We have, $f(x)=x(x-1)^2$ in $[0,1]$.

(i) Since, $f(x)=x(x-1)^2$ is a polynomial function.

So, it is continuous in $[0,1]$.

$$\begin{aligned} &\text { (ii) Now, }\\ &\begin{aligned} f^{\prime}(x) & =x \cdot \frac{d}{d x}(x-1)^2+(x-1)^2 \frac{d}{d x} x \\ & =x \cdot 2(x-1) \cdot 1+(x-1)^2 \\ & =2 x^2-2 x+x^2+1-2 x \\ & =3 x^2-4 x+1 \text { which exists in }(0,1) \end{aligned} \end{aligned}$$

So, $f(x)$ is differentiable in $(0,1)$. (iii) Now, $f(0)=0$ and $f(1)=0 \Rightarrow f(0)=f(1)$

$f$ satisfies the above conditions of Rolle's theorem.

Hence, by Rolle's theorem $\exists c \in(0,1)$ such that

$$\begin{aligned} f^{\prime}(c) & =0 \\ \Rightarrow\quad 3 c^2-4 c+1 & =0 \\ \Rightarrow\quad 3 c^2-3 c-c+1 & =0 \\ \Rightarrow\quad 3 c(c-1)-1(c-1) & =0 \\ \Rightarrow\quad (3 c-1)(c-1) & =0 \\ \Rightarrow\quad c & =\frac{1}{3}, 1 \Rightarrow \frac{1}{3} \in(0,1) \end{aligned}$$

Thus, we see that there exists a real number c in the open interval $(0,1)$. Hence, Rolle's theorem has been verified.

We have, $f(x)=\sin ^4 x+\cos ^4 x$ in $\left[0, \frac{\pi}{2}\right]\quad\text{.... (i)}$

(i) $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$

[since, $\sin ^4 x$ and $\cos ^4 x$ are continuous functions and we know that, if $g$ and $h$ be continuous functions, then $(g+h)$ is a continuous function.]

$$\begin{aligned} &\begin{aligned} \text { (ii) }\quad f^{\prime}(x) & =4(\sin x)^3 \cdot \cos x+4(\cos x)^3 \cdot(-\sin x) \\ & =4 \sin ^3 x \cdot \cos x-4 \sin x \cdot \cos ^3 x \\ & =4 \sin x \cos x\left(\sin ^2 x-\cos ^2 x\right) \text { which exists in }\left(0, \frac{\pi}{2}\right)\quad\text{.... (ii)} \end{aligned} \end{aligned}$$

Hence, $f(x)$ is differentiable in $\left(0, \frac{\pi}{2}\right)$.

(iii) Also, $f(0)=0+1=1$ and $f\left(\frac{\pi}{2}\right)=1+0=1$

$$\Rightarrow \quad f(0)=f\left(\frac{\pi}{2}\right)$$

Conditions of Rolle's theorem are satisfied.

Hence, there exists at least one $c\in\left(0, \frac{\pi}{2}\right)$ such that $f^{\prime}(c)=0$.

$$\begin{aligned} &\begin{array}{lr} \therefore & 4 \sin c \cos c\left(\sin ^2 c-\cos ^2 c\right)=0 \\ \Rightarrow & 4 \operatorname{sinc} \cos c(-\cos 2 c)=0 \\ \Rightarrow & -2 \sin 2 c \cdot \cos 2 c=0 \\ \Rightarrow & -\sin 4 c=0 \\ \Rightarrow & \sin 4 c=0 \\ \Rightarrow & 4 c=\pi \\ \Rightarrow & c=\frac{\pi}{4} \\ \text { and } & \frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right) \end{array}\\ &\text { Hence, Rolle's theorem has been verified. } \end{aligned}$$

We have, $$f(x)=\log \left(x^2+2\right)-\log 3$$

(i) Logarithmic functions are continuous in their domain. Hence, $f(x)=\log \left(x^2+2\right)-\log 3$ is continuous in $[-1,1]$.

$$\begin{aligned} \text{(ii)}\quad f^{\prime}(x) & =\frac{1}{x^2+2} \cdot 2 x-0 \\ & =\frac{2 x}{x^2+2}, \text { which exists in }(-1,1) \end{aligned}$$

Hence, $f(x)$ is differentiable in $(-1,1)$.

$$\begin{aligned} & \text { (iii) } f(-1)=\log \left[(-1)^2+2\right]-\log 3=\log 3-\log 3=0 \text { and } \\ & f(1)=\log \left(1^2+2\right)-\log 3=\log 3-\log 3=0 \\ & \Rightarrow \quad f(-1)=f(1) \end{aligned}$$

Conditions of Rolle's theorem are satisfied.

Hence, there exists a real number c such that

$$\begin{aligned} f^{\prime}(c) =0 \\ \Rightarrow \quad \frac{2 c}{c^2+2} & =0 \\ \Rightarrow \quad c =0 \in(-1,1) \end{aligned}$$

Hence, Rolle's theorem has been verified.

We have, $$f(x)=x(x+3) \mathrm{e}^{-x / 2}$$

(i) $f(x)$ is a continuous function. [since, it is a combination of polynomial functions $x(x+3)$ and an exponential function $\mathrm{e}^{-x / 2}$ which are continuous functions]

So, $f(x)=x(x+3) \mathrm{e}^{-x / 2}$ is continuous in $[-3,0]$.

(ii) $\therefore \quad \quad f(x)=\left(x^2+3 x\right) \cdot \frac{d}{d x} \mathrm{e}^{-x / 2}+\mathrm{e}^{-x / 2} \cdot \frac{d}{d x}\left(x^2+3 x\right)$

$$\begin{aligned} & =\left(x^2+3 x\right) \cdot \mathrm{e}^{-x / 2} \cdot\left(-\frac{1}{2}\right)+\mathrm{e}^{-x / 2} \cdot(2 x+3) \\ & =\mathrm{e}^{-x / 2}\left[2 x+3-\frac{1}{2} \cdot\left(x^2+3 x\right)\right] \\ & =\mathrm{e}^{-x / 2}\left[\frac{4 x+6-x^2-3 x}{2}\right] \\ & =\mathrm{e}^{-x / 2} \cdot \frac{1}{2}\left[-x^2+x+6\right] \\ & =\frac{-1}{2} \mathrm{e}^{-x / 2}\left[x^2-x-6\right] \\ & =\frac{-1}{2} \mathrm{e}^{-x / 2}\left[x^2-3 x+2 x-6\right] \\ & =\frac{-1}{2} \mathrm{e}^{-x / 2}[(x+2)(x-3)] \text { which exists in }(-3,0) . \end{aligned}$$

Hence, $f(x)$ is differentiable in $(-3,0)$. (iii) $\therefore$ $$f(-3)=-3(-3+3) e^{-3 / 2}=0$$

and $$f(0)=0(0+3) e^{-0 / 2}=0$$

$$\Rightarrow \quad f(-3)=f(0)$$

Since, conditions of Rolle's theorem are satisfied.

Hence, there exists a real number $c$ such that $f(c)=0$

$$\begin{array}{ll} \Rightarrow & -\frac{1}{2} e^{-c / 2}(c+2)(c-3)=0 \\ \Rightarrow & c=-2,3 \text { where }-2 \in(-3,0) \end{array}$$

Therefore, Rolle's theorem has been verified.

We have, $f(x)=\sqrt{4-x^2}=\left(4-x^2\right)^{1 / 2}$

(i) $f(x)=\sqrt{4-x^2}$ is a continuous function.

[since every polynomial function is a continuous function]

Hence, $f(x)$ is continuous in $[-2,2]$.

(ii) $f^{\prime}(x)=\frac{1}{2}\left(4-x^2\right)^{-1 / 2} \cdot(-2 x)$

$=-x \cdot \frac{1}{\sqrt{4-x^2}}$, which exists everywhere except at $x= \pm 2$.

Hence, $f(x)$ is differentiable in $(-2,2)$.

$$\begin{aligned} \text{(iii)}\quad f(-2)=\sqrt{(4-4)}=0 \text { and } f(2)=\sqrt{(4-4)} & =0 \\ \Rightarrow \quad f(-2) & =f(2) \end{aligned}$$

conditions of Rolle's theorem are satisfied.

Hence, there exists a real number $c$ such that $f^{\prime}(c)=0$.

$$\begin{aligned} &\begin{aligned} & \Rightarrow \quad-c \frac{1}{\sqrt{4-c^2}}=0 \\ & \Rightarrow \quad c=0 \in(-2,2) \end{aligned}\\ &\text { Hence, Rolle's theorem has been verified. } \end{aligned}$$