Ex 5.14 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution:

Since \( f(x) = x^{2} – 5x + 6 \) is a polynomial, it is continuous and differentiable for all \( x \).

First, check the values at the endpoints: \[ f(1) = 1^2 – 5 \cdot 1 + 6 = 1 – 5 + 6 = 2, \] \[ f(4) = 4^2 – 5 \cdot 4 + 6 = 16 – 20 + 6 = 2. \] Since \( f(1) = f(4) \), the conditions of Rolle’s theorem are satisfied.

Next, compute the derivative: \[ f'(x) = 2x – 5. \] According to Rolle’s theorem, there exists a \( c \in (1,4) \) such that \[ f'(c) = 0. \] Setting the derivative equal to zero gives: \[ 2c – 5 = 0 \quad \Rightarrow \quad c = \frac{5}{2}. \] The value \( c = \frac{5}{2} \) lies in the interval \( (1,4) \).

Solution:

Since \( f(x) = x^{2} + 2 \) is a polynomial, it is continuous and differentiable for all \( x \).

Evaluate the function at the endpoints: \[ f(-2) = (-2)^{2} + 2 = 4 + 2 = 6, \] \[ f(2) = 2^{2} + 2 = 4 + 2 = 6. \] Thus, \( f(-2) = f(2) \), satisfying the condition of Rolle’s theorem.

Next, compute the derivative: \[ f'(x) = 2x. \] Setting the derivative equal to zero gives: \[ 2x = 0 \quad \Rightarrow \quad x = 0. \] The value \( x = 0 \) lies within the interval \( (-2,2) \).

Solution:

The function \( f(x) = x^{2} + 2x – 8 \) is a polynomial, hence it is continuous and differentiable for all \( x \).

First, evaluate the function at the endpoints: \[ f(-4) = (-4)^{2} + 2(-4) – 8 = 16 – 8 – 8 = 0, \] \[ f(2) = (2)^{2} + 2(2) – 8 = 4 + 4 – 8 = 0. \] Since \( f(-4) = f(2) = 0 \), the conditions of Rolle’s theorem are satisfied.

Next, compute the derivative: \[ f'(x) = 2x + 2. \] Setting \( f'(x) = 0 \) gives: \[ 2x + 2 = 0 \quad \Rightarrow \quad 2x = -2 \quad \Rightarrow \quad x = -1. \] The value \( x = -1 \) lies within the interval \( (-4,2) \), fulfilling the requirement of Rolle’s theorem.

Solution:

The function \( f(x) = x^{3} – 3x \) is a polynomial and is therefore continuous and differentiable for all \( x \).

First, evaluate the function at the endpoints: \[ f(-\sqrt{3}) = (-\sqrt{3})^{3} – 3(-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0, \] \[ f(0) = 0^{3} – 3 \cdot 0 = 0. \] Since \( f(-\sqrt{3}) = f(0) = 0 \), the conditions of Rolle’s theorem are satisfied.

Next, compute the derivative: \[ f'(x) = 3x^{2} – 3. \] Setting the derivative equal to zero: \[ 3x^{2} – 3 = 0 \quad \Rightarrow \quad x^{2} = 1 \quad \Rightarrow \quad x = \pm 1. \] Among the solutions, \( x = -1 \) lies in the interval \( \left(-\sqrt{3}, 0\right) \) (since \(-\sqrt{3} \approx -1.732\) and \(-1 > -1.732\)).

Solution:

The function \( f(x) = x^{3} + 3x^{2} – 24x – 80 \) is a polynomial, so it is continuous and differentiable for all \( x \).

Evaluate \( f(x) \) at the endpoints: \[ f(-4) = (-4)^{3} + 3(-4)^{2} – 24(-4) – 80 = -64 + 48 + 96 – 80 = 0, \] \[ f(5) = (5)^{3} + 3(5)^{2} – 24(5) – 80 = 125 + 75 – 120 – 80 = 0. \] Since \( f(-4) = f(5) \), the conditions of Rolle’s theorem are satisfied.

Next, compute the derivative: \[ f'(x) = 3x^{2} + 6x – 24. \] Set \( f'(x) = 0 \): \[ 3x^{2} + 6x – 24 = 0 \quad \Rightarrow \quad x^{2} + 2x – 8 = 0. \] Solve the quadratic equation: \[ x = \frac{-2 \pm \sqrt{(2)^{2} – 4(1)(-8)}}{2} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}. \] This gives the solutions: \[ x = \frac{4}{2} = 2 \quad \text{or} \quad x = \frac{-8}{2} = -4. \] Since \( x = -4 \) is an endpoint, the \( c \) in the open interval \( (-4,5) \) is \( c = 2 \).

Solution:

The function \( f(x) = x(x-1)^2 \) is a polynomial and thus continuous and differentiable for all \( x \).

Evaluate the function at the endpoints: \[ f(0) = 0 \cdot (0-1)^2 = 0, \quad f(1) = 1 \cdot (1-1)^2 = 0. \] Since \( f(0) = f(1) = 0 \), the conditions of Rolle’s theorem are satisfied.

Next, compute the derivative using the product rule: \[ f'(x) = (x-1)^2 + x \cdot 2(x-1) = (x-1)^2 + 2x(x-1). \] Factor out the common term \( (x-1) \): \[ f'(x) = (x-1)\left[(x-1)+2x\right] = (x-1)(3x-1). \] Setting \( f'(x) = 0 \), we have: \[ (x-1)(3x-1)=0. \] This yields: \[ x = 1 \quad \text{or} \quad x = \frac{1}{3}. \] Since \( x = 1 \) is an endpoint, the value of \( c \) within \( (0,1) \) is \( \frac{1}{3} \).

Solution:

The function \[ f(x) = (x-1)(x-2)(x-3) \] is a cubic polynomial and hence continuous and differentiable for all \( x \).

First, evaluate the function at the endpoints: \[ f(1) = (1-1)(1-2)(1-3) = 0, \] \[ f(3) = (3-1)(3-2)(3-3) = 0. \] Since \( f(1) = f(3) \), the conditions of Rolle’s theorem are satisfied.

Next, expand the function: \[ f(x) = (x-1)(x-2)(x-3) = x^{3} – 6x^{2} + 11x – 6. \] Compute the derivative: \[ f'(x) = 3x^{2} – 12x + 11. \] Set \( f'(x) = 0 \): \[ 3x^{2} – 12x + 11 = 0. \] Solve using the quadratic formula: \[ x = \frac{12 \pm \sqrt{12^{2} – 4 \cdot 3 \cdot 11}}{2 \cdot 3} = \frac{12 \pm \sqrt{144 – 132}}{6} = \frac{12 \pm \sqrt{12}}{6}. \] Note that: \[ \sqrt{12} = 2\sqrt{3}, \] so \[ x = \frac{12 \pm 2\sqrt{3}}{6} = 2 \pm \frac{\sqrt{3}}{3}. \] Recognizing that \[ \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}, \] the two values for \( c \) are: \[ c = 2 + \frac{1}{\sqrt{3}} \quad \text{and} \quad c = 2 – \frac{1}{\sqrt{3}}. \] Both values lie in the interval \( (1,3) \).

Solution:

The function \[ f(x)=\sqrt{4-x^{2}} \] is continuous on \( [-2,2] \) (its domain) and differentiable on \( (-2,2) \).

Evaluate the function at the endpoints: \[ f(-2)=\sqrt{4-(-2)^{2}}=\sqrt{4-4}=0, \] \[ f(2)=\sqrt{4-2^{2}}=\sqrt{4-4}=0. \] Since \( f(-2)=f(2)=0 \), the conditions of Rolle’s theorem are satisfied.

Next, compute the derivative of \( f(x) \). We have: \[ f(x)=\sqrt{4-x^{2}}=(4-x^{2})^{1/2}. \] Using the chain rule, the derivative is: \[ f'(x)=\frac{1}{2}(4-x^{2})^{-1/2} \cdot (-2x)=-\frac{x}{\sqrt{4-x^{2}}}. \] Set \( f'(x)=0 \): \[ -\frac{x}{\sqrt{4-x^{2}}}=0. \] The fraction is zero when the numerator is zero, hence \( x=0 \). Since \( 0 \) lies in the open interval \( (-2,2) \), it is the required value of \( c \) as per Rolle’s theorem.

Solution:

The function \( f(x)=\cos 2x \) is continuous and differentiable for all \( x \).

Evaluate \( f(x) \) at the endpoints: \[ f\left(-\frac{\pi}{4}\right)=\cos\left(-\frac{\pi}{2}\right)=0, \] \[ f\left(\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{2}\right)=0. \] Since \( f\left(-\frac{\pi}{4}\right)=f\left(\frac{\pi}{4}\right) \), the conditions of Rolle’s theorem are satisfied.

Next, differentiate \( f(x) \): \[ f'(x)=-2\sin 2x. \] Set \( f'(x)=0 \): \[ -2\sin 2x=0 \quad \Longrightarrow \quad \sin 2x=0. \] The general solution is \( 2x=n\pi \), for \( n\in \mathbb{Z} \). Within the interval \( \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \), the only solution is when \( n=0 \), i.e., \[ 2x=0 \quad \Longrightarrow \quad x=0. \]

Solution:

The function \( f(x) = \sin x – 1 \) is continuous and differentiable for all \( x \).

Evaluate the function at the endpoints: \[ f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) – 1 = 1 – 1 = 0, \] \[ f\left(\frac{5\pi}{2}\right) = \sin\left(\frac{5\pi}{2}\right) – 1 = 1 – 1 = 0. \] As \( f\left(\frac{\pi}{2}\right) = f\left(\frac{5\pi}{2}\right) \), the conditions of Rolle’s theorem are satisfied.

Next, compute the derivative: \[ f'(x) = \cos x. \] To find \( c \), set the derivative equal to zero: \[ \cos x = 0. \] The general solution for \( \cos x = 0 \) is \[ x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}. \] In the interval \( \left(\frac{\pi}{2}, \frac{5\pi}{2}\right) \), the solution is: \[ x = \frac{\pi}{2} + \pi = \frac{3\pi}{2}. \] Thus, \( c = \frac{3\pi}{2} \) is the required point.

Solution:

The function \( f(x)=\sin 2x \) is continuous and differentiable for all \( x \).

Evaluate the function at the endpoints: \[ f(0)=\sin(0)=0, \] \[ f\left(\frac{\pi}{2}\right)=\sin\left(\pi\right)=0. \] Since \( f(0)=f\left(\frac{\pi}{2}\right) \), the conditions of Rolle’s theorem are met.

Next, differentiate \( f(x) \): \[ f'(x)=2\cos 2x. \] Set the derivative equal to zero: \[ 2\cos 2x=0 \quad \Rightarrow \quad \cos 2x=0. \] The general solution for \( \cos 2x=0 \) is: \[ 2x=\frac{\pi}{2}+k\pi,\quad k\in \mathbb{Z}. \] For \( k=0 \), we have: \[ 2x=\frac{\pi}{2} \quad \Rightarrow \quad x=\frac{\pi}{4}. \] Since \( \frac{\pi}{4} \) lies in \( \left(0,\frac{\pi}{2}\right) \), it is the required value.

Solution: We first note that the function \[ f(x)=\sin 3x \] is continuous on the closed interval \([0,\pi]\) and differentiable on the open interval \((0,\pi)\). Furthermore, the endpoint values are \[ f(0)=\sin(0)=0 \quad \text{and} \quad f(\pi)=\sin(3\pi)=0. \] Since \( f(0)=f(\pi) \), all the hypotheses of Rolle’s theorem are satisfied. By Rolle’s theorem, there exists at least one point \( c \) in \((0,\pi)\) such that \[ f'(c)=0. \] Differentiating \( f(x) \) with respect to \( x \), we have \[ f'(x)=\frac{d}{dx}(\sin 3x)=3\cos 3x. \] Setting the derivative equal to zero, \[ 3\cos 3c=0 \quad \Longrightarrow \quad \cos 3c=0. \] The cosine function equals zero when its argument is of the form \[ 3c = \frac{\pi}{2}+ k\pi,\quad k\in\mathbb{Z}. \] Hence, we can express \( c \) as: \[ c=\frac{\pi}{6}+ \frac{k\pi}{3}. \] We now determine the integer values of \( k \) for which \( c \) lies in the interval \((0,\pi)\): \[ \begin{aligned} &\text{For } k = 0: && c=\frac{\pi}{6} && \in (0,\pi),\\[6mm] &\text{For } k = 1: && c=\frac{\pi}{6}+ \frac{\pi}{3}=\frac{\pi}{2} && \in (0,\pi),\\[6mm] &\text{For } k = 2: && c=\frac{\pi}{6}+ \frac{2\pi}{3}=\frac{5\pi}{6} && \in (0,\pi). \end{aligned} \] For \( k = 3 \), \[ c=\frac{\pi}{6}+ \pi = \frac{7\pi}{6}, \] which exceeds \(\pi\) and is therefore not acceptable.

Solution: The function \[ f(x)=e^{x}\sin x \] is continuous on the closed interval \([0,\pi]\) and differentiable on the open interval \((0,\pi)\). Evaluating the endpoints, we have: \[ f(0)=e^{0}\sin 0=1\cdot 0=0,\quad f(\pi)=e^{\pi}\sin \pi=e^{\pi}\cdot 0=0. \] Thus, \( f(0)=f(\pi)=0 \). By Rolle’s theorem, there exists at least one point \( c \) in \((0,\pi)\) such that: \[ f'(c)=0. \] Differentiating \( f(x) \) using the product rule: \[ f'(x)=\frac{d}{dx}\left(e^{x}\sin x\right)=e^{x}\sin x+e^{x}\cos x=e^{x}(\sin x+\cos x). \] Since \( e^{x}\neq 0 \) for all \( x \), setting \( f'(x)=0 \) yields: \[ \sin x+\cos x=0. \] Rearranging, we get: \[ \sin x=-\cos x. \] Dividing both sides by \(\cos x\) (noting \(\cos x\neq0\) except possibly at isolated points which can be verified separately), we obtain: \[ \tan x=-1. \] The general solution for \( \tan x=-1 \) is: \[ x=-\frac{\pi}{4}+n\pi,\quad n\in\mathbb{Z}. \] Restricting \( x \) to the interval \((0,\pi)\), we choose \( n=1 \): \[ c=-\frac{\pi}{4}+\pi=\frac{3\pi}{4}. \]

Solution: The function \[ f(x)=e^{x}\cos x \] is continuous on the closed interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and differentiable on the open interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Evaluating the endpoints: \[ f\left(-\frac{\pi}{2}\right)=e^{-\frac{\pi}{2}}\cos\left(-\frac{\pi}{2}\right)=e^{-\frac{\pi}{2}}\cdot 0=0,\quad f\left(\frac{\pi}{2}\right)=e^{\frac{\pi}{2}}\cos\left(\frac{\pi}{2}\right)=e^{\frac{\pi}{2}}\cdot 0=0. \] Thus, \( f\left(-\frac{\pi}{2}\right)=f\left(\frac{\pi}{2}\right)=0 \). By Rolle’s theorem, there exists at least one point \( c \) in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) such that \[ f'(c)=0. \] Differentiating \( f(x) \) using the product rule: \[ f'(x)=\frac{d}{dx}\left(e^{x}\cos x\right)=e^{x}\cos x – e^{x}\sin x=e^{x}(\cos x-\sin x). \] Since \( e^{x}\neq 0 \) for all \( x \), setting \( f'(x)=0 \) yields: \[ \cos x-\sin x=0, \] or equivalently, \[ \cos x=\sin x. \] Dividing both sides by \(\cos x\) (for \( \cos x\neq 0 \)), we obtain: \[ \tan x=1. \] The general solution is: \[ x=\frac{\pi}{4}+n\pi,\quad n\in\mathbb{Z}. \] Within the interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), the only valid value is: \[ c=\frac{\pi}{4}. \]

Solution: First, we verify that \( f(x)=e^{-x}\sin x \) is continuous on \([0,\pi]\) and differentiable on \((0,\pi)\). Compute the endpoint values: \[ f(0)=e^{-0}\sin0=1\cdot 0=0, \] \[ f(\pi)=e^{-\pi}\sin\pi=e^{-\pi}\cdot 0=0. \] Since \( f(0)=f(\pi)=0 \), the conditions of Rolle’s theorem are satisfied. According to Rolle’s theorem, there exists at least one point \( c \) in \((0,\pi)\) such that: \[ f'(c)=0. \] Differentiate \( f(x)=e^{-x}\sin x \) using the product rule: \[ f'(x)=\frac{d}{dx}\left(e^{-x}\sin x\right)=-e^{-x}\sin x+e^{-x}\cos x=e^{-x}(\cos x-\sin x). \] Since \( e^{-x} \neq 0 \) for all \( x \), setting \( f'(x)=0 \) requires: \[ \cos x-\sin x=0. \] This leads to: \[ \cos x=\sin x. \] Dividing both sides by \(\cos x\) (assuming \( \cos x \neq 0 \)): \[ \tan x=1. \] The general solution for this equation is: \[ x=\frac{\pi}{4}+n\pi,\quad n\in\mathbb{Z}. \] Restricting \( x \) to the interval \( (0, \pi) \), we choose \( n=0 \): \[ c=\frac{\pi}{4}. \]

Solution: First, we verify the endpoint values. Define \[ f(x)=e^{2x}(\sin 2x-\cos 2x). \] At \( x=\frac{\pi}{8} \): \[ f\left(\frac{\pi}{8}\right)=e^{\frac{\pi}{4}}\left(\sin\frac{\pi}{4}-\cos\frac{\pi}{4}\right) =e^{\frac{\pi}{4}}\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\right)=0. \] At \( x=\frac{5\pi}{8} \): \[ f\left(\frac{5\pi}{8}\right)=e^{\frac{5\pi}{4}}\left(\sin\frac{5\pi}{4}-\cos\frac{5\pi}{4}\right) =e^{\frac{5\pi}{4}}\left(-\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}\right)=0. \] Thus, \( f\left(\frac{\pi}{8}\right)=f\left(\frac{5\pi}{8}\right)=0 \). Since \( f(x) \) is the product of exponential and trigonometric functions, it is continuous on \(\left[\frac{\pi}{8}, \frac{5\pi}{8}\right]\) and differentiable on \(\left(\frac{\pi}{8}, \frac{5\pi}{8}\right)\). Hence, by Rolle’s theorem, there exists a point \( c \) in \(\left(\frac{\pi}{8}, \frac{5\pi}{8}\right)\) such that: \[ f'(c)=0. \] We differentiate \( f(x) \) using the product rule. Write: \[ f(x)=e^{2x} \cdot (\sin 2x-\cos 2x). \] The derivative is: \[ f'(x)=\frac{d}{dx}(e^{2x})(\sin 2x-\cos 2x)+e^{2x}\frac{d}{dx}(\sin 2x-\cos 2x). \] Compute: \[ \frac{d}{dx}(e^{2x})=2e^{2x}, \] and \[ \frac{d}{dx}(\sin 2x-\cos 2x)=2\cos 2x+2\sin 2x. \] Thus, \[ f'(x)=2e^{2x}(\sin 2x-\cos 2x)+e^{2x}[2\cos 2x+2\sin 2x]. \] Factor out \(2e^{2x}\): \[ f'(x)=2e^{2x}\left[(\sin 2x-\cos 2x)+(\cos 2x+\sin 2x)\right] =2e^{2x}\left[2\sin 2x\right] =4e^{2x}\sin 2x. \] Setting \( f'(x)=0 \), we have \[ 4e^{2x}\sin 2x=0. \] Since \( 4e^{2x}\neq 0 \) for any \( x \), we must have: \[ \sin 2x=0. \] The general solution is: \[ 2x=n\pi,\quad n\in\mathbb{Z} \quad \Longrightarrow \quad x=\frac{n\pi}{2}. \] We require \( x \) to be in \(\left(\frac{\pi}{8}, \frac{5\pi}{8}\right)\). For \( n=1 \): \[ x=\frac{\pi}{2}, \] which indeed lies in the interval.

Solution: First, we check the continuity and differentiability of the function \[ f(x)=\log\left(x^{2}+2\right)-\log3. \] The function \( x^{2}+2 \) is a polynomial and is continuous for all \( x \). Since the logarithm function \( \text{log} \) is continuous for positive arguments and \( x^{2}+2\geq 2 > 0 \) for all \( x \), the function \( f(x) \) is continuous on \([-1,1]\). Also, \( f(x) \) is differentiable on \((-1,1)\). Next, we verify the endpoint values: \[ f(-1)=\log\left((-1)^{2}+2\right)-\log3=\log\left(1+2\right)-\log3=\log3-\log3=0, \] \[ f(1)=\log\left(1^{2}+2\right)-\log3=\log\left(1+2\right)-\log3=\log3-\log3=0. \] Since \( f(-1)=f(1)=0 \), the condition \( f(a)=f(b) \) is satisfied. By Rolle’s Theorem, there exists at least one point \( c \) in \( (-1,1) \) such that \[ f'(c)=0. \] We now differentiate \( f(x) \): \[ f(x)=\log\left(x^{2}+2\right)-\log3. \] Note that the derivative of the constant \( \log3 \) is 0. Using the chain rule, the derivative of \( \log\left(x^{2}+2\right) \) is: \[ \frac{d}{dx}\left[\log\left(x^{2}+2\right)\right]=\frac{1}{x^{2}+2}\cdot 2x=\frac{2x}{x^{2}+2}. \] Thus, \[ f'(x)=\frac{2x}{x^{2}+2}. \] Setting \( f'(x)=0 \): \[ \frac{2x}{x^{2}+2}=0 \quad \Longrightarrow \quad 2x=0 \quad \Longrightarrow \quad x=0. \] Hence, \( c=0 \) is the value in \( (-1,1) \) where the tangent is horizontal. Finally, the corresponding \( y \)-coordinate is: \[ f(0)=\log\left(0^{2}+2\right)-\log3=\log2-\log3=\log\left(\frac{2}{3}\right). \]

Solution: \(\underline{\text{(i) For } y=x^2 \text{ in } [-2,2]:}\) The function \[ f(x)=x^2 \] is continuous on \([-2,2]\) and differentiable on \((-2,2)\). Compute the endpoint values: \[ f(-2)=(-2)^2=4,\quad f(2)=2^2=4. \] Since \(f(-2)=f(2)=4\), Rolle’s theorem guarantees the existence of at least one point \(c\) in \((-2,2)\) where \(f'(c)=0\). Differentiate: \[ f'(x)=2x. \] Setting \( f'(c)=0 \) gives: \[ 2c=0 \quad \Longrightarrow \quad c=0. \] The corresponding point on the curve is: \[ (0,\,0^2)=(0,0). \] \(\underline{\text{(ii) For } y=-1+\cos x \text{ on } [0,2\pi]:}\) The function \[ f(x)=-1+\cos x \] is continuous on \([0,2\pi]\) and differentiable on \((0,2\pi)\). Evaluate the endpoints: \[ f(0)=-1+\cos 0=-1+1=0, \quad f(2\pi)=-1+\cos(2\pi)=-1+1=0. \] Since \(f(0)=f(2\pi)=0\), Rolle’s theorem applies. Differentiate: \[ f'(x)=-\sin x. \] Setting \(f'(c)=0\) yields: \[ -\sin c=0 \quad \Longrightarrow \quad \sin c=0. \] The solutions for \(\sin c=0\) in the open interval \((0,2\pi)\) is: \[ c=\pi. \] The corresponding point on the curve is: \[ \left(\pi,\,-1+\cos\pi\right)=\left(\pi,\,-1+(-1)\right)=\left(\pi,\,-2\right). \]

Solution: Since \( f(x)=x^3+ax^2+bx \) is a polynomial, it is continuous on \([1,2]\) and differentiable on \((1,2)\). For Rolle’s theorem to be applicable, the function must satisfy: \[ f(1)=f(2). \] We calculate: \[ f(1)=1^3+a(1)^2+b(1)=1+a+b, \] \[ f(2)=2^3+a(2)^2+b(2)=8+4a+2b. \] Setting \( f(1)=f(2) \): \[ 1+a+b=8+4a+2b. \] Simplifying: \[ 1+a+b-8-4a-2b=0, \] \[ -7-3a-b=0, \] which yields \[ 3a+b=-7. \tag{1} \] Next, by Rolle’s theorem, there exists a point \( c \) in \( (1,2) \) where \( f'(c)=0 \). We are given that this point is \( c=\frac{4}{3} \). First, differentiate \( f(x) \): \[ f'(x)=3x^2+2ax+b. \] Now, set \( f’\left(\frac{4}{3}\right)=0 \): \[ 3\left(\frac{4}{3}\right)^2+2a\left(\frac{4}{3}\right)+b=0. \] Compute \( \left(\frac{4}{3}\right)^2=\frac{16}{9} \), so: \[ 3\cdot\frac{16}{9}+\frac{8a}{3}+b=0, \] \[ \frac{16}{3}+\frac{8a}{3}+b=0. \] Multiplying both sides by 3 to eliminate the denominators: \[ 16+8a+3b=0. \tag{2} \] We now have the system of equations: \[ \begin{cases} 3a+b=-7, \\ 8a+3b=-16. \end{cases} \] Multiply the first equation by 3: \[ 9a+3b=-21. \] Subtract equation (2) from this result: \[ (9a+3b)-(8a+3b)= -21-(-16), \] \[ a=-5. \] Substitute \( a=-5 \) into equation (1): \[ 3(-5)+b=-7, \] \[ -15+b=-7 \quad \Longrightarrow \quad b=8. \]

Solution: To verify Rolle’s theorem for a function \( f(x) \) on \([a,b]\), we check the following three conditions:

1. \( f(x) \) is continuous on \([a,b]\).
2. \( f(x) \) is differentiable on \((a,b)\).
3. \( f(a) = f(b) \).

For \( f(x) = [x] \) (greatest integer function) on the interval \([-2,2]\):

• The function \( [x] \) has jump discontinuities at all integer points. In particular, there is a jump at \( x=0 \). Hence, \( f(x) \) is not continuous on \([-2,2]\).
• Due to these discontinuities, the function also fails the differentiability condition at those integer points.

Because it is not continuous on \([-2,2]\), Rolle’s theorem cannot be applied. Regarding the converse of Rolle’s theorem: even if there exist points in \((-2,2)\) where \( f'(x)=0 \) for some differentiable segments, it does not guarantee the function meets all the hypotheses of Rolle’s theorem. Therefore, the converse does not hold in this situation.

Solution: For Rolle’s theorem to be applicable on an interval \([a,b]\), the function \( f(x) \) must satisfy the following conditions:

1. \( f(x) \) is continuous on \([a,b]\).
2. \( f(x) \) is differentiable on \((a,b)\).
3. \( f(a)=f(b) \).

Consider the function \( f(x)=x^{1/3} \) on the interval \([-1,1]\): \[ f(x)=x^{1/3} \] \(\textbf{Continuity:}\) The function \( x^{1/3} \) is continuous on \(\mathbb{R}\), hence it is continuous on \([-1,1]\). \(\textbf{Differentiability:}\) The derivative of \( f(x) \) is given by \[ f'(x)=\frac{1}{3} x^{-2/3}. \] This derivative is not defined at \( x=0 \) because \( x^{-2/3} \) becomes unbounded as \( x \to 0 \). Thus, \( f(x) \) is not differentiable on the entire open interval \((-1,1)\). Since the function fails to be differentiable on \((-1,1)\), Rolle’s theorem is not applicable for \( f(x)=x^{1/3} \) in \([-1,1]\).

Solution: To apply Rolle’s theorem on an interval \([a,b]\), the function \( f(x) \) must satisfy these conditions:

1. \( f(x) \) is continuous on \([a,b]\).
2. \( f(x) \) is differentiable on \((a,b)\).
3. \( f(a)=f(b) \).

Consider the function: \[ f(x)=x^{2/3}. \] \(\textbf{Continuity:}\) \( f(x) \) is continuous on \(\mathbb{R}\), hence it is continuous on \([-2,2]\). \(\textbf{Differentiability:}\) Differentiate \( f(x) \): \[ f'(x)=\frac{2}{3}x^{-1/3}. \] The derivative \( f'(x) \) is undefined at \( x=0 \), as \( x^{-1/3} \) becomes unbounded. Therefore, \( f(x) \) is not differentiable on the entire open interval \((-2,2)\). Since the differentiability condition is not met, Rolle’s theorem is not applicable.

Solution: For applying Rolle’s theorem on an interval \([a,b]\), the function \( f(x) \) must satisfy:

1. \( f(x) \) is continuous on \([a,b]\).
2. \( f(x) \) is differentiable on \((a,b)\).
3. \( f(a)=f(b) \).

Consider the function: \[ f(x)=2+(x-1)^{2/3}. \] \(\textbf{Continuity:}\) The function \( (x-1)^{2/3} \) is continuous for all \( x \) and hence \( f(x) \) is continuous on \([0,2]\). \(\textbf{Differentiability:}\) To check the differentiability, differentiate \( f(x) \): \[ f'(x)=\frac{2}{3}(x-1)^{-1/3}. \] The derivative \( f'(x) \) is undefined at \( x=1 \) because \( (x-1)^{-1/3} \) becomes unbounded as \( x \to 1 \). Therefore, \( f(x) \) is not differentiable on the entire open interval \((0,2)\). Since the differentiability condition is not met, Rolle’s theorem is not applicable.

Solution: For Rolle’s theorem to apply on an interval \([a,b]\), the function \( f(x) \) must satisfy:

1. \( f(x) \) is continuous on \([a,b]\).
2. \( f(x) \) is differentiable on \((a,b)\).
3. \( f(a)=f(b) \).

Consider the function: \[ f(x)=1+|x-2|. \] \(\textbf{Continuity:}\) The function \( |x-2| \) is continuous everywhere, so \( f(x) \) is continuous on \([0,4]\). \(\textbf{Differentiability:}\) The absolute value function \( |x-2| \) is not differentiable at \( x=2 \) because of the sharp corner at that point. Hence, \( f(x) \) is not differentiable on the entire open interval \((0,4)\). Since the differentiability condition is not met, Rolle’s theorem is not applicable.

Solution: For Rolle’s theorem to apply on an interval \([a,b]\), the function \( f(x) \) must satisfy:

1. \( f(x) \) is continuous on \([a,b]\).
2. \( f(x) \) is differentiable on \((a,b)\).
3. \( f(a)=f(b) \).

Consider the function: \[ f(x)=\tan x = \frac{\sin x}{\cos x}. \] \(\textbf{Continuity:}\) Although \(\tan x\) is continuous on intervals where \(\cos x \neq 0\), within \([0, \pi]\) the function is discontinuous at \( x=\frac{\pi}{2} \) since \( \cos \left(\frac{\pi}{2}\right)=0 \). \(\textbf{Differentiability:}\) Due to the discontinuity at \( x=\frac{\pi}{2} \), the function cannot be differentiable on the entire open interval \((0,\pi)\). Since the continuity and differentiability conditions are not met, Rolle’s theorem is not applicable for \( f(x)=\tan x \) in \([0,\pi]\).

Solution: For Rolle’s theorem to apply on an interval \([a,b]\), the function \( f(x) \) must satisfy:

1. \( f(x) \) is continuous on \([a,b]\).
2. \( f(x) \) is differentiable on \((a,b)\).
3. \( f(a)=f(b) \).

Consider the function: \[ f(x)=\sec x=\frac{1}{\cos x}. \] \(\textbf{Continuity:}\) The function \( \sec x \) is continuous wherever \(\cos x\neq 0\). However, in the interval \([0,2\pi]\), \(\cos x=0\) at \( x=\frac{\pi}{2} \) (and at \( x=\frac{3\pi}{2} \)), leading to discontinuities. \(\textbf{Differentiability:}\) Due to these discontinuities, \( f(x) \) is not differentiable on the entire open interval \((0,2\pi)\). Since the continuity condition is not met, Rolle’s theorem is not applicable for \( f(x)=\sec x \) in \([0,2\pi]\).

Solution: To apply Rolle’s theorem on an interval \([a,b]\), the function \( f(x) \) must satisfy:

1. \( f(x) \) is continuous on \([a,b]\).
2. \( f(x) \) is differentiable on \((a,b)\).
3. \( f(a)=f(b) \).

Consider the function: \[ f(x)=\frac{x(x-2)}{x-1}. \] Notice that the function is undefined at \( x=1 \) because the denominator becomes zero, which causes a discontinuity in the interval \([0,2]\). Since \( f(x) \) is discontinuous at \( x=1 \), it fails the necessary conditions for Rolle’s theorem.

Solution: For Rolle’s theorem to be applicable on an interval \([a,b]\), the function \( f(x) \) must satisfy:

1. \( f(x) \) is continuous on \([a,b] \),
2. \( f(x) \) is differentiable on \((a,b) \),
3. \( f(a) = f(b) \).

Let us verify these conditions for \[ f(x) = x^2 + 1 \quad \text{on} \quad [-1, 2]. \]

• \( f(x) \) is a polynomial function, hence continuous and differentiable everywhere.
• Check values at the endpoints: \[ f(-1) = (-1)^2 + 1 = 2, \quad f(2) = 2^2 + 1 = 5. \]

Since \( f(-1) \ne f(2) \), the condition \( f(a) = f(b) \) is not satisfied. Therefore, Rolle’s theorem is not applicable.