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

(A) Let $f(x) = ax^3 + bx^2 + cx$.
Since $f(x)$ is a polynomial,it is continuous and differentiable on $\mathbb{R}$.
We observe that $f(0) = a(0)^3 + b(0)^2 + c(0) = 0$.
Given $a + b + c = 0$,we have $f(1) = a(1)^3 + b(1)^2 + c(1) = a + b + c = 0$.
Since $f(0) = f(1) = 0$,by Rolle's Theorem,there exists at least one $\alpha \in (0, 1)$ such that $f'(\alpha) = 0$.
Calculating the derivative,$f'(x) = 3ax^2 + 2bx + c$.
Thus,$f'(\alpha) = 3a\alpha^2 + 2b\alpha + c = 0$ for some $\alpha \in (0, 1)$.
Therefore,the quadratic equation $3ax^2 + 2bx + c = 0$ has at least one root in $[0, 1]$.

(A) Let $f(x) = ax^2 + bx + c$.
Define $F(x) = \int_{0}^{x} f(t) dt = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx$.
Clearly,$F(0) = 0$.
Also,$F(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6}$.
Given $2a + 3b + 6c = 0$,we have $F(1) = 0$.
Since $F(0) = F(1) = 0$ and $F(x)$ is a polynomial (and thus continuous and differentiable),by Rolle's Theorem,there exists at least one $x \in (0, 1)$ such that $F'(x) = 0$.
Since $F'(x) = f(x) = ax^2 + bx + c$,there exists at least one root of $ax^2 + bx + c = 0$ in the interval $(0, 1)$.

(D) Let $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x$.
Since $f(0) = 0$ and $f(\alpha) = 0$,where $\alpha > 0$,the function $f(x)$ satisfies the conditions of Rolle's Theorem on the interval $[0, \alpha]$.
According to Rolle's Theorem,there exists at least one root of the derivative $f'(x) = 0$ in the open interval $(0, \alpha)$.
The derivative is $f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1$.
Therefore,the equation $n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1 = 0$ must have at least one positive root smaller than $\alpha$.

(C) Given that $f$ is differentiable for all $x$,we can apply the Lagrange's Mean Value Theorem on the interval $[1, 6]$.
According to the theorem,there exists some $c \in (1, 6)$ such that $\frac{f(6) - f(1)}{6 - 1} = f'(c)$.
We are given $f'(x) \ge 2$ for all $x \in [1, 6]$,so $f'(c) \ge 2$.
Substituting the values,we get $\frac{f(6) - (-2)}{5} \ge 2$.
$\frac{f(6) + 2}{5} \ge 2$.
$f(6) + 2 \ge 10$.
$f(6) \ge 8$.

(B) Let $g(x) = f(x) - x^2$.
Since $f(1) = 1, f(2) = 4, f(3) = 9$,we have $g(1) = 1 - 1^2 = 0$,$g(2) = 4 - 2^2 = 0$,and $g(3) = 9 - 3^2 = 0$.
Thus,$g(x)$ has at least $3$ real roots at $x = 1, 2, 3$.
By Rolle's Theorem,$g'(x)$ has at least one root in $(1, 2)$ and at least one root in $(2, 3)$,meaning $g'(x)$ has at least $2$ roots in $(1, 3)$.
Applying Rolle's Theorem again to $g'(x)$,$g''(x)$ must have at least $1$ root in $(1, 3)$.
Since $g(x) = f(x) - x^2$,we have $g'(x) = f'(x) - 2x$ and $g''(x) = f''(x) - 2$.
Since $g''(c) = 0$ for some $c \in (1, 3)$,it follows that $f''(c) - 2 = 0$,or $f''(c) = 2$ for at least one $c \in (1, 3)$.

(A) By the Mean Value Theorem,there exists a $c \in (1, 5)$ such that $f'(c) = \frac{f(5) - f(1)}{5 - 1}$.
Given $f'(x) \ge 9$ for all $x \in (1, 5)$,we have $f'(c) \ge 9$.
Substituting the values,we get $\frac{f(5) - (-3)}{4} \ge 9$.
$f(5) + 3 \ge 36$.
$f(5) \ge 33$.

(D) Given $f(x) = x^2 - 2x + 4$.
First,find the derivative $f'(x) = 2x - 2$.
At $x = c$,$f'(c) = 2c - 2$.
Now,calculate $f(5)$ and $f(1)$:
$f(5) = 5^2 - 2(5) + 4 = 25 - 10 + 4 = 19$.
$f(1) = 1^2 - 2(1) + 4 = 1 - 2 + 4 = 3$.
Substitute these values into the given equation $\frac{f(5) - f(1)}{5 - 1} = f'(c)$:
$\frac{19 - 3}{5 - 1} = 2c - 2$.
$\frac{16}{4} = 2c - 2$.
$4 = 2c - 2$.
$2c = 6$.
$c = 3$.

(D) Rolle's theorem requires three conditions for a function $f(x)$ on an interval $[a, b]$:
$1$. $f(x)$ must be continuous on $[a, b]$.
$2$. $f(x)$ must be differentiable on $(a, b)$.
$3$. $f(a) = f(b)$.
Given $f(x) = \frac{x^2 - 3x}{x - 1}$.
The function is undefined at $x = 1$. Therefore,it is discontinuous at $x = 1$.
For any interval $[a, b]$ containing $x = 1$,the function fails the continuity condition.
In the given options:
- For $[0, 3]$,$1 \in (0, 3)$,so it is not continuous.
- For $[-3, 0]$,the function is continuous and differentiable,but $f(-3) = \frac{9+9}{-4} = -4.5$ and $f(0) = 0$. Since $f(-3) \neq f(0)$,Rolle's theorem does not apply.
- For $[1.5, 3]$,the function is continuous and differentiable,but $f(1.5) = \frac{2.25 - 4.5}{0.5} = -4.5$ and $f(3) = \frac{9 - 9}{2} = 0$. Since $f(1.5) \neq f(3)$,Rolle's theorem does not apply.
Thus,there is no interval among the choices where all conditions are satisfied.

(B) According to the Lagrange's Mean Value Theorem,there exists a point $c \in (a, b)$ such that $\frac{f(b) - f(a)}{b - a} = f'(c)$.
Given $f(x) = e^x$,$a = 0$,and $b = 1$.
First,calculate $f(a) = e^0 = 1$ and $f(b) = e^1 = e$.
The derivative is $f'(x) = e^x$,so $f'(c) = e^c$.
Substituting these values into the formula:
$\frac{e - 1}{1 - 0} = e^c$
$e - 1 = e^c$
Taking the natural logarithm on both sides:
$\log(e - 1) = c$.
Thus,the value of $c$ is $\log(e - 1)$.

(B) The function is defined as $f(x) = \begin{cases} -x, & \text{if } -1 \le x < 0 \\ x, & \text{if } 0 \le x \le 1 \end{cases}$.
First,we check the conditions for Rolle's theorem:
$1$. $f(x)$ is continuous on $[-1, 1]$.
$2$. $f(-1) = |-1| = 1$ and $f(1) = |1| = 1$,so $f(-1) = f(1)$.
$3$. We check for differentiability at $x = 0$:
Right-hand derivative: $Rf'(0) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$.
Left-hand derivative: $Lf'(0) = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$.
Since $Rf'(0) \neq Lf'(0)$,the function is not differentiable at $x = 0$,which lies in the interval $(-1, 1)$.
Therefore,Rolle's theorem is not applicable because the function is not differentiable on $(-1, 1)$.

(C) The Mean Value Theorem states that there exists a point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Given $f(x) = \cos x$ on $[0, \frac{\pi}{2}]$,we have $a = 0$ and $b = \frac{\pi}{2}$.
$f(a) = f(0) = \cos(0) = 1$.
$f(b) = f(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0$.
Now,calculate the slope: $\frac{f(b) - f(a)}{b - a} = \frac{0 - 1}{\frac{\pi}{2} - 0} = -\frac{2}{\pi}$.
Since $f'(x) = -\sin x$,we set $f'(c) = -\sin c = -\frac{2}{\pi}$.
Therefore,$\sin c = \frac{2}{\pi}$,which implies $c = \sin^{-1}\left(\frac{2}{\pi}\right)$.

(A) Given $f(x) = \frac{1}{x}$.
Then $f'(x) = -\frac{1}{x^2}$.
According to the Mean Value Theorem,$f'(x_1) = \frac{f(b) - f(a)}{b - a}$.
Substituting the values,we get $-\frac{1}{x_1^2} = \frac{\frac{1}{b} - \frac{1}{a}}{b - a}$.
Simplifying the right side: $\frac{\frac{a - b}{ab}}{b - a} = \frac{-(b - a)}{ab(b - a)} = -\frac{1}{ab}$.
Thus,$-\frac{1}{x_1^2} = -\frac{1}{ab}$,which implies $x_1^2 = ab$.
Since $a < x_1 < b$,we have $x_1 = \sqrt{ab}$.

(C) To determine $c$ in Rolle's theorem,we solve $f'(c) = 0$.
Given $f(x) = (x^2 + 3x)e^{-(1/2)x}$.
Using the product rule,$f'(x) = (2x + 3)e^{-(1/2)x} + (x^2 + 3x)e^{-(1/2)x} \cdot (-1/2)$.
$f'(x) = e^{-(1/2)x} \left[ 2x + 3 - \frac{1}{2}x^2 - \frac{3}{2}x \right]$.
$f'(x) = e^{-(1/2)x} \left[ -\frac{1}{2}x^2 + \frac{1}{2}x + 3 \right]$.
Setting $f'(c) = 0$,we get $-\frac{1}{2}c^2 + \frac{1}{2}c + 3 = 0$.
Multiplying by $-2$,we get $c^2 - c - 6 = 0$.
$(c - 3)(c + 2) = 0$.
Thus,$c = 3$ or $c = -2$.
Since $c$ must lie in the interval $(-3, 0)$,we reject $c = 3$.
Therefore,$c = -2$.

(B) For Rolle's theorem to be applicable to a function $f(x)$ on an interval $[a, b]$,the following conditions must be met:
$1$. $f(x)$ must be continuous on $[a, b]$.
$2$. $f(x)$ must be differentiable on $(a, b)$.
$3$. $f(a) = f(b)$.
Given $f(x) = x^2 - 4$,which is a polynomial function,it is continuous and differentiable everywhere.
We check the condition $f(a) = f(b)$ for the given options:
For option $(b)$,the interval is $[-2, 2]$.
$f(-2) = (-2)^2 - 4 = 4 - 4 = 0$.
$f(2) = (2)^2 - 4 = 4 - 4 = 0$.
Since $f(-2) = f(2)$,Rolle's theorem is applicable on the interval $[-2, 2]$.

(B) The Mean Value Theorem states that there exists at least one $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Given $f(x) = x + \frac{1}{x}$ on $[1, 3]$,we have $a = 1$ and $b = 3$.
$f(1) = 1 + \frac{1}{1} = 2$.
$f(3) = 3 + \frac{1}{3} = \frac{10}{3}$.
$f'(x) = 1 - \frac{1}{x^2}$.
Applying the theorem: $1 - \frac{1}{c^2} = \frac{\frac{10}{3} - 2}{3 - 1}$.
$1 - \frac{1}{c^2} = \frac{\frac{4}{3}}{2} = \frac{2}{3}$.
$\frac{1}{c^2} = 1 - \frac{2}{3} = \frac{1}{3}$.
$c^2 = 3 \Rightarrow c = \sqrt{3}$ (since $c \in (1, 3)$).

(C) According to the Mean Value Theorem $(MVT)$,if a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$,then there exists at least one point ${x_1}$ in the open interval $(a, b)$ such that:
$f'({x_1}) = \frac{f(b) - f(a)}{b - a}$
By the definition of the Mean Value Theorem,the point ${x_1}$ must lie strictly between $a$ and $b$.
Therefore,$a < {x_1} < b$.
Thus,the correct option is $C$.

(B) By the Mean Value Theorem,there exists some $c \in (0, 2)$ such that $\frac{f(2) - f(0)}{2 - 0} = f'(c)$.
Given $f(0) = 0$,we have $\frac{f(2)}{2} = f'(c)$.
For any $x \in [0, 2]$,applying the Mean Value Theorem on $[0, x]$,there exists $c_x \in (0, x)$ such that $\frac{f(x) - f(0)}{x - 0} = f'(c_x)$.
Since $f(0) = 0$,this simplifies to $\frac{f(x)}{x} = f'(c_x)$,which implies $f(x) = x \cdot f'(c_x)$.
Taking the absolute value,$|f(x)| = |x| \cdot |f'(c_x)|$.
Given $|f'(x)| \le \frac{1}{2}$ for all $x \in [0, 2]$,we have $|f(x)| \le |x| \cdot \frac{1}{2}$.
Since the maximum value of $x$ in $[0, 2]$ is $2$,we get $|f(x)| \le 2 \cdot \frac{1}{2} = 1$.
Thus,$|f(x)| \le 1$.

(A) For Rolle's theorem to be satisfied on $[1, 3]$,we must have $f(1) = f(3)$.
First,calculate $f(1) = (1)^3 - 6(1)^2 + a(1) + b = 1 - 6 + a + b = a + b - 5$.
Next,calculate $f(3) = (3)^3 - 6(3)^2 + a(3) + b = 27 - 54 + 3a + b = 3a + b - 27$.
Equating $f(1) = f(3)$,we get $a + b - 5 = 3a + b - 27$.
Subtracting $b$ from both sides: $a - 5 = 3a - 27$.
Rearranging terms: $27 - 5 = 3a - a$,which gives $22 = 2a$,so $a = 11$.
Rolle's theorem also requires $f'(c) = 0$ for some $c \in (1, 3)$.
$f'(x) = 3x^2 - 12x + a = 3x^2 - 12x + 11$.
Setting $f'(c) = 0$: $3c^2 - 12c + 11 = 0$.
The roots are $c = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{12 \pm \sqrt{12}}{6} = 2 \pm \frac{\sqrt{3}}{3}$.
Since $2 - \frac{\sqrt{3}}{3} \approx 2 - 0.577 = 1.423$,which is in $(1, 3)$,the condition is satisfied for $a = 11$.
Since $b$ is not uniquely determined by Rolle's theorem conditions alone,we check the options. Option $A$ provides $a = 11$ and $b = -6$.

(A) Given the function $f(x) = e^{-2x} \sin 2x$.
First,we find the derivative $f'(x)$ using the product rule:
$f'(x) = \frac{d}{dx}(e^{-2x}) \cdot \sin 2x + e^{-2x} \cdot \frac{d}{dx}(\sin 2x)$
$f'(x) = -2e^{-2x} \sin 2x + e^{-2x} \cdot 2 \cos 2x$
$f'(x) = 2e^{-2x} (\cos 2x - \sin 2x)$.
According to Rolle's theorem,there exists a $c \in (0, \pi/2)$ such that $f'(c) = 0$.
Setting $f'(c) = 0$:
$2e^{-2c} (\cos 2c - \sin 2c) = 0$.
Since $2e^{-2c} \neq 0$ for any real $c$,we must have:
$\cos 2c - \sin 2c = 0$
$\cos 2c = \sin 2c$
$\tan 2c = 1$.
Since $c \in (0, \pi/2)$,then $2c \in (0, \pi)$.
The value of $2c$ for which $\tan 2c = 1$ is $2c = \pi/4$.
Therefore,$c = \pi/8$.

(B) According to the Fundamental Theorem of Calculus,$\int_1^2 f'(x) dx = [f(x)]_1^2 = f(2) - f(1)$.
Since $f(x)$ satisfies the conditions of Rolle's theorem on the interval $[1, 2]$,one of the necessary conditions is that $f(1) = f(2)$.
Therefore,$f(2) - f(1) = 0$.
Hence,$\int_1^2 f'(x) dx = 0$.

(D) Given the function $f(x) = x^3 - 6x^2 + ax + b$.
First,we find the derivative $f'(x) = 3x^2 - 12x + a$.
According to Rolle's theorem,there exists a point $c \in (1, 3)$ such that $f'(c) = 0$.
We are given that $f'\left( \frac{2\sqrt{3} + 1}{\sqrt{3}} \right) = 0$,which simplifies to $f'\left( 2 + \frac{1}{\sqrt{3}} \right) = 0$.
Substituting $x = 2 + \frac{1}{\sqrt{3}}$ into $f'(x) = 0$:
$3\left( 2 + \frac{1}{\sqrt{3}} \right)^2 - 12\left( 2 + \frac{1}{\sqrt{3}} \right) + a = 0$.
Expanding the terms:
$3\left( 4 + \frac{1}{3} + \frac{4}{\sqrt{3}} \right) - 24 - \frac{12}{\sqrt{3}} + a = 0$.
$12 + 1 + 4\sqrt{3} - 24 - 4\sqrt{3} + a = 0$.
$13 - 24 + a = 0$.
$-11 + a = 0$,which gives $a = 11$.

(A) Given $f(x) = \sqrt{x}$,$a = 4$,and $b = 9$.
First,calculate $f(a)$ and $f(b)$:
$f(4) = \sqrt{4} = 2$
$f(9) = \sqrt{9} = 3$
Next,find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$
According to the Mean Value Theorem:
$f'(c) = \frac{f(b) - f(a)}{b - a}$
Substitute the values:
$\frac{1}{2\sqrt{c}} = \frac{3 - 2}{9 - 4}$
$\frac{1}{2\sqrt{c}} = \frac{1}{5}$
$2\sqrt{c} = 5$
$\sqrt{c} = \frac{5}{2} = 2.5$
$c = (2.5)^2 = 6.25$
Thus,the value of $c$ is $6.25$.

(A) Given the curve $f(x) = x^3$ on the interval $[-2, 2]$.
According to the Mean Value Theorem,there exists at least one point $c \in (-2, 2)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Here,$a = -2$ and $b = 2$.
$f(2) = 2^3 = 8$ and $f(-2) = (-2)^3 = -8$.
The slope of the secant line is $\frac{f(2) - f(-2)}{2 - (-2)} = \frac{8 - (-8)}{4} = \frac{16}{4} = 4$.
The derivative is $f'(x) = 3x^2$.
Setting $f'(c) = 4$,we get $3c^2 = 4$.
$c^2 = \frac{4}{3}$,which implies $c = \pm \frac{2}{\sqrt{3}}$.

(B) According to the Mean Value Theorem,there exists a point $c \in (3, 4)$ such that $f'(c) = \frac{f(4) - f(3)}{4 - 3}$.
First,calculate the slope of the chord joining $(3, 0)$ and $(4, 1)$:
$m = \frac{1 - 0}{4 - 3} = 1$.
Now,find the derivative of $f(x) = (x - 3)^2$:
$f'(x) = 2(x - 3)$.
Set the derivative equal to the slope of the chord:
$2(c - 3) = 1 \Rightarrow c - 3 = \frac{1}{2} \Rightarrow c = \frac{7}{2}$.
Now,find the corresponding $y$-coordinate by substituting $x = \frac{7}{2}$ into the function:
$y = \left( \frac{7}{2} - 3 \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Thus,the point is $\left( \frac{7}{2}, \frac{1}{4} \right)$.

(D) Given the function $f(x) = \sqrt{x}$ on the interval $[4, 9]$.
First,calculate $f(a)$ and $f(b)$:
$f(4) = \sqrt{4} = 2$
$f(9) = \sqrt{9} = 3$
Next,find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2\sqrt{x}}$
According to the Mean Value Theorem,there exists a $c \in (4, 9)$ such that:
$f'(c) = \frac{f(b) - f(a)}{b - a}$
Substitute the values:
$\frac{1}{2\sqrt{c}} = \frac{3 - 2}{9 - 4}$
$\frac{1}{2\sqrt{c}} = \frac{1}{5}$
$2\sqrt{c} = 5$
$\sqrt{c} = 2.5$
$c = (2.5)^2 = 6.25$
Thus,the value of $c$ is $6.25$.

(D) Lagrange's Mean Value Theorem $(LMVT)$ states that for a function $f(x)$ to be applicable on $[a, b]$,it must be continuous on $[a, b]$ and differentiable on $(a, b)$.
Checking option $(d)$: $f(x) = |x|$.
This function is continuous on $[0, 1]$,but it is not differentiable at $x = 0$,which lies within the interval $[0, 1]$.
Therefore,$LMVT$ is not applicable to $f(x) = |x|$ on $[0, 1]$.

(B) According to Lagrange's Mean Value Theorem,there exists a point $c \in (1, 2)$ such that $f'(c) = \frac{f(2) - f(1)}{2 - 1}$.
Given $f(x) = x^3 - 6ax^2 + 5x$,we have:
$f(2) = 2^3 - 6a(2^2) + 5(2) = 8 - 24a + 10 = 18 - 24a$.
$f(1) = 1^3 - 6a(1^2) + 5(1) = 1 - 6a + 5 = 6 - 6a$.
Thus,the slope of the chord is $\frac{f(2) - f(1)}{2 - 1} = (18 - 24a) - (6 - 6a) = 12 - 18a$.
The derivative is $f'(x) = 3x^2 - 12ax + 5$.
Since the tangent at $x = \frac{7}{4}$ is parallel to the chord,$f'(\frac{7}{4}) = 12 - 18a$.
$3(\frac{7}{4})^2 - 12a(\frac{7}{4}) + 5 = 12 - 18a$.
$3(\frac{49}{16}) - 21a + 5 = 12 - 18a$.
$\frac{147}{16} + 5 - 12 = 21a - 18a$.
$\frac{147 - 112}{16} = 3a$.
$\frac{35}{16} = 3a$.
$a = \frac{35}{48}$.

(D) For Rolle's theorem to be applicable to $f(x)$ on $[0, 1]$,the following conditions must be satisfied:
$1$. $f(x)$ must be continuous on $[0, 1]$.
$2$. $f(x)$ must be differentiable on $(0, 1)$.
$3$. $f(0) = f(1)$.
Checking condition $3$: $f(1) = 1^\alpha \ln(1) = 0$ and $f(0) = 0$. Thus,$f(0) = f(1) = 0$ holds for any $\alpha$.
Checking condition $1$ (Continuity at $x=0$): We require $\lim_{x \to 0^+} x^\alpha \ln x = 0$. This limit exists and equals $0$ if and only if $\alpha > 0$.
Checking condition $2$ (Differentiability on $(0, 1)$): $f'(x) = \alpha x^{\alpha-1} \ln x + x^{\alpha-1} = x^{\alpha-1}(\alpha \ln x + 1)$. This is defined for all $x \in (0, 1)$ for any $\alpha$.
Among the given options,only $\alpha = 0.5$ satisfies the condition $\alpha > 0$.

(A) Given $g(x) = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx$. Note that $g'(x) = ax^2 + bx + c$.
Calculate $g(0) = 0$ and $g(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6}$.
Since $2a + 3b + 6c = 0$,we have $g(1) = 0$.
Since $g(0) = 0$ and $g(1) = 0$,and $g(x)$ is a polynomial (which is continuous and differentiable everywhere),Rolle's Theorem applies to $g(x)$ on $[0, 1]$.
Thus,Statement-$2$ is true.
By Rolle's Theorem,there exists at least one $c_0 \in (0, 1)$ such that $g'(c_0) = 0$.
Since $g'(x) = ax^2 + bx + c$,this means $ac_0^2 + bc_0 + c = 0$.
Therefore,the quadratic equation $ax^2 + bx + c = 0$ has at least one root in $(0, 1)$.
Thus,Statement-$1$ is true and Statement-$2$ is the correct explanation for Statement-$1$.

(B) Given function is $f(x) = \log(\sin x)$.
According to Lagrange's Mean Value Theorem $(LMVT)$,there exists a point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Here,$a = \frac{\pi}{6}$ and $b = \frac{5\pi}{6}$.
$f(a) = \log(\sin \frac{\pi}{6}) = \log(\frac{1}{2})$ and $f(b) = \log(\sin \frac{5\pi}{6}) = \log(\frac{1}{2})$.
Thus,$f'(c) = \frac{\log(1/2) - \log(1/2)}{\frac{5\pi}{6} - \frac{\pi}{6}} = 0$.
Now,$f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.
So,$\cot c = 0$.
Since $c \in (\frac{\pi}{6}, \frac{5\pi}{6})$,the value of $c$ is $\frac{\pi}{2}$.

(C) Let $f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$.
Since $f(x)$ is a polynomial,it is continuous on the closed interval $[\alpha, \beta]$ and differentiable on the open interval $(\alpha, \beta)$.
Given that $\alpha$ and $\beta$ are roots of $f(x) = 0$,we have $f(\alpha) = 0$ and $f(\beta) = 0$.
According to Rolle's Theorem,there exists at least one point $c \in (\alpha, \beta)$ such that $f'(c) = 0$.
The derivative of the polynomial is $f'(x) = na_nx^{n-1} + (n - 1)a_{n-1}x^{n-2} + \dots + a_1$.
Therefore,the equation $f'(x) = 0$ has at least one root in the interval $(\alpha, \beta)$.

(B) Given that $f(x)$ is continuous on $[2, 4]$ and differentiable on $(2, 4)$.
By Lagrange's Mean Value Theorem,there exists at least one $c \in (2, 4)$ such that $f'(c) = \frac{f(4) - f(2)}{4 - 2}$.
We are given $f'(x) \geq 6$ for all $x \in [2, 4]$,so $f'(c) \geq 6$.
Substituting the values,we get $\frac{f(4) - (-4)}{2} \geq 6$.
$f(4) + 4 \geq 12$.
$f(4) \geq 8$.

(A) The Mean Value Theorem states that for a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$,there exists at least one point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Given $f(x) = \log_e x$,$a = 1$,and $b = 3$.
The derivative is $f'(x) = \frac{1}{x}$.
Applying the formula:
$f'(c) = \frac{f(3) - f(1)}{3 - 1}$
$\frac{1}{c} = \frac{\log_e 3 - \log_e 1}{2}$
Since $\log_e 1 = 0$,we have:
$\frac{1}{c} = \frac{\log_e 3}{2}$
$c = \frac{2}{\log_e 3} = 2 \log_3 e$.
However,checking the options provided,the expression for $f'(c)$ itself is $\frac{1}{2} \log_e 3$. If the question asks for the value of $f'(c)$,the answer is $\frac{1}{2} \log_e 3$. If it asks for $c$,$c = \frac{2}{\log_e 3}$.

(D) For Rolle's theorem to be applicable on $[0, 1]$,the function $f(x)$ must be continuous on $[0, 1]$ and differentiable on $(0, 1)$,with $f(0) = f(1)$.
Given $f(0) = 0$ and $f(1) = 1^{\alpha} \log(1) = 0$,the condition $f(0) = f(1)$ is satisfied for any $\alpha$.
For continuity at $x = 0$,we must have $\lim_{x \to 0^+} f(x) = f(0) = 0$.
$\lim_{x \to 0^+} x^{\alpha} \log x = \lim_{x \to 0^+} \frac{\log x}{x^{-\alpha}}$.
Using $L$'$H$ôpital's rule,this limit is $\lim_{x \to 0^+} \frac{1/x}{-\alpha x^{-\alpha-1}} = \lim_{x \to 0^+} \frac{x^{\alpha}}{-\alpha} = 0$ if $\alpha > 0$.
Checking the options,for $\alpha = 1/2$,the limit is $0$,which satisfies the continuity condition.
Thus,$\alpha = 1/2$ is the correct value.

(D) Let $f(x) = 2x^3 + 3x + K$.
For $f(x)$ to have two real roots in the interval $[0, 1]$,by Rolle's Theorem,there must exist at least one point $c \in (0, 1)$ such that $f'(c) = 0$.
Calculating the derivative: $f'(x) = 6x^2 + 3$.
Since $6x^2 + 3 > 0$ for all real $x$,$f'(x)$ is never zero.
Therefore,$f(x)$ is a strictly increasing function on the interval $[0, 1]$.
$A$ strictly increasing function can have at most one real root in any given interval.
Thus,it is impossible for the equation $2x^3 + 3x + K = 0$ to have two real roots in the interval $[0, 1]$ for any real value of $K$.

(A) By the Lagrange's Mean Value Theorem $(LMVT)$,there exists a $c \in (1, 6)$ such that $f'(c) = \frac{f(6) - f(1)}{6 - 1}$.
Given $f'(x) \geq 2$ for all $x \in [1, 6]$,we have $f'(c) \geq 2$.
Substituting the values,we get $\frac{f(6) - (-2)}{5} \geq 2$.
$f(6) + 2 \geq 10$.
$f(6) \geq 8$.

(C) For Rolle's theorem to hold on $[1, 3]$,we must have $f(1) = f(3)$.
$f(1) = 1^3 - 6(1)^2 + a(1) + b = a + b - 5$.
$f(3) = 3^3 - 6(3)^2 + a(3) + b = 27 - 54 + 3a + b = 3a + b - 27$.
Equating $f(1) = f(3)$ gives $a + b - 5 = 3a + b - 27$,which simplifies to $2a = 22$,so $a = 11$.
Since $b$ cancels out in the equation $f(1) = f(3)$,$b$ can be any real number $(b \in R)$.
Also,$f'(x) = 3x^2 - 12x + a = 3x^2 - 12x + 11$.
By Rolle's theorem,$f'(c) = 0$ for $c = \frac{2\sqrt{3} + 1}{\sqrt{3}} = 2 + \frac{1}{\sqrt{3}}$.
$f'(c) = 3(2 + \frac{1}{\sqrt{3}})^2 - 12(2 + \frac{1}{\sqrt{3}}) + 11 = 3(4 + \frac{4}{\sqrt{3}} + \frac{1}{3}) - 24 - \frac{12}{\sqrt{3}} + 11 = 12 + 4\sqrt{3} + 1 - 24 - 4\sqrt{3} + 11 = 0$.
Thus,the condition $f'(c) = 0$ is satisfied for $a = 11$ and any $b \in R$.

(A) Let $f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x$.
We are given that $f(0) = 0$ and $f(\alpha) = 0$.
Since $f(x)$ is a polynomial,it is continuous on $[0, \alpha]$ and differentiable on $(0, \alpha)$.
By Rolle's Theorem,there exists at least one point $c \in (0, \alpha)$ such that $f'(c) = 0$.
Note that $f'(x) = na_nx^{n-1} + (n-1)a_{n-1}x^{n-2} + \dots + a_1$.
Therefore,the equation $f'(x) = 0$ has at least one positive root in the interval $(0, \alpha)$,which means the root is less than $\alpha$.

(A) Let $f(x) = ax^3 + bx^2 + cx + d$.
Then $f'(x) = 3ax^2 + 2bx + c$.
We are given $a + b + c = 0$.
Evaluate $f(0) = a(0)^3 + b(0)^2 + c(0) + d = d$.
Evaluate $f(1) = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d$.
Since $a + b + c = 0$,we have $f(1) = 0 + d = d$.
Thus,$f(0) = f(1) = d$.
Since $f(x)$ is a polynomial,it is continuous on $[0, 1]$ and differentiable on $(0, 1)$.
By Rolle's Theorem,there exists at least one $c \in (0, 1)$ such that $f'(c) = 0$.
Therefore,the equation $3ax^2 + 2bx + c = 0$ has at least one root in the interval $(0, 1)$.

(A) Let $f(x) = ax^3 + bx^2 + cx$.
Since $f(x)$ is a polynomial,it is continuous on $[0, 1]$ and differentiable on $(0, 1)$.
Calculate the values at the endpoints:
$f(0) = a(0)^3 + b(0)^2 + c(0) = 0$.
$f(1) = a(1)^3 + b(1)^2 + c(1) = a + b + c$.
Given that $a + b + c = 0$,we have $f(1) = 0$.
Since $f(0) = f(1) = 0$,by Rolle's Theorem,there exists at least one $c_0 \in (0, 1)$ such that $f'(c_0) = 0$.
$f'(x) = 3ax^2 + 2bx + c$.
Thus,the equation $3ax^2 + 2bx + c = 0$ has at least one root in the interval $(0, 1)$.

(C) Let $f(x) = ax^4 + bx^3 + cx^2 + dx + e$.
Then,$f'(x) = 4ax^3 + 3bx^2 + 2cx + d$.
We are given that $f'(x) = 0$ has at least one root.
Consider the values of $f(x)$ at $x=0$ and $x=3$:
$f(0) = a(0)^4 + b(0)^3 + c(0)^2 + d(0) + e = e$.
$f(3) = a(3)^4 + b(3)^3 + c(3)^2 + d(3) + e = 81a + 27b + 9c + 3d + e$.
We can rewrite $f(3)$ as $f(3) = 3(27a + 9b + 3c + d) + e$.
Since it is given that $27a + 9b + 3c + d = 0$,we have $f(3) = 3(0) + e = e$.
Thus,$f(0) = f(3) = e$.
Since $f(x)$ is a polynomial,it is continuous on $[0, 3]$ and differentiable on $(0, 3)$.
By Rolle's Theorem,there exists at least one $c \in (0, 3)$ such that $f'(c) = 0$.
Therefore,the equation $4ax^3 + 3bx^2 + 2cx + d = 0$ has at least one root in the interval $(0, 3)$.

(A) Let $f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$.
Then $f'(x) = ax^2 + bx + c$.
We observe that $f(0) = 0$.
Also,$f(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6}$.
Given $2a + 3b + 6c = 0$,we have $f(1) = \frac{0}{6} = 0$.
Since $f(x)$ is a polynomial,it is continuous on $[0, 1]$ and differentiable on $(0, 1)$.
Since $f(0) = f(1) = 0$,by Rolle's Theorem,there exists at least one $c_0 \in (0, 1)$ such that $f'(c_0) = 0$.
Thus,the equation $ax^2 + bx + c = 0$ has at least one root in the interval $(0, 1)$.

(C) Given that $f(x)$ is differentiable on $[2, 5]$.
By Lagrange's Mean Value Theorem $(LMVT)$,there exists at least one $c \in (2, 5)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Here,$a = 2$ and $b = 5$.
$f'(c) = \frac{f(5) - f(2)}{5 - 2} = \frac{1/2 - 1/5}{3}$.
$f'(c) = \frac{(5 - 2)/10}{3} = \frac{3/10}{3} = \frac{1}{10}$.

(C) For $x \in [2, 5]$,the function is defined as $f(x) = (x - 2) - (x - 5) = 3$.
Since $f(x) = 3$ is a constant function on the interval $[2, 5]$,its derivative $f'(x) = 0$ for all $x \in (2, 5)$.
Thus,$f'(4) = 0$,so Statement-$1$ is true.
For Statement-$2$,$f(x) = |x - 2| + |x - 5|$ is a sum of continuous functions,hence continuous on $[2, 5]$.
On the interval $(2, 5)$,$f(x) = 3$,which is a constant function and therefore differentiable.
Also,$f(2) = |2 - 2| + |2 - 5| = 3$ and $f(5) = |5 - 2| + |5 - 5| = 3$,so $f(2) = f(5)$.
Statement-$2$ is true.
Since $f(x)$ is constant on $[2, 5]$,its derivative is zero everywhere in $(2, 5)$,which directly explains why $f'(4) = 0$. Thus,Statement-$2$ is the correct explanation for Statement-$1$.

(C) Given the condition $2a + 3b + 6c = 0$.
Dividing by $6$,we get $\frac{a}{3} + \frac{b}{2} + c = 0$.
Let $f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$. Note that $f(x)$ is a polynomial function,so it is continuous on $[0, 1]$ and differentiable on $(0, 1)$.
Calculate the values at the endpoints:
$f(0) = 0$.
$f(1) = \frac{a}{3} + \frac{b}{2} + c = 0$.
Since $f(0) = f(1) = 0$,by Rolle's Theorem,there exists at least one $c_0 \in (0, 1)$ such that $f'(c_0) = 0$.
$f'(x) = ax^2 + bx + c$.
Therefore,the equation $ax^2 + bx + c = 0$ has at least one root in $(0, 1)$.

(D) The Mean Value Theorem $(MVT)$ requires a function to be continuous on $[a, b]$ and differentiable on $(a, b)$.
For option $D$,$f(x) = |x|$ is not differentiable at $x = 0$. Since $0 \in [0, 1]$,the function $f(x) = |x|$ is not differentiable on the interval $(0, 1)$.
Therefore,the Mean Value Theorem cannot be applied to $f(x) = |x|$ on $[0, 1]$ because it fails the differentiability condition at $x = 0$.

(A) Since $f(x)$ satisfies Rolle's theorem on $[1, 3]$,we have $f(1) = f(3)$.
$a(1)^3 + b(1)^2 + 11(1) - 6 = a(3)^3 + b(3)^2 + 11(3) - 6$
$a + b + 5 = 27a + 9b + 27$
$26a + 8b + 22 = 0 \implies 13a + 4b + 11 = 0 \dots (i)$
Now,$f'(x) = 3ax^2 + 2bx + 11$.
Given $f'\left( 2 + \frac{1}{\sqrt{3}} \right) = 0$,we substitute $x = 2 + \frac{1}{\sqrt{3}}$:
$3a\left( 2 + \frac{1}{\sqrt{3}} \right)^2 + 2b\left( 2 + \frac{1}{\sqrt{3}} \right) + 11 = 0$
$3a\left( 4 + \frac{1}{3} + \frac{4}{\sqrt{3}} \right) + 2b\left( 2 + \frac{1}{\sqrt{3}} \right) + 11 = 0$
$3a\left( \frac{13}{3} + \frac{4}{\sqrt{3}} \right) + 2b\left( 2 + \frac{1}{\sqrt{3}} \right) + 11 = 0$
$13a + 4\sqrt{3}a + 4b + \frac{2b}{\sqrt{3}} + 11 = 0$
Since $13a + 4b + 11 = 0$ from $(i)$,we have $4\sqrt{3}a + \frac{2b}{\sqrt{3}} = 0$.
$4\sqrt{3}a = -\frac{2b}{\sqrt{3}} \implies 12a = -2b \implies b = -6a$.
Substitute $b = -6a$ into $(i)$:
$13a + 4(-6a) + 11 = 0$
$13a - 24a + 11 = 0 \implies -11a = -11 \implies a = 1$.
Then $b = -6(1) = -6$.

(A) According to the Mean Value Theorem,there exists a value $c \in (1, 3)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Given $f(x) = \log_e x$,we have $f'(x) = \frac{1}{x}$.
Here,$a = 1$ and $b = 3$.
So,$f'(c) = \frac{f(3) - f(1)}{3 - 1} = \frac{\log_e 3 - \log_e 1}{2}$.
Since $\log_e 1 = 0$,we get $f'(c) = \frac{\log_e 3}{2}$.
Substituting $f'(c) = \frac{1}{c}$,we have $\frac{1}{c} = \frac{\log_e 3}{2}$.
Therefore,$c = \frac{2}{\log_e 3} = 2 \log_3 e$.

(D) The function is defined as $f(x) = |x - 2| + |x - 5|$.
We can express this piecewise as:
$f(x) = \begin{cases} -(x-2) - (x-5) = -2x + 7 & \text{if } x < 2 \\ (x-2) - (x-5) = 3 & \text{if } 2 \le x < 5 \\ (x-2) + (x-5) = 2x - 7 & \text{if } x \ge 5 \end{cases}$
For Statement-$1$:
In the interval $(2, 5)$,$f(x) = 3$. Therefore,$f'(x) = 0$ for all $x \in (2, 5)$.
Since $4 \in (2, 5)$,$f'(4) = 0$. Thus,Statement-$1$ is true.
For Statement-$2$:
$f(x)$ is a sum of continuous functions,so it is continuous everywhere. In $[2, 5]$,$f(x) = 3$,which is a constant function,so it is differentiable in $(2, 5)$.
Also,$f(2) = |2-2| + |2-5| = 3$ and $f(5) = |5-2| + |5-5| = 3$. Thus,$f(2) = f(5)$.
Statement-$2$ is true.
Statement-$2$ describes the conditions of Rolle's Theorem,which implies the existence of a point $c \in (2, 5)$ such that $f'(c) = 0$. Since $f(x)$ is constant on $[2, 5]$,$f'(x) = 0$ for all $x \in (2, 5)$,which confirms Statement-$1$. Thus,Statement-$2$ is a correct explanation for Statement-$1$.