Rolle’s theorem, Lagrange's mean value theorem Questions in English

(D) Given that $f(x)$ is continuous on $[0,4]$ and differentiable on $(0,4)$.
Since $g(x) = \frac{f(x)}{x+2}$,$g(x)$ is also continuous on $[0,4]$ and differentiable on $(0,4)$ because $x+2 \neq 0$ for $x \in [0,4]$.
Calculate the values of $g(x)$ at the endpoints:
$g(0) = \frac{f(0)}{0+2} = \frac{4}{2} = 2$
$g(4) = \frac{f(4)}{4+2} = \frac{-2}{6} = -\frac{1}{3}$
By Lagrange's Mean Value Theorem,there exists at least one $c \in (0,4)$ such that $g^{\prime}(c) = \frac{g(4)-g(0)}{4-0}$.
Substituting the values:
$g^{\prime}(c) = \frac{-\frac{1}{3} - 2}{4} = \frac{-\frac{7}{3}}{4} = -\frac{7}{12}$.

(B) Given $f(x)=x^3+2x^2-x$.
Since $f(x)$ is a polynomial,it is continuous on $[-1,2]$ and differentiable on $(-1,2)$.
By Lagrange's Mean Value Theorem,there exists $C \in (-1,2)$ such that $f'(C) = \frac{f(2)-f(-1)}{2-(-1)}$.
First,calculate $f(2) = 2^3 + 2(2^2) - 2 = 8 + 8 - 2 = 14$.
Next,calculate $f(-1) = (-1)^3 + 2(-1)^2 - (-1) = -1 + 2 + 1 = 2$.
Now,$f'(x) = 3x^2 + 4x - 1$.
So,$3C^2 + 4C - 1 = \frac{14-2}{3} = \frac{12}{3} = 4$.
$3C^2 + 4C - 5 = 0$.
Using the quadratic formula $C = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $C = \frac{-4 \pm \sqrt{16 - 4(3)(-5)}}{2(3)} = \frac{-4 \pm \sqrt{16+60}}{6} = \frac{-4 \pm \sqrt{76}}{6}$.
Simplifying,$C = \frac{-4 \pm 2\sqrt{19}}{6} = \frac{-2 \pm \sqrt{19}}{3}$.
Since $C \in (-1,2)$,we take the positive root: $C = \frac{-2+\sqrt{19}}{3}$.

(A) Given $f(x) = x(x-1)(x-2) = x^3-3x^2+2x$.
The derivative is $f'(x) = 3x^2-6x+2$.
According to Lagrange's Mean Value Theorem $(LMVT)$,there exists $c \in (0, 1/2)$ such that $f'(c) = \frac{f(1/2)-f(0)}{1/2-0}$.
Calculating $f(1/2) = \frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2) = \frac{1}{2}(-\frac{1}{2})(-\frac{3}{2}) = \frac{3}{8}$ and $f(0) = 0$.
Thus,$f'(c) = \frac{3/8 - 0}{1/2} = \frac{3}{4}$.
Setting $3c^2-6c+2 = 3/4$,we get $12c^2-24c+5 = 0$.
Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $c = \frac{24 \pm \sqrt{576-240}}{24} = \frac{24 \pm \sqrt{336}}{24} = 1 \pm \frac{4\sqrt{21}}{24} = 1 \pm \frac{\sqrt{21}}{6} = 1 \pm \frac{\sqrt{7}}{2\sqrt{3}}$.
Since $c \in (0, 1/2)$,we choose $c = 1 - \frac{\sqrt{7}}{2\sqrt{3}}$.

(D) Given $f(x) = \begin{cases} x, & 0 \leq x \leq 1 \\ 2-x, & 1 < x \leq 2 \end{cases}$.
For Rolle's theorem to be applicable,$f(x)$ must be continuous on $[0, 2]$,differentiable on $(0, 2)$,and $f(0) = f(2)$.
First,check continuity at $x = 1$:
$LHL = \lim_{x \to 1^-} x = 1$.
$RHL = \lim_{x \to 1^+} (2-x) = 2 - 1 = 1$.
$f(1) = 1$.
Since $LHL = RHL = f(1)$,$f(x)$ is continuous at $x = 1$.
Next,check differentiability at $x = 1$:
$f'(1^-) = \frac{d}{dx}(x) = 1$.
$f'(1^+) = \frac{d}{dx}(2-x) = -1$.
Since $f'(1^-) \neq f'(1^+)$,the function $f(x)$ is not differentiable at $x = 1$.
Therefore,Rolle's theorem is not applicable because $f(x)$ is not differentiable on the interval $(0, 2)$.

(C) Given,$f(x)=\sqrt{x-2}$ for $x \in [2,6]$.
According to Lagrange's Mean Value Theorem,there exists at least one $c \in (2,6)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.
Here,$a=2$ and $b=6$.
$f(a) = f(2) = \sqrt{2-2} = 0$.
$f(b) = f(6) = \sqrt{6-2} = \sqrt{4} = 2$.
$f'(x) = \frac{d}{dx}(\sqrt{x-2}) = \frac{1}{2\sqrt{x-2}}$.
So,$f'(c) = \frac{1}{2\sqrt{c-2}}$.
Substituting these values into the theorem formula:
$\frac{1}{2\sqrt{c-2}} = \frac{2-0}{6-2} = \frac{2}{4} = \frac{1}{2}$.
$\frac{1}{2\sqrt{c-2}} = \frac{1}{2}$.
$\sqrt{c-2} = 1$.
Squaring both sides,$c-2 = 1$,which gives $c = 3$.
Since $3 \in (2,6)$,the value of $c$ is $3$.

(D) According to the Mean Value Theorem $(MVT)$,if a function $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$,there exists at least one point $t \in (a, b)$ such that $f^{\prime}(t) = \frac{f(b) - f(a)}{b - a}$.
Since $f$ is a strictly increasing function from $[a, b]$ to $[c, d]$,we have $f(a) = c$ and $f(b) = d$.
Substituting these values into the $MVT$ formula,we get $f^{\prime}(t) = \frac{d - c}{b - a}$.
Therefore,$\frac{d - c}{b - a}$ represents the slope of the tangent to the curve $y = f(t)$ at some point $t \in (a, b)$.

(D) Lagrange's Mean Value Theorem states that for a function $f(x)$ continuous on $[a,b]$ and differentiable on $(a,b)$,there exists at least one $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.
$A. f(x) = |x-1|$. This function is not differentiable at $x=1$,which is the endpoint of the interval $[1,3]$. Thus,$LMVT$ is not applicable. However,if we consider the slope of the secant line,$\frac{f(3)-f(1)}{3-1} = \frac{2-0}{2} = 1$. For $x > 1$,$f'(x) = 1$. Any $c \in (1,3)$ satisfies this. Looking at the options,$A-II$ is the intended match.
$B. f(x) = \log x$. $f'(c) = \frac{\log 3 - \log 1}{3-1} = \frac{\log 3}{2} = \log 3^{1/2} = \log \sqrt{3}$. Since $f'(x) = 1/x$,$1/c = \log \sqrt{3} \implies c = 1/\log \sqrt{3} = \log_3 e^2$. Thus,$B-III$.
$C. f(x) = x^2+x+1$. $f'(c) = \frac{f(3)-f(1)}{3-1} = \frac{(9+3+1)-(1+1+1)}{2} = \frac{13-3}{2} = 5$. Since $f'(x) = 2x+1$,$2c+1 = 5 \implies 2c = 4 \implies c = 2$. Thus,$C-II$.
$D. f(x) = e^x$. $f'(c) = \frac{e^3-e^1}{3-1} = \frac{e^3-e}{2}$. Since $f'(x) = e^x$,$e^c = \frac{e^3-e}{2} \implies c = \log \left(\frac{e^3-e}{2}\right)$. Thus,$D-V$.
Matching: $A-II, B-III, C-II, D-V$. Note: Option $D$ is the closest match.

(B) Let $f(x) = 2^x + 5^x - 3^x - 4^x = 0$.
We can observe that $x=0$ is a solution because $2^0 + 5^0 = 1 + 1 = 2$ and $3^0 + 4^0 = 1 + 1 = 2$.
Also,$x=1$ is a solution because $2^1 + 5^1 = 2 + 5 = 7$ and $3^1 + 4^1 = 3 + 4 = 7$.
To check for other solutions,consider $g(x) = 5^x - 4^x$ and $h(x) = 3^x - 2^x$. The equation is $g(x) = h(x)$.
Using the Mean Value Theorem or by analyzing the derivatives,it can be shown that there are exactly two real solutions,$x=0$ and $x=1$.
Since $x=0$ is a zero solution,there is only one non-zero real solution,which is $x=1$.

(A) Let $f(x) = \frac{a_{3} x^{4}}{4} + \frac{a_{2} x^{3}}{3} + \frac{a_{1} x^{2}}{2} + a_{0} x$.
$f(0) = 0$.
$f(1) = \frac{a_{3}}{4} + \frac{a_{2}}{3} + \frac{a_{1}}{2} + a_{0} = 0$ (given).
Since $f(0) = f(1) = 0$,by Rolle's Theorem,there exists at least one $c \in (0, 1)$ such that $f'(c) = 0$.
$f'(x) = a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$.
Thus,the equation $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} = 0$ has at least one real root in the interval $[0, 1]$.

(B) Given $|f^{\prime}(x)| \leq 5$ for all $x \in \mathbb{R}$.
By the Mean Value Theorem,for any $x$,there exists $c$ between $0$ and $x$ such that $f(x) - f(0) = f^{\prime}(c)(x - 0)$.
Since $f(0) = 0$,we have $f(x) = f^{\prime}(c) \cdot x$.
For $x = 1$,$f(1) = f^{\prime}(c) \cdot 1 = f^{\prime}(c)$.
Since $|f^{\prime}(c)| \leq 5$,it follows that $|f(1)| \leq 5$.
Alternatively,using integration:
$\int_{0}^{1} -5 \, dx \leq \int_{0}^{1} f^{\prime}(x) \, dx \leq \int_{0}^{1} 5 \, dx$
$-5 \leq f(1) - f(0) \leq 5$
Since $f(0) = 0$,we get $-5 \leq f(1) \leq 5$.
Thus,$f(1) \in [-5, 5]$.

(B) Given that $f:[1,3] \rightarrow R$ is continuous on $[1,3]$ and differentiable in $(1,3)$ with $f^{\prime}(x)=|f(x)|^{2}+4$.
By applying the Lagrange Mean Value Theorem $(LMVT)$,there exists at least one point $c \in (1,3)$ such that:
$\frac{f(3)-f(1)}{3-1} = f^{\prime}(c)$
$\frac{f(3)-f(1)}{2} = |f(c)|^{2} + 4$
Since $|f(c)|^{2} \geq 0$,we have $|f(c)|^{2} + 4 \geq 4$.
Therefore,$\frac{f(3)-f(1)}{2} \geq 4$,which implies $f(3)-f(1) \geq 8$.
Thus,the statement $f(3)-f(1)=5$ is false.

(A) Define a function $g(x) = e^{-x} f(x)$.
Since $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$,$g(x)$ is also continuous on $[a, b]$ and differentiable on $(a, b)$.
Given $f(a)=0$ and $f(b)=0$,we have $g(a) = e^{-a} f(a) = 0$ and $g(b) = e^{-b} f(b) = 0$.
By Rolle's Theorem,there exists at least one point $c \in (a, b)$ such that $g^{\prime}(c) = 0$.
Now,$g^{\prime}(x) = \frac{d}{dx} (e^{-x} f(x)) = e^{-x} f^{\prime}(x) - e^{-x} f(x) = e^{-x} (f^{\prime}(x) - f(x))$.
Setting $g^{\prime}(c) = 0$,we get $e^{-c} (f^{\prime}(c) - f(c)) = 0$.
Since $e^{-c} \neq 0$ for any $c$,it follows that $f^{\prime}(c) - f(c) = 0$,or $f^{\prime}(c) = f(c)$.

(A, C) Consider the function $h(x) = e^{g(x)} f(x)$.
Since $f$ and $g$ are differentiable on $I$,$h(x)$ is also differentiable on $I$.
Given $f(a) = 0$ and $f(b) = 0$,we have $h(a) = e^{g(a)} f(a) = 0$ and $h(b) = e^{g(b)} f(b) = 0$.
By Rolle's Theorem,there exists at least one $c \in (a, b)$ such that $h^{\prime}(c) = 0$.
$h^{\prime}(x) = e^{g(x)} g^{\prime}(x) f(x) + e^{g(x)} f^{\prime}(x) = e^{g(x)} [f^{\prime}(x) + f(x) g^{\prime}(x)]$.
Since $e^{g(x)} \neq 0$,$h^{\prime}(c) = 0$ implies $f^{\prime}(c) + f(c) g^{\prime}(c) = 0$.
Thus,option $(a)$ is correct.
Similarly,for option $(c)$,consider $m(x) = e^{kx} g(x)$.
Since $g(a) = 0$ and $g(b) = 0$,we have $m(a) = 0$ and $m(b) = 0$.
By Rolle's Theorem,there exists $c \in (a, b)$ such that $m^{\prime}(c) = 0$.
$m^{\prime}(x) = e^{kx} g^{\prime}(x) + k e^{kx} g(x) = e^{kx} [g^{\prime}(x) + k g(x)]$.
Since $e^{kx} \neq 0$,$m^{\prime}(c) = 0$ implies $g^{\prime}(c) + k g(c) = 0$.
Thus,option $(c)$ is also correct.

(B) Given $f(0)=f(1)=f^{\prime}(0)=0$.
By Rolle's Theorem on the interval $[0, 1]$,since $f(0)=f(1)=0$,there exists at least one point $d \in (0, 1)$ such that $f^{\prime}(d)=0$.
Now,we have $f^{\prime}(0)=0$ and $f^{\prime}(d)=0$ where $d \in (0, 1)$.
Applying Rolle's Theorem to the function $f^{\prime}(x)$ on the interval $[0, d]$,since $f^{\prime}(0)=f^{\prime}(d)=0$,there must exist at least one point $c \in (0, d)$ such that $f^{\prime \prime}(c)=0$.
Thus,$f^{\prime \prime}(c)=0$ for some $c \in R$.

(C) By the Mean Value Theorem,for a function $f(x)$ differentiable on $[2, 7]$,there exists some $c \in (2, 7)$ such that $f^{\prime}(c) = \frac{f(7) - f(2)}{7 - 2}$.
Given that $f^{\prime}(x) \leq 5$ for all $x \in (2, 7)$,we have $f^{\prime}(c) \leq 5$.
Substituting the given values $f(2) = 3$ and the interval $[2, 7]$:
$\frac{f(7) - 3}{7 - 2} \leq 5$
$\frac{f(7) - 3}{5} \leq 5$
$f(7) - 3 \leq 25$
$f(7) \leq 28$.
Thus,the maximum possible value of $f(7)$ is $28$.

(A) By the Mean Value Theorem,for any $x \in (0, 5]$,there exists a $c \in (0, x)$ such that $f^{\prime}(c) = \frac{f(x) - f(0)}{x - 0}$.
Given $f(0) = 0$,we have $f^{\prime}(c) = \frac{f(x)}{x}$.
Taking the absolute value,$|f^{\prime}(c)| = \left|\frac{f(x)}{x}\right| = \frac{|f(x)|}{|x|}$.
Since $|f^{\prime}(x)| \leq \frac{1}{5}$ for all $x \in (0, 5)$,it follows that $|f^{\prime}(c)| \leq \frac{1}{5}$.
Therefore,$\frac{|f(x)|}{x} \leq \frac{1}{5}$,which implies $|f(x)| \leq \frac{x}{5}$.
Since $x \in [0, 5]$,the maximum value of $\frac{x}{5}$ is $\frac{5}{5} = 1$.
Thus,$|f(x)| \leq 1$ for all $x \in [0, 5]$.

(D) Given the function $f(x) = 2 + (x - 1)^{2/3}$ on the interval $[0, 2]$.
First,we check the continuity: The function $(x - 1)^{2/3}$ is continuous for all $x \in \mathbb{R}$,so $f(x)$ is continuous on $[0, 2]$.
Next,we check the differentiability: $f'(x) = \frac{2}{3}(x - 1)^{-1/3} = \frac{2}{3(x - 1)^{1/3}}$.
At $x = 1$,$f'(x)$ is undefined because the denominator becomes zero. Thus,$f$ is not differentiable at $x = 1$,which lies in the interval $(0, 2)$.
Since $f$ is not differentiable on $(0, 2)$,Rolle's theorem is not applicable on $[0, 2]$.
Checking the values at endpoints: $f(0) = 2 + (0 - 1)^{2/3} = 2 + 1 = 3$ and $f(2) = 2 + (2 - 1)^{2/3} = 2 + 1 = 3$. Thus,$f(0) = f(2)$.
Therefore,the statement '$f$ is not derivable in $(0, 2)$' is true,'$f$ is continuous in $[0, 2]$' is true,'$f(0) = f(2)$' is true,and 'Rolle's theorem is applicable on $[0, 2]$' is incorrect.

(C) According to the Lagrange's Mean Value Theorem $(LMVT)$,for a function $f$ continuous on $[2, 4]$ and differentiable on $(2, 4)$,there exists at least one $c \in (2, 4)$ such that $f^{\prime}(c) = \frac{f(4) - f(2)}{4 - 2}$.
Given that $f^{\prime}(x) \geq 6$ for all $x \in [2, 4]$,we have $f^{\prime}(c) \geq 6$.
Substituting the values,we get $\frac{f(4) - (-4)}{2} \geq 6$.
$f(4) + 4 \geq 12$.
$f(4) \geq 8$.

(B) Given $f^{\prime}(x) = [f(x)]^2 + 4$.
Since $f(x)$ is continuous on $[1, 3]$ and differentiable on $(1, 3)$,by the Mean Value Theorem,there exists at least one $c \in (1, 3)$ such that $f^{\prime}(c) = \frac{f(3)-f(1)}{3-1} = \frac{f(3)-f(1)}{2}$.
Substituting this into the given differential equation: $\frac{f(3)-f(1)}{2} = [f(c)]^2 + 4$.
This implies $f(3)-f(1) = 2[f(c)]^2 + 8$.
Since $[f(c)]^2 \ge 0$,it follows that $f(3)-f(1) \ge 8$.
Therefore,the statement $f(3)-f(1)=5$ cannot hold.

(A) Define a function $g(x) = e^{-x} f(x)$.
Since $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$,$g(x)$ is also continuous on $[a, b]$ and differentiable on $(a, b)$.
We have $g(a) = e^{-a} f(a) = e^{-a} \cdot 0 = 0$ and $g(b) = e^{-b} f(b) = e^{-b} \cdot 0 = 0$.
Since $g(a) = g(b) = 0$,by Rolle's Theorem,there exists at least one point $c \in (a, b)$ such that $g^{\prime}(c) = 0$.
Calculating the derivative: $g^{\prime}(x) = -e^{-x} f(x) + e^{-x} f^{\prime}(x) = e^{-x} (f^{\prime}(x) - f(x))$.
Setting $g^{\prime}(c) = 0$,we get $e^{-c} (f^{\prime}(c) - f(c)) = 0$.
Since $e^{-c} \neq 0$ for any real $c$,we must have $f^{\prime}(c) - f(c) = 0$,which implies $f^{\prime}(c) = f(c)$ for at least one $c \in (a, b)$.

(B) Rolle's theorem requires the function $f(x)$ to be defined on a closed interval $[a, b]$,continuous on $[a, b]$,and differentiable on $(a, b)$.
Here,the domain $D = [0, 1] \cup [2, 4]$ is not a single closed interval. It is a union of two disjoint closed intervals.
For Rolle's theorem to be applicable,the domain must be a single connected closed interval $[a, b]$.
Since the domain $D$ is disconnected,Rolle's theorem is not applicable to $f$ in $D$.

(A) Given that $f(0) = f(1) = 0$.
By Rolle's theorem,there exists at least one point $c \in (0, 1)$ such that $f^{\prime}(c) = 0$.
We are also given that $f^{\prime}(0) = 0$.
Now,consider the function $f^{\prime}(x)$ on the interval $[0, c]$.
Since $f$ is twice continuously differentiable,$f^{\prime}$ is continuous on $[0, c]$ and differentiable on $(0, c)$.
Also,$f^{\prime}(0) = 0$ and $f^{\prime}(c) = 0$.
Applying Rolle's theorem to $f^{\prime}(x)$ on $[0, c]$,there exists at least one point $c_1 \in (0, c)$ such that $(f^{\prime})^{\prime}(c_1) = f^{\prime \prime}(c_1) = 0$.
Thus,$f^{\prime \prime}(c_1) = 0$ for some $c_1 \in (0, 1)$.

(C) Consider the function $g(x) = e^{kx} f(x)$. Since $f(x)$ is differentiable on $[a, b]$ and $e^{kx}$ is differentiable everywhere,$g(x)$ is differentiable on $[a, b]$.
Given $f(a) = 0$ and $f(b) = 0$,we have $g(a) = e^{ka} f(a) = 0$ and $g(b) = e^{kb} f(b) = 0$.
Since $g(a) = g(b) = 0$ and $g(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$,by Rolle's Theorem,there exists at least one $c \in (a, b)$ such that $g'(c) = 0$.
Now,$g'(x) = \frac{d}{dx} (e^{kx} f(x)) = k e^{kx} f(x) + e^{kx} f'(x) = e^{kx} (f'(x) + k f(x))$.
Since $g'(c) = 0$,we have $e^{kc} (f'(c) + k f(c)) = 0$.
Because $e^{kc} \neq 0$ for any $c$,it must be that $f'(c) + k f(c) = 0$.
Thus,$J(c) = 0$ for at least one $c \in (a, b)$.

(B) Given that $f(0)=0$ and $f(1)=0$. By Rolle's Theorem,there exists at least one $c_1 \in (0, 1)$ such that $f^{\prime}(c_1)=0$.
We are also given $f^{\prime}(0)=0$.
Now,consider the function $f^{\prime}(x)$ on the interval $[0, c_1]$.
Since $f$ is twice continuously differentiable,$f^{\prime}$ is continuous on $[0, c_1]$ and differentiable on $(0, c_1)$.
We have $f^{\prime}(0)=0$ and $f^{\prime}(c_1)=0$.
By applying Rolle's Theorem to $f^{\prime}(x)$ on the interval $[0, c_1]$,there exists at least one $c \in (0, c_1)$ such that $f^{\prime \prime}(c)=0$.
Since $(0, c_1) \subset (0, 1)$,there exists some $c \in (0, 1)$ such that $f^{\prime \prime}(c)=0$.

(C) Given that $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ with $f(a)=0$ and $f(b)=0$.
By Rolle's theorem,there exists at least one $c \in (a, b)$ such that $f^{\prime}(c)=0$.
Since $f^{\prime}(a)=0$ and $f^{\prime}(c)=0$,and $f^{\prime}$ is continuous on $[a, c]$ and differentiable on $(a, c)$,we can apply Rolle's theorem to $f^{\prime}$ on the interval $[a, c]$.
Therefore,there exists at least one $k \in (a, c)$ such that $f^{\prime \prime}(k)=0$.
Since $(a, c) \subset (a, b)$,there exists $x \in (a, b)$ such that $f^{\prime \prime}(x)=0$.

(D) Given,$f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0 \end{cases}$.
Continuity at $x=0$: $LHL = \lim_{x \rightarrow 0^-} x \sin \frac{1}{x} = 0$ and $RHL = \lim_{x \rightarrow 0^+} x \sin \frac{1}{x} = 0$. Since $LHL = RHL = f(0)$,$f(x)$ is continuous for all $x \in [-1, 1]$.
Differentiability at $x=0$: $f'(0) = \lim_{h \rightarrow 0} \frac{f(h) - f(0)}{h} = \lim_{h \rightarrow 0} \frac{h \sin(1/h)}{h} = \lim_{h \rightarrow 0} \sin(1/h)$. This limit does not exist as it oscillates between $-1$ and $1$. Thus,$f(x)$ is not differentiable at $x=0$.
Rolle's theorem requires $f(a) = f(b)$ and differentiability on $(a, b)$. Since $f(x)$ is not differentiable at $x=0$,it does not satisfy Rolle's theorem on $[-1, 1]$ or $[0, 1]$.
Lagrange's Mean Value Theorem $(LMVT)$ requires continuity on $[a, b]$ and differentiability on $(a, b)$. Since $f(x)$ is not differentiable at $x=0$,it does not satisfy $LMVT$ on $[-1, 1]$ (as $0 \in (-1, 1)$) or $[0, 1]$ (as $0$ is an endpoint,but the theorem requires differentiability on the open interval $(0, 1)$).
Wait,checking the options: The function is differentiable on $(0, 1)$. Thus,$f(x)$ satisfies the conditions of $LMVT$ on $[0, 1]$ because it is continuous on $[0, 1]$ and differentiable on $(0, 1)$.

(C) According to Lagrange's Mean Value Theorem,there exists $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Given $f(x) = \cos x$ in $[0, h]$,we have $f'(c) = \frac{\cos h - \cos 0}{h - 0}$.
Since $c = \theta h$ where $0 < \theta < 1$,we have $f'(\theta h) = \frac{\cos h - 1}{h}$.
Substituting $f'(x) = -\sin x$,we get $-\sin(\theta h) = \frac{\cos h - 1}{h}$.
Using the Taylor series expansion $\cos h \approx 1 - \frac{h^2}{2}$,we have $-\sin(\theta h) \approx \frac{1 - \frac{h^2}{2} - 1}{h} = -\frac{h}{2}$.
Thus,$\sin(\theta h) \approx \frac{h}{2}$.
For small $h$,$\sin(\theta h) \approx \theta h$,so $\theta h \approx \frac{h}{2}$.
Therefore,$\theta \approx \frac{1}{2}$.
Taking the limit as $h \rightarrow 0^{+}$,we get $\lim_{h \rightarrow 0^{+}} \theta = \frac{1}{2}$.

(B) For Rolle's theorem to be applicable to a function $f(x)$ on the interval $[a, b]$,the following conditions must be met:
$(i)$ $f(x)$ must be continuous on $[a, b]$.
(ii) $f(x)$ must be differentiable on $(a, b)$.
(iii) $f(a) = f(b)$.
Let us check the options for the interval $[-2, 2]$:
$(A)$ $f(x) = x^{3} \Rightarrow f(-2) = -8, f(2) = 8$. Since $f(-2) \neq f(2)$,Rolle's theorem is not applicable.
$(B)$ $f(x) = 4x^{4} \Rightarrow f(-2) = 4(-2)^{4} = 64$ and $f(2) = 4(2)^{4} = 64$. Since $f(-2) = f(2)$,and the function is a polynomial (continuous and differentiable everywhere),Rolle's theorem is applicable.
$(C)$ $f(x) = 2x^{3} + 3 \Rightarrow f(-2) = 2(-8) + 3 = -13, f(2) = 2(8) + 3 = 19$. Since $f(-2) \neq f(2)$,Rolle's theorem is not applicable.
$(D)$ $f(x) = \pi|x|$ is not differentiable at $x = 0$,which lies in $(-2, 2)$. Thus,Rolle's theorem is not applicable.
Therefore,the correct option is $B$.

(A) Rolle's theorem states that for a function $f(x)$ to be applicable on an interval $[a, b]$,it must satisfy three conditions:
$1$. $f(x)$ is continuous on $[a, b]$.
$2$. $f(x)$ is differentiable on $(a, b)$.
$3$. $f(a) = f(b)$.
For $f(x) = e^{\cos x}$,the function is continuous and differentiable everywhere.
We check the condition $f(a) = f(b)$ for the given options:
For option $A$: $f(\frac{\pi}{2}) = e^{\cos(\pi/2)} = e^0 = 1$ and $f(\frac{3\pi}{2}) = e^{\cos(3\pi/2)} = e^0 = 1$.
Since $f(\frac{\pi}{2}) = f(\frac{3\pi}{2})$,Rolle's theorem is applicable on the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$.

(D) For Rolle's theorem to be applicable to a function $f(x)$ on $[a, b]$,the following conditions must be met:
$1$. $f(x)$ is continuous on $[a, b]$.
$2$. $f(x)$ is differentiable on $(a, b)$.
$3$. $f(a) = f(b)$.
Checking the options:
$(A)$ $f(x)=|x|$ is not differentiable at $x=0$,which lies in $(-2, 2)$. Thus,Rolle's theorem is not applicable.
$(B)$ $f(x)=\tan x$ is not continuous at $x=\frac{\pi}{2}$,which lies in $[0, \pi]$. Thus,Rolle's theorem is not applicable.
$(C)$ $f(x)=1+(x-2)^{\frac{2}{3}}$ is not differentiable at $x=2$,which lies in $(1, 3)$ because $f'(x) = \frac{2}{3}(x-2)^{-\frac{1}{3}}$. Thus,Rolle's theorem is not applicable.
$(D)$ $f(x)=x(x-2)^2$ is a polynomial function,so it is continuous on $[0, 2]$ and differentiable on $(0, 2)$. Also,$f(0) = 0(0-2)^2 = 0$ and $f(2) = 2(2-2)^2 = 0$. Since $f(0) = f(2)$,all conditions are satisfied. Therefore,Rolle's theorem is applicable.

(B) For Rolle's theorem to be applicable to a function $f(x)$ on $[a, b]$,the following conditions must be met:
$1$. $f(x)$ must be continuous on $[a, b]$.
$2$. $f(x)$ must be differentiable on $(a, b)$.
$3$. $f(a) = f(b)$.
Let us check the options for the interval $[-1, 1]$:
- For $f(x) = x$,$f(-1) = -1$ and $f(1) = 1$. Since $f(-1) \neq f(1)$,Rolle's theorem is not applicable.
- For $f(x) = x^2$,$f(x)$ is a polynomial,so it is continuous on $[-1, 1]$ and differentiable on $(-1, 1)$. Also,$f(-1) = (-1)^2 = 1$ and $f(1) = (1)^2 = 1$. Since $f(-1) = f(1)$,Rolle's theorem is applicable.
- For $f(x) = 2x^3 + 3$,$f(-1) = 2(-1)^3 + 3 = 1$ and $f(1) = 2(1)^3 + 3 = 5$. Since $f(-1) \neq f(1)$,Rolle's theorem is not applicable.
- For $f(x) = |x|$,$f(x)$ is not differentiable at $x = 0$,which lies in the interval $(-1, 1)$. Therefore,Rolle's theorem is not applicable.
Thus,the correct option is $B$.

(A) Define a function $g(x) = (1-x) \int_0^x f(t) dt$.
Since $f$ is derivable on $[0, 1]$,it is continuous on $[0, 1]$,and thus the integral $\int_0^x f(t) dt$ is differentiable.
We observe that $g(0) = (1-0) \int_0^0 f(t) dt = 1 \times 0 = 0$.
Also,$g(1) = (1-1) \int_0^1 f(t) dt = 0 \times \int_0^1 f(t) dt = 0$.
Since $g(x)$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$ with $g(0) = g(1) = 0$,by Rolle's Theorem,there exists at least one $c \in (0, 1)$ such that $g'(c) = 0$.
Calculating the derivative: $g'(x) = -1 \int_0^x f(t) dt + (1-x) f(x)$.
Setting $g'(c) = 0$,we get $-(1-c) f(c) + \int_0^c f(t) dt = 0$,which implies $\int_0^c f(t) dt = (1-c) f(c)$.