Differentiability Questions in English

(C) The function $f(x)$ can be rewritten by expanding the absolute value as:
$f(x) = \begin{cases} \frac{1}{2}(-x - 1), & x < -1 \\ \tan^{-1}x, & -1 \le x \le 1 \\ \frac{1}{2}(x - 1), & x > 1 \end{cases}$
To find the domain of the derivative $f'(x)$,we check the differentiability at the transition points $x = -1$ and $x = 1$.
For $x < -1$,$f'(x) = -\frac{1}{2}$.
For $-1 < x < 1$,$f'(x) = \frac{1}{1+x^2}$.
For $x > 1$,$f'(x) = \frac{1}{2}$.
At $x = -1$:
Left-hand derivative $f'(-1^-) = -\frac{1}{2}$.
Right-hand derivative $f'(-1^+) = \frac{1}{1+(-1)^2} = \frac{1}{2}$.
Since $f'(-1^-) \neq f'(-1^+)$,the function is not differentiable at $x = -1$.
At $x = 1$:
Left-hand derivative $f'(1^-) = \frac{1}{1+(1)^2} = \frac{1}{2}$.
Right-hand derivative $f'(1^+) = \frac{1}{2}$.
Since $f'(1^-) = f'(1^+)$,the function is differentiable at $x = 1$.
Thus,the derivative $f'(x)$ exists for all $x \in R$ except $x = -1$.
Therefore,the domain of $f'(x)$ is $R - \{-1\}$.

(D) The correct statement is $(d)$.
By the fundamental theorem of calculus,if a function $f(x)$ is differentiable at a point $x = c$,then it must be continuous at that point $x = c$.
However,the converse is not necessarily true; a continuous function may not be differentiable (e.g.,$f(x) = |x|$ at $x = 0$).
Therefore,$(d)$ is the only universally true statement among the given options.

(D) To find $f'(2)$,we must check the left-hand derivative $(Lf'(2))$ and the right-hand derivative $(Rf'(2))$ at $x = 2$.
$1$. Left-hand derivative $(Lf'(2))$:
$Lf'(2) = \lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h}$
Since $x < 2$,$f(x) = x + 1$. Thus,$f(2+h) = (2+h) + 1 = 3+h$ and $f(2) = 2(2) - 1 = 3$.
$Lf'(2) = \lim_{h \to 0^-} \frac{(3+h) - 3}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1$.
$2$. Right-hand derivative $(Rf'(2))$:
$Rf'(2) = \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h}$
Since $x \ge 2$,$f(x) = 2x - 1$. Thus,$f(2+h) = 2(2+h) - 1 = 3+2h$ and $f(2) = 3$.
$Rf'(2) = \lim_{h \to 0^+} \frac{(3+2h) - 3}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2$.
Since $Lf'(2) \neq Rf'(2)$,the derivative $f'(2)$ does not exist.

(D) $\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} f(3 - h) = \mathop {\lim }\limits_{h \to 0} |3 - h - 3| = 0$
$\mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} f(3 + h) = \mathop {\lim }\limits_{h \to 0} |3 + h - 3| = 0$
$\because \mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ + }} f(x) = f(3)$
Hence,$f$ is continuous at $x = 3$.
Now,$L f'(3) = \mathop {\lim }\limits_{h \to 0} \frac{f(3 - h) - f(3)}{-h}$
$= \mathop {\lim }\limits_{h \to 0} \frac{|3 - h - 3| - 0}{-h} = \mathop {\lim }\limits_{h \to 0} \frac{h}{-h} = -1$
$R f'(3) = \mathop {\lim }\limits_{h \to 0} \frac{f(3 + h) - f(3)}{h}$
$= \mathop {\lim }\limits_{h \to 0} \frac{|3 + h - 3| - 0}{h} = \mathop {\lim }\limits_{h \to 0} \frac{h}{h} = 1$
$\because L f'(3) \neq R f'(3)$
Hence,$f$ is not differentiable at $x = 3$.
Trick: By observing the graph,the function is continuous,but the tangent is not defined at $x = 3$ due to the sharp corner.

(B) We check the continuity at $x = 1$:
$f(1) = 1 - 1 = 0$
$LHL = \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} ((1 - h) - 1) = 0$
$RHL = \lim_{x \to 1^+} f(x) = \lim_{h \to 0} f(1 + h) = \lim_{h \to 0} ((1 + h)^3 - 1) = 0$
Since $LHL = RHL = f(1)$,the function is continuous at $x = 1$.
Now,we check the differentiability at $x = 1$:
$Lf'(1) = \lim_{h \to 0} \frac{f(1 - h) - f(1)}{-h} = \lim_{h \to 0} \frac{(1 - h - 1) - 0}{-h} = \lim_{h \to 0} \frac{-h}{-h} = 1$
$Rf'(1) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0} \frac{((1 + h)^3 - 1) - 0}{h} = \lim_{h \to 0} \frac{1 + 3h + 3h^2 + h^3 - 1}{h} = \lim_{h \to 0} (3 + 3h + h^2) = 3$
Since $Lf'(1) \neq Rf'(1)$,the function is not differentiable at $x = 1$.
Thus,the function is continuous and not differentiable at $x = 1$.

(B) Given that $f(x)$ is differentiable at $x = 0$.
Since every differentiable function is continuous,$f(x)$ must be continuous at $x = 0$.
Thus,$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
$\lim_{x \to 0^-} (e^x + ax) = e^0 + a(0) = 1$.
$\lim_{x \to 0^+} b(x - 1)^2 = b(0 - 1)^2 = b$.
Equating these,we get $b = 1$.
Since $f(x)$ is differentiable at $x = 0$,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$ at $x = 0$.
$LHD = \frac{d}{dx}(e^x + ax) = e^x + a$. At $x = 0$,$LHD = e^0 + a = 1 + a$.
$RHD = \frac{d}{dx}(b(x - 1)^2) = 2b(x - 1)$. At $x = 0$,$RHD = 2b(0 - 1) = -2b$.
Setting $LHD = RHD$,we have $1 + a = -2b$.
Substituting $b = 1$,we get $1 + a = -2(1) \implies 1 + a = -2 \implies a = -3$.
Therefore,$(a, b) = (-3, 1)$.

(D) The function $f(x) = |\sin x|$ is continuous for all real numbers $x$.
However,it is not differentiable at points where the graph has sharp corners (cusps).
From the graph of $f(x) = |\sin x|$,we can observe that the function has sharp corners at $x = 0, \pm \pi, \pm 2\pi, \dots$.
These points are of the form $x = k\pi$,where $k$ is any integer.
Therefore,the function is not differentiable at $x = k\pi$ for any integer $k$.

(C) We have,$f(x) = \begin{cases} e^x, & x < 0 \\ e^{-x}, & x \ge 0 \end{cases}$
Clearly,$f(x)$ is continuous and differentiable for all $x \neq 0$.
Now,$\lim_{x \to 0^-} f(x) = \lim_{x \to 0} e^x = 1$ and $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{-x} = 1$.
Also,$f(0) = e^0 = 1$. Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$,the function is continuous at $x = 0$.
Now,check for differentiability at $x = 0$:
Left-hand derivative $(LHD)$ at $x = 0$ is $\frac{d}{dx}(e^x) |_{x=0} = e^0 = 1$.
Right-hand derivative $(RHD)$ at $x = 0$ is $\frac{d}{dx}(e^{-x}) |_{x=0} = -e^0 = -1$.
Since $LHD \neq RHD$ at $x = 0$,the function is not differentiable at $x = 0$.
Thus,$f(x) = e^{-|x|}$ is continuous but not differentiable at $x = 0$.

(A) The left-hand derivative at $x = k$ is defined as $f'(k^-) = \lim_{h \to 0^+} \frac{f(k - h) - f(k)}{-h}$.
Given $f(x) = [x]\sin(\pi x)$,we have $f(k) = [k]\sin(\pi k) = k \times 0 = 0$.
For small $h > 0$,$[k - h] = k - 1$.
Thus,$f(k - h) = (k - 1)\sin(\pi(k - h)) = (k - 1)\sin(\pi k - \pi h) = (k - 1)(\sin(\pi k)\cos(\pi h) - \cos(\pi k)\sin(\pi h))$.
Since $\sin(\pi k) = 0$ and $\cos(\pi k) = (-1)^k$,we get $f(k - h) = (k - 1)(0 - (-1)^k \sin(\pi h)) = -(k - 1)(-1)^k \sin(\pi h) = (k - 1)(-1)^{k+1} \sin(\pi h)$.
Substituting this into the limit: $f'(k^-) = \lim_{h \to 0^+} \frac{(k - 1)(-1)^{k+1} \sin(\pi h) - 0}{-h}$.
Using $\lim_{h \to 0} \frac{\sin(\pi h)}{\pi h} = 1$,we get $f'(k^-) = (k - 1)(-1)^{k+1} \times (-\pi) = (k - 1)(-1)^{k+1} (-1) \pi = (k - 1)(-1)^{k+2} \pi = (k - 1)(-1)^k \pi$.

(D) To check if $f'(2)$ exists,we calculate the right-hand derivative $(Rf'(2))$ and the left-hand derivative $(Lf'(2))$ at $x = 2$.
First,find $f(2)$ using the definition $f(x) = 2x - 1$ for $x \ge 2$:
$f(2) = 2(2) - 1 = 3$.
Right-hand derivative $(Rf'(2))$:
$Rf'(2) = \lim_{h \to 0^+} \frac{f(2 + h) - f(2)}{h} = \lim_{h \to 0^+} \frac{2(2 + h) - 1 - 3}{h} = \lim_{h \to 0^+} \frac{4 + 2h - 1 - 3}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2$.
Left-hand derivative $(Lf'(2))$:
$Lf'(2) = \lim_{h \to 0^+} \frac{f(2 - h) - f(2)}{-h} = \lim_{h \to 0^+} \frac{(2 - h) + 1 - 3}{-h} = \lim_{h \to 0^+} \frac{3 - h - 3}{-h} = \lim_{h \to 0^+} \frac{-h}{-h} = 1$.
Since $Rf'(2) \neq Lf'(2)$,the derivative $f'(2)$ does not exist.

(B) polynomial function is continuous everywhere,so option $A$ is true.
Every differentiable function is continuous,so option $C$ is true.
The exponential function $e^x$ is continuous for all $x \in \mathbb{R}$,so option $D$ is true.
However,a continuous function is not necessarily differentiable (e.g.,$f(x) = |x|$ is continuous at $x = 0$ but not differentiable at $x = 0$).
Therefore,statement $B$ is not true.

(D) To check continuity at $x = 0$:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin \frac{1}{x}$.
Since $-1 \le \sin \frac{1}{x} \le 1$,we have $-x^2 \le x^2 \sin \frac{1}{x} \le x^2$.
By the squeeze theorem,$\lim_{x \to 0} f(x) = 0 = f(0)$. Thus,$f(x)$ is continuous at $x = 0$.
To check differentiability at $x = 0$:
$f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin \frac{1}{h}$.
Since $\lim_{h \to 0} h = 0$ and $\sin(1/h)$ is bounded,the limit is $0$.
Thus,$f'(0) = 0$,so the function is differentiable at $x = 0$.

(B) Given $f(x) = \frac{x - 1}{2x^2 - 7x + 5}$ for $x \neq 1$.
We can simplify the expression for $x \neq 1$ by factoring the denominator:
$2x^2 - 7x + 5 = 2x^2 - 2x - 5x + 5 = 2x(x - 1) - 5(x - 1) = (2x - 5)(x - 1)$.
So,for $x \neq 1$,$f(x) = \frac{x - 1}{(2x - 5)(x - 1)} = \frac{1}{2x - 5}$.
By the definition of the derivative at $x = 1$:
$f'(1) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h}$.
Substituting $f(1 + h) = \frac{1}{2(1 + h) - 5} = \frac{1}{2h - 3}$ and $f(1) = -\frac{1}{3}$:
$f'(1) = \lim_{h \to 0} \frac{\frac{1}{2h - 3} - (-\frac{1}{3})}{h} = \lim_{h \to 0} \frac{\frac{1}{2h - 3} + \frac{1}{3}}{h}$.
$f'(1) = \lim_{h \to 0} \frac{3 + (2h - 3)}{3h(2h - 3)} = \lim_{h \to 0} \frac{2h}{3h(2h - 3)}$.
$f'(1) = \lim_{h \to 0} \frac{2}{3(2h - 3)} = \frac{2}{3(0 - 3)} = \frac{2}{-9} = -\frac{2}{9}$.

(D) For $f(x)$ to be differentiable at $x = 1$,it must first be continuous at $x = 1$.
For continuity,$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.
$\lim_{x \to 1^-} (mx^2) = m(1)^2 = m$.
$\lim_{x \to 1^+} (2x) = 2(1) = 2$.
Thus,for continuity,$m = 2$.
Now,check for differentiability at $x = 1$ with $m = 2$:
Left-hand derivative $L f'(1) = \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} = \lim_{h \to 0} \frac{2(1-h)^2 - 2}{-h} = \lim_{h \to 0} \frac{2(1 + h^2 - 2h) - 2}{-h} = \lim_{h \to 0} \frac{2h^2 - 4h}{-h} = \lim_{h \to 0} (4 - 2h) = 4$.
Right-hand derivative $R f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{2(1+h) - 2}{h} = \lim_{h \to 0} \frac{2h}{h} = 2$.
Since $L f'(1) \neq R f'(1)$ $(4 \neq 2)$,the function is not differentiable at $x = 1$ for any value of $m$.

(D) To find $(g \circ f)'(0)$,we must check the left-hand derivative and right-hand derivative at $x = 0$.
For the right-hand derivative $(g \circ f)'(0^+)$:
$(g \circ f)(x) = g(f(x)) = g(\sin x) = e^{\sin x}$ for $x \ge 0$.
$(g \circ f)'(x) = e^{\sin x} \cdot \cos x$.
At $x = 0$,$(g \circ f)'(0^+) = e^{\sin 0} \cdot \cos 0 = e^0 \cdot 1 = 1$.
For the left-hand derivative $(g \circ f)'(0^-)$:
$(g \circ f)(x) = g(f(x)) = g(1 - \cos x) = e^{1 - \cos x}$ for $x \le 0$.
$(g \circ f)'(x) = e^{1 - \cos x} \cdot \sin x$.
At $x = 0$,$(g \circ f)'(0^-) = e^{1 - \cos 0} \cdot \sin 0 = e^0 \cdot 0 = 0$.
Since the left-hand derivative $(0)$ is not equal to the right-hand derivative $(1)$,the derivative $(g \circ f)'(0)$ does not exist.
Therefore,the correct option is $D$.

(D) Given the condition $|f(x) - f(y)| \le (x - y)^2$ for all $x, y \in R$.
Dividing both sides by $|x - y|$ (where $x \neq y$),we get:
$\left| \frac{f(x) - f(y)}{x - y} \right| \le |x - y|$
Taking the limit as $x \to y$ on both sides:
$\lim_{x \to y} \left| \frac{f(x) - f(y)}{x - y} \right| \le \lim_{x \to y} |x - y|$
This implies $|f'(y)| \le 0$.
Since the absolute value cannot be negative,we must have $|f'(y)| = 0$,which means $f'(y) = 0$ for all $y \in R$.
If the derivative of a function is zero everywhere,the function must be a constant.
Thus,$f(x) = C$ for some constant $C$.
Given $f(0) = 0$,we have $C = 0$.
Therefore,$f(x) = 0$ for all $x \in R$.
Hence,$f(1) = 0$.

(B) The function is defined as $f(x) = ||x| - 1|$.
We can break this down based on the sign of the inner expression:
$f(x) = \begin{cases} |x| - 1, & |x| - 1 \ge 0 \\ -(|x| - 1), & |x| - 1 < 0 \end{cases}$
$f(x) = \begin{cases} |x| - 1, & |x| \ge 1 \\ 1 - |x|, & |x| < 1 \end{cases}$
This can be further expanded for different intervals of $x$:
$f(x) = \begin{cases} -x - 1, & x \le -1 \\ x + 1, & -1 < x < 0 \\ 1 - x, & 0 \le x < 1 \\ x - 1, & x \ge 1 \end{cases}$
$A$ function is not differentiable at points where there are sharp corners (cusps) in its graph.
By observing the graph,the function has sharp corners at $x = -1$,$x = 0$,and $x = 1$.
Therefore,$f(x)$ is not differentiable at $x \in \{-1, 0, 1\}$.

(B) Given that $f\left( \frac{1}{n} \right) = 0$ for all $n \in \{1, 2, 3, \dots\}$.
Since $f(x)$ is continuous (as it is differentiable),we have $f(0) = \lim_{n \to \infty} f\left( \frac{1}{n} \right) = 0$.
Now,consider the sequence $x_n = \frac{1}{n}$. We have $f(x_n) = 0$ and $f(0) = 0$.
By the definition of the derivative at $x = 0$:
$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{n \to \infty} \frac{f(1/n) - f(0)}{1/n - 0} = \lim_{n \to \infty} \frac{0 - 0}{1/n} = 0$.
Thus,$f(0) = 0$ and $f'(0) = 0$.

(D) To check the differentiability of $f(x)$ at $x = 0$,we calculate the Left-Hand Derivative $(LHD)$ and the Right-Hand Derivative $(RHD)$.
For $x < 0$,$f(x) = 1$. Therefore,the $LHD$ at $x = 0$ is:
$LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{1 - 1}{h} = 0$.
For $x \ge 0$,$f(x) = 1 + \sin x$. Therefore,the $RHD$ at $x = 0$ is:
$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{(1 + \sin h) - 1}{h} = \lim_{h \to 0^+} \frac{\sin h}{h} = 1$.
Since $LHD \neq RHD$ (i.e.,$0 \neq 1$),the derivative $f'(0)$ does not exist.

(D) The function $f(x) = (x^2 - 1)|(x - 1)(x - 2)| + \cos(|x|)$.
First,consider the term $\cos(|x|)$. The function $|x|$ is not differentiable at $x = 0$,so $\cos(|x|)$ is not differentiable at $x = 0$.
Next,consider the term $(x^2 - 1)|(x - 1)(x - 2)|$. The expression $|(x - 1)(x - 2)|$ is not differentiable at the roots of $(x - 1)(x - 2) = 0$,which are $x = 1$ and $x = 2$.
At $x = 1$,$f(x) = (x^2 - 1)|(x - 1)(x - 2)| + \cos(|x|)$. Since $(x^2 - 1)$ has a factor $(x - 1)$,the product $(x^2 - 1)|(x - 1)(x - 2)|$ becomes differentiable at $x = 1$ because the zero of the polynomial cancels the non-differentiability of the absolute value.
At $x = 2$,the term $(x^2 - 1)$ is $3 \neq 0$. Thus,the non-differentiability of $|(x - 1)(x - 2)|$ at $x = 2$ persists.
Checking $x = 2$:
$Lf'(2) = \lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h} = -3 - \sin(2)$
$Rf'(2) = \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h} = 3 - \sin(2)$
Since $Lf'(2) \neq Rf'(2)$,the function is not differentiable at $x = 2$.

(C) To determine which function is continuous for all real values of $x$ and differentiable at $x = 0$:
$1$. For $f(x) = |x|$,the function is continuous for all $x \in \mathbb{R}$,but it is not differentiable at $x = 0$ because the left-hand derivative is $-1$ and the right-hand derivative is $1$.
$2$. For $f(x) = \log x$,the function is not defined for $x \le 0$,so it is not continuous for all real values of $x$.
$3$. For $f(x) = \sin x$,the function is continuous for all $x \in \mathbb{R}$. Its derivative is $f'(x) = \cos x$. At $x = 0$,$f'(0) = \cos(0) = 1$,which is defined. Thus,it is differentiable at $x = 0$.
$4$. For $f(x) = x^{\frac{1}{2}}$,the function is not defined for $x < 0$,so it is not continuous for all real values of $x$.
Therefore,the correct option is $C$.

(C) The correct answer is $C$.
Statement $A$ is true: Differentiability implies continuity.
Statement $B$ is true: If $f'(x) = 0$ for all $x$ in an interval,then $f(x)$ is a constant function.
Statement $D$ is true: The derivative of a constant function is always $0$.
Statement $C$ is false: $A$ function can have a maximum or minimum at a point where it is not differentiable (e.g.,$f(x) = |x|$ at $x = 0$ has a minimum,but it is not differentiable at $x = 0$).

(D) Given $f(x) = \begin{cases} x + 2, & -1 < x < 3 \\ 5, & x = 3 \\ 8 - x, & x > 3 \end{cases}$ and $f(3) = 5$.
To check differentiability at $x = 3$,we calculate the Left Hand Derivative $(LHD)$ and Right Hand Derivative $(RHD)$.
$LHD = \lim_{x \to 3^-} \frac{f(x) - f(3)}{x - 3} = \lim_{h \to 0} \frac{f(3 - h) - f(3)}{-h}$
$= \lim_{h \to 0} \frac{(3 - h + 2) - 5}{-h} = \lim_{h \to 0} \frac{5 - h - 5}{-h} = \lim_{h \to 0} \frac{-h}{-h} = 1$.
$RHD = \lim_{x \to 3^+} \frac{f(x) - f(3)}{x - 3} = \lim_{h \to 0} \frac{f(3 + h) - f(3)}{h}$
$= \lim_{h \to 0} \frac{(8 - (3 + h)) - 5}{h} = \lim_{h \to 0} \frac{8 - 3 - h - 5}{h} = \lim_{h \to 0} \frac{-h}{h} = -1$.
Since $LHD \neq RHD$,the derivative $f'(3)$ does not exist.

(D) To find $f'(0)$,we must check the left-hand derivative $(Lf'(0))$ and the right-hand derivative $(Rf'(0))$.
$Rf'(0) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{(1 + \sin h) - 1}{h} = \lim_{h \to 0^+} \frac{\sin h}{h} = 1$.
$Lf'(0) = \lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{1 - 1}{-h} = 0$.
Since $Lf'(0) \neq Rf'(0)$,the derivative $f'(0)$ does not exist.

(C) For $f(x)$ to possess a derivative at $x = 0$,it must be continuous at $x = 0$.
For continuity at $x = 0$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$
$\lim_{x \to 0^-} (ax^2 + b) = b$
$\lim_{x \to 0^+} (x^2) = 0$
$f(0) = b$
Thus,$b = 0$.
Now,for differentiability at $x = 0$,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$:
$LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{a(h)^2 + b - b}{h} = \lim_{h \to 0^-} (ah) = 0$
$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{(h)^2 - 0}{h} = \lim_{h \to 0^+} (h) = 0$
Since $LHD = RHD = 0$ for any value of $a$,the function is differentiable at $x = 0$ for all $a \in R$ provided $b = 0$.

(A) Given the function $f(x) = \frac{x}{1 + |x|}$.
We analyze the differentiability of $f(x)$ at $x = 0$ and other points.
For $x > 0$,$f(x) = \frac{x}{1 + x}$,which is a rational function with a non-zero denominator,hence differentiable.
For $x < 0$,$f(x) = \frac{x}{1 - x}$,which is also a rational function with a non-zero denominator,hence differentiable.
Now,check differentiability at $x = 0$ using the definition of the derivative:
$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h - 0} = \lim_{h \to 0} \frac{\frac{h}{1 + |h|} - 0}{h} = \lim_{h \to 0} \frac{1}{1 + |h|}$.
As $h \to 0$,$|h| \to 0$,so the limit is $\frac{1}{1 + 0} = 1$.
Since the limit exists and is finite,the function $f(x)$ is differentiable at $x = 0$.
Thus,the function is differentiable for all $x \in ( - \infty, \infty )$.

(B) Let $f(x) = \sin^{-1}\left(\frac{2x}{1 + x^2}\right)$.
We know that $\frac{2x}{1 + x^2} = \sin(2\tan^{-1}x)$.
Thus,$y = \sin^{-1}(\sin(2\tan^{-1}x))$.
Case $1$: If $|x| \le 1$,then $-\frac{\pi}{2} \le 2\tan^{-1}x \le \frac{\pi}{2}$,so $y = 2\tan^{-1}x$. The derivative is $y' = \frac{2}{1 + x^2}$.
Case $2$: If $x > 1$,then $2\tan^{-1}x > \frac{\pi}{2}$,so $y = \pi - 2\tan^{-1}x$. The derivative is $y' = -\frac{2}{1 + x^2}$.
Case $3$: If $x < -1$,then $2\tan^{-1}x < -\frac{\pi}{2}$,so $y = -\pi - 2\tan^{-1}x$. The derivative is $y' = -\frac{2}{1 + x^2}$.
At $x = 1$,the left-hand derivative is $\frac{2}{1 + 1^2} = 1$ and the right-hand derivative is $-\frac{2}{1 + 1^2} = -1$.
At $x = -1$,the left-hand derivative is $-\frac{2}{1 + (-1)^2} = -1$ and the right-hand derivative is $\frac{2}{1 + (-1)^2} = 1$.
Since the left-hand and right-hand derivatives are not equal at $x = 1$ and $x = -1$,the function is not differentiable at these points.

(C) The function $f(x) = x(\sqrt{x} - \sqrt{x + 1})$ is defined for $x \ge 0$ because of the term $\sqrt{x}$.
For $x < 0$,the term $\sqrt{x}$ is not defined in the set of real numbers.
Since the function is not defined in the neighborhood $(-\delta, 0)$ for any $\delta > 0$,it cannot be continuous at $x = 0$.
By definition,a function must be continuous at a point to be differentiable at that point.
Therefore,$f(x)$ is not differentiable at $x = 0$.

(D) Given $f(x) = \text{sgn}(x^3)$.
We know that $\text{sgn}(x^3) = \text{sgn}(x)$ because the sign of $x^3$ is the same as the sign of $x$.
Thus,$f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases}$.
Since $\lim_{x \to 0^-} f(x) = -1$ and $\lim_{x \to 0^+} f(x) = 1$,the function is discontinuous at $x = 0$.
$A$ function that is not continuous at a point cannot be derivable at that point.
Therefore,$f$ is not derivable at $x = 0$.

(A) Given $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$.
Then $g(x) = x \cdot f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$.
To check the differentiability of $g(x)$ at $x = 0$:
$g'(0) = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin(1/h)$.
Since $\lim_{h \to 0} h = 0$ and $\sin(1/h)$ is bounded in $[-1, 1]$,by the squeeze theorem,$g'(0) = 0$.
Thus,$g(x)$ is differentiable at $x = 0$.
Now,for $x \ne 0$,$g'(x) = \frac{d}{dx} (x^2 \sin \frac{1}{x}) = 2x \sin \frac{1}{x} + x^2 \cos \frac{1}{x} \cdot (-\frac{1}{x^2}) = 2x \sin \frac{1}{x} - \cos \frac{1}{x}$.
As $x \to 0$,$2x \sin \frac{1}{x} \to 0$,but $\lim_{x \to 0} \cos \frac{1}{x}$ does not exist.
Therefore,$\lim_{x \to 0} g'(x)$ does not exist,which means $g'(x)$ is not continuous at $x = 0$.

(C) Given $f(x) = \max \{(1 - x), (1 + x), 2\}.$
We can define $f(x)$ piecewise by analyzing the intervals:
If $x > 1$,then $1 + x > 2$ and $1 + x > 1 - x$,so $f(x) = 1 + x.$
If $- 1 \le x \le 1$,then $2 \ge 1 + x$ and $2 \ge 1 - x$,so $f(x) = 2.$
If $x < - 1$,then $1 - x > 2$ and $1 - x > 1 + x$,so $f(x) = 1 - x.$
Thus,$f(x) = \begin{cases} 1 - x, & x < - 1 \\ 2, & - 1 \le x \le 1 \\ 1 + x, & x > 1 \end{cases}$
Since $f(x)$ is a polynomial function in each interval and the pieces match at the boundaries ($f(-1) = 2$ and $f(1) = 2$),the function is continuous everywhere.
Checking differentiability at $x = - 1$:
Left-hand derivative: $\frac{d}{dx}(1 - x) = - 1.$
Right-hand derivative: $\frac{d}{dx}(2) = 0.$
Since $- 1 \neq 0$,it is not differentiable at $x = - 1.$
Checking differentiability at $x = 1$:
Left-hand derivative: $\frac{d}{dx}(2) = 0.$
Right-hand derivative: $\frac{d}{dx}(1 + x) = 1.$
Since $0 \neq 1$,it is not differentiable at $x = 1.$
Therefore,the function is differentiable at all points except at $x = 1$ and $x = - 1$.

(A) We have $f(x) = |x| + |x - 1|$.
By definition of absolute value,we can write $f(x)$ as:
$f(x) = \begin{cases} -2x + 1, & x < 0 \\ 1, & 0 \le x < 1 \\ 2x - 1, & x \ge 1 \end{cases}$
For continuity at $x = 1$:
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (1) = 1$
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 2(1) - 1 = 1$
$f(1) = 2(1) - 1 = 1$
Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$,the function is continuous at $x = 1$.
For differentiability at $x = 1$:
Left-hand derivative $(LHD)$ at $x = 1$ is $\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \frac{d}{dx}(1) = 0$.
Right-hand derivative $(RHD)$ at $x = 1$ is $\lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \frac{d}{dx}(2x - 1) = 2$.
Since $LHD \neq RHD$,the function is not differentiable at $x = 1$.

(D) Given $f(x) = |x|.$ We check the differentiability at $x = 0$ by calculating the left-hand derivative $(Lf'(0))$ and the right-hand derivative $(Rf'(0))$.
$Rf'(0) = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{|h| - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1.$
$Lf'(0) = \lim_{h \to 0^+} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{|-h| - 0}{-h} = \lim_{h \to 0^+} \frac{h}{-h} = -1.$
Since $Rf'(0) \neq Lf'(0),$ the derivative $f'(0)$ does not exist.
Therefore,the correct option is $D$.

(D) The function is defined as $y = f(x) = 1 - |x|$.
We can express this as a piecewise function:
$f(x) = \begin{cases} 1 - x, & x \ge 0 \\ 1 + x, & x < 0 \end{cases}$
To check for differentiability at $x = 0$,we find the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$:
$LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{(1 + h) - 1}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1$
$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{(1 - h) - 1}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1$
Since $LHD \neq RHD$,the derivative of the function at $x = 0$ does not exist.

(A) Given the function $f(x) = |x|^3$.
We can write this as $f(x) = x^3$ for $x \ge 0$ and $f(x) = -x^3$ for $x < 0$.
To find the derivative at $x = 0$,we calculate the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$.
$LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{|h|^3 - 0}{h} = \lim_{h \to 0^-} \frac{-h^3}{h} = \lim_{h \to 0^-} (-h^2) = 0$.
$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{|h|^3 - 0}{h} = \lim_{h \to 0^+} \frac{h^3}{h} = \lim_{h \to 0^+} (h^2) = 0$.
Since $LHD = RHD = 0$,the derivative $f'(0)$ exists and is equal to $0$.

(C) Let $y = \sec^{-1} \left( \frac{2x}{1 + x^2} \right) + \sin^{-1} \left( \frac{x - 1}{x + 1} \right)$.
For the term $\sec^{-1} \left( \frac{2x}{1 + x^2} \right)$,the domain of $\sec^{-1}(u)$ is $|u| \ge 1$.
Thus,we must have $\left| \frac{2x}{1 + x^2} \right| \ge 1$.
This implies $2|x| \ge 1 + x^2$,which simplifies to $x^2 - 2|x| + 1 \le 0$,or $(|x| - 1)^2 \le 0$.
Since a square cannot be negative,this is only possible if $|x| - 1 = 0$,which means $x = 1$ or $x = -1$.
Since the function $y$ is defined only at discrete points $x = 1$ and $x = -1$,it is not defined on any interval.
Therefore,the derivative $\frac{dy}{dx}$ does not exist.

(B) Critical points are the points where $f'(x) = 0$ or where $f(x)$ is not differentiable.
For $x < 0$,$f'(x) = 1 \neq 0$.
For $x > 0$,$f'(x) = -3 \neq 0$.
Now,check for differentiability at $x = 0$:
Left-hand derivative $(LHD)$ at $x = 0$: $\lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{(1+h) - 2}{h} = \lim_{h \to 0^-} \frac{h-1}{h} = \infty$.
Since the derivative does not exist at $x = 0$,$x = 0$ is a critical point.

(C) The function is defined as $f(x) = |x|$.
For $x < 0$,$f(x) = -x$,and for $x > 0$,$f(x) = x$.
At $x = 0$,$f(0) = 0$.
Since $|x| \geq 0$ for all real $x$,the value $f(0) = 0$ is the smallest possible value of the function.
Therefore,$f(x)$ has a global minimum at $x = 0$.
Note that $f'(0)$ does not exist because the left-hand derivative is $-1$ and the right-hand derivative is $1$.

(A) The function is defined as $f(x) = \frac{x}{1 + |x|}$.
We can write this as a piecewise function:
$f(x) = \begin{cases} \frac{x}{1 - x}, & x < 0 \\ \frac{x}{1 + x}, & x \geq 0 \end{cases}$
Now,we find the derivative $f'(x)$ for both cases:
For $x < 0$,$f'(x) = \frac{(1-x)(1) - x(-1)}{(1-x)^2} = \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2}$.
For $x > 0$,$f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1+x-x}{(1+x)^2} = \frac{1}{(1+x)^2}$.
At $x = 0$,we check the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$:
$LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{\frac{h}{1-h} - 0}{h} = \lim_{h \to 0^-} \frac{1}{1-h} = 1$.
$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\frac{h}{1+h} - 0}{h} = \lim_{h \to 0^+} \frac{1}{1+h} = 1$.
Since $LHD = RHD = 1$,the function is differentiable at $x = 0$ as well.
Therefore,the function is differentiable for all $x \in ( - \infty, \infty )$.

(C) The function is defined as $f(x) = \text{Min}\{x + 1, |x| + 1\}$.
We can analyze this by considering two cases for $x$:
Case $1$: $x \ge 0$. Then $|x| = x$,so $f(x) = \text{Min}\{x + 1, x + 1\} = x + 1$.
Case $2$: $x < 0$. Then $|x| = -x$,so $f(x) = \text{Min}\{x + 1, -x + 1\}$.
For $x < 0$,$x + 1 < -x + 1$ is equivalent to $2x < 0$,which is true for all $x < 0$.
Thus,$f(x) = x + 1$ for all $x < 0$.
Combining these,$f(x) = x + 1$ for all $x \in R$.
Since $f(x) = x + 1$ is a linear polynomial,it is differentiable everywhere in $R$.
Therefore,option $C$ is correct.

(D) To check differentiability at $x = 0$:
$f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{(h-1) \sin(\frac{1}{h-1}) - (-1) \sin(-1)}{h}$
$= \lim_{h \to 0} \frac{(h-1) \sin(\frac{1}{h-1}) - \sin(1)}{h}$.
As $h \to 0$,the limit is $\frac{-\sin(-1) - \sin(1)}{0}$,which is not defined. Thus,$f$ is differentiable at $x = 0$.
To check differentiability at $x = 1$:
$f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{h \sin(\frac{1}{h}) - 0}{h} = \lim_{h \to 0} \sin(\frac{1}{h})$.
Since $\lim_{h \to 0} \sin(\frac{1}{h})$ oscillates between $-1$ and $1$,the limit does not exist. Thus,$f$ is not differentiable at $x = 1$.
Therefore,$f$ is differentiable at $x = 0$ but not at $x = 1$.

(D) We have $f(x) = x|x|$ and $g(x) = \sin x$.
Therefore,$(gof)(x) = \sin(x|x|) = \begin{cases} -\sin(x^2), & x < 0 \\ \sin(x^2), & x \ge 0 \end{cases}$.
$LHD$ of $gof$ at $x=0$: $\lim_{x \to 0^-} \frac{-\sin(x^2) - 0}{x - 0} = \lim_{x \to 0^-} -x \left(\frac{\sin(x^2)}{x^2}\right) = 0$.
$RHD$ of $gof$ at $x=0$: $\lim_{x \to 0^+} \frac{\sin(x^2) - 0}{x - 0} = \lim_{x \to 0^+} x \left(\frac{\sin(x^2)}{x^2}\right) = 0$.
Since $LHD = RHD = 0$,$gof$ is differentiable at $x=0$ and $(gof)'(0) = 0$.
The derivative is $(gof)'(x) = \begin{cases} -2x\cos(x^2), & x < 0 \\ 2x\cos(x^2), & x \ge 0 \end{cases}$.
Since $\lim_{x \to 0^-} (gof)'(x) = 0$ and $\lim_{x \to 0^+} (gof)'(x) = 0$,the derivative is continuous at $x=0$. Thus,Statement-$1$ is true.
Now,check for second differentiability at $x=0$:
$LHD$ of $(gof)'$ at $x=0$: $\lim_{x \to 0^-} \frac{-2x\cos(x^2) - 0}{x - 0} = -2\cos(0) = -2$.
$RHD$ of $(gof)'$ at $x=0$: $\lim_{x \to 0^+} \frac{2x\cos(x^2) - 0}{x - 0} = 2\cos(0) = 2$.
Since $LHD \neq RHD$,$gof$ is not twice differentiable at $x=0$. Thus,Statement-$2$ is false.

(B) Since $g(x)$ is differentiable,it must be continuous at $x=3$.
For continuity at $x=3$,$\lim_{x \to 3^-} g(x) = \lim_{x \to 3^+} g(x) = g(3)$.
$\lim_{x \to 3^-} k\sqrt{x+1} = k\sqrt{3+1} = 2k$.
$\lim_{x \to 3^+} (mx+2) = 3m+2$.
Thus,$2k = 3m+2$ --- $(i)$.
For differentiability at $x=3$,$g'(3^-) = g'(3^+)$.
$g'(x) = \begin{cases} \frac{k}{2\sqrt{x+1}}, & 0 < x < 3 \\ m, & 3 < x < 5 \end{cases}$.
$g'(3^-) = \frac{k}{2\sqrt{3+1}} = \frac{k}{4}$.
$g'(3^+) = m$.
So,$\frac{k}{4} = m \Rightarrow k = 4m$ --- $(ii)$.
Substituting $(ii)$ into $(i)$: $2(4m) = 3m+2 \Rightarrow 8m = 3m+2 \Rightarrow 5m = 2 \Rightarrow m = \frac{2}{5}$.
Then $k = 4(\frac{2}{5}) = \frac{8}{5}$.
Therefore,$k+m = \frac{8}{5} + \frac{2}{5} = \frac{10}{5} = 2$.

(D) The function is given by $f(x) = |x-\pi|(e^{|x|}-1)\sin|x|$.
We check the differentiability of $f(x)$ at the points where the absolute value functions might not be differentiable,namely $x=0$ and $x=\pi$.
At $x=\pi$:
$f(\pi) = 0$.
$RHD = \lim_{h \to 0^+} \frac{|\pi+h-\pi|(e^{|\pi+h|}-1)\sin|\pi+h| - 0}{h} = \lim_{h \to 0^+} \frac{h(e^{\pi+h}-1)\sin(\pi+h)}{h} = (e^{\pi}-1)\sin(\pi) = 0$.
$LHD = \lim_{h \to 0^+} \frac{|\pi-h-\pi|(e^{|\pi-h|}-1)\sin|\pi-h| - 0}{-h} = \lim_{h \to 0^+} \frac{h(e^{\pi-h}-1)\sin(\pi-h)}{-h} = -(e^{\pi}-1)\sin(\pi) = 0$.
Since $RHD = LHD = 0$,$f(x)$ is differentiable at $x=\pi$.
At $x=0$:
$f(0) = 0$.
$RHD = \lim_{h \to 0^+} \frac{|h-\pi|(e^{|h|}-1)\sin|h| - 0}{h} = \lim_{h \to 0^+} \frac{|h-\pi|(e^h-1)\sin(h)}{h} = |-\pi| \cdot (1) \cdot (0) = 0$.
$LHD = \lim_{h \to 0^+} \frac{|-h-\pi|(e^{|-h|}-1)\sin|-h| - 0}{-h} = \lim_{h \to 0^+} \frac{|-h-\pi|(e^h-1)\sin(h)}{-h} = |-\pi| \cdot (1) \cdot (0) = 0$.
Since $RHD = LHD = 0$,$f(x)$ is differentiable at $x=0$.
Since $f(x)$ is differentiable at all points,the set $S$ of points where $f(x)$ is not differentiable is empty,i.e.,$S = \emptyset$.

(D) To find the points where $f(x) = \max(x, x^3)$ is not differentiable,we analyze the behavior of $x$ and $x^3$ in different intervals:
$1$. If $x < -1$,then $x^3 < x$,so $f(x) = x$.
$2$. If $-1 < x < 0$,then $x^3 > x$,so $f(x) = x^3$.
$3$. If $0 < x < 1$,then $x > x^3$,so $f(x) = x$.
$4$. If $x > 1$,then $x^3 > x$,so $f(x) = x^3$.
At the transition points $x = -1, 0, 1$,we check the left-hand and right-hand derivatives:
At $x = -1$: $f'( -1^-) = 1$ and $f'( -1^+) = 3(-1)^2 = 3$. Since $1 \neq 3$,it is not differentiable.
At $x = 0$: $f'( 0^-) = 3(0)^2 = 0$ and $f'( 0^+) = 1$. Since $0 \neq 1$,it is not differentiable.
At $x = 1$: $f'( 1^-) = 1$ and $f'( 1^+) = 3(1)^2 = 3$. Since $1 \neq 3$,it is not differentiable.
Thus,the set of points where $f(x)$ is not differentiable is $\{ -1, 0, 1\}$.

(D) Let $f(x) = \sin(|x|) - |x|$. We check the differentiability at $x = 0$ by calculating the Left Hand Derivative $(LHD)$ and Right Hand Derivative $(RHD)$.
For $x < 0$,$|x| = -x$,so $f(x) = \sin(-x) - (-x) = -\sin x + x$. The derivative is $f'(x) = -\cos x + 1$. Thus,$LHD = \lim_{x \to 0^-} (-\cos x + 1) = -\cos(0) + 1 = -1 + 1 = 0$.
For $x > 0$,$|x| = x$,so $f(x) = \sin x - x$. The derivative is $f'(x) = \cos x - 1$. Thus,$RHD = \lim_{x \to 0^+} (\cos x - 1) = \cos(0) - 1 = 1 - 1 = 0$.
Since $LHD = RHD = 0$,the function $\sin(|x|) - |x|$ is differentiable at $x = 0$.
For the other options,$\cos(|x|)$ is differentiable at $x=0$ but $|x|$ is not,so their sum or difference is not differentiable at $x=0$. Similarly,$\sin(|x|) + |x|$ has $LHD = -2$ and $RHD = 2$,so it is not differentiable.

(B) By the definition of the derivative at $x = 0$:
$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{g(h) \cos(\frac{1}{h}) - 0}{h} = \lim_{h \to 0} \frac{g(h)}{h} \cos(\frac{1}{h})$.
Since $g(x)$ is an even function,$g(-x) = g(x)$.
Given $g(x)$ is differentiable at $x = 0$ and passes through the origin,we have $g(0) = 0$.
Thus,$g'(0) = \lim_{h \to 0} \frac{g(h) - g(0)}{h} = \lim_{h \to 0} \frac{g(h)}{h}$.
Since $g(x)$ is even,$g'(0) = \lim_{h \to 0} \frac{g(h) - g(0)}{h} = \lim_{h \to 0} \frac{g(-h) - g(0)}{-h} = -g'(0)$,which implies $g'(0) = 0$.
Therefore,$f'(0) = \lim_{h \to 0} \frac{g(h)}{h} \cos(\frac{1}{h}) = g'(0) \times (\text{a bounded value}) = 0 \times \text{bounded value} = 0$.

(C) The function $h(x) = \max\{f(x), g(x)\}$ can be written as $h(x) = \frac{f(x) + g(x) + |f(x) - g(x)|}{2}$.
Since $f(x)$ and $g(x)$ are differentiable at $x = x_0$,their sum and difference are also differentiable.
The differentiability of $h(x)$ depends on the term $|f(x) - g(x)|$.
If $f(x_0) \neq g(x_0)$,then by the continuity of $f$ and $g$,there exists a neighborhood around $x_0$ where $f(x) - g(x)$ maintains a constant sign,making $|f(x) - g(x)|$ differentiable at $x_0$.
If $f(x_0) = g(x_0)$,the function $h(x)$ is differentiable at $x_0$ if and only if $f'(x_0) = g'(x_0)$.
Therefore,$h(x)$ is guaranteed to be differentiable at $x_0$ if $f(x_0) \neq g(x_0)$.

(C) To check for differentiability at $x = 0$,we calculate the derivative using the limit definition $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$.
$(1)$ For $f(x)$:
$f'(0) = \lim_{h \to 0} \frac{h \sin(1/h) - 0}{h} = \lim_{h \to 0} \sin(1/h)$.
Since $\lim_{h \to 0} \sin(1/h)$ oscillates between $-1$ and $1$,the limit does not exist. Thus,$f(x)$ is not differentiable at $x = 0$.
$(2)$ For $g(x)$:
$g'(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin(1/h)$.
Since $|h \sin(1/h)| \le |h|$,by the Squeeze Theorem,the limit is $0$. Thus,$g(x)$ is differentiable at $x = 0$.
$(3)$ For $h(x) = |x|^3$:
$h'(0) = \lim_{h \to 0} \frac{|h|^3 - 0}{h} = \lim_{h \to 0} \frac{|h|^3}{h}$.
For $h > 0$,$\frac{h^3}{h} = h^2 \to 0$. For $h < 0$,$\frac{(-h)^3}{h} = -h^2 \to 0$.
Since the limit exists and is $0$,$h(x)$ is differentiable at $x = 0$.
Therefore,$g$ and $h$ are differentiable at $x = 0$. The correct option is $C$.

(D) For $g(x)$ to be differentiable at $x = 0$,it must first be continuous at $x = 0$.
For continuity: $\lim_{x \to 0^-} g(x) = \lim_{x \to 0^+} g(x) = g(0)$.
$\lim_{x \to 0^-} (x + b) = b$ and $\lim_{x \to 0^+} \cos x = \cos(0) = 1$.
Thus,$b = 1$.
Now,check for differentiability using the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$ at $x = 0$:
$RHD = g'(0^+) = \lim_{h \to 0^+} \frac{\cos(0 + h) - \cos(0)}{h} = \lim_{h \to 0^+} \frac{\cos h - 1}{h} = 0$.
$LHD = g'(0^-) = \lim_{h \to 0^-} \frac{(0 + h + b) - (0 + b)}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1$.
Since $LHD \neq RHD$ $(1 \neq 0)$,the function $g(x)$ cannot be made differentiable at $x = 0$ for any value of $b$.