Continuity Questions in English

(D) For $f(x)$ to be continuous on $\mathbb{R}$,it must be continuous at $x=0, x=1,$ and $x=2$.
At $x=0$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) \implies \sin(0) = 0^2 + a^2 \implies a^2 = 0 \implies a = 0$.
At $x=1$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \implies 1^2 + a^2 = b(1) + 2 \implies 1 + 0 = b + 2 \implies b = -1$.
At $x=2$: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \implies b(2) + 2 = 0 \implies 2(-1) + 2 = 0$,which is consistent.
Thus,$a = 0$ and $b = -1$.
Calculating $a+b+ab = 0 + (-1) + (0)(-1) = -1$.

(D) 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,calculate the Left Hand Limit $(LHL)$:
$\lim_{x \to 0^-} (1+|\sin x|)^{\frac{p}{|\sin x|}} = \lim_{x \to 0^-} (1-\sin x)^{\frac{p}{-\sin x}}$ (since $|\sin x| = -\sin x$ for $x < 0$).
Let $h = -x$,as $x \to 0^-$,$h \to 0^+$. Then $\sin x = -\sin h$.
$\lim_{h \to 0^+} (1+\sin h)^{\frac{p}{\sin h}} = e^{\lim_{h \to 0^+} \sin h \cdot \frac{p}{\sin h}} = e^p$.
Next,calculate the Right Hand Limit $(RHL)$:
$\lim_{x \to 0^+} e^{\frac{\sin 2x}{\sin 3x}} = e^{\lim_{x \to 0^+} \frac{\sin 2x}{\sin 3x}} = e^{\lim_{x \to 0^+} \frac{\sin 2x}{2x} \cdot \frac{3x}{\sin 3x} \cdot \frac{2}{3}} = e^{2/3}$.
Since $f(0) = q$,we equate the limits:
$e^p = q = e^{2/3}$.
Thus,$p = 2/3$ and $q = e^{2/3}$.
Wait,checking the $LHL$ again: $(1+|\sin x|)^{\frac{p}{|\sin x|}} = (1-\sin x)^{\frac{p}{-\sin x}} = ((1-\sin x)^{\frac{1}{-\sin x}})^p = e^p$.
If the original question implies $f(x) = (1+|\sin x|)^{\frac{p}{\sin x}}$,then $LHL$ $= e^{-p}$.
Equating $e^{-p} = e^{2/3} = q$,we get $p = -2/3$ and $q = e^{2/3}$.

(A) For the function $f(x)$ to be continuous at $x=0$,we must have $f(0) = \lim_{x \rightarrow 0} f(x)$.
Given $f(x) = (x+1)^{\cot x}$.
Taking the limit as $x \rightarrow 0$:
$\lim_{x \rightarrow 0} (x+1)^{\cot x} = e^{\lim_{x \rightarrow 0} \cot x \cdot \ln(1+x)}$.
Using the standard limit $\lim_{x \rightarrow 0} \frac{\ln(1+x)}{x} = 1$,we can rewrite the exponent as:
$\lim_{x \rightarrow 0} \frac{\ln(1+x)}{\tan x} = \lim_{x \rightarrow 0} \left( \frac{\ln(1+x)}{x} \cdot \frac{x}{\tan x} \right)$.
Since $\lim_{x \rightarrow 0} \frac{\ln(1+x)}{x} = 1$ and $\lim_{x \rightarrow 0} \frac{x}{\tan x} = 1$,the limit of the exponent is $1 \cdot 1 = 1$.
Therefore,$f(0) = e^1 = e$.

(B) Given that $f(x)$ is continuous at $x = 0$,the condition $\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x)$ must hold.
First,calculate the Left-Hand Limit $(LHL)$:
$\text{LHL} = \lim_{x \to 0^-} \left( \beta + \left[ \frac{\sin x - x}{x^3} \right] \right)$.
Using the Taylor series expansion,$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$,so $\frac{\sin x - x}{x^3} = \frac{-x^3/6 + x^5/120}{x^3} = -\frac{1}{6} + \frac{x^2}{120}$.
As $x \to 0^-$,$\frac{\sin x - x}{x^3} \to -\frac{1}{6}$. Thus,$\text{LHL} = \beta + [-1/6] = \beta - 1$.
Next,calculate the Right-Hand Limit $(RHL)$:
$\text{RHL} = \lim_{x \to 0^+} \left( \alpha + \frac{\sin [x]}{x} \right)$.
For $0 < x < 1$,$[x] = 0$,so $\sin [x] = \sin 0 = 0$.
Thus,$\text{RHL} = \lim_{x \to 0^+} (\alpha + 0) = \alpha$.
Given $f(0) = 2$,we have $\beta - 1 = 2 = \alpha$.
So,$\beta = 3$ and $\alpha = 2$.
Therefore,$\beta - \alpha = 3 - 2 = 1$.

(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0)$.
First,consider the $LHL$: $\lim_{x \rightarrow 0^-} \frac{a+2 \cos x}{x^2}$. For this limit to exist and be finite,the numerator must approach $0$ as $x \rightarrow 0$. Thus,$a + 2 \cos(0) = 0 \Rightarrow a + 2 = 0 \Rightarrow a = -2$.
Substituting $a = -2$ into the limit: $\lim_{x \rightarrow 0^-} \frac{-2 + 2 \cos x}{x^2} = \lim_{x \rightarrow 0^-} \frac{-2(1 - \cos x)}{x^2} = \lim_{x \rightarrow 0^-} \frac{-2(2 \sin^2(x/2))}{x^2} = -2 \lim_{x \rightarrow 0^-} \frac{\sin^2(x/2)}{(x/2)^2 \times 2} = -2 \times \frac{1}{2} = -1$.
Now,consider the $RHL$: $\lim_{x \rightarrow 0^+} b \tan \frac{\pi}{[x+4]}$. As $x \rightarrow 0^+$,$[x+4] = 4$.
So,$RHL = b \tan \frac{\pi}{4} = b(1) = b$.
Since $LHL = RHL$,we have $-1 = b$.
Thus,the ordered pair $(a, b)$ is $(-2, -1)$.

(B) For the function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x \rightarrow 0$ must exist and be equal to $f(0)$.
$f(0) = k$.
Now,we calculate the limit:
$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{1 + 3 x^2 - \cos 2 x}{x^2}$
Using the trigonometric identity $\cos 2x = 1 - 2 \sin^2 x$:
$\lim_{x \rightarrow 0} \frac{1 + 3 x^2 - (1 - 2 \sin^2 x)}{x^2}$
$= \lim_{x \rightarrow 0} \frac{1 + 3 x^2 - 1 + 2 \sin^2 x}{x^2}$
$= \lim_{x \rightarrow 0} \frac{3 x^2 + 2 \sin^2 x}{x^2}$
$= \lim_{x \rightarrow 0} \left( \frac{3 x^2}{x^2} + 2 \frac{\sin^2 x}{x^2} \right)$
$= \lim_{x \rightarrow 0} \left( 3 + 2 \left( \frac{\sin x}{x} \right)^2 \right)$
Since $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have:
$= 3 + 2(1)^2 = 3 + 2 = 5$.
Since the function is continuous at $x = 0$,$f(0) = \lim_{x \rightarrow 0} f(x)$,therefore $k = 5$.

(D) For a function $f(x)$ to be continuous at $x=0$,the condition $\lim_{x \rightarrow 0} f(x) = f(0)$ must hold.
Given $f(x) = \frac{2 \sin x - \sin 2x}{2x \cos x}$ for $x \neq 0$.
We know that $\sin 2x = 2 \sin x \cos x$.
Substituting this into the expression:
$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{2 \sin x - 2 \sin x \cos x}{2x \cos x}$
$= \lim_{x \rightarrow 0} \frac{2 \sin x (1 - \cos x)}{2x \cos x}$
$= \lim_{x \rightarrow 0} \left( \frac{\sin x}{x} \right) \cdot \left( \frac{1 - \cos x}{\cos x} \right)$
Since $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim_{x \rightarrow 0} \frac{1 - \cos x}{\cos x} = \frac{1 - 1}{1} = 0$.
Therefore,$\lim_{x \rightarrow 0} f(x) = 1 \cdot 0 = 0$.
Since $f(0) = a$,for continuity,we must have $a = 0$.

(C) Given that $f(x) = \begin{cases} \frac{x + 2}{x^2 + 3 x + 2}, & x \in R - \{-1, -2\} \\ -1, & x = -2 \\ 0, & x = -1 \end{cases}$
For $x \in R - \{-1, -2\}$,$f(x) = \frac{x + 2}{(x + 1)(x + 2)} = \frac{1}{x + 1}$.
Now,we check the continuity at $x = -2$ and $x = -1$.
At $x = -2$:
$\lim_{x \rightarrow -2} f(x) = \lim_{x \rightarrow -2} \frac{1}{x + 1} = \frac{1}{-2 + 1} = -1$.
Since $f(-2) = -1$,we have $\lim_{x \rightarrow -2} f(x) = f(-2)$,so $f$ is continuous at $x = -2$.
At $x = -1$:
$\lim_{x \rightarrow -1^-} f(x) = \lim_{x \rightarrow -1^-} \frac{1}{x + 1} = -\infty$ and $\lim_{x \rightarrow -1^+} f(x) = \lim_{x \rightarrow -1^+} \frac{1}{x + 1} = \infty$.
Since the limit does not exist at $x = -1$,$f$ is discontinuous at $x = -1$.
Therefore,$f$ is continuous on the set $R - \{-1\}$.

(B) For $f(x)$ to be continuous at $x = 0$,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$ and the value of the function $f(0)$.
$LHL$ = $\lim_{x \to 0^-} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$
Rationalizing the numerator:
$LHL$ = $\lim_{x \to 0^-} \frac{(\sqrt{1+kx}-\sqrt{1-kx})(\sqrt{1+kx}+\sqrt{1-kx})}{x(\sqrt{1+kx}+\sqrt{1-kx})}$
$LHL$ = $\lim_{x \to 0^-} \frac{(1+kx)-(1-kx)}{x(\sqrt{1+kx}+\sqrt{1-kx})} = \lim_{x \to 0^-} \frac{2kx}{x(\sqrt{1+kx}+\sqrt{1-kx})} = \frac{2k}{1+1} = k$
$RHL$ = $\lim_{x \to 0^+} (2x^2+3x-2) = 2(0)^2+3(0)-2 = -2$
Since the function is continuous,$LHL$ = $RHL$,therefore $k = -2$.

(A) For the function $f(x)$ to be continuous at $x=0$,the left-hand limit,right-hand limit,and the value of the function at $x=0$ must be equal.
$1$. The value of the function at $x=0$ is $f(0) = a^2 \cos^2(0) + b^2 \sin^2(0) = a^2(1) + b^2(0) = a^2$.
$2$. The right-hand limit $(RHL)$ as $x \rightarrow 0^+$ is $\lim_{x \rightarrow 0^+} e^{ax+b} = e^{a(0)+b} = e^b$.
$3$. The left-hand limit $(LHL)$ as $x \rightarrow 0^-$ is $\lim_{x \rightarrow 0^-} (a^2 \cos^2 x + b^2 \sin^2 x) = a^2(1) + b^2(0) = a^2$.
For continuity,$LHL = RHL = f(0)$,so $e^b = a^2$.
Taking the natural logarithm on both sides,we get $\ln(e^b) = \ln(a^2)$,which simplifies to $b = 2 \ln |a|$ or $b = 2 \log |a|$.

(C) The function $f(x) = x - [x]$ is the fractional part function,denoted as $\{x\}$.
We check the continuity of $f(x)$ at any integer $n \in Z$.
The left-hand limit is:
$\lim_{x \rightarrow n^-} f(x) = \lim_{h \rightarrow 0} f(n-h) = \lim_{h \rightarrow 0} ((n-h) - [n-h]) = \lim_{h \rightarrow 0} ((n-h) - (n-1)) = \lim_{h \rightarrow 0} (1-h) = 1$.
The right-hand limit is:
$\lim_{x \rightarrow n^+} f(x) = \lim_{h \rightarrow 0} f(n+h) = \lim_{h \rightarrow 0} ((n+h) - [n+h]) = \lim_{h \rightarrow 0} ((n+h) - n) = \lim_{h \rightarrow 0} h = 0$.
The value of the function at $x=n$ is:
$f(n) = n - [n] = n - n = 0$.
Since $\lim_{x \rightarrow n^-} f(x) \neq \lim_{x \rightarrow n^+} f(x)$,the function $f(x)$ is discontinuous at every integer $n \in Z$.
Therefore,the set of points of discontinuity is $Z$.

(B) For $f(x)$: $\lim_{x \to 0} f(x) = \lim_{x \to 0} x \sin(1/x) = 0$,which equals $f(0)$,so $f(x)$ is continuous at $x = 0$.
For differentiability at $x = 0$,$f'(0) = \lim_{h \to 0} \frac{h \sin(1/h) - 0}{h} = \lim_{h \to 0} \sin(1/h)$,which does not exist. Thus,$(i)$ is true.
For $g(x) = x f(x) = x^2 \sin(1/x)$ for $x \neq 0$ and $g(0) = 0$.
$g'(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin(1/h) = 0$. So $g(x)$ is differentiable at $x = 0$.
For $x \neq 0$,$g'(x) = 2x \sin(1/x) - \cos(1/x)$.
As $x \to 0$,$\lim_{x \to 0} g'(x) = \lim_{x \to 0} (2x \sin(1/x) - \cos(1/x))$,which does not exist because $\cos(1/x)$ oscillates.
Since $\lim_{x \to 0} g'(x) \neq g'(0)$,$g'(x)$ is not continuous at $x = 0$. Thus,$(ii)$ is true.

(A) Given $f(x) = \begin{cases} 1 + 6x - 3x^2, & x \leq 1 \\ x + \log_2(b^2 + 7), & x > 1 \end{cases}$.
For $f(1)$ to be the maximum value,$f(x)$ must be less than or equal to $f(1)$ for all $x$.
First,calculate $f(1) = 1 + 6(1) - 3(1)^2 = 1 + 6 - 3 = 4$.
For $x > 1$,we require $f(x) \leq 4$,which means $x + \log_2(b^2 + 7) \leq 4$.
Since this must hold for all $x > 1$,we look at the limit as $x \to 1^+$,which gives $1 + \log_2(b^2 + 7) \leq 4$.
$\log_2(b^2 + 7) \leq 3$.
$b^2 + 7 \leq 2^3 = 8$.
$b^2 \leq 1$.
Therefore,$-1 \leq b \leq 1$,so $b \in [-1, 1]$.

(D) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
Using the identity $1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})$,we have:
$\lim_{x \to 0} \frac{1 - \cos(1 - \cos x)}{x^4} = \lim_{x \to 0} \frac{2 \sin^2(\frac{1 - \cos x}{2})}{x^4}$
Since $1 - \cos x = 2 \sin^2(\frac{x}{2})$,this becomes:
$\lim_{x \to 0} \frac{2 \sin^2(\sin^2(\frac{x}{2}))}{x^4}$
Using the limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,we approximate $\sin \theta \approx \theta$ for small $\theta$:
$\lim_{x \to 0} \frac{2 (\sin^2(\frac{x}{2}))^2}{x^4} = \lim_{x \to 0} \frac{2 (\frac{x}{2})^4}{x^4} = \lim_{x \to 0} \frac{2 \cdot \frac{x^4}{16}}{x^4} = \frac{2}{16} = \frac{1}{8}$.

(D) Given the interval $x \in (-2 \pi, 0)$.
As $x$ approaches $0$ from the left,$x^3$ approaches $0$ from the negative side.
Therefore,$\frac{1}{x^3}$ approaches $-\infty$.
Since the function $f(x) = \sin \left(\frac{1}{x^3}\right)$ oscillates between $-1$ and $1$ as its argument $\frac{1}{x^3}$ approaches $-\infty$,the function will cross the $x$-axis infinitely many times in any neighborhood of $0$.
Thus,the function changes sign infinitely many times in the interval $(-2 \pi, 0)$.

(D) continuous function $f: I \rightarrow \mathbb{R}$ defined on an interval $I$ that takes only irrational values must be a constant function.
If $f(x)$ were not constant,by the Intermediate Value Theorem,it would take all values between any two values it assumes. Since the set of rational numbers is dense in $\mathbb{R}$,any non-constant continuous function must take rational values.
Given $f(\sqrt{2})=\sqrt{2}$,and since $f(x)$ is constant,we have $f(x)=\sqrt{2}$ for all $x \in [-2,2]$.
Therefore,$f(\sqrt{2}-1)=\sqrt{2}$.

(A) Let $g(x) = x \log_e x + x - 2$.
We are looking for roots of the equation $g(x) = 0$.
Evaluate $g(x)$ at the endpoints of the interval $(1, 2)$:
$g(1) = 1 \cdot \log_e(1) + 1 - 2 = 0 + 1 - 2 = -1$.
$g(2) = 2 \cdot \log_e(2) + 2 - 2 = 2 \log_e(2) \approx 2(0.693) = 1.386$.
Since $g(1) = -1 < 0$ and $g(2) = 2 \log_e(2) > 0$,and $g(x)$ is a continuous function on the interval $[1, 2]$,by the Intermediate Value Theorem,there must exist at least one root $c \in (1, 2)$ such that $g(c) = 0$.

(D) The function $f(x) = x - [x]$ is known as the fractional part function,denoted as $\{x\}$.
For any integer $n \in Z$,we evaluate the left-hand limit and right-hand limit:
$\lim_{x \to n^-} f(x) = \lim_{x \to n^-} (x - [x]) = n - (n - 1) = 1$.
$\lim_{x \to n^+} f(x) = \lim_{x \to n^+} (x - [x]) = n - n = 0$.
Since the left-hand limit is not equal to the right-hand limit at any integer $n$,the function is discontinuous at all integers.
Therefore,the set of points of discontinuity is $Z$.

(C) The function is defined as $f(x) = \begin{cases} x+1, & -1 \leq x \leq 0 \\ -x, & 0 < x \leq 1 \end{cases}$.
Check continuity at $x=0$:
Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x+1) = 1$.
Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (-x) = 0$.
Since $\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$,the function is discontinuous at $x=0$.
Now,analyze the range of $f(x)$ on $[-1, 1]$:
For $x \in [-1, 0]$,$f(x) = x+1$. The range is $[0, 1]$.
For $x \in (0, 1]$,$f(x) = -x$. The range is $[-1, 0)$.
Combining these,the range of $f(x)$ is $[-1, 0) \cup [0, 1] = [-1, 1]$.
The maximum value is $1$ (attained at $x=0$) and the minimum value is $-1$ (attained at $x=1$).
Thus,$f(x)$ is discontinuous in $[-1, 1]$ but still has a maximum and minimum value.

(D) The function $f(x) = [x^2] \sin(\pi x)$ is a product of two functions: $g(x) = [x^2]$ and $h(x) = \sin(\pi x)$.
$g(x) = [x^2]$ is discontinuous at all points where $x^2$ is an integer,i.e.,$x = \sqrt{n}$ for $n \in \{1, 2, 3, \dots\}$.
For the product $f(x) = g(x)h(x)$ to be continuous at a point $x_0$ where $g(x)$ is discontinuous,$h(x_0)$ must be $0$.
Here,$h(x) = \sin(\pi x) = 0$ when $x$ is an integer.
If $x^2 = n$ (where $n$ is an integer) and $x$ is an integer,then $x = \sqrt{n}$ must be an integer,which means $n$ is a perfect square.
If $x^2 = n$ but $x$ is not an integer,then $\sin(\pi x) \neq 0$,so the function $f(x)$ remains discontinuous at these points.
Therefore,$f(x)$ is continuous only at points where $x^2$ is an integer and $\sin(\pi x) = 0$,which occurs when $x$ is an integer.
Since the options provided do not correctly describe the set of points of continuity,the correct choice is $D$.

(D) For the function to be continuous at $x = 0$,we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
First,calculate the left-hand limit $(LHL)$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin(a+1)x + \sin x}{x} = \lim_{x \to 0^-} \left( \frac{\sin(a+1)x}{x} + \frac{\sin x}{x} \right) = (a+1) + 1 = a+2$.
Next,calculate the right-hand limit $(RHL)$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x+bx^2} - \sqrt{x}}{bx^{1/2}} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1+bx} - 1)}{bx^{1/2}} = \lim_{x \to 0^+} \frac{\sqrt{1+bx} - 1}{b}$.
Using the binomial expansion $(1+bx)^{1/2} \approx 1 + \frac{1}{2}bx$,we get:
$\lim_{x \to 0^+} \frac{1 + \frac{1}{2}bx - 1}{b} = \lim_{x \to 0^+} \frac{\frac{1}{2}bx}{b} = \lim_{x \to 0^+} \frac{x}{2} = 0$.
Since $f(0) = c$,for continuity we require $a+2 = 0$ and $c = 0$.
Thus,$a = -2$ and $c = 0$. Since the denominator in the $RHL$ expression contains $b$,we must have $b \neq 0$ for the function to be defined for $x > 0$.
Therefore,$a = -2, b \in \mathbb{R} - \{0\}, c = 0$.

(A, D) Given $f(x) = \begin{cases} 0, & -1 \leq x < 0 \\ 1, & x = 0 \\ 2, & 0 < x \leq 1 \end{cases}$.
For $-1 \leq x \leq 0$,$F(x) = \int_{-1}^{x} 0 \, dt = 0$.
For $0 < x \leq 1$,$F(x) = \int_{-1}^{0} 0 \, dt + \int_{0}^{x} 2 \, dt = 0 + [2t]_{0}^{x} = 2x$.
So,$F(x) = \begin{cases} 0, & -1 \leq x \leq 0 \\ 2x, & 0 < x \leq 1 \end{cases}$.
Checking continuity at $x = 0$:
$\lim_{x \to 0^-} F(x) = 0$ and $\lim_{x \to 0^+} F(x) = 2(0) = 0$.
Since $\lim_{x \to 0^-} F(x) = \lim_{x \to 0^+} F(x) = F(0) = 0$,$F(x)$ is continuous in $[-1, 1]$.
Checking differentiability at $x = 0$:
Left-hand derivative $LHD = \lim_{h \to 0^-} \frac{F(0+h) - F(0)}{h} = \lim_{h \to 0^-} \frac{0 - 0}{h} = 0$.
Right-hand derivative $RHD = \lim_{h \to 0^+} \frac{F(0+h) - F(0)}{h} = \lim_{h \to 0^+} \frac{2(0+h) - 0}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2$.
Since $LHD \neq RHD$,$F'(x)$ does not exist at $x = 0$. Thus,options $A$ and $D$ are correct.

(B) For $f(x)$ to be continuous,it must be continuous at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.
At $x = -\frac{\pi}{2}$:
$LHL = \lim_{x \to -\frac{\pi}{2}^-} (-2 \sin x) = -2 \sin(-\frac{\pi}{2}) = -2(-1) = 2$.
$RHL = \lim_{x \to -\frac{\pi}{2}^+} (A \sin x + B) = A \sin(-\frac{\pi}{2}) + B = -A + B$.
For continuity,$LHL = RHL \implies -A + B = 2$ (Equation $i$).
At $x = \frac{\pi}{2}$:
$LHL = \lim_{x \to \frac{\pi}{2}^-} (A \sin x + B) = A \sin(\frac{\pi}{2}) + B = A + B$.
$RHL = \lim_{x \to \frac{\pi}{2}^+} (\cos x) = \cos(\frac{\pi}{2}) = 0$.
For continuity,$LHL = RHL \implies A + B = 0$ (Equation $ii$).
Adding Equation $i$ and Equation $ii$: $(-A + B) + (A + B) = 2 + 0 \implies 2B = 2 \implies B = 1$.
Substituting $B = 1$ into Equation $ii$: $A + 1 = 0 \implies A = -1$.
Thus,$f$ is continuous for $A = -1$ and $B = 1$.

(C) For the function $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \to 0} f(x) = f(0)$.
Given $f(x) = \frac{\sin [-x^2]}{[-x^2]}$ for $x \neq 0$.
As $x \to 0$,$x^2$ approaches $0$ from the positive side,so $-x^2$ approaches $0$ from the negative side (i.e.,$-x^2 \in (-1, 0)$).
Therefore,the greatest integer function $[-x^2]$ will take the value $-1$ as $x \to 0$.
Thus,$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin [-x^2]}{[-x^2]} = \frac{\sin(-1)}{-1} = \frac{-\sin(1)}{-1} = \sin(1)$.
Since $f(0) = \alpha$,for continuity,we must have $\alpha = \sin(1)$.

(A) Given $f(2x - 1) = f(x)$.
For any $x$,we can write $f(x) = f(2x - 1) = f(2(2x - 1) - 1) = f(4x - 3) = f(2^n x - (2^n - 1))$.
As $n \rightarrow \infty$,$2^n x - 2^n + 1 = 2^n(x - 1) + 1$.
If $x \neq 1$,then $2^n(x - 1) + 1 \rightarrow \pm \infty$.
Since $f$ is continuous at $x = 1$,we consider the limit as $x \rightarrow 1$. Let $x_n$ be a sequence such that $x_n \rightarrow 1$. Then $f(x_n) = f(2x_n - 1)$.
By iterating the relation,$f(x) = f(1)$ for all $x$ in the domain because the sequence $x_{n+1} = \frac{x_n + 1}{2}$ converges to $1$.
Since $f(1) = 1$,it follows that $f(x) = 1$ for all $x \in R$.
Thus,$f(2) = 1$ and $f$ is a constant function,which is continuous everywhere.

(A) For the function $f(x)$ to be continuous at $x=2$,the limit of $f(x)$ as $x \to 2$ must exist and be equal to $f(2)$.
Given $f(2) = 2$.
We evaluate the limit: $\lim_{x \to 2} \frac{x^2-(A+2)x+A}{x-2}$.
For the limit to exist,the numerator must be zero at $x=2$ because the denominator is zero.
Substituting $x=2$ in the numerator: $2^2 - (A+2)(2) + A = 4 - 2A - 4 + A = -A$.
Setting $-A = 0$,we get $A = 0$.
Now,check the limit with $A=0$: $\lim_{x \to 2} \frac{x^2-2x}{x-2} = \lim_{x \to 2} \frac{x(x-2)}{x-2} = \lim_{x \to 2} x = 2$.
Since the limit $2$ equals $f(2) = 2$,the function is continuous at $x=2$ when $A=0$.

(A) For $f(x)$ to be continuous at $x = 2$,the limit of $f(x)$ as $x \to 2$ must equal $f(2)$.
Given $f(x) = [x] + [-x]$ for $x \neq 2$.
We know that for any integer $n$,$[n] + [-n] = 0$,but for any non-integer $x$,$[x] + [-x] = -1$.
As $x \to 2$,$x$ takes values close to $2$ but not equal to $2$. Since $x$ is not an integer in the neighborhood of $2$,we have $[x] + [-x] = -1$.
Therefore,$\lim_{x \to 2} f(x) = -1$.
Since the function is continuous at $x = 2$,we must have $f(2) = \lim_{x \to 2} f(x)$.
Thus,$\lambda = -1$.

(A) The given function is $f(x) = x + |x|$.
We can define the absolute value function $|x|$ as:
$|x| = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases}$
Therefore,the function $f(x)$ can be written as:
$f(x) = \begin{cases} x + x, & x \geq 0 \\ x - x, & x < 0 \end{cases} = \begin{cases} 2x, & x \geq 0 \\ 0, & x < 0 \end{cases}$
Now,we check the continuity at $x = 0$:
Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (0) = 0$
Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2x) = 2(0) = 0$
Value of the function: $f(0) = 2(0) = 0$
Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0$,the function is continuous at $x = 0$.
Since $f(x) = 2x$ is a polynomial for $x > 0$ and $f(x) = 0$ is a constant function for $x < 0$,the function is continuous for all $x \in (-\infty, \infty)$.

(B) Given $f(x) = 1$ if $x \in \mathbb{Q}$ and $f(x) = 0$ if $x \notin \mathbb{Q}$.
Given $g(x) = 0$ if $x \in \mathbb{Q}$ and $g(x) = 1$ if $x \notin \mathbb{Q}$.
Consider the function $h(x) = f(x) + g(x)$.
For any $x \in [0,1]$,if $x$ is rational,$h(x) = f(x) + g(x) = 1 + 0 = 1$.
If $x$ is irrational,$h(x) = f(x) + g(x) = 0 + 1 = 1$.
Thus,$h(x) = 1$ for all $x \in [0,1]$,which is a constant function.
$A$ constant function is continuous and differentiable everywhere in its domain.
Therefore,$f+g$ is continuous at $x = \frac{2}{3}$.
Since $f$ and $g$ are Dirichlet-type functions,they are discontinuous at every point in $[0,1]$.
Thus,option $B$ is correct.

(D) function $f(x)$ defined as $f(x) = \begin{cases} g(x), & x \in \mathbb{Q} \\ h(x), & x \notin \mathbb{Q} \end{cases}$ is continuous at $x = a$ if and only if $g(a) = h(a)$.
Here,$g(x) = \sin |x|$ and $h(x) = 0$.
For continuity,we require $\sin |x| = 0$.
This occurs when $|x| = n\pi$ for some integer $n$,which implies $x = n\pi$ for $n \in \mathbb{Z}$.
At any point $x = k\pi$ (where $k \in \mathbb{Z}$),$f(x) = \sin |k\pi| = 0$.
For any other point $x \neq k\pi$,$\sin |x| \neq 0$,so the function is discontinuous because the values for rational and irrational numbers do not coincide.
Thus,$f$ is continuous only at $x = k\pi$ and discontinuous everywhere else.
Therefore,option $D$ is correct.

(A) Given,$f(x)=\frac{\tan \{\pi[x-\frac{\pi}{2}]\}}{2+[x]^{2}}$.
Since $[x-\frac{\pi}{2}]$ is an integer for all $x$,let $[x-\frac{\pi}{2}] = k$,where $k \in \mathbb{Z}$.
Then the numerator becomes $\tan(\pi k)$,which is equal to $0$ for all integers $k$.
Since the denominator $2+[x]^{2}$ is always $\geq 2$ and never zero,the function simplifies to $f(x) = \frac{0}{2+[x]^{2}} = 0$ for all $x \in \mathbb{R}$.
$A$ constant function $f(x) = 0$ is continuous and differentiable for all values of $x$.
Therefore,the function is continuous for all values of $x$.

(B, C) The function $f(x) = \frac{x^3}{4} - \sin(\pi x) + 3$ is continuous on the interval $[-2, 2]$.
Evaluating at the endpoints:
$f(-2) = \frac{(-2)^3}{4} - \sin(-2\pi) + 3 = \frac{-8}{4} - 0 + 3 = -2 + 3 = 1$.
$f(2) = \frac{2^3}{4} - \sin(2\pi) + 3 = \frac{8}{4} - 0 + 3 = 2 + 3 = 5$.
By the Intermediate Value Theorem,since $f(x)$ is continuous on $[-2, 2]$,it must attain every value in the interval $[f(-2), f(2)]$,which is $[1, 5]$.
Since $2 \frac{1}{3} = \frac{7}{3} \approx 2.33$ and $3 \frac{1}{4} = 3.25$ both lie within the interval $[1, 5]$,the function $f(x)$ attains both these values.
Thus,both options $B$ and $C$ are correct.

(C, D) Given the function $f(x) = x^3$ on the interval $x \in [-1, 1]$.
$1$. Check for extrema: $f'(x) = 3x^2$. Setting $f'(x) = 0$ gives $x = 0$. Since $f'(x) \ge 0$ for all $x \in [-1, 1]$,the function is strictly increasing. Thus,it has no local minimum or maximum at $x = 0$. The absolute minimum is at $x = -1$ $(f(-1) = -1)$ and the absolute maximum is at $x = 1$ $(f(1) = 1)$. So,options $A$ and $B$ are incorrect.
$2$. Continuity: The function $f(x) = x^3$ is a polynomial function,which is continuous for all real numbers,including the interval $[-1, 1]$. Thus,option $C$ is correct.
$3$. Boundedness: Since $f(x)$ is continuous on the closed interval $[-1, 1]$,by the Extreme Value Theorem,it is bounded. Specifically,$-1 \le f(x) \le 1$ for all $x \in [-1, 1]$. Thus,option $D$ is correct.
Therefore,the correct statements are $C$ and $D$.

(A) Given $f(x) = x^2 + x \sin x - \cos x$.
We find the derivative: $f'(x) = 2x + (\sin x + x \cos x) + \sin x = 2x + 2 \sin x + x \cos x = x(2 + \cos x) + 2 \sin x$.
For $x > 0$,$f'(x) = x(2 + \cos x) + 2 \sin x$. Since $2 + \cos x > 0$ and $x > 0$,and $2 \sin x$ is not always negative,we analyze the behavior. Alternatively,note that $f(0) = -1$.
As $x \to \infty$,$f(x) \to \infty$. Since $f(0) = -1 < 0$ and $f(x)$ is continuous,by the Intermediate Value Theorem,there exists at least one root in $(0, \infty)$.
Similarly,as $x \to -\infty$,$f(x) \to \infty$. Since $f(0) = -1 < 0$,there exists at least one root in $(-\infty, 0)$.
Thus,$f(x) = 0$ has at least two real roots (one positive and one negative),which implies it has at least one real root.

(A) Given $f(x) = \lim_{\theta \rightarrow 0} \frac{\cos \pi x - x^{(2/\theta)} \sin(x-1)}{1 + x^{(2/\theta)}(x-1)}$.
Case $1$: $|x| < 1$. As $\theta \rightarrow 0$,$x^{(2/\theta)} \rightarrow 0$. Thus,$f(x) = \cos \pi x$.
Case $2$: $|x| > 1$. As $\theta \rightarrow 0$,$x^{(2/\theta)} \rightarrow \infty$. Dividing numerator and denominator by $x^{(2/\theta)}$,we get $f(x) = \frac{-\sin(x-1)}{x-1}$.
At $x=1$: $LHL = \lim_{x \rightarrow 1^-} \cos \pi x = -1$. $RHL = \lim_{x \rightarrow 1^+} \frac{-\sin(x-1)}{x-1} = -1$. Since $LHL = RHL = f(1) = -1$,$f(x)$ is continuous at $x=1$. Statement $(I)$ is False.
At $x=-1$: $LHL = \lim_{x \rightarrow -1^-} \frac{-\sin(x-1)}{x-1} = \frac{-\sin(-2)}{-2} = \frac{-\sin 2}{2}$. $RHL = \lim_{x \rightarrow -1^+} \cos \pi x = \cos(-\pi) = -1$. Since $LHL \neq RHL$,$f(x)$ is discontinuous at $x=-1$. Statement $(II)$ is False.
Therefore,neither $(I)$ nor $(II)$ is true.

(A) For the function to be continuous at $x=0$,$f(0) = \lim_{x \to 0} f(x)$.
$f(0) = \lim_{x \to 0} \frac{e^{\tan x} - e^x + \ln(\sec x + \tan x) - x}{\tan x - x}$.
Using the Taylor series expansions near $x=0$:
$e^{\tan x} = 1 + (x + \frac{x^3}{3}) + \frac{(x + \frac{x^3}{3})^2}{2} + \frac{(x + \frac{x^3}{3})^3}{6} + \dots = 1 + x + \frac{x^2}{2} + \frac{2x^3}{3} + O(x^4)$.
$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$.
$\ln(\sec x + \tan x) = x + \frac{x^3}{6} + O(x^5)$.
Substituting these into the numerator:
$N = (1 + x + \frac{x^2}{2} + \frac{2x^3}{3}) - (1 + x + \frac{x^2}{2} + \frac{x^3}{6}) + (x + \frac{x^3}{6}) - x = \frac{2x^3}{3} + O(x^4)$.
The denominator is $\tan x - x = (x + \frac{x^3}{3} + O(x^5)) - x = \frac{x^3}{3} + O(x^5)$.
Thus,$f(0) = \lim_{x \to 0} \frac{\frac{2x^3}{3}}{\frac{x^3}{3}} = \frac{2}{1} = 2$.

(C) For the function to be continuous at $x = 0$,we must have $f(0) = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x)$.
Given $f(0) = a$.
For $x > 0$,$f(x) = \frac{\sin x - \frac{1}{2} \sin 2x}{x^3} = \frac{\sin x - \sin x \cos x}{x^3} = \frac{\sin x(1 - \cos x)}{x^3} = \frac{\sin x}{x} \cdot \frac{2 \sin^2(x/2)}{x^2} = \frac{\sin x}{x} \cdot \frac{2 \sin^2(x/2)}{4(x/2)^2} = 1 \cdot \frac{1}{2} = \frac{1}{2}$.
Thus,$a = \frac{1}{2}$.
For $x < 0$,$f(x) = b^2 \sin \left(\frac{\pi}{2} \left[\frac{\pi}{2}(\cos x + \sin x) \cos x\right]\right)$. As $x \to 0^-$,$\cos x \to 1$ and $\sin x \to 0$. The expression inside the greatest integer function becomes $\frac{\pi}{2}(1 + 0)(1) = \frac{\pi}{2} \approx 1.57$. Since $x < 0$,$\cos x < 1$ and $\sin x < 0$,so the value is slightly less than $\frac{\pi}{2}$. Thus,$[\frac{\pi}{2}(\cos x + \sin x) \cos x] = 1$.
So,$\lim_{x \to 0^-} f(x) = b^2 \sin(\frac{\pi}{2} \cdot 1) = b^2$.
Equating the limits,$b^2 = \frac{1}{2}$.
Therefore,$a^2 + b^2 = (\frac{1}{2})^2 + \frac{1}{2} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$.

(C) The function $g(x) = |x|[x^2]$ is discontinuous where $[x^2]$ is discontinuous,provided $|x| \neq 0$.
$[x^2]$ is discontinuous at $x^2 \in \mathbb{Z}$.
For $x \in (-2, 2)$,$x^2 \in [0, 4)$.
The integers in $[0, 4)$ are $0, 1, 2, 3$.
Thus,$x^2 = 1, 2, 3$ gives $x = \pm 1, \pm \sqrt{2}, \pm \sqrt{3}$.
At $x=0$,$g(x) = |x|[x^2] = 0 \cdot [0] = 0$,and $\lim_{x \to 0} |x|[x^2] = 0$,so $g(x)$ is continuous at $x=0$.
Thus,$S = \{-1, 1, -\sqrt{2}, \sqrt{2}, -\sqrt{3}, \sqrt{3}\}$.
Now,$f(x) = \min \{\sqrt{2}x, x^2\}$.
$f(-1) = \min \{-\sqrt{2}, 1\} = -\sqrt{2}$.
$f(1) = \min \{\sqrt{2}, 1\} = 1$.
$f(-\sqrt{2}) = \min \{-2, 2\} = -2$.
$f(\sqrt{2}) = \min \{2, 2\} = 2$.
$f(-\sqrt{3}) = \min \{-\sqrt{6}, 3\} = -\sqrt{6}$.
$f(\sqrt{3}) = \min \{\sqrt{6}, 3\} = \sqrt{6}$.
Summing these values: $\sum_{x \in S} f(x) = -\sqrt{2} + 1 - 2 + 2 - \sqrt{6} + \sqrt{6} = 1 - \sqrt{2}$.

(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = b$.
First,consider the right-hand limit $(x > 0)$:
$\lim_{x \to 0^+} \frac{ax + x^2 - 2\sin x \cos x}{x} = \lim_{x \to 0^+} \left( a + x - \frac{\sin(2x)}{x} \right) = a + 0 - 2 = a - 2$.
Next,consider the left-hand limit $(x < 0)$:
Let $x = -h$ where $h > 0$. As $x \to 0^-$,$h \to 0^+$.
$\lim_{h \to 0^+} \frac{a|-h| + (-h)^2 - 2\sin|-h|\cos|-h|}{-h} = \lim_{h \to 0^+} \frac{ah + h^2 - 2\sin h \cos h}{-h} = \lim_{h \to 0^+} \left( -a - h + \frac{\sin(2h)}{h} \right) = -a - 0 + 2 = -a + 2$.
For continuity,$a - 2 = -a + 2 = b$.
From $a - 2 = -a + 2$,we get $2a = 4$,so $a = 2$.
Substituting $a = 2$ into $b = a - 2$,we get $b = 2 - 2 = 0$.
Therefore,$a + b = 2 + 0 = 2$.

(B) For $f(x)$ to be continuous at $x = -\frac{3}{2}$,the limit $\lim_{x \to -\frac{3}{2}} f(x)$ must exist and equal $f(-\frac{3}{2}) = b$.
Since the denominator $(2x-1)(2x+3)$ approaches $0$ as $x \to -\frac{3}{2}$,the numerator $ax^2 + 2ax + 3$ must also approach $0$.
$a(-\frac{3}{2})^2 + 2a(-\frac{3}{2}) + 3 = 0$ $\Rightarrow \frac{9a}{4} - 3a + 3 = 0$ $\Rightarrow -\frac{3a}{4} = -3$ $\Rightarrow a = 4$.
Substituting $a=4$,$f(x) = \frac{4x^2+8x+3}{(2x-1)(2x+3)} = \frac{(2x+1)(2x+3)}{(2x-1)(2x+3)} = \frac{2x+1}{2x-1}$ for $x \neq -\frac{3}{2}, \frac{1}{2}$.
Now,$f(f(x)) = f\left(\frac{2x+1}{2x-1}\right) = \frac{2(\frac{2x+1}{2x-1}) + 1}{2(\frac{2x+1}{2x-1}) - 1} = \frac{4x+2+2x-1}{4x+2-2x+1} = \frac{6x+1}{2x+3}$.
Given $f(f(x)) = \frac{7}{5}$,we have $\frac{6x+1}{2x+3} = \frac{7}{5}$.
$5(6x+1) = 7(2x+3)$ $\Rightarrow 30x + 5 = 14x + 21$ $\Rightarrow 16x = 16$ $\Rightarrow x = 1$.

(B) For $f(x)$ to be continuous at $x=2$,the condition $\lim_{x \to 2} f(x) = f(2) = k$ must be satisfied.
First,we factor the numerator $x^3+x^2-16x+20$. Since $x=2$ is a root (as $2^3+2^2-16(2)+20 = 8+4-32+20 = 0$),we divide by $(x-2)$:
$x^3+x^2-16x+20 = (x-2)(x^2+3x-10)$.
Further factoring the quadratic term:
$(x^2+3x-10) = (x-2)(x+5)$.
So,$x^3+x^2-16x+20 = (x-2)^2(x+5)$.
For $x \neq 2$,$f(x) = \frac{(x-2)^2(x+5)}{(x-2)^2} = x+5$.
Now,calculating the limit: $\lim_{x \to 2} f(x) = \lim_{x \to 2} (x+5) = 2+5 = 7$.
Since the function is continuous at $x=2$,$k = 7$.

(C) For a function $f(x)$ to be continuous at $x = a$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function at $x = a$ must be equal.
Here,$a = \pi$.
$LHL$: $\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx + 1) = k\pi + 1$.
$RHL$: $\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (\cos x) = \cos(\pi) = -1$.
Since the function is continuous at $x = \pi$,we have $LHL$ = $RHL$.
Therefore,$k\pi + 1 = -1$.
$k\pi = -2$.
$k = -\frac{2}{\pi}$.

(A) First,analyze the continuity of $f(x)$ at $x=0$.
$f(0^-) = \lim_{x \to 0^-} (x^3+8) = 8$.
$f(0^+) = \lim_{x \to 0^+} (x^2-4) = -4$.
Since $f(0^-) \neq f(0^+)$,$f(x)$ is discontinuous at $x=0$.
Now,consider the composite function $g(f(x))$.
At $x=0$,the left-hand limit is $g(f(0^-)) = g(8)$. Since $8 \ge 0$,$g(8) = (8+4)^{1/2} = \sqrt{12} = 2\sqrt{3}$.
The right-hand limit is $g(f(0^+)) = g(-4)$. Since $-4 < 0$,$g(-4) = (-4-8)^{1/3} = (-12)^{1/3} = -\sqrt[3]{12}$.
Since $g(f(0^-)) \neq g(f(0^+))$,$g(f(x))$ is discontinuous at $x=0$.
Next,check for points where $f(x)$ takes values that make $g(f(x))$ discontinuous. $g(u)$ is continuous for all $u$ in its domain.
We check if $f(x)$ crosses the boundary $u=0$ where $g(u)$ changes definition.
For $x < 0$,$f(x) = x^3+8 = 0 \implies x = -2$. At $x=-2$,$f(-2)=0$. $g(f(-2)) = g(0) = (0+4)^{1/2} = 2$. The limit exists and is continuous.
For $x \ge 0$,$f(x) = x^2-4 = 0 \implies x = 2$. At $x=2$,$f(2)=0$. $g(f(2)) = g(0) = 2$. The limit exists and is continuous.
Thus,the only point of discontinuity is $x=0$. The number of points is $1$.

(D) Let $g(x) = x^2 - x - 0.5$. The function $f(x) = [g(x)]$ is discontinuous at points where $g(x)$ takes an integer value.
We need to find the number of points $x \in [2, 4]$ such that $g(x) = k$,where $k$ is an integer.
First,evaluate the range of $g(x)$ on the interval $[2, 4]$.
$g(2) = 2^2 - 2 - 0.5 = 1.5$.
$g(4) = 4^2 - 4 - 0.5 = 11.5$.
Since $g'(x) = 2x - 1$,for $x \in [2, 4]$,$g'(x) > 0$,so $g(x)$ is strictly increasing.
As $x$ varies from $2$ to $4$,$g(x)$ takes all values in the interval $[1.5, 11.5]$.
The integer values $k$ that $g(x)$ takes in this interval are $k \in \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$.
For each such integer $k$,there exists exactly one $x \in [2, 4]$ such that $g(x) = k$ because $g(x)$ is strictly increasing.
The number of such integers is $11 - 2 + 1 = 10$.
Thus,there are $10$ points of discontinuity in the interval $[2, 4]$.