limit $(1)$,the function is discontinuous at $x = 2$.
Therefore,$x = 2$ is the only point of discontinuity.

(D) The given function is $f(x) = \begin{cases} |x| + 3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \ge 3 \end{cases}$.
Case $I$: If $c < -3$,then $f(c) = -c + 3$. The limit $\lim_{x \to c} f(x) = \lim_{x \to c} (-x + 3) = -c + 3 = f(c)$. Thus,$f$ is continuous for all $x < -3$.
Case $II$: If $c = -3$,then $f(-3) = |-3| + 3 = 6$. The left-hand limit is $\lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} (-x + 3) = -(-3) + 3 = 6$. The right-hand limit is $\lim_{x \to -3^+} f(x) = \lim_{x \to -3^+} (-2x) = -2(-3) = 6$. Since $\lim_{x \to -3^-} f(x) = \lim_{x \to -3^+} f(x) = f(-3)$,$f$ is continuous at $x = -3$.
Case $III$: If $-3 < c < 3$,then $f(c) = -2c$. The limit $\lim_{x \to c} f(x) = \lim_{x \to c} (-2x) = -2c = f(c)$. Thus,$f$ is continuous for all $x \in (-3, 3)$.
Case $IV$: If $c = 3$,then $f(3) = 6(3) + 2 = 20$. The left-hand limit is $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (-2x) = -2(3) = -6$. The right-hand limit is $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (6x + 2) = 6(3) + 2 = 20$. Since the left-hand limit $\neq$ right-hand limit,$f$ is discontinuous at $x = 3$.
Case $V$: If $c > 3$,then $f(c) = 6c + 2$. The limit $\lim_{x \to c} f(x) = \lim_{x \to c} (6x + 2) = 6c + 2 = f(c)$. Thus,$f$ is continuous for all $x > 3$.
Therefore,the only point of discontinuity is $x = 3$.

(D) The given function $f$ is $f(x) = \begin{cases} \frac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$
It is known that $x < 0 \implies |x| = -x$ and $x > 0 \implies |x| = x$.
Therefore,the function can be rewritten as:
$f(x) = \begin{cases} -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \\ 1, & \text{if } x > 0 \end{cases}$
The function $f$ is defined for all real numbers. Let $c$ be any real number.
Case $I$: If $c < 0$,then $f(c) = -1$.
$\lim_{x \to c} f(x) = \lim_{x \to c} (-1) = -1 = f(c)$.
Thus,$f$ is continuous for all $x < 0$.
Case $II$: If $c = 0$,the left-hand limit is $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-1) = -1$.
The right-hand limit is $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1) = 1$.
Since the left-hand limit $\neq$ right-hand limit,$f$ is discontinuous at $x = 0$.
Case $III$: If $c > 0$,then $f(c) = 1$.
$\lim_{x \to c} f(x) = \lim_{x \to c} (1) = 1 = f(c)$.
Thus,$f$ is continuous for all $x > 0$.
Conclusion: $x = 0$ is the only point of discontinuity.

(D) The given function is $f(x) = \begin{cases} \frac{x}{|x|}, & \text{if } x < 0 \\ -1, & \text{if } x \ge 0 \end{cases}$.
For $x < 0$,we have $|x| = -x$. Therefore,for $x < 0$,$f(x) = \frac{x}{-x} = -1$.
Thus,the function can be written as $f(x) = -1$ for all $x \in \mathbb{R}$.
Let $c$ be any real number. Then,$\lim_{x \to c} f(x) = \lim_{x \to c} (-1) = -1$.
Also,$f(c) = -1$ for any $c \in \mathbb{R}$.
Since $\lim_{x \to c} f(x) = f(c)$ for all $c \in \mathbb{R}$,the function $f(x)$ is continuous everywhere.
Hence,the function has no points of discontinuity.

(NONE) The given function is $f(x) = \begin{cases} x + 1, & \text{if } x \ge 1 \\ x^2 + 1, & \text{if } x < 1 \end{cases}$.
The function $f$ is defined for all real numbers.
Case $I$: If $c < 1$,then $f(c) = c^2 + 1$. The limit is $\lim_{x \to c} f(x) = \lim_{x \to c} (x^2 + 1) = c^2 + 1$. Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous for all $x < 1$.
Case $II$: If $c = 1$,then $f(1) = 1 + 1 = 2$.
The left-hand limit is $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 1) = 1^2 + 1 = 2$.
The right-hand limit is $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + 1) = 1 + 1 = 2$.
Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 2$,$f$ is continuous at $x = 1$.
Case $III$: If $c > 1$,then $f(c) = c + 1$. The limit is $\lim_{x \to c} f(x) = \lim_{x \to c} (x + 1) = c + 1$. Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous for all $x > 1$.
Conclusion: The function $f$ has no points of discontinuity and is a continuous function.

(NONE) The given function $f$ is $f(x) = \begin{cases} x^3 - 3, & \text{if } x \le 2 \\ x^2 + 1, & \text{if } x > 2 \end{cases}$
The function $f$ is defined for all real numbers.
Case $I$: If $c < 2,$ then $f(c) = c^3 - 3.$
$\lim_{x \to c} f(x) = \lim_{x \to c} (x^3 - 3) = c^3 - 3 = f(c).$
Thus,$f$ is continuous for all $x < 2.$
Case $II$: If $c = 2,$ then $f(2) = 2^3 - 3 = 5.$
Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^3 - 3) = 2^3 - 3 = 5.$
Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 + 1) = 2^2 + 1 = 5.$
Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 5,$ the function is continuous at $x = 2.$
Case $III$: If $c > 2,$ then $f(c) = c^2 + 1.$
$\lim_{x \to c} f(x) = \lim_{x \to c} (x^2 + 1) = c^2 + 1 = f(c).$
Thus,$f$ is continuous for all $x > 2.$
Conclusion: The function $f$ is continuous at every point on the real line. Therefore,there are no points of discontinuity.

(A) The given function is $f(x) = \begin{cases} x^{10} - 1, & \text{if } x \le 1 \\ x^2, & \text{if } x > 1 \end{cases}$.
The function $f$ is defined for all real numbers.
Case $I$: If $c < 1$,then $f(c) = c^{10} - 1$. The limit $\lim_{x \to c} f(x) = \lim_{x \to c} (x^{10} - 1) = c^{10} - 1 = f(c)$. Thus,$f$ is continuous for all $x < 1$.
Case $II$: If $c = 1$,we check the limits at $x = 1$.
The left-hand limit is $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10} - 1) = 1^{10} - 1 = 0$.
The right-hand limit is $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2) = 1^2 = 1$.
Since the left-hand limit $(0)$ is not equal to the right-hand limit $(1)$,the function $f$ is discontinuous at $x = 1$.
Case $III$: If $c > 1$,then $f(c) = c^2$. The limit $\lim_{x \to c} f(x) = \lim_{x \to c} (x^2) = c^2 = f(c)$. Thus,$f$ is continuous for all $x > 1$.
Conclusion: The only point of discontinuity is $x = 1$.

(B) The given function is $f(x) = \begin{cases} x + 5, & \text{if } x \le 1 \\ x - 5, & \text{if } x > 1 \end{cases}$.
The function $f$ is defined for all real numbers.
Let $c$ be any real number.
Case $I$: If $c < 1$,then $f(c) = c + 5$ and $\lim_{x \to c} f(x) = \lim_{x \to c} (x + 5) = c + 5$. Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous for all $x < 1$.
Case $II$: If $c = 1$,then $f(1) = 1 + 5 = 6$. The left-hand limit is $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 5) = 6$. The right-hand limit is $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - 5) = -4$. Since the left-hand limit $\neq$ right-hand limit,$f$ is not continuous at $x = 1$.
Case $III$: If $c > 1$,then $f(c) = c - 5$ and $\lim_{x \to c} f(x) = \lim_{x \to c} (x - 5) = c - 5$. Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous for all $x > 1$.
Conclusion: The function $f$ is not continuous at $x = 1$.

(D) The given function is $f(x) = \begin{cases} 3, & \text{if } 0 \le x \le 1 \\ 4, & \text{if } 1 < x < 3 \\ 5, & \text{if } 3 \le x \le 10 \end{cases}$
To check the continuity at $x=3,$ we evaluate the left-hand limit,right-hand limit,and the value of the function at $x=3.$
$1$. Value of the function at $x=3$:
$f(3) = 5$ (from the third part of the definition).
$2$. Left-hand limit at $x=3$:
$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (4) = 4$.
$3$. Right-hand limit at $x=3$:
$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (5) = 5$.
Since the left-hand limit $(\lim_{x \to 3^-} f(x) = 4)$ is not equal to the right-hand limit $(\lim_{x \to 3^+} f(x) = 5)$,the limit $\lim_{x \to 3} f(x)$ does not exist.
Therefore,the function $f$ is not continuous at $x=3.$

(A) The given function is $f(x) = \begin{cases} 2x, & \text{if } x < 0 \\ 0, & \text{if } 0 \le x \le 1 \\ 4x, & \text{if } x > 1 \end{cases}$.
To check the continuity at $x=3$,we evaluate the limit of the function as $x \to 3$ and compare it with the value of the function at $x=3$.
Since $3 > 1$,the function is defined by $f(x) = 4x$ in the neighborhood of $x=3$.
$1$. Value of the function at $x=3$:
$f(3) = 4(3) = 12$.
$2$. Limit of the function as $x \to 3$:
$\lim_{x \to 3} f(x) = \lim_{x \to 3} (4x) = 4(3) = 12$.
Since $\lim_{x \to 3} f(x) = f(3) = 12$,the function $f$ is continuous at $x=3$.

(A) The given function is $f(x) = \begin{cases} -2, & \text{if } x \le -1 \\ 2x, & \text{if } -1 < x \le 1 \\ 2, & \text{if } x > 1 \end{cases}$.
To check the continuity at $x=3$,we observe the definition of the function for $x > 1$.
For any $x > 1$,$f(x) = 2$.
At $x=3$,the value of the function is $f(3) = 2$.
The limit of the function as $x$ approaches $3$ is:
$\lim_{x \to 3} f(x) = \lim_{x \to 3} (2) = 2$.
Since $\lim_{x \to 3} f(x) = f(3) = 2$,the function $f$ is continuous at $x=3$.

(D) The given function $f$ is defined as $f(x) = \begin{cases} ax + 1, & \text{if } x \le 3 \\ bx + 3, & \text{if } x > 3 \end{cases}$.
For the function $f$ to be continuous at $x = 3$,the left-hand limit,right-hand limit,and the value of the function at $x = 3$ must be equal:
$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$.
$1$. Left-hand limit at $x = 3$:
$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (ax + 1) = 3a + 1$.
$2$. Right-hand limit at $x = 3$:
$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (bx + 3) = 3b + 3$.
$3$. Value of the function at $x = 3$:
$f(3) = 3a + 1$.
Equating the limits:
$3a + 1 = 3b + 3$
$3a = 3b + 2$
$a = b + \frac{2}{3}$.
Thus,the required relationship is $a = b + \frac{2}{3}$.

(NONE) 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.
Left-hand limit at $x=0$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \lambda(x^2 - 2x) = \lambda(0^2 - 2(0)) = 0$
Right-hand limit at $x=0$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (4x + 1) = 4(0) + 1 = 1$
Value of the function at $x=0$:
$f(0) = \lambda(0^2 - 2(0)) = 0$
Since the left-hand limit $(0)$ is not equal to the right-hand limit $(1)$,the function is discontinuous at $x=0$ for any value of $\lambda$.
At $x=1$,the function is defined by $f(x) = 4x + 1$,which is a polynomial function. Polynomial functions are continuous everywhere in their domain. Thus,$f(x)$ is continuous at $x=1$ for any value of $\lambda$.

The given function is $g(x)=x-[x]$.
It is evident that $g$ is defined at all integral points.
Let $n$ be an integer.
Then $g(n)=n-[n]=n-n=0$.
The left-hand limit of $g$ at $x=n$ is:
$\lim_{x \to n^-} g(x) = \lim_{x \to n^-} (x-[x]) = \lim_{x \to n^-} (x) - \lim_{x \to n^-} [x] = n - (n-1) = 1$.
The right-hand limit of $g$ at $x=n$ is:
$\lim_{x \to n^+} g(x) = \lim_{x \to n^+} (x-[x]) = \lim_{x \to n^+} (x) - \lim_{x \to n^+} [x] = n - n = 0$.
It is observed that the left-hand limit and right-hand limit of $g$ at $x=n$ do not coincide (since $1 \neq 0$).
Therefore,$g$ is not continuous at $x=n$.
Hence,$g$ is discontinuous at all integral points.

(A) The given function is $f(x) = x^{2} - \sin x + 5$.
First,we evaluate the function at $x = \pi$:
$f(\pi) = \pi^{2} - \sin \pi + 5 = \pi^{2} - 0 + 5 = \pi^{2} + 5$.
Next,we calculate the limit of the function as $x \to \pi$:
$\lim_{x \to \pi} f(x) = \lim_{x \to \pi} (x^{2} - \sin x + 5)$.
Using the property of limits for sums and differences:
$\lim_{x \to \pi} f(x) = \lim_{x \to \pi} x^{2} - \lim_{x \to \pi} \sin x + \lim_{x \to \pi} 5$.
Substituting $x = \pi$:
$\lim_{x \to \pi} f(x) = \pi^{2} - \sin \pi + 5 = \pi^{2} - 0 + 5 = \pi^{2} + 5$.
Since $\lim_{x \to \pi} f(x) = f(\pi) = \pi^{2} + 5$,the function $f(x)$ is continuous at $x = \pi$.

(A) It is known that if $g$ and $h$ are two continuous functions,then $g+h$,$g-h$,and $g \cdot h$ are also continuous.
First,we prove that $g(x) = \sin x$ and $h(x) = \cos x$ are continuous functions.
For $g(x) = \sin x$:
$g(x)$ is defined for every real number. Let $c$ be a real number. Put $x = c + h$. As $x \to c$,$h \to 0$.
$g(c) = \sin c$.
$\lim_{x \to c} g(x) = \lim_{h \to 0} \sin(c + h) = \lim_{h \to 0} (\sin c \cos h + \cos c \sin h) = \sin c(1) + \cos c(0) = \sin c$.
Since $\lim_{x \to c} g(x) = g(c)$,$g(x)$ is continuous.
For $h(x) = \cos x$:
$h(x)$ is defined for every real number. Let $c$ be a real number. Put $x = c + h$. As $x \to c$,$h \to 0$.
$h(c) = \cos c$.
$\lim_{x \to c} h(x) = \lim_{h \to 0} \cos(c + h) = \lim_{h \to 0} (\cos c \cos h - \sin c \sin h) = \cos c(1) - \sin c(0) = \cos c$.
Since $\lim_{x \to c} h(x) = h(c)$,$h(x)$ is continuous.
Conclusion:
a) $f(x) = g(x) + h(x) = \sin x + \cos x$ is continuous.
b) $f(x) = g(x) - h(x) = \sin x - \cos x$ is continuous.
c) $f(x) = g(x) \cdot h(x) = \sin x \cos x$ is continuous.

(N/A) It is known that if $g$ and $h$ are two continuous functions,then:
$i.$ $\frac{h(x)}{g(x)}, g(x) \neq 0$ is continuous.
$ii.$ $\frac{1}{g(x)}, g(x) \neq 0$ is continuous.
$iii.$ $\frac{1}{h(x)}, h(x) \neq 0$ is continuous.
First,we prove that $g(x) = \sin x$ and $h(x) = \cos x$ are continuous functions.
For $g(x) = \sin x$,let $c$ be a real number. Put $x = c + h$. As $x \to c$,$h \to 0$.
$\lim_{x \to c} \sin x = \lim_{h \to 0} \sin(c + h) = \lim_{h \to 0} [\sin c \cos h + \cos c \sin h] = \sin c(1) + \cos c(0) = \sin c = g(c)$.
Thus,$g(x) = \sin x$ is continuous for all $x \in \mathbb{R}$.
Similarly,for $h(x) = \cos x$,$\lim_{x \to c} \cos x = \lim_{h \to 0} \cos(c + h) = \lim_{h \to 0} [\cos c \cos h - \sin c \sin h] = \cos c(1) - \sin c(0) = \cos c = h(c)$.
Thus,$h(x) = \cos x$ is continuous for all $x \in \mathbb{R}$.
Now,for the other functions:
$1.$ $\cos x$ is continuous for all $x \in \mathbb{R}$.
$2.$ $\csc x = \frac{1}{\sin x}$ is continuous where $\sin x \neq 0$,i.e.,$x \neq n\pi, n \in \mathbb{Z}$.
$3.$ $\sec x = \frac{1}{\cos x}$ is continuous where $\cos x \neq 0$,i.e.,$x \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z}$.
$4.$ $\cot x = \frac{\cos x}{\sin x}$ is continuous where $\sin x \neq 0$,i.e.,$x \neq n\pi, n \in \mathbb{Z}$.

(NONE) The given function $f$ is $f(x) = \begin{cases} \frac{\sin x}{x}, & \text{if } x < 0 \\ x + 1, & \text{if } x \ge 0 \end{cases}$
It is evident that $f$ is defined at all points of the real line.
Let $c$ be a real number.
Case $I$: If $c < 0,$ then $f(c) = \frac{\sin c}{c}$ and $\lim_{x \to c} f(x) = \lim_{x \to c} \left( \frac{\sin x}{x} \right) = \frac{\sin c}{c}.$
Since $\lim_{x \to c} f(x) = f(c),$ $f$ is continuous at all points $x < 0.$
Case $II$: If $c > 0,$ then $f(c) = c + 1$ and $\lim_{x \to c} f(x) = \lim_{x \to c} (x + 1) = c + 1.$
Since $\lim_{x \to c} f(x) = f(c),$ $f$ is continuous at all points $x > 0.$
Case $III$: If $c = 0,$ then $f(0) = 0 + 1 = 1.$
The left-hand limit at $x = 0$ is $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin x}{x} = 1.$
The right-hand limit at $x = 0$ is $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1) = 1.$
Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 1,$ $f$ is continuous at $x = 0.$
From the above observations,$f$ is continuous at all points of the real line. Thus,$f$ has no points of discontinuity.

(A) The given function $f$ is $f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$
It is evident that $f$ is defined at all points of the real line.
Let $c$ be a real number.
Case $I$: If $c \neq 0$,then $f(c) = c^2 \sin \frac{1}{c}$.
$\lim_{x \to c} f(x) = \lim_{x \to c} (x^2 \sin \frac{1}{x}) = (\lim_{x \to c} x^2) (\lim_{x \to c} \sin \frac{1}{x}) = c^2 \sin \frac{1}{c}$.
Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous at all points $x \neq 0$.
Case $II$: If $c = 0$,then $f(0) = 0$.
We know that $-1 \leq \sin \frac{1}{x} \leq 1$ for $x \neq 0$.
Multiplying by $x^2$,we get $-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$.
By the Squeeze Theorem,since $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} (x^2) = 0$,it follows that $\lim_{x \to 0} (x^2 \sin \frac{1}{x}) = 0$.
Thus,$\lim_{x \to 0} f(x) = 0 = f(0)$.
Therefore,$f$ is continuous at $x = 0$.
From the above observations,$f$ is continuous at every point of the real line. Thus,$f$ is a continuous function.

(A) The given function $f$ is defined as $f(x) = \begin{cases} \sin x - \cos x, & \text{if } x \neq 0 \\ -1, & \text{if } x = 0 \end{cases}$.
It is evident that $f$ is defined at all points of the real line.
Let $c$ be a real number.
Case $I$: If $c \neq 0$,then $f(c) = \sin c - \cos c$.
$\lim_{x \to c} f(x) = \lim_{x \to c} (\sin x - \cos x) = \sin c - \cos c$.
Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous at all points $x \neq 0$.
Case $II$: If $c = 0$,then $f(0) = -1$.
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0} (\sin x - \cos x) = \sin 0 - \cos 0 = 0 - 1 = -1$.
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0} (\sin x - \cos x) = \sin 0 - \cos 0 = 0 - 1 = -1$.
Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = -1$,$f$ is continuous at $x = 0$.
From the above observations,$f$ is continuous at every point of the real line.

(A) The given function $f$ is defined as:
$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$
Since the function $f$ is continuous at $x = \frac{\pi}{2}$,the limit of $f(x)$ as $x \to \frac{\pi}{2}$ must be equal to $f\left(\frac{\pi}{2}\right)$.
Given $f\left(\frac{\pi}{2}\right) = 3$.
Now,calculate the limit:
$\lim_{x \to \frac{\pi}{2}} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x}$
Let $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}$,$h \to 0$.
$\lim_{h \to 0} \frac{k \cos(\frac{\pi}{2} + h)}{\pi - 2(\frac{\pi}{2} + h)} = \lim_{h \to 0} \frac{k(-\sin h)}{\pi - \pi - 2h} = \lim_{h \to 0} \frac{-k \sin h}{-2h} = \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h}$
Since $\lim_{h \to 0} \frac{\sin h}{h} = 1$,we have:
$\lim_{x \to \frac{\pi}{2}} f(x) = \frac{k}{2} \cdot 1 = \frac{k}{2}$
Equating the limit to the function value:
$\frac{k}{2} = 3 \implies k = 6$.
Therefore,the required value of $k$ is $6$.

(A) For the function $f(x)$ to be continuous at $x=2$,the left-hand limit,right-hand limit,and the value of the function at $x=2$ must be equal.
$1$. The value of the function at $x=2$ is $f(2) = k(2)^2 = 4k$.
$2$. The left-hand limit at $x=2$ is $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (kx^2) = k(2)^2 = 4k$.
$3$. The right-hand limit at $x=2$ is $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3) = 3$.
For continuity,$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$.
Therefore,$4k = 3$.
Solving for $k$,we get $k = 3/4$.

(A) For the function $f(x)$ to be continuous at $x = \pi$,the left-hand limit,right-hand limit,and the value of the function at $x = \pi$ must be equal.
$1$. Value of the function at $x = \pi$:
$f(\pi) = k(\pi) + 1 = k\pi + 1$.
$2$. Left-hand limit $(LHL)$ at $x = \pi$:
$\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx + 1) = k\pi + 1$.
$3$. Right-hand limit $(RHL)$ at $x = \pi$:
$\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} \cos x = \cos(\pi) = -1$.
For continuity,$LHL$ = $RHL$ = $f(\pi)$:
$k\pi + 1 = -1$.
Solving for $k$:
$k\pi = -1 - 1$
$k\pi = -2$
$k = -\frac{2}{\pi}$.
Thus,the required value of $k$ is $-\frac{2}{\pi}$.

(A) The given function is $f(x) = \begin{cases} kx + 1, & \text{if } x \le 5 \\ 3x - 5, & \text{if } x > 5 \end{cases}$.
For the function $f$ to be continuous at $x = 5$,the left-hand limit,right-hand limit,and the value of the function at $x = 5$ must be equal.
First,we find the value of the function at $x = 5$: $f(5) = k(5) + 1 = 5k + 1$.
Next,we calculate the left-hand limit: $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (kx + 1) = 5k + 1$.
Then,we calculate the right-hand limit: $\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (3x - 5) = 3(5) - 5 = 15 - 5 = 10$.
For continuity,$\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5)$.
So,$5k + 1 = 10$.
$5k = 9$.
$k = 9/5$.

(A) The given function $f$ is $f(x) = \begin{cases} 5, & \text{if } x \le 2 \\ ax + b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \ge 10 \end{cases}$
Since $f$ is a continuous function,it must be continuous at all points,including $x=2$ and $x=10$.
For continuity at $x=2$:
$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$
$5 = 2a + b = 5 \implies 2a + b = 5$ $(1)$
For continuity at $x=10$:
$\lim_{x \to 10^-} f(x) = \lim_{x \to 10^+} f(x) = f(10)$
$10a + b = 21 = 21 \implies 10a + b = 21$ $(2)$
Subtracting equation $(1)$ from $(2)$:
$(10a + b) - (2a + b) = 21 - 5$
$8a = 16 \implies a = 2$
Substituting $a=2$ into equation $(1)$:
$2(2) + b = 5$
$4 + b = 5 \implies b = 1$
Thus,the values are $a=2$ and $b=1$.

(N/A) The given function is $f(x)=\cos \left(x^{2}\right)$.
This function $f$ is defined for every real number and $f$ can be written as the composition of two functions as $f=g \circ h$,where $g(x)=\cos x$ and $h(x)=x^{2}$.
$[\because (g \circ h)(x)=g(h(x))=g(x^{2})=\cos(x^{2})=f(x)]$.
It has to be first proved that $g(x)=\cos x$ and $h(x)=x^{2}$ are continuous functions.
It is evident that $g$ is defined for every real number. Let $c$ be a real number. Then $g(c)=\cos c$.
Put $x=c+h$. If $x \to c$,then $h \to 0$.
$\lim_{x \to c} g(x) = \lim_{x \to c} \cos x = \lim_{h \to 0} \cos(c+h) = \lim_{h \to 0} [\cos c \cos h - \sin c \sin h]$.
$= \cos c \lim_{h \to 0} \cos h - \sin c \lim_{h \to 0} \sin h = \cos c \times 1 - \sin c \times 0 = \cos c$.
$\therefore \lim_{x \to c} g(x) = g(c)$. Therefore,$g(x)=\cos x$ is a continuous function.
Now,$h(x)=x^{2}$. Clearly,$h$ is defined for every real number. Let $k$ be a real number,then $h(k)=k^{2}$.
$\lim_{x \to k} h(x) = \lim_{x \to k} x^{2} = k^{2} = h(k)$. Therefore,$h$ is a continuous function.
It is known that if $g$ and $h$ are continuous functions,then their composition $(g \circ h)$ is also continuous.
Since $g(x)=\cos x$ and $h(x)=x^{2}$ are continuous,their composition $f(x)=(g \circ h)(x)=\cos(x^{2})$ is a continuous function.

(N/A) The given function is $f(x)=|\cos x|$.
This function $f$ is defined for every real number and can be written as the composition of two functions $f=g \circ h$,where $g(x)=|x|$ and $h(x)=\cos x$.
$[\because (g \circ h)(x) = g(h(x)) = g(\cos x) = |\cos x| = f(x)]$
First,we prove that $g(x)=|x|$ and $h(x)=\cos x$ are continuous functions.
$g(x)=|x|$ can be written as:
$g(x) = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \ge 0 \end{cases}$
Clearly,$g$ is defined for all real numbers. Let $c$ be a real number.
Case $I$: If $c < 0$,then $g(c)=-c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} (-x) = -c = g(c)$. Thus,$g$ is continuous for $x < 0$.
Case $II$: If $c > 0$,then $g(c)=c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} x = c = g(c)$. Thus,$g$ is continuous for $x > 0$.
Case $III$: If $c=0$,then $g(0)=0$. $\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (-x) = 0$ and $\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (x) = 0$. Since $\lim_{x \to 0^-} g(x) = \lim_{x \to 0^+} g(x) = g(0)$,$g$ is continuous at $x=0$.
Thus,$g(x)=|x|$ is continuous everywhere.
Now,$h(x)=\cos x$ is defined for every real number. Let $c$ be a real number. Put $x=c+h$. As $x \to c$,$h \to 0$.
$\lim_{x \to c} h(x) = \lim_{h \to 0} \cos(c+h) = \lim_{h \to 0} (\cos c \cos h - \sin c \sin h) = \cos c(1) - \sin c(0) = \cos c = h(c)$.
Thus,$h(x)=\cos x$ is continuous everywhere.
Since the composition of two continuous functions is continuous,$f(x) = (g \circ h)(x) = |\cos x|$ is a continuous function.

(N/A) Let $f(x) = \sin |x|$.
This function $f$ is defined for every real number and $f$ can be written as the composition of two functions $f = h \circ g$,where $g(x) = |x|$ and $h(x) = \sin x$.
$[\because (h \circ g)(x) = h(g(x)) = h(|x|) = \sin |x| = f(x)]$
First,we prove that $g(x) = |x|$ and $h(x) = \sin x$ are continuous functions.
$g(x) = |x|$ can be written as:
$g(x) = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \ge 0 \end{cases}$
Clearly,$g$ is defined for all real numbers. Let $c$ be a real number.
Case $I$: If $c < 0$,$g(c) = -c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} (-x) = -c$. Thus,$\lim_{x \to c} g(x) = g(c)$. So,$g$ is continuous for $x < 0$.
Case $II$: If $c > 0$,$g(c) = c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} (x) = c$. Thus,$\lim_{x \to c} g(x) = g(c)$. So,$g$ is continuous for $x > 0$.
Case $III$: If $c = 0$,$g(0) = 0$. $\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (-x) = 0$ and $\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (x) = 0$. Since $\lim_{x \to 0^-} g(x) = \lim_{x \to 0^+} g(x) = g(0)$,$g$ is continuous at $x = 0$.
Thus,$g(x) = |x|$ is continuous everywhere.
Now,$h(x) = \sin x$ is defined for every real number. Let $c$ be a real number. Put $x = c + k$. If $x \to c$,then $k \to 0$.
$\lim_{x \to c} h(x) = \lim_{k \to 0} \sin(c + k) = \lim_{k \to 0} (\sin c \cos k + \cos c \sin k) = \sin c(1) + \cos c(0) = \sin c = h(c)$.
Thus,$h(x) = \sin x$ is continuous everywhere.
Since the composition of two continuous functions is continuous,$f(x) = (h \circ g)(x) = \sin |x|$ is a continuous function.

(NONE) The given function is $f(x) = |x| - |x + 1|$.
Define two functions $g$ and $h$ as follows:
$g(x) = |x|$ and $h(x) = |x + 1|$.
Therefore,$f = g - h$.
First,let us check the continuity of $g$ and $h$.
The function $g(x) = |x|$ can be written as:
$g(x) = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \ge 0 \end{cases}$
It is clear that $g$ is defined for all real numbers.
Let $c$ be a real number.
Case $I$: If $c < 0$,then $g(c) = -c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} (-x) = -c$. Thus,$g$ is continuous for $x < 0$.
Case $II$: If $c > 0$,then $g(c) = c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} (x) = c$. Thus,$g$ is continuous for $x > 0$.
Case $III$: If $c = 0$,then $g(0) = 0$. The left-hand limit is $\lim_{x \to 0^-} (-x) = 0$ and the right-hand limit is $\lim_{x \to 0^+} (x) = 0$. Thus,$g$ is continuous at $x = 0$.
Thus,$g$ is continuous at every point.
Similarly,$h(x) = |x + 1|$ can be written as:
$h(x) = \begin{cases} -(x + 1), & \text{if } x < -1 \\ x + 1, & \text{if } x \ge -1 \end{cases}$
It can be proved in the same way that $h$ is continuous for every real number.
The difference of two continuous functions is also a continuous function. Therefore,$f = g - h$ is also continuous at every point.
Thus,$f$ has no points of discontinuity.

(A) The function is continuous at $x = 1$ if $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.
First,calculate the left-hand limit:
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1} (a + bx) = a + b$.
Next,calculate the right-hand limit:
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1} (b - ax) = b - a$.
Given $f(1) = 4$,we have the system of equations:
$a + b = 4$
$b - a = 4$
Adding the two equations: $(a + b) + (b - a) = 4 + 4$ $\Rightarrow 2b = 8$ $\Rightarrow b = 4$.
Substituting $b = 4$ into $a + b = 4$: $a + 4 = 4 \Rightarrow a = 0$.
Therefore,the values are $a = 0$ and $b = 4$.

(A) For $\lim_{x \to 0} f(x)$ to exist,the left-hand limit and right-hand limit at $x = 0$ must be equal.
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0} (mx^2 + n) = m(0)^2 + n = n$
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0} (nx + m) = n(0) + m = m$
Thus,$\lim_{x \to 0} f(x)$ exists if $m = n$.
For $\lim_{x \to 1} f(x)$ to exist,the left-hand limit and right-hand limit at $x = 1$ must be equal.
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1} (nx + m) = n(1) + m = n + m$
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1} (nx^3 + m) = n(1)^3 + m = n + m$
Since $n + m = n + m$,the limit $\lim_{x \to 1} f(x)$ exists for all integers $m$ and $n$.
Therefore,the condition for both limits to exist is $m = n$ for any integer values of $m$ and $n$.

(A) For $f(x)$ to be twice differentiable,it must be continuous and differentiable at $x = \pi$.
$1$. Continuity at $x = \pi$:
$f(\pi^{-}) = f(\pi) = f(\pi^{+})$
$k_{1}(\pi-\pi)^{2} - 1 = k_{2} \cos(\pi)$
$-1 = -k_{2} \implies k_{2} = 1$.
$2$. First derivative $f'(x)$:
$f'(x) = \begin{cases} 2k_{1}(x-\pi), & x \leq \pi \\ -k_{2} \sin x, & x > \pi \end{cases}$
At $x = \pi$,$f'(\pi^{-}) = 2k_{1}(\pi-\pi) = 0$ and $f'(\pi^{+}) = -k_{2} \sin(\pi) = 0$.
Since $0 = 0$,the function is differentiable at $x = \pi$ for any $k_{1}, k_{2}$.
$3$. Second derivative $f''(x)$:
$f''(x) = \begin{cases} 2k_{1}, & x \leq \pi \\ -k_{2} \cos x, & x > \pi \end{cases}$
For $f''(x)$ to be continuous at $x = \pi$:
$f''(\pi^{-}) = f''(\pi^{+})$
$2k_{1} = -k_{2} \cos(\pi)$
$2k_{1} = -k_{2}(-1) = k_{2}$
Since $k_{2} = 1$,we have $2k_{1} = 1 \implies k_{1} = \frac{1}{2}$.
Thus,$(k_{1}, k_{2}) = (\frac{1}{2}, 1)$.

(A) The function is defined as $f(x) = x \cdot \left[ \frac{x}{2} \right]$ for $x \in (-10, 10)$.
The greatest integer function $[t]$ is discontinuous at all integer values of $t$.
Here,$t = \frac{x}{2}$. Thus,$f(x)$ is potentially discontinuous when $\frac{x}{2} = k$,where $k \in \mathbb{Z}$.
Given $-10 < x < 10$,we have $-5 < \frac{x}{2} < 5$.
The possible integer values for $k = \frac{x}{2}$ are $\{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$.
Let us check the continuity at these points:
$1$. For $k \neq 0$ (i.e.,$x \neq 0$),the function $f(x) = x \cdot \left[ \frac{x}{2} \right]$ is discontinuous because the greatest integer function $\left[ \frac{x}{2} \right]$ jumps at these points,and $x$ is non-zero.
$2$. For $k = 0$,$x = 0$. We check the continuity at $x = 0$:
$f(0) = 0 \cdot [0] = 0$.
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x \cdot [0] = 0$.
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x \cdot [-1] = 0$.
Since $f(0) = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = 0$,the function is continuous at $x = 0$.
Therefore,the points of discontinuity are $\frac{x}{2} \in \{-4, -3, -2, -1, 1, 2, 3, 4\}$,which corresponds to $x \in \{-8, -6, -4, -2, 2, 4, 6, 8\}$.
There are $8$ such points.

(A) For the function to be continuous at $x = 0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
Using the formula $\cos A - \cos B = -2 \sin(\frac{A+B}{2}) \sin(\frac{A-B}{2})$,we get:
$\lim_{x \to 0} \frac{-2 \sin(\frac{\sin x + x}{2}) \sin(\frac{\sin x - x}{2})}{x^4} = \frac{1}{k}$
$\lim_{x \to 0} \frac{2 \sin(\frac{\sin x + x}{2}) \sin(\frac{x - \sin x}{2})}{x^4} = \frac{1}{k}$
Using $\sin \theta \approx \theta$ as $\theta \to 0$ and the Taylor series expansion $\sin x = x - \frac{x^3}{6} + \dots$,we have:
$\lim_{x \to 0} 2 \left( \frac{\sin x + x}{2x} \right) \left( \frac{x - \sin x}{2x^3} \right) = \frac{1}{k}$
$\lim_{x \to 0} 2 \left( \frac{x + x}{2x} \right) \left( \frac{x - (x - \frac{x^3}{6})}{2x^3} \right) = \frac{1}{k}$
$2 \times (1) \times \frac{x^3/6}{2x^3} = \frac{1}{k}$
$2 \times 1 \times \frac{1}{12} = \frac{1}{k}$
$\frac{1}{6} = \frac{1}{k} \Rightarrow k = 6$.

(C) For the function to be continuous at $x=0$,we must have $\operatorname{Lim}_{x \rightarrow 0^{+}} f(x) = \operatorname{Lim}_{x \rightarrow 0^{-}} f(x) = f(0) = \alpha$.
First,consider the Right Hand Limit $(RHL)$ as $x \rightarrow 0^{+}$:
$\operatorname{Lim}_{x \rightarrow 0^{+}} \frac{\cos^{-1}(1-x^2) \sin^{-1}(1-x)}{x(1-x)(1+x)} = \operatorname{Lim}_{x \rightarrow 0^{+}} \frac{\cos^{-1}(1-x^2)}{x} \cdot \frac{\sin^{-1}(1-x)}{1-x^2} = \frac{\pi}{2} \cdot \operatorname{Lim}_{x \rightarrow 0^{+}} \frac{\cos^{-1}(1-x^2)}{x}$.
Let $1-x^2 = \cos \theta$,then as $x \rightarrow 0^{+}$,$\theta \rightarrow 0^{+}$.
$\operatorname{Lim}_{\theta \rightarrow 0^{+}} \frac{\theta}{\sqrt{1-\cos \theta}} = \operatorname{Lim}_{\theta \rightarrow 0^{+}} \frac{\theta}{\sqrt{2} \sin(\theta/2)} = \frac{1}{\sqrt{2}} \cdot 2 = \sqrt{2}$.
So,$RHL = \frac{\pi}{2} \cdot \sqrt{2} = \frac{\pi}{\sqrt{2}}$.
Now,consider the Left Hand Limit $(LHL)$ as $x \rightarrow 0^{-}$:
For $x \in (-1, 0)$,$\{x\} = x+1$.
$\operatorname{Lim}_{x \rightarrow 0^{-}} \frac{\cos^{-1}(1-(x+1)^2) \sin^{-1}(1-(x+1))}{(x+1)-(x+1)^3} = \operatorname{Lim}_{x \rightarrow 0^{-}} \frac{\cos^{-1}(1-(x+1)^2) \sin^{-1}(-x)}{(x+1)(1-(x+1)^2)} = \operatorname{Lim}_{x \rightarrow 0^{-}} \frac{\cos^{-1}(1-(x+1)^2) \cdot (-x)}{(x+1)(-x)(2+x)} = \frac{\cos^{-1}(0)}{1 \cdot 2} = \frac{\pi/2}{2} = \frac{\pi}{4}$.
Since $RHL = \frac{\pi}{\sqrt{2}}$ and $LHL = \frac{\pi}{4}$,the limits are not equal.
Therefore,the function is not continuous at $x=0$ for any value of $\alpha$.

(D) Since $f(x)$ is continuous at $x=0$,we must have $\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{+}} f(x) = f(0)$.
First,find $f(0) = b$.
Next,calculate the left-hand limit:
$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} \frac{\sin(a+1)x + \sin 2x}{2x} = \lim_{x \rightarrow 0^{-}} \left( \frac{\sin(a+1)x}{2x} + \frac{\sin 2x}{2x} \right) = \frac{a+1}{2} + \frac{2}{2} = \frac{a+1}{2} + 1$.
Now,calculate the right-hand limit:
$\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} \frac{\sqrt{x+bx^3} - \sqrt{x}}{bx^{5/2}} = \lim_{x \rightarrow 0^{+}} \frac{\sqrt{x}(\sqrt{1+bx^2} - 1)}{bx^{5/2}} = \lim_{x \rightarrow 0^{+}} \frac{\sqrt{1+bx^2} - 1}{bx^2}$.
Using the expansion $\sqrt{1+u} \approx 1 + \frac{u}{2}$,we get $\lim_{x \rightarrow 0^{+}} \frac{1 + \frac{bx^2}{2} - 1}{bx^2} = \frac{1}{2}$.
Equating the limits:
$b = \frac{1}{2}$ and $\frac{a+1}{2} + 1 = \frac{1}{2}$.
From $\frac{a+1}{2} = -\frac{1}{2}$,we get $a+1 = -1$,so $a = -2$.
Therefore,$a+b = -2 + \frac{1}{2} = -\frac{3}{2}$.

(C) Let $x = n$,where $n \in Z$.
First,consider the value of the function at $x = n$:
$f(n) = [n-1] \cos \left(\frac{2n-1}{2}\right) \pi = (n-1) \cos \left(n\pi - \frac{\pi}{2}\right) = (n-1) \cdot 0 = 0$.
Now,calculate the Left Hand Limit $(LHL)$:
$LHL = \lim_{x \rightarrow n^-} [x-1] \cos \left(\frac{2x-1}{2}\right) \pi$.
As $x \rightarrow n^-$,$[x-1] = n-2$.
$LHL = (n-2) \cos \left(n\pi - \frac{\pi}{2}\right) = (n-2) \cdot 0 = 0$.
Now,calculate the Right Hand Limit $(RHL)$:
$RHL = \lim_{x \rightarrow n^+} [x-1] \cos \left(\frac{2x-1}{2}\right) \pi$.
As $x \rightarrow n^+$,$[x-1] = n-1$.
$RHL = (n-1) \cos \left(n\pi - \frac{\pi}{2}\right) = (n-1) \cdot 0 = 0$.
Since $LHL = RHL = f(n) = 0$ for all $n \in Z$,the function is continuous at all integral values of $x$.
For non-integral values of $x$,the function is a product of a constant (locally) and a continuous trigonometric function,hence it is continuous.
Therefore,$f(x)$ is continuous for every real $x$.

(B) $f(x)$ is continuous on $R$,so it must be continuous at $x = 1$ and $x = -1$.
For continuity at $x = 1$:
$\lim_{x \rightarrow 1^-} f(x) = f(1) = \lim_{x \rightarrow 1^+} f(x)$
$|a(1)^2 + 1 + b| = \lim_{x \rightarrow 1^+} \sin(\pi x)$
$|a + 1 + b| = \sin(\pi) = 0$
$a + b + 1 = 0 \Rightarrow a + b = -1$ ... $(1)$
For continuity at $x = -1$:
$\lim_{x \rightarrow -1^-} f(x) = f(-1) = \lim_{x \rightarrow -1^+} f(x)$
$\lim_{x \rightarrow -1^-} 2 \sin \left(-\frac{\pi x}{2}\right) = |a(-1)^2 + (-1) + b|$
$2 \sin \left(\frac{\pi}{2}\right) = |a - 1 + b|$
$2(1) = |a + b - 1|$
$|a + b - 1| = 2$
This implies $a + b - 1 = 2$ or $a + b - 1 = -2$.
$a + b = 3$ or $a + b = -1$.
Since we already established $a + b = -1$ from the continuity at $x = 1$,the value of $a + b$ is $-1$.

(B) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
First,find $f(0)$ and the left-hand limit:
$f(0) = a \sin \frac{\pi}{2}(0-1) = a \sin(-\frac{\pi}{2}) = -a$.
Next,find the right-hand limit:
$\lim_{x \to 0^+} \frac{\tan 2x - \sin 2x}{bx^3} = \lim_{x \to 0^+} \frac{\tan 2x(1 - \cos 2x)}{bx^3} = \lim_{x \to 0^+} \frac{\tan 2x \cdot 2 \sin^2 x}{bx^3}$.
Using standard limits $\tan \theta \approx \theta$ and $\sin \theta \approx \theta$ as $\theta \to 0$:
$\lim_{x \to 0^+} \frac{(2x) \cdot 2(x^2)}{bx^3} = \frac{4x^3}{bx^3} = \frac{4}{b}$.
Equating the limits:
$-a = \frac{4}{b} \implies -ab = 4$.
Finally,calculate $10 - ab$:
$10 - ab = 10 + 4 = 14$.

(A) Given $f(x) = x - [x]$ and $g(x) = 1 - (x - [x])$. Let ${x} = x - [x]$. Then $f(x) = {x}$ and $g(x) = 1 - {x}$.
$h(x) = \min \{ {x}, 1 - {x} \}$.
For $x \in [n, n+1)$,${x} = x - n$. So $h(x) = \min \{ x-n, 1-(x-n) \} = \min \{ x-n, 1-x+n \}$.
The graphs of $f(x)$ and $g(x)$ intersect when ${x} = 1 - {x}$,which implies $2{x} = 1$,or ${x} = 0.5$.
In each interval $[n, n+1)$,the function $h(x)$ increases from $0$ to $0.5$ and then decreases from $0.5$ to $0$.
Since $h(x)$ is continuous everywhere and the graph shows sharp corners at $x = n$ (where ${x}=0$) and at $x = n + 0.5$ (where ${x}=0.5$),we count the points of non-differentiability in $(-2, 2)$.
The points are $x = -1.5, -1, -0.5, 0, 0.5, 1, 1.5$. There are $7$ such points.
Since $7 > 4$,option $A$ is correct.

(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{+}} f(x) = f(0) = k$.
First,calculate the $RHL$:
$\lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2}+1}-1} = \lim_{x \to 0^{+}} \frac{\cos(2x) - 1}{\sqrt{x^{2}+1}-1}$.
Using $\cos(2x) - 1 = -2\sin^{2} x$,we get:
$\lim_{x \to 0^{+}} \frac{-2\sin^{2} x}{\sqrt{x^{2}+1}-1} \times \frac{\sqrt{x^{2}+1}+1}{\sqrt{x^{2}+1}+1} = \lim_{x \to 0^{+}} \frac{-2\sin^{2} x (\sqrt{x^{2}+1}+1)}{x^{2}} = -2(1)^{2}(1+1) = -4$.
So,$k = -4$.
Next,calculate the $LHL$:
$\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} \frac{1}{x} \ln\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right) = \lim_{x \to 0^{-}} \left[ \frac{\ln(1+\frac{x}{a})}{x} - \frac{\ln(1-\frac{x}{b})}{x} \right]$.
Using $\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1$,we get:
$\frac{1}{a} - (-\frac{1}{b}) = \frac{1}{a} + \frac{1}{b}$.
Since $LHL = k$,we have $\frac{1}{a} + \frac{1}{b} = -4$.
Finally,calculate $\frac{1}{a} + \frac{1}{b} + \frac{4}{k} = -4 + \frac{4}{-4} = -4 - 1 = -5$.