Differentiability Questions in English

(A) Consider the function $f(x) = |x| + |x - 1|$.
Since the modulus function is continuous everywhere and the sum of two continuous functions is also continuous,$f(x)$ is continuous for all $x \in \mathbb{R}$.
To check the differentiability of $f(x)$,we examine the points where the expression inside the modulus changes sign,which are $x = 0$ and $x = 1$.
At $x = 0$:
Left-hand derivative $(LHD)$ $= \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac{(|x| + |x - 1|) - 1}{x} = \lim_{x \to 0^-} \frac{(-x - x + 1) - 1}{x} = \lim_{x \to 0^-} \frac{-2x}{x} = -2$.
Right-hand derivative $(RHD)$ $= \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac{(|x| + |x - 1|) - 1}{x} = \lim_{x \to 0^+} \frac{(x - x + 1) - 1}{x} = \lim_{x \to 0^+} \frac{0}{x} = 0$.
Since $LHD \neq RHD$,$f(x)$ is not differentiable at $x = 0$.
At $x = 1$:
$LHD$ $= \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^-} \frac{(|x| + |x - 1|) - 1}{x - 1} = \lim_{x \to 1^-} \frac{(x - x + 1) - 1}{x - 1} = \lim_{x \to 1^-} \frac{0}{x - 1} = 0$.
$RHD$ $= \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^+} \frac{(|x| + |x - 1|) - 1}{x - 1} = \lim_{x \to 1^+} \frac{(x + x - 1) - 1}{x - 1} = \lim_{x \to 1^+} \frac{2x - 2}{x - 1} = 2$.
Since $LHD \neq RHD$,$f(x)$ is not differentiable at $x = 1$.
Thus,$f(x) = |x| + |x - 1|$ is continuous everywhere but not differentiable at exactly two points,$x = 0$ and $x = 1$.

(A) The function is defined as $f(x) = \begin{cases} x^2, & x < 0 \\ x, & 0 \leq x \leq 1 \\ x^2, & x > 1 \end{cases}$.
To check for differentiability at $x = 0$:
Left-hand derivative $(LHD)$ at $x = 0$ is $\frac{d}{dx}(x^2) = 2x = 0$.
Right-hand derivative $(RHD)$ at $x = 0$ is $\frac{d}{dx}(x) = 1$.
Since $LHD \neq RHD$,$f$ is not differentiable at $x = 0$.
To check for differentiability at $x = 1$:
Left-hand derivative $(LHD)$ at $x = 1$ is $\frac{d}{dx}(x) = 1$.
Right-hand derivative $(RHD)$ at $x = 1$ is $\frac{d}{dx}(x^2) = 2x = 2$.
Since $LHD \neq RHD$,$f$ is not differentiable at $x = 1$.
Thus,the set of points where $f$ is not differentiable is $S = \{0, 1\}$.

(A) To find $f''(0)$,we first find $f'(x)$.
For $x < 0$,$f'(x) = 5x^4 \sin(1/x) - x^3 \cos(1/x) + 10x$. Thus,$f'(0) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} (h^4 \sin(1/h) + 5h) = 0$.
For $x > 0$,$f'(x) = 5x^4 \cos(1/x) + x^3 \sin(1/h) + 2\lambda x$. Thus,$f'(0) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} (h^4 \cos(1/h) + \lambda h) = 0$.
Now,$f''(0) = \lim_{h \to 0} \frac{f'(h) - f'(0)}{h}$.
Left-hand derivative $(LHD)$ at $x=0$: $\lim_{h \to 0^-} \frac{5h^4 \sin(1/h) - h^3 \cos(1/h) + 10h - 0}{h} = \lim_{h \to 0^-} (5h^3 \sin(1/h) - h^2 \cos(1/h) + 10) = 10$.
Right-hand derivative $(RHD)$ at $x=0$: $\lim_{h \to 0^+} \frac{5h^4 \cos(1/h) + h^3 \sin(1/h) + 2\lambda h - 0}{h} = \lim_{h \to 0^+} (5h^3 \cos(1/h) + h^2 \sin(1/h) + 2\lambda) = 2\lambda$.
For $f''(0)$ to exist,$LHD$ = $RHD$,so $2\lambda = 10$,which implies $\lambda = 5$.

(D) Given $f(x) = \begin{cases} \frac{1}{|x|} & ; |x| \geq 1 \\ ax^2 + b & ; |x| < 1 \end{cases}$.
For $x \geq 1$,$f(x) = \frac{1}{x}$. For $x \leq -1$,$f(x) = -\frac{1}{x}$.
Since $f(x)$ is differentiable at every point,it must be continuous at $x = 1$.
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \implies a(1)^2 + b = \frac{1}{1} \implies a + b = 1 \quad \dots(1)$.
Also,$f(x)$ must be differentiable at $x = 1$.
$f'(x) = \begin{cases} -\frac{1}{x^2} & ; x > 1 \\ 2ax & ; -1 < x < 1 \end{cases}$.
Equating the derivatives at $x = 1$:
$\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x) \implies 2a(1) = -\frac{1}{(1)^2} \implies 2a = -1 \implies a = -\frac{1}{2}$.
Substituting $a = -\frac{1}{2}$ into equation $(1)$:
$-\frac{1}{2} + b = 1 \implies b = 1 + \frac{1}{2} = \frac{3}{2}$.
Thus,the values are $a = -\frac{1}{2}$ and $b = \frac{3}{2}$.

(C) Given function: $f(x) = |2x+1| - 3|x+2| + |x^2+x-2|$
Factorize the quadratic term: $|x^2+x-2| = |(x+2)(x-1)| = |x+2||x-1|$
Substitute back into the function: $f(x) = |2x+1| - 3|x+2| + |x+2||x-1|$
$f(x) = |2x+1| + |x+2|(|x-1| - 3)$
The points where the expression inside the absolute value becomes zero are $x = -1/2$,$x = -2$,and $x = 1$.
Let $g(x) = |x-1| - 3$. The function $f(x)$ is non-differentiable at points where the argument of an absolute value is zero,provided the function does not have a smooth turning point there.
$1$. At $x = -1/2$,$|2x+1|$ is non-differentiable,and the other terms are smooth. Thus,$x = -1/2$ is a point of non-differentiability.
$2$. At $x = 1$,$|x-1|$ is non-differentiable,and $|x+2| = 3 \neq 0$. Thus,$x = 1$ is a point of non-differentiability.
$3$. At $x = -2$,we check the behavior: $f(x) = |2x+1| + |x+2|(|x-1|-3)$. Near $x = -2$,$|x-1| = -(x-1) = 1-x$. So $f(x) \approx |2x+1| + |x+2|(1-x-3) = |2x+1| + |x+2|(-x-2) = |2x+1| - |x+2|(x+2) = |2x+1| - (x+2)^2$. Since $(x+2)^2$ is differentiable at $x = -2$,the non-differentiability of $|x+2|$ is cancelled out by the factor $(x+2)$.
Therefore,the points of non-differentiability are $x = -1/2$ and $x = 1$.
The number of points of non-differentiability is $2$.

(C) For $x \in [-2, 2]$,$f(x) = \min \{|x|, 2-x^2\}$.
Solving $|x| = 2-x^2$,we get $x^2 + |x| - 2 = 0$,which gives $(|x|+2)(|x|-1) = 0$. Since $|x| \geq 0$,we have $|x| = 1$,so $x = \pm 1$.
Thus,$f(x) = |x|$ for $x \in [-1, 1]$ and $f(x) = 2-x^2$ for $x \in [-2, -1) \cup (1, 2]$.
Points of non-differentiability in $(-2, 2)$ are $x = -1, 0, 1$ (due to $|x|$ and the intersection of curves).
For $2 < |x| \leq 3$,$f(x) = [|x|]$.
For $x \in (2, 3]$,$f(x) = [x] = 2$ for $x \in (2, 3)$ and $f(3) = 3$. This is discontinuous at $x = 3$ (not in $(-3, 3)$) and $x = 2$ (jump discontinuity).
For $x \in [-3, -2)$,$f(x) = [|x|] = [|x|]$. For $x \in (-3, -2)$,$f(x) = 2$. At $x = -2$,$f(-2) = 2$. At $x = -3$,$f(-3) = 3$.
Checking points: $x = -2, -1, 0, 1, 2$. At these $5$ points,the function is either discontinuous or has a sharp corner.
Therefore,the number of points of non-differentiability is $5$.

(C) Given $f(x)=|x^{2}-2 x-3| \cdot e^{|9 x^{2}-12 x+4|}$.
We can factorize the expressions inside the absolute values:
$x^{2}-2 x-3 = (x-3)(x+1)$
$9 x^{2}-12 x+4 = (3 x-2)^{2}$
So,$f(x)=|(x-3)(x+1)| \cdot e^{(3 x-2)^{2}}$.
Note that $e^{(3 x-2)^{2}}$ is a smooth,differentiable function for all $x \in \mathbb{R}$.
The non-differentiability of $f(x)$ depends solely on the term $|(x-3)(x+1)|$.
$A$ function of the form $|g(x)|$ is non-differentiable at the roots of $g(x)$ where the sign changes.
Here,$g(x) = (x-3)(x+1)$ changes sign at $x=3$ and $x=-1$.
At $x=3$ and $x=-1$,the function $|(x-3)(x+1)|$ has sharp corners (cusps).
Therefore,$f(x)$ is not differentiable at exactly two points: $x=3$ and $x=-1$.

(C) Let $f(t) = t^3 - 6t^2 + 9t - 3$.
Then $f'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$.
The critical points are $t=1$ and $t=3$.
$f(0) = -3$,$f(1) = 1 - 6 + 9 - 3 = 1$,and $f(3) = 27 - 54 + 27 - 3 = -3$.
For $0 \leq x \leq 1$,$f(t)$ is increasing,so $\max_{0 \leq t \leq x} f(t) = f(x)$.
For $1 < x \leq 3$,the maximum value of $f(t)$ on $[0, x]$ is $f(1) = 1$.
Thus,$g(x) = \begin{cases} x^3 - 6x^2 + 9x - 3 & , 0 \leq x \leq 1 \\ 1 & , 1 < x \leq 3 \\ 4 - x & , 3 < x \leq 4 \end{cases}$.
Now check differentiability:
At $x=1$: $g'(1^-) = f'(1) = 0$ and $g'(1^+) = 0$. So $g(x)$ is differentiable at $x=1$.
At $x=3$: $g(3^-) = 1$ and $g(3^+) = 4-3 = 1$. $g(x)$ is continuous at $x=3$.
$g'(3^-) = 0$ and $g'(3^+) = -1$. Since $g'(3^-) \neq g'(3^+)$,$g(x)$ is not differentiable at $x=3$.
Thus,there is only $1$ point in $(0,4)$ where $g(x)$ is not differentiable.

(C) First,analyze the condition $8x^2 - 6x + 1 < 0$. Solving $(2x - 1)(4x - 1) < 0$,we get $x \in (1/4, 1/2)$.
For $x \in (1/4, 1/2)$,$f(x) = [4x^2 - 8x + 5]$. Let $g(x) = 4x^2 - 8x + 5 = 4(x-1)^2 + 1$. In $(1/4, 1/2)$,$g(x)$ decreases from $g(1/4) = 4(1/16) - 2 + 5 = 13/4 = 3.25$ to $g(1/2) = 4(1/4) - 4 + 5 = 2$.
Thus,$f(x) = [g(x)]$ takes values $3$ for $x \in (1/4, x_1)$ where $g(x_1) = 3$,and $2$ for $x \in [x_1, 1/2)$.
At $x = 1/4$,$g(1/4) = 3.25$. The function is continuous but has a sharp corner (non-differentiable).
At $x = x_1$,$g(x_1) = 3$. The function jumps from $3$ to $2$,so it is discontinuous and non-differentiable.
At $x = 1/2$,$g(1/2) = 2$. The function jumps from $2$ to $g(1/2) = 2$ (continuous),but the derivative changes abruptly,making it non-differentiable.
Thus,the points of non-differentiability are $x = 1/4, x_1, 1/2$. The total number of points is $3$.

(B) The function is given by $f(x) = 4|2x + 3| + 9[x + \frac{1}{2}] - 12[x + 20]$.
$1$. The term $4|2x + 3|$ is non-differentiable at $2x + 3 = 0$,which gives $x = -\frac{3}{2}$. This is $1$ point.
$2$. The term $9[x + \frac{1}{2}]$ is non-differentiable when $x + \frac{1}{2} = k$ for any integer $k$. In the interval $(-20, 20)$,$x + \frac{1}{2} \in (-19.5, 20.5)$. The integers $k$ are $\{-19, -18, \dots, 20\}$. There are $20 - (-19) + 1 = 40$ points. However,we must check if any of these coincide with $x = -\frac{3}{2}$. For $x = -\frac{3}{2}$,$x + \frac{1}{2} = -1$,which is an integer. So,one point is common.
$3$. The term $-12[x + 20]$ is non-differentiable when $x + 20 = k$ for any integer $k$. In the interval $(-20, 20)$,$x + 20 \in (0, 40)$. The integers $k$ are $\{1, 2, \dots, 39\}$. There are $39$ points.
$4$. Total points of non-differentiability = (Points from $|2x+3|$) + (Points from $[x+1/2]$) + (Points from $[x+20]$) - (Common points).
Points from $|2x+3|$: $x = -1.5$ ($1$ point).
Points from $[x+1/2]$: $x \in \{-19.5, -18.5, \dots, 19.5\}$ ($39$ points).
Points from $[x+20]$: $x \in \{-19, -18, \dots, 19\}$ ($39$ points).
Since the sets of points are disjoint,the total number of points is $1 + 39 + 39 = 79$.

(B) Given function: $f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x^2-5x+4|$.
Factorize the quadratic term: $x^2-5x+4 = (x-1)(x-4)$.
So,$f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x-1||x-4|$.
Factor out $|x-1|$: $f(x) = |x-1| [\cos |x-2| \sin |x-1| + (x-3)|x-4|]$.
Let $g(x) = |x-1|$ and $h(x) = \cos |x-2| \sin |x-1| + (x-3)|x-4|$.
The function $f(x) = g(x) \cdot h(x)$ is non-differentiable where $g(x)$ is non-differentiable,provided $h(x) \neq 0$ at those points.
$g(x) = |x-1|$ is non-differentiable at $x = 1$.
Check $h(1) = \cos |1-2| \sin |1-1| + (1-3)|1-4| = \cos(1) \cdot 0 + (-2) \cdot |-3| = -6 \neq 0$.
Thus,$f(x)$ is non-differentiable at $x = 1$.
Now check the term $|x-4|$ in $h(x)$. The function $f(x)$ involves $|x-4|$ multiplied by $(x-3)$.
At $x = 4$,$f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x-1||x-4|$.
Near $x = 4$,the term $|x-1| \cos |x-2| \sin |x-1|$ is differentiable.
The term $(x-3)|x-1||x-4|$ is of the form $k(x)|x-4|$,where $k(4) = (4-3)|4-1| = 3 \neq 0$.
Since $k(4) \neq 0$,the function is non-differentiable at $x = 4$.
Therefore,the function is non-differentiable at $x = 1$ and $x = 4$. The total number of points is $2$.

(D) Given the function $f(x) = \begin{cases} \frac{\sin(x^2)}{x} & x \neq 0 \\ 0 & x = 0 \end{cases}$.
First,check for continuity at $x=0$:
$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{\sin(x^2)}{x} = \lim_{x \rightarrow 0} x \cdot \frac{\sin(x^2)}{x^2} = 0 \cdot 1 = 0 = f(0)$.
Since the limit equals the function value,$f(x)$ is continuous at $x=0$.
Next,check for differentiability at $x=0$:
$f'(0) = \lim_{h \rightarrow 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \rightarrow 0} \frac{\frac{\sin(h^2)}{h} - 0}{h} = \lim_{h \rightarrow 0} \frac{\sin(h^2)}{h^2} = 1$.
Since the limit exists,$f(x)$ is differentiable at $x=0$ and $f'(0) = 1$.
Now,find $f'(x)$ for $x \neq 0$:
$f'(x) = \frac{d}{dx} \left( \frac{\sin(x^2)}{x} \right) = \frac{x \cdot \cos(x^2) \cdot 2x - \sin(x^2) \cdot 1}{x^2} = 2\cos(x^2) - \frac{\sin(x^2)}{x^2}$.
Check if $f'(x)$ is continuous at $x=0$:
$\lim_{x \rightarrow 0} f'(x) = \lim_{x \rightarrow 0} \left( 2\cos(x^2) - \frac{\sin(x^2)}{x^2} \right) = 2(1) - 1 = 1$.
Since $\lim_{x \rightarrow 0} f'(x) = f'(0) = 1$,the derivative is continuous at $x=0$.
Thus,$f$ is differentiable and the derivative is continuous.

(B) Given $f(x) = x |\sin x|$.
At $x = n\pi$ (where $n \in Z$),we check the differentiability using the limit definition:
$f'(n\pi) = \lim_{x \to n\pi} \frac{f(x) - f(n\pi)}{x - n\pi} = \lim_{x \to n\pi} \frac{x |\sin x| - 0}{x - n\pi} = \lim_{x \to n\pi} \frac{x |\sin x|}{x - n\pi}$.
Let $x = n\pi + h$,where $h \to 0$.
Then $f'(n\pi) = \lim_{h \to 0} \frac{(n\pi + h) |\sin(n\pi + h)|}{h} = \lim_{h \to 0} \frac{(n\pi + h) |(-1)^n \sin h|}{h} = \lim_{h \to 0} (n\pi + h) \frac{|\sin h|}{|h|} \cdot |h| \cdot \frac{(-1)^n}{h}$.
Since $\frac{|\sin h|}{h} \to 1$ as $h \to 0$,the limit becomes $\lim_{h \to 0} (n\pi + h) \cdot 1 \cdot \frac{|h|}{h} \cdot (-1)^n$.
For $n = 0$,$f'(0) = \lim_{h \to 0} h \cdot \frac{|h|}{h} = \lim_{h \to 0} |h| = 0$. Thus,$f(x)$ is differentiable at $x = 0$.
For $n \neq 0$,the limit $\lim_{h \to 0} n\pi \cdot (-1)^n \cdot \frac{|h|}{h}$ does not exist because the left-hand limit is $-n\pi(-1)^n$ and the right-hand limit is $n\pi(-1)^n$.
Therefore,$f(x)$ is differentiable for all $x$ except at $x = n\pi$ for $n = \pm 1, \pm 2, \pm 3, \dots$.

(C) To check the differentiability of $f(x)$ at $x=0$,we examine the limit of the difference quotient:
$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \left| \cos \left(\frac{\pi}{h}\right) \right| - 0}{h} = \lim_{h \to 0} h \left| \cos \left(\frac{\pi}{h}\right) \right|$.
Since $|\cos(\frac{\pi}{h})| \leq 1$,by the Squeeze Theorem,$\lim_{h \to 0} h \left| \cos \left(\frac{\pi}{h}\right) \right| = 0$. Thus,$f$ is differentiable at $x=0$.
For $x \neq 0$,$f(x) = x^2 |\cos(\frac{\pi}{x})|$. The function $|\cos(\frac{\pi}{x})|$ is not differentiable where the argument of the absolute value is zero,i.e.,$\cos(\frac{\pi}{x}) = 0$.
$\cos(\frac{\pi}{x}) = 0 \implies \frac{\pi}{x} = (2n+1)\frac{\pi}{2}$ for $n \in Z$.
$x = \frac{2}{2n+1}$.
Thus,$f$ is not differentiable at points where $\cos(\frac{\pi}{x}) = 0$,which are $x = \frac{2}{2n+1}$ for $n \in Z$.

(B) Since $f(x)$ is differentiable on $R$,it must be continuous at $x = 0$.
For continuity at $x = 0$:
$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0)$
$\lim_{x \rightarrow 0^-} \frac{\sin x^2}{x} = \lim_{x \rightarrow 0^+} (x^2 + ax + b)$
Using the limit $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we have $\lim_{x \rightarrow 0^-} (x \cdot \frac{\sin x^2}{x^2}) = 0 \cdot 1 = 0$.
Thus,$0 = b$,so $b = 0$.
For differentiability at $x = 0$,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$:
$LHD = \lim_{h \rightarrow 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \rightarrow 0^+} \frac{\frac{\sin(-h)^2}{-h} - 0}{-h} = \lim_{h \rightarrow 0^+} \frac{\sin h^2}{h^2} = 1$.
$RHD = \lim_{h \rightarrow 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \rightarrow 0^+} \frac{h^2 + ah + b - b}{h} = \lim_{h \rightarrow 0^+} (h + a) = a$.
Equating $LHD$ and $RHD$,we get $a = 1$.
Therefore,$a = 1$ and $b = 0$.

(D) The function is given by $f(x) = [a + 13 \sin x]$ for $x \in (0, \pi)$.
Since $a$ is an integer,we can write $f(x) = a + [13 \sin x]$.
The function $f(x)$ is not differentiable at points where the argument of the greatest integer function,$13 \sin x$,is an integer.
For $x \in (0, \pi)$,the range of $13 \sin x$ is $(0, 13]$.
The values of $13 \sin x$ are integers at $13 \sin x = 1, 2, 3, \dots, 13$.
For each integer $k \in \{1, 2, \dots, 12\}$,there are $2$ values of $x$ in $(0, \pi)$ such that $13 \sin x = k$.
For $k = 13$,there is only $1$ value of $x$ in $(0, \pi)$,which is $x = \frac{\pi}{2}$.
Thus,the total number of points where the function is not differentiable is $2 \times 12 + 1 = 25$.

(A) Given $f(x) = \begin{cases} \frac{1}{x} & , x \geq 2 \\ ax^2 + 2b & , -2 < x < 2 \\ -\frac{1}{x} & , x \leq -2 \end{cases}$.
For $f(x)$ to be differentiable on $\mathbb{R}$,it must be continuous and differentiable at $x = 2$ and $x = -2$.
At $x = 2$,continuity implies $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^-} f(x) = f(2)$.
$\frac{1}{2} = a(2)^2 + 2b \Rightarrow 4a + 2b = \frac{1}{2} \Rightarrow 8a + 4b = 1$.
For differentiability at $x = 2$,$f'(2^+) = f'(2^-)$.
$f'(x) = -\frac{1}{x^2}$ for $x > 2$ and $f'(x) = 2ax$ for $-2 < x < 2$.
$-\frac{1}{2^2} = 2a(2) \Rightarrow -\frac{1}{4} = 4a \Rightarrow a = -\frac{1}{16}$.
Substituting $a = -\frac{1}{16}$ into $8a + 4b = 1$:
$8(-\frac{1}{16}) + 4b = 1 \Rightarrow -\frac{1}{2} + 4b = 1 \Rightarrow 4b = \frac{3}{2} \Rightarrow b = \frac{3}{8}$.
Now,calculate $48(a+b)$:
$48(-\frac{1}{16} + \frac{3}{8}) = 48(\frac{-1+6}{16}) = 48(\frac{5}{16}) = 3 \times 5 = 15$.

(C) Given the function $f(x) = e^{-\left|\ln x\right|}$ for $x \in (0, \infty)$.
We can rewrite the function by considering the definition of the absolute value:
$f(x) = \begin{cases} e^{-(-\ln x)} = e^{\ln x} = x, & 0 < x < 1 \\ e^{-\ln x} = \frac{1}{x}, & x \geq 1 \end{cases}$
Now,let's check for continuity at $x = 1$:
Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$
Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1}{x} = 1$
Value at $x = 1$: $f(1) = \frac{1}{1} = 1$
Since the limits are equal,the function is continuous everywhere in its domain. Thus,$m = 0$.
Now,let's check for differentiability at $x = 1$:
Left-hand derivative: $f'(1^-) = \frac{d}{dx}(x) \big|_{x=1} = 1$
Right-hand derivative: $f'(1^+) = \frac{d}{dx}(\frac{1}{x}) \big|_{x=1} = -\frac{1}{x^2} \big|_{x=1} = -1$
Since $f'(1^-) \neq f'(1^+)$,the function is not differentiable at $x = 1$. Thus,$n = 1$.
Therefore,$m + n = 0 + 1 = 1$.

(D) Given $f(x) = |2x^2 + 5|x| - 3|$.
Since $f(x)$ is a composition of continuous functions (polynomials and absolute value),it is continuous everywhere. Thus,the number of points of discontinuity is $m = 0$.
To find the points of non-differentiability,we analyze $g(x) = 2x^2 + 5|x| - 3$.
For $x \ge 0$,$g(x) = 2x^2 + 5x - 3 = (2x - 1)(x + 3)$. The roots are $x = 1/2$ and $x = -3$. Since we consider $x \ge 0$,the root is $x = 1/2$.
For $x < 0$,$g(x) = 2x^2 - 5x - 3 = (2x + 1)(x - 3)$. The roots are $x = -1/2$ and $x = 3$. Since we consider $x < 0$,the root is $x = -1/2$.
Also,the function $f(x)$ involves $|x|$,which is non-differentiable at $x = 0$.
Thus,$f(x)$ is non-differentiable at the roots of $2x^2 + 5|x| - 3 = 0$ (where the graph touches the x-axis) and at $x = 0$ (due to $|x|$).
The points are $x = 1/2$,$x = -1/2$,and $x = 0$.
So,the number of points of non-differentiability is $n = 3$.
Therefore,$m + n = 0 + 3 = 3$.

(C) Given $f(x) = a \cos (|x^3-x|) + b|x| \sin (|x^3+x|)$.
Since $\cos(\theta) = \cos(-\theta)$,we have $\cos(|x^3-x|) = \cos(x^3-x)$.
Also,$|x| \sin(|x^3+x|) = x \sin(x^3+x)$ for all $x \in R$ because if $x \ge 0$,$|x|=x$ and $|x^3+x|=x^3+x$,and if $x < 0$,$|x|=-x$ and $|x^3+x|=-(x^3+x)$,so $-x \sin(-(x^3+x)) = -x(-\sin(x^3+x)) = x \sin(x^3+x)$.
Thus,$f(x) = a \cos(x^3-x) + b x \sin(x^3+x)$ for all $x \in R$.
Since $\cos(x^3-x)$ and $x \sin(x^3+x)$ are differentiable functions on $R$,$f(x)$ is differentiable for all $x \in R$ for any $a, b \in R$.
Therefore,$f$ is differentiable at $x=0$ and $x=1$ for any $a, b \in R$.
Checking the options:
$(A)$ Differentiable at $x=0$ if $a=0, b=1$ (True).
$(B)$ Differentiable at $x=1$ if $a=1, b=0$ (True).
$(C)$ Not differentiable at $x=0$ if $a=1, b=0$ (False).
$(D)$ Not differentiable at $x=1$ if $a=1, b=1$ (False).
Thus,options $A$ and $B$ are correct.

(B) $(i)$ Check for differentiability at $x=0$:
$LHD = f'(0^-) = \lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{(-h)^2 |\cos(-\pi/h)| - 0}{-h} = \lim_{h \to 0^+} -h |\cos(\pi/h)| = 0$.
$RHD = f'(0^+) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2 |\cos(\pi/h)| - 0}{h} = \lim_{h \to 0^+} h |\cos(\pi/h)| = 0$.
Since $LHD = RHD = 0$,$f(x)$ is differentiable at $x=0$.
$(ii)$ Check for differentiability at $x=2$:
$f(2) = 2^2 |\cos(\pi/2)| = 0$.
$RHD = f'(2^+) = \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0^+} \frac{(2+h)^2 |\cos(\pi/(2+h))|}{h}$.
Since $\cos(\pi/(2+h)) > 0$ for small $h > 0$,$|\cos(\pi/(2+h))| = \cos(\pi/(2+h)) = \sin(\pi/2 - \pi/(2+h)) = \sin(\pi h / (2(2+h)))$.
$RHD = \lim_{h \to 0^+} \frac{(2+h)^2 \sin(\pi h / (2(2+h)))}{h} = 4 \cdot \frac{\pi}{4} = \pi$.
$LHD = f'(2^-) = \lim_{h \to 0^+} \frac{f(2-h) - f(2)}{-h} = \lim_{h \to 0^+} \frac{(2-h)^2 |\cos(\pi/(2-h))|}{-h}$.
Since $\cos(\pi/(2-h)) < 0$ for small $h > 0$,$|\cos(\pi/(2-h))| = -\cos(\pi/(2-h)) = -\sin(\pi/2 - \pi/(2-h)) = -\sin(-\pi h / (2(2-h))) = \sin(\pi h / (2(2-h)))$.
$LHD = \lim_{h \to 0^+} \frac{(2-h)^2 \sin(\pi h / (2(2-h)))}{-h} = 4 \cdot (-\pi/4) = -\pi$.
Since $LHD \neq RHD$,$f(x)$ is not differentiable at $x=2$.

(D) Given $f(x) = x - x^2 + (x - 1) \sin x = -(x - 1)^2 + (x - 1) \sin x = (x - 1) [-(x - 1) + \sin x]$.
Note that $f(1) = 0$.
Also,$f'(x) = 1 - 2x + \sin x + (x - 1) \cos x$. Thus,$f'(1) = 1 - 2 + \sin 1 + 0 = \sin 1 - 1$.
Let $h(x) = (fg)(x) = f(x)g(x)$.
For differentiability at $x=1$,we check $h'(1) = \lim_{k \to 0} \frac{f(1+k)g(1+k) - f(1)g(1)}{k} = \lim_{k \to 0} \frac{f(1+k)g(1+k)}{k}$.
Since $f(1+k) = k(-k + \sin(1+k))$,the limit becomes $\lim_{k \to 0} \frac{k(-k + \sin(1+k))g(1+k)}{k} = \lim_{k \to 0} (-k + \sin(1+k))g(1+k) = \sin(1) \cdot g(1)$.
If $g$ is continuous at $x=1$,$g(1+k) \to g(1)$,so the limit exists. Thus,$(A)$ is true.
If $g$ is differentiable at $x=1$,it is continuous,so $(C)$ is true.
If $fg$ is differentiable,it does not imply $g$ is continuous or differentiable (e.g.,if $f(x)=0$ at $x=1$,$fg$ could be differentiable even if $g$ is not). Thus,$(B)$ and $(D)$ are false.
Therefore,the correct statements are $(A)$ and $(C)$.

(B) Given $f(x+y) = f(x) + f(y) + f(x)f(y)$. Adding $1$ to both sides gives $1 + f(x+y) = (1 + f(x))(1 + f(y))$.
Let $h(x) = 1 + f(x)$. Then $h(x+y) = h(x)h(y)$.
Since $f(x) = xg(x)$ and $\lim_{x \to 0} g(x) = 1$,we have $\lim_{x \to 0} f(x) = 0$,so $h(0) = 1 + f(0) = 1$.
For $h(x+y) = h(x)h(y)$,the solution is $h(x) = e^{cx}$.
Thus $1 + f(x) = e^{cx}$,or $f(x) = e^{cx} - 1$.
Given $f(x) = xg(x)$,we have $g(x) = \frac{e^{cx}-1}{x}$.
$\lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{e^{cx}-1}{x} = c$.
Since $\lim_{x \to 0} g(x) = 1$,we have $c = 1$.
Therefore,$f(x) = e^x - 1$.
$f'(x) = e^x$,which is defined for all $x \in R$,so $(A)$ is $TRUE$.
$f'(0) = e^0 = 1$,so $(D)$ is $TRUE$.
$f'(1) = e^1 = e \neq 1$,so $(C)$ is $FALSE$.
For $g(x) = \frac{e^x-1}{x}$ for $x \neq 0$ and $g(0)=1$,$g$ is differentiable at all $x \neq 0$. At $x=0$,$g'(0) = \lim_{h \to 0} \frac{\frac{e^h-1}{h}-1}{h} = \lim_{h \to 0} \frac{e^h-1-h}{h^2} = \frac{1}{2}$. Thus $g$ is differentiable everywhere,so $(B)$ is $TRUE$.

(B) Given $f(x) = |x| + 1$ and $g(x) = x^2 + 1$.
For $x \leq 0$,$f(x) = -x + 1$ and $g(x) = x^2 + 1$. We define $h(x) = \max\{-x+1, x^2+1\}$.
Since $x^2+1 \geq -x+1$ for $x \in [-1, 0]$ (as $x^2+x \geq 0$),$h(x) = x^2+1$ for $x \in [-1, 0]$ and $h(x) = -x+1$ for $x < -1$.
For $x > 0$,$f(x) = x + 1$ and $g(x) = x^2 + 1$. We define $h(x) = \min\{x+1, x^2+1\}$.
Since $x^2+1 \leq x+1$ for $x \in [0, 1]$ (as $x^2-x \leq 0$),$h(x) = x^2+1$ for $x \in [0, 1]$ and $h(x) = x+1$ for $x > 1$.
Thus,$h(x) = \begin{cases} -x+1 & x < -1 \\ x^2+1 & -1 \leq x \leq 1 \\ x+1 & x > 1 \end{cases}$.
Checking differentiability:
At $x = -1$: $h(-1) = 2$. Left derivative is $-1$,right derivative is $2(-1) = -2$. Not differentiable.
At $x = 1$: $h(1) = 2$. Left derivative is $2(1) = 2$,right derivative is $1$. Not differentiable.
At $x = 0$: $h(0) = 1$. Left derivative is $2(0) = 0$,right derivative is $2(0) = 0$. Differentiable.
There are $2$ points of non-differentiability.

(D) Differentiability of $f(x)$ at $x=0$:
$\text{LHD} = \lim_{\delta \rightarrow 0^+} \frac{f(0-\delta) - f(0)}{-\delta} = \lim_{\delta \rightarrow 0^+} \frac{\frac{-\delta}{|-\delta|} g(-\delta) - 0}{-\delta} = \lim_{\delta \rightarrow 0^+} \frac{-g(-\delta)}{-\delta} = g^{\prime}(0) = 0$.
$\text{RHD} = \lim_{\delta \rightarrow 0^+} \frac{f(0+\delta) - f(0)}{\delta} = \lim_{\delta \rightarrow 0^+} \frac{\frac{\delta}{|\delta|} g(\delta) - 0}{\delta} = \lim_{\delta \rightarrow 0^+} \frac{g(\delta)}{\delta} = g^{\prime}(0) = 0$.
Since $\text{LHD} = \text{RHD} = 0$,$f(x)$ is differentiable at $x=0$.
Differentiability of $h(x) = e^{|x|}$ at $x=0$:
$h(x)$ is not differentiable at $x=0$ because $|x|$ is not differentiable at $x=0$.
Differentiability of $f(h(x))$ at $x=0$:
Since $h(x) = e^{|x|} > 0$ for all $x$,$f(h(x)) = \frac{h(x)}{|h(x)|} g(h(x)) = 1 \cdot g(e^{|x|}) = g(e^{|x|})$.
$\text{LHD} = \lim_{\delta \rightarrow 0^+} \frac{g(e^{|-\delta|}) - g(e^0)}{-\delta} = \lim_{\delta \rightarrow 0^+} \frac{g(e^{\delta}) - g(1)}{-\delta} = -g^{\prime}(1)$.
$\text{RHD} = \lim_{\delta \rightarrow 0^+} \frac{g(e^{\delta}) - g(1)}{\delta} = g^{\prime}(1)$.
Since $g^{\prime}(1) \neq 0$,$f(h(x))$ is not differentiable at $x=0$.
Differentiability of $h(f(x))$ at $x=0$:
$h(f(x)) = e^{|f(x)|} = e^{|\frac{x}{|x|} g(x)|} = e^{|g(x)|}$.
$\text{LHD} = \lim_{\delta \rightarrow 0^+} \frac{e^{|g(-\delta)|} - e^{|g(0)|}}{-\delta} = \lim_{\delta \rightarrow 0^+} \frac{e^{|g(-\delta)|} - 1}{|g(-\delta)|} \cdot \frac{|g(-\delta)|}{-\delta} = 1 \cdot 0 = 0$.
$\text{RHD} = \lim_{\delta \rightarrow 0^+} \frac{e^{|g(\delta)|} - 1}{\delta} = \lim_{\delta \rightarrow 0^+} \frac{e^{|g(\delta)|} - 1}{|g(\delta)|} \cdot \frac{|g(\delta)|}{\delta} = 1 \cdot 0 = 0$.
Thus,$h(f(x))$ is differentiable at $x=0$.

(C) The function $f(x) = (x^2 - 1)|x^2 - ax + 2| + \cos|x|$ is non-differentiable where the expression inside the modulus is zero,provided the expression does not have a double root at that point.
Since $\cos|x|$ is differentiable everywhere,we only need to check $g(x) = x^2 - ax + 2$.
Given that $x = \alpha = 2$ is a point of non-differentiability,$g(2) = 0$.
$2^2 - a(2) + 2 = 0 \implies 6 - 2a = 0 \implies a = 3$.
Substituting $a = 3$,we get $g(x) = x^2 - 3x + 2 = (x - 1)(x - 2)$.
The roots are $x = 1$ and $x = 2$.
At $x = 1$,$f(x) = (x^2 - 1)|(x - 1)(x - 2)| + \cos|x|$. Since $(x^2 - 1) = (x - 1)(x + 1)$,the term becomes $(x - 1)^2(x + 1)|x - 2|$,which is differentiable at $x = 1$.
Thus,$x = 1$ is not a point of non-differentiability.
However,the problem states there are two points of non-differentiability. This implies $x^2 - ax + 2$ must have a root that makes the function non-differentiable. If $a=3$,we only have one point $x=2$. For two points,the quadratic must have two distinct roots,and the factor $(x^2-1)$ must not cancel the non-differentiability. Re-evaluating,if $\beta = -1$,then $(x^2-1) = (x-1)(x+1)$. If $x=-1$ is a root,the function is differentiable. The only way to have two points is if the roots of $x^2-ax+2$ are $\alpha=2$ and $\beta$ such that $\beta \neq \pm 1$. Given the structure,$\beta = 1$ is a root,but it is cancelled. If we assume the question implies $\beta$ is the other root,$\beta = 1$. The distance from $(2, 1)$ to $12x + 5y + 10 = 0$ is $d = \frac{|12(2) + 5(1) + 10|}{\sqrt{12^2 + 5^2}} = \frac{|24 + 5 + 10|}{13} = \frac{39}{13} = 3$.

(C) First,check differentiability at $x=0$:
$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{2-2h^2-h^2 \sin(1/h) - 2}{h} = \lim_{h \to 0} (-2h - h \sin(1/h)) = 0$.
Since the limit exists,$f$ is differentiable at $x=0$. Thus,$(A)$ is false.
For $x \neq 0$,$f'(x) = -4x - 2x \sin(1/h) - x^2 \cos(1/h) (-1/x^2) = -4x - 2x \sin(1/x) + \cos(1/x)$.
As $x \to 0$,$f'(x)$ oscillates due to the $\cos(1/x)$ term.
For any $\delta > 0$,in the interval $(0, \delta)$ or $(-\delta, 0)$,$f'(x)$ takes both positive and negative values because $\cos(1/x)$ oscillates between $-1$ and $1$.
Therefore,$f$ is neither increasing nor decreasing on any interval $(0, \delta)$ or $(-\delta, 0)$.
This makes $(B)$ false and $(C)$ true.
Finally,$f(0)=2$ and for small $h \neq 0$,$f(h) = 2 - 2h^2 - h^2 \sin(1/h) = 2 - h^2(2 + \sin(1/h))$. Since $2 + \sin(1/h) > 0$,$f(h) < 2$ for small $h$. Thus,$x=0$ is a local maximum,making $(D)$ false.

(A) Given,$f(x) = \begin{cases} e^{-x}, & x \geq 0 \\ e^{x}, & x < 0 \end{cases}$
$LHL = \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0} e^{x} = 1$
$RHL = \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0} e^{-x} = 1$
Also,$f(0) = e^{0} = 1$
Since $LHL = RHL = f(0)$,the function is continuous for every value of $x$.
Now,we check for differentiability at $x = 0$:
$LHD = \left(\frac{d}{dx} e^{x}\right)_{x = 0} = \left[e^{x}\right]_{x = 0} = 1$
$RHD = \left(\frac{d}{dx} e^{-x}\right)_{x = 0} = \left[-e^{-x}\right]_{x = 0} = -1$
Since $LHD \neq RHD$,$f(x)$ is not differentiable at $x = 0$.
Therefore,$f(x) = e^{-|x|}$ is continuous everywhere but not differentiable at $x = 0$.

(B) Given $f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x|$.
Since $|x|$ is not differentiable at $x = 0$,we check the differentiability of $f(x)$ at $x = 0$.
For $x \ge 0$,$f(x) = \sin x - x + 2(x - \pi) \cos x$.
For $x < 0$,$f(x) = \sin(-x) - (-x) + 2(x - \pi) \cos(-x) = -\sin x + x + 2(x - \pi) \cos x$.
Now,we find the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$ at $x = 0$.
$f'(x)$ for $x > 0$ is $\cos x - 1 + 2 \cos x - 2(x - \pi) \sin x = 3 \cos x - 1 - 2(x - \pi) \sin x$.
$f'(0^+) = 3(1) - 1 - 2(-\pi)(0) = 2$.
$f'(x)$ for $x < 0$ is $-\cos x + 1 + 2 \cos x - 2(x - \pi) \sin x = \cos x + 1 - 2(x - \pi) \sin x$.
$f'(0^-) = 1 + 1 - 2(-\pi)(0) = 2$.
Since $f'(0^+) = f'(0^-) = 2$,the function is differentiable at $x = 0$.
Since the function is composed of differentiable functions everywhere else,$f(x)$ is differentiable for all $x \in \mathbb{R}$.
Therefore,the set $K$ of points where $f(x)$ is not differentiable is an empty set.

(A) To check the differentiability of $f(x) = |x - \pi|(e^{|x|} - 1) \sin |x|$ at $x = \pi$ and $x = 0$:
$1$. Differentiability at $x = \pi$:
$f(\pi) = 0$.
$f'(x) = \frac{d}{dx} [|x - \pi|(e^{|x|} - 1) \sin |x|]$.
Since $|x - \pi|$ is differentiable at $x = \pi$ only if the other factors are zero,note that at $x = \pi$,$(e^{|\pi|} - 1) \sin |\pi| = (e^{\pi} - 1) \cdot 0 = 0$.
Using the definition of the derivative,$\lim_{h \to 0} \frac{f(\pi + h) - f(\pi)}{h} = \lim_{h \to 0} \frac{|h|(e^{|\pi + h|} - 1) \sin |\pi + h|}{h} = \lim_{h \to 0} \frac{|h|}{h} (e^{\pi} - 1) \sin \pi = 0$.
Thus,$f(x)$ is differentiable at $x = \pi$.
$2$. Differentiability at $x = 0$:
$f(0) = 0$.
$\lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{|h - \pi|(e^{|h|} - 1) \sin |h|}{h}$.
As $h \to 0$,$e^{|h|} - 1 \approx |h|$ and $\sin |h| \approx |h|$.
So,the limit becomes $\lim_{h \to 0} \frac{|-\pi| \cdot |h| \cdot |h|}{h} = \lim_{h \to 0} \pi \cdot \frac{h^2}{h} = \lim_{h \to 0} \pi h = 0$.
Thus,$f(x)$ is differentiable at $x = 0$.
Since the function is differentiable everywhere,the set $S$ is an empty set $\phi$.

(B) The function $f(x) = |x-1| e^{x}$ is a product of two functions: $g(x) = |x-1|$ and $h(x) = e^{x}$.
We know that $e^{x}$ is differentiable for all $x \in R$.
The function $g(x) = |x-1|$ is continuous everywhere but is not differentiable at $x = 1$ because the left-hand derivative and right-hand derivative at $x = 1$ are not equal.
Specifically,at $x = 1$,the left-hand derivative is $-e^{1} = -e$ and the right-hand derivative is $e^{1} = e$.
Since the product of a non-differentiable function and a non-zero differentiable function at a point is non-differentiable at that point,$f(x)$ is not differentiable at $x = 1$.
Therefore,the set of points where $f(x)$ is differentiable is $R - \{1\}$.

(B) Given,$f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$.
On differentiating with respect to $x$,we get:
$f^{\prime}(x) = \frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}}} \times \frac{d}{d x}\left(\frac{2 x}{1+x^{2}}\right)$
$f^{\prime}(x) = \frac{1+x^{2}}{\sqrt{(1-x^{2})^{2}}} \times \frac{2(1-x^{2})}{(1+x^{2})^{2}}$
$f^{\prime}(x) = \frac{2}{1+x^{2}} \times \frac{1-x^{2}}{|1-x^{2}|}$
This simplifies to:
$f^{\prime}(x) = \begin{cases} \frac{2}{1+x^{2}}, & \text{if } |x| < 1 \\ -\frac{2}{1+x^{2}}, & \text{if } |x| > 1 \end{cases}$
At $x = 1$ and $x = -1$,the derivative does not exist because the left-hand derivative and right-hand derivative are not equal.
Therefore,$f(x)$ is differentiable on $R - \{-1, 1\}$.

(C) $\text{Given } f(x) = \begin{cases} \frac{1}{2}(-x-1), & x < -1 \\ \tan^{-1} x, & -1 \le x \le 1 \\ \frac{1}{2}(x-1), & x > 1 \end{cases}$
$\text{The derivative } f'(x) \text{ is defined as:}$
$f'(x) = \begin{cases} -\frac{1}{2}, & x < -1 \\ \frac{1}{1+x^2}, & -1 < x < 1 \\ \frac{1}{2}, & x > 1 \end{cases}$
$\text{At } x = -1: \text{ Left-hand derivative } = -\frac{1}{2}, \text{ Right-hand derivative } = \frac{1}{1+(-1)^2} = \frac{1}{2}. \text{ Since } -\frac{1}{2} \neq \frac{1}{2}, f(x) \text{ is not differentiable at } x = -1.$
$\text{At } x = 1: \text{ Left-hand derivative } = \frac{1}{1+1^2} = \frac{1}{2}, \text{ Right-hand derivative } = \frac{1}{2}. \text{ However, check continuity: } f(1) = \tan^{-1}(1) = \frac{\pi}{4}, \text{ but } \lim_{x \to 1^+} f(x) = \frac{1}{2}(1-1) = 0. \text{ Since } f(x) \text{ is discontinuous at } x = 1, \text{ it is not differentiable at } x = 1.$
$\text{Thus, the domain of } f'(x) \text{ is } R - \{-1, 1\}.$

(D) Given,$f(x)=|x-3|$.
We can redefine the function as:
$f(x) = \begin{cases} 3-x, & x < 3 \\ 0, & x=3 \\ x-3, & x > 3 \end{cases}$
To check the differentiability at $x=3$,we find the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$:
$LHD = \lim_{h \to 0^-} \frac{f(3+h)-f(3)}{h} = \lim_{h \to 0^-} \frac{|3+h-3|-0}{h} = \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$.
$RHD = \lim_{h \to 0^+} \frac{f(3+h)-f(3)}{h} = \lim_{h \to 0^+} \frac{|3+h-3|-0}{h} = \lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$.
Since $LHD \neq RHD$,the derivative $f^{\prime}(3)$ does not exist.

(A) The function is defined as $f(x) = \frac{x}{1+|x|}$.
We can express this piecewise as:
$f(x) = \begin{cases} \frac{x}{1-x}, & x < 0 \\ \frac{x}{1+x}, & x \geq 0 \end{cases}$
Now,we check the differentiability at $x = 0$:
Left-hand derivative $(LHD)$ at $x = 0$:
$LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{\frac{h}{1-h} - 0}{h} = \lim_{h \to 0^-} \frac{1}{1-h} = 1$
Right-hand derivative $(RHD)$ at $x = 0$:
$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\frac{h}{1+h} - 0}{h} = \lim_{h \to 0^+} \frac{1}{1+h} = 1$
Since $LHD = RHD = 1$,the function is differentiable at $x = 0$.
For $x \neq 0$,the function is a rational function with a non-zero denominator,so it is differentiable everywhere.
Thus,the derivative exists for all $x \in (-\infty, \infty)$.

(C) We have $f(x) = [x]$.
By the definition of the right-hand derivative at $x = 1$:
$f^{\prime}(1^{+}) = \lim_{h \to 0^{+}} \frac{f(1+h) - f(1)}{h}$.
Since $h$ is a small positive value,$1+h$ is slightly greater than $1$,so $[1+h] = 1$.
Also,$[1] = 1$.
Substituting these values:
$f^{\prime}(1^{+}) = \lim_{h \to 0^{+}} \frac{1 - 1}{h} = \lim_{h \to 0^{+}} \frac{0}{h} = 0$.

(A) Given $f(x) = [x] \sin(\pi x)$.
For $x = k$,where $k \in \mathbb{Z}$,the left-hand derivative $(LHD)$ is defined as:
$LHD = \lim_{h \to 0^+} \frac{f(k-h) - f(k)}{-h}$
Since $k$ is an integer,$[k-h] = k-1$ for small $h > 0$.
Also,$f(k) = [k] \sin(k\pi) = k \cdot 0 = 0$.
Substituting these into the limit:
$LHD = \lim_{h \to 0^+} \frac{(k-1) \sin(\pi(k-h)) - 0}{-h}$
Using the identity $\sin(k\pi - \pi h) = \sin(k\pi)\cos(\pi h) - \cos(k\pi)\sin(\pi h)$:
Since $\sin(k\pi) = 0$ and $\cos(k\pi) = (-1)^k$,we have $\sin(k\pi - \pi h) = -(-1)^k \sin(\pi h) = (-1)^{k+1} \sin(\pi h)$.
$LHD = \lim_{h \to 0^+} \frac{(k-1) (-1)^{k+1} \sin(\pi h)}{-h}$
Using $\lim_{h \to 0} \frac{\sin(\pi h)}{h} = \pi$:
$LHD = (k-1) (-1)^{k+1} (-\pi) = (k-1) (-1)^{k+1+1} \pi = (-1)^k (k-1) \pi$.

(C) The graph represents the function $f(x) = |x-1|$.
At $x=1$,the graph is continuous because there is no break in the curve.
However,at $x=1$,there is a sharp corner (or cusp),which means the left-hand derivative and right-hand derivative are not equal.
Specifically,the left-hand derivative is $-1$ and the right-hand derivative is $1$.
Therefore,the function is continuous but not differentiable at $x=1$.

(B) The given function is $f(x) = |\cos x|$.
We know that the function $g(x) = \cos x$ is continuous and differentiable for all $x \in R$.
The modulus function $h(x) = |x|$ is continuous everywhere but not differentiable at $x = 0$.
Therefore,the composite function $f(x) = |\cos x|$ is continuous for all $x \in R$.
However,$f(x)$ will not be differentiable where the expression inside the modulus is zero,i.e.,at $\cos x = 0$.
This occurs at $x = (2n + 1) \frac{\pi}{2}$ for all $n \in Z$.
At these points,the graph of $f(x)$ has sharp corners (cusps),as shown in the graph.
Thus,$f(x)$ is everywhere continuous but not differentiable at odd multiples of $\frac{\pi}{2}$.

(A) Given,$f(x) = \begin{cases} 2a - x & \text{when } -a < x < a \\ 3x - 2a & \text{when } a \leq x \end{cases}$.
First,check for continuity at $x = a$:
Left-hand limit: $\lim_{x \to a^-} f(x) = \lim_{x \to a^-} (2a - x) = 2a - a = a$.
Right-hand limit: $\lim_{x \to a^+} f(x) = \lim_{x \to a^+} (3x - 2a) = 3a - 2a = a$.
Value of function: $f(a) = 3(a) - 2a = a$.
Since $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$,the function is continuous at $x = a$.
Now,check for differentiability at $x = a$:
Left-hand derivative: $Lf'(a) = \frac{d}{dx}(2a - x) = -1$.
Right-hand derivative: $Rf'(a) = \frac{d}{dx}(3x - 2a) = 3$.
Since $Lf'(a) \neq Rf'(a)$,the function $f(x)$ is not differentiable at $x = a$.

(B) Given,$f(x) = \begin{cases} 2a-x & \text{in } -a < x < a \\ 3x-2a & \text{in } a \leq x \end{cases}$
Check continuity at $x=a$:
$LHL = \lim_{x \to a^-} f(x) = \lim_{x \to a} (2a-x) = 2a-a = a$
$RHL = \lim_{x \to a^+} f(x) = \lim_{x \to a} (3x-2a) = 3a-2a = a$
$f(a) = 3(a)-2a = a$
Since $LHL = RHL = f(a)$,the function is continuous at $x=a$.
Check differentiability at $x=a$:
$LHD = \lim_{h \to 0} \frac{f(a-h)-f(a)}{-h} = \lim_{h \to 0} \frac{2a-(a-h)-a}{-h} = \lim_{h \to 0} \frac{h}{-h} = -1$
$RHD = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{3(a+h)-2a-a}{h} = \lim_{h \to 0} \frac{3h}{h} = 3$
Since $LHD \neq RHD$,the function is not differentiable at $x=a$.
Thus,option $B$ is correct.

(C) For $f(x)$ to be differentiable at $x = 0$,it must first be continuous at $x = 0$.
Left-hand limit $(LHL)$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (e^x + ax) = e^0 + a(0) = 1$.
Right-hand limit $(RHL)$: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} b(x - 1)^2 = b(0 - 1)^2 = b$.
Since $f(x)$ is continuous,$LHL = RHL$,so $b = 1$.
Now,for differentiability,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$ at $x = 0$.
$LHD = \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} (e^x + a) = e^0 + a = 1 + a$.
$RHD = \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} 2b(x - 1) = 2b(0 - 1) = -2b$.
Equating $LHD$ and $RHD$: $1 + a = -2b$.
Substituting $b = 1$: $1 + a = -2(1) \Rightarrow 1 + a = -2 \Rightarrow a = -3$.
Thus,$a = -3$ and $b = 1$.

(B) Given $f(x)=\begin{cases} x^{3}-1, & 1 < x < \infty \\ x-1, & -\infty < x \leq 1 \end{cases}$
First,we check the continuity at $x=1$.
$RHL = \lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1^{+}} (x^{3}-1) = 1^{3}-1 = 0$
$LHL = \lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (x-1) = 1-1 = 0$
$f(1) = 1-1 = 0$
Since $LHL = RHL = f(1)$,the function is continuous at $x=1$.
Next,we check the differentiability at $x=1$.
$f'(x) = \begin{cases} 3x^{2}, & 1 < x < \infty \\ 1, & -\infty < x < 1 \end{cases}$
$LHD = \lim_{x \rightarrow 1^{-}} f'(x) = 1$
$RHD = \lim_{x \rightarrow 1^{+}} f'(x) = 3(1)^{2} = 3$
Since $LHD \neq RHD$,the function is non-differentiable at $x=1$.
Therefore,the function is continuous and non-differentiable.

(B) We need to evaluate the limit: $\lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x}$.
Given $f(0) = 0$,the expression becomes $\lim_{x \rightarrow 0} \frac{x \left(1 + \frac{1}{2} \sin (\log x^2) \right) - 0}{x}$.
Simplifying the expression by canceling $x$ (since $x \neq 0$ as $x \rightarrow 0$):
$= \lim_{x \rightarrow 0} \left(1 + \frac{1}{2} \sin (\log x^2) \right)$.
As $x \rightarrow 0$,$x^2 \rightarrow 0^+$,so $\log x^2 \rightarrow -\infty$.
The function $\sin(\log x^2)$ oscillates between $-1$ and $1$ as $x \rightarrow 0$.
Therefore,the limit $\lim_{x \rightarrow 0} \sin(\log x^2)$ does not exist.
Consequently,$\lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x}$ does not exist.

(C) The function $f(x) = |x-3| + |x+5|$ is differentiable everywhere except at the points where the expressions inside the absolute values are zero,which are $x = 3$ and $x = -5$.
Thus,the set $A$ is $\mathbb{R} \setminus \{-5, 3\}$.
We are looking for real numbers $x$ such that $x \in (-\infty, -3) \cup (5, \infty)$ and $x \notin A$.
Since $A = \mathbb{R} \setminus \{-5, 3\}$,the condition $x \notin A$ implies $x \in \{-5, 3\}$.
Checking the intervals:
For $x = -5$,$-5$ is not in $(-\infty, -3)$ because the interval is open at $-3$ and $-5 < -3$,but wait,$-5$ is in $(-\infty, -3)$.
For $x = 3$,$3$ is not in $(-\infty, -3) \cup (5, \infty)$ because $3$ lies between $-3$ and $5$.
Thus,the only value in the set $\{-5, 3\}$ that lies in $(-\infty, -3) \cup (5, \infty)$ is $-5$.
Therefore,the number of such real numbers is $1$.

(A) The function $f(x)$ is defined as:
$f(x) = \begin{cases} \frac{1}{2}(-x-1), & \text{if } x < -1 \\ \tan^{-1} x, & \text{if } -1 \leq x \leq 1 \\ \frac{1}{2}(x-1), & \text{if } x > 1 \end{cases}$
Check continuity at $x = -1$:
$\lim_{x \to -1^-} f(x) = \frac{1}{2}(1-1) = 0$
$f(-1) = \tan^{-1}(-1) = -\frac{\pi}{4}$
Since $\lim_{x \to -1^-} f(x) \neq f(-1)$,$f(x)$ is discontinuous at $x = -1$.
Check continuity at $x = 1$:
$\lim_{x \to 1^+} f(x) = \frac{1}{2}(1-1) = 0$
$f(1) = \tan^{-1}(1) = \frac{\pi}{4}$
Since $\lim_{x \to 1^+} f(x) \neq f(1)$,$f(x)$ is discontinuous at $x = 1$.
$A$ function must be continuous to be differentiable. Since $f(x)$ is discontinuous at $x = -1$ and $x = 1$,it is not differentiable at these points.
Thus,the domain of $f'(x)$ is $R - \{-1, 1\}$.

(B) Given,$f: R \rightarrow R$ is a continuous function such that $|f(x)-f(y)| \leq 10|x-y|^{201}$.
Dividing both sides by $|x-y|$ (where $x \neq y$):
$\frac{|f(x)-f(y)|}{|x-y|} \leq 10|x-y|^{200}$.
Taking the limit as $y \rightarrow x$:
$\lim_{y \rightarrow x} \left| \frac{f(x)-f(y)}{x-y} \right| \leq \lim_{y \rightarrow x} 10|x-y|^{200}$.
$|f'(x)| \leq 0$.
Since the absolute value cannot be negative,we must have $|f'(x)| = 0$,which implies $f'(x) = 0$ for all $x \in R$.
Therefore,$f(x) = C$ (a constant function).
Since $f(x)$ is a constant function,$f(2019) = f(2020) = f(2021) = f(2022) = C$.
Checking the options:
$f(2019) + f(2022) = C + C = 2C$.
$2f(2021) = 2C$.
Thus,$f(2019) + f(2022) = 2f(2021)$ is true.
Hence,option $B$ is correct.

(C) Given,$f(x) = \begin{cases} x^{\alpha} \sin \left( \frac{1}{x} \right) ; & x \neq 0 \\ 0 ; & x = 0 \end{cases}$
For continuity at $x = 0$,$\lim_{x \rightarrow 0} f(x) = f(0) = 0$.
$\lim_{x \rightarrow 0} x^{\alpha} \sin \left( \frac{1}{x} \right) = 0$ only if $\alpha > 0$.
Thus,$f(x)$ is continuous for $\alpha > 0$.
For differentiability at $x = 0$,$f^{\prime}(0) = \lim_{h \rightarrow 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \rightarrow 0} \frac{h^{\alpha} \sin \left( \frac{1}{h} \right) - 0}{h} = \lim_{h \rightarrow 0} h^{\alpha - 1} \sin \left( \frac{1}{h} \right)$.
This limit exists and is equal to $0$ if $\alpha - 1 > 0$,i.e.,$\alpha > 1$.
Therefore,$f(x)$ is continuous and differentiable for $\alpha > 1$.

(A) Since $f(x)$ is differentiable $\forall x \in R$,it must be continuous $\forall x \in R$.
For continuity at $x = 2$:
$\lim_{x \rightarrow 2^+} f(x) = \lim_{x \rightarrow 2^-} f(x)$
$\Rightarrow \lim_{x \rightarrow 2} (bx^3 + 1) = \lim_{x \rightarrow 2} (3x - 3)$
$\Rightarrow 8b + 1 = 3(2) - 3 = 3$
$\Rightarrow 8b = 2 \Rightarrow b = \frac{1}{4}$.
For continuity at $x = 1$:
$\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^+} f(x)$
$\Rightarrow \lim_{x \rightarrow 1} (ax^2 + bx - \frac{13}{8}) = \lim_{x \rightarrow 1} (3x - 3)$
$\Rightarrow a + b - \frac{13}{8} = 3(1) - 3 = 0$
$\Rightarrow a + \frac{1}{4} - \frac{13}{8} = 0$
$\Rightarrow a = \frac{13}{8} - \frac{2}{8} = \frac{11}{8}$.
Therefore,$a - b = \frac{11}{8} - \frac{1}{4} = \frac{11}{8} - \frac{2}{8} = \frac{9}{8}$.

(D) For $x > 1$,$|x| = x$ and $|x - 1| = x - 1$.
Then $f(x) = (x - (x - 1))^2 = (1)^2 = 1$.
For $0 < x < 1$,$|x| = x$ and $|x - 1| = -(x - 1) = 1 - x$.
Then $f(x) = (x - (1 - x))^2 = (2x - 1)^2 = 4x^2 - 4x + 1$.
The left-hand derivative at $x = 1$ is $LHD = \lim_{h \rightarrow 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \rightarrow 0^-} \frac{(2(1+h) - 1)^2 - 1}{h} = \lim_{h \rightarrow 0^-} \frac{(2h + 1)^2 - 1}{h} = \lim_{h \rightarrow 0^-} \frac{4h^2 + 4h}{h} = 4$.
The right-hand derivative at $x = 1$ is $RHD = \lim_{h \rightarrow 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \rightarrow 0^+} \frac{1 - 1}{h} = 0$.
Since $LHD \neq RHD$,the derivative at $x = 1$ does not exist.
Thus,Assertion $(A)$ is false.
The Reason $(R)$ is the standard definition of the derivative,which is true.
Therefore,$(A)$ is false and $(R)$ is true.