Question

Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = cos 2 (x − π/4) on [0, π/2]

(ii) f(x) = sin 2x on [0, π/2]

(iii) f(x) = cos 2x on [−π/4, π/4]

(iv) f(x) = e^x sin x on [0, π]

(v) f(x) = e^x cos x on [−π/2, π/2]

(vi) f(x) = cos 2x on [0, π]

(vii) f(x) = $\frac{\sin x}{e^x}$ on 0 ≤ x ≤ π

(viii) f(x) = sin 3x on [0, π]

(ix) f(x) = ${e^{1-x}}^2$ on [−1, 1]

(x) f(x) = log (x² + 2) − log 3 on [−1, 1]

(xi) f(x) = sin x + cos x on [0, π/2]

(xii) f(x) = 2 sin x + sin 2x on [0, π]

(xiii) $f\left(x\right)=\frac{x}{2}-\sin \frac{\mathrm{\pi }x}{6}\mathrm{on}[-1,0]$

(xiv) $f\left(x\right)=\frac{6x}{\mathrm{\pi }}-4{\sin }^{2}x\mathrm{on}[0,\mathrm{\pi }/6]$

(xv) f(x) = 4^{sin x} on [0, π]

(xvi) f(x) = x² − 5x + 4 on [1, 4]

(xvii) f(x) = sin⁴ x + cos⁴ x on $\left[0,\frac{\mathrm{\pi }}{2}\right]$

(xviii) f(x) = sin x − sin 2x on [0, π]

Answer 1

(i) The given function is $f\left(x\right)=\cos 2\left(x-\frac{\mathrm{\pi }}{4}\right)=\cos \left(2x-\frac{\mathrm{\pi }}{2}\right)=\sin 2x$.

Since $\sin 2x$ is everywhere continuous and differentiable.

Therefore, $\sin 2x$ is continuous on $\left[0,\frac{\mathrm{\pi }}{2}\right]$ and differentiable on $\left(0,\frac{\mathrm{\pi }}{2}\right)$.

Also,
$f\left(\frac{\mathrm{\pi }}{2}\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(0,\frac{\mathrm{\pi }}{2}\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=\sin 2x \\ \Rightarrow f\text{'}\left(x\right)=2\cos 2x \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow 2\cos 2x=0 \\ \Rightarrow \cos 2x=0 \\ \Rightarrow x=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Thus, $c=\frac{\mathrm{\pi }}{4}\in \left(0,\frac{\mathrm{\pi }}{2}\right)$ such that $f\text{'}\left(c\right)=0$.

Hence, Rolle's theorem is verified.

(ii) The given function is $f\left(x\right)=\sin 2x$.

Since $\sin 2x$ is everywhere continuous and differentiable.

Therefore, $\sin 2x$ is continuous on $\left[0,\frac{\mathrm{\pi }}{2}\right]$ and differentiable on $\left(0,\frac{\mathrm{\pi }}{2}\right)$.

Also,
$f\left(\frac{\mathrm{\pi }}{2}\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(0,\frac{\mathrm{\pi }}{2}\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=\sin 2x \\ \Rightarrow f\text{'}\left(x\right)=2\cos 2x \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow 2\cos 2x=0 \\ \Rightarrow \cos 2x=0 \\ \Rightarrow x=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Thus, $c=\frac{\mathrm{\pi }}{4}\in \left(0,\frac{\mathrm{\pi }}{2}\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.


(iii)

The given function is $f\left(x\right)=\cos 2x$.

Since $\cos 2x$ is everywhere continuous and differentiable, $\cos 2x$ is continuous on $\left[\frac{-\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right]$ and differentiable on $\left(\frac{-\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right)$.

Also,
$f\left(\frac{\mathrm{\pi }}{4}\right)=f\left(\frac{-\mathrm{\pi }}{4}\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(\frac{-\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=\cos 2x \\ \Rightarrow f\text{'}\left(x\right)=-2\sin 2x \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow -2\sin 2x=0 \\ \Rightarrow \sin 2x=0 \\ \Rightarrow \sin 2x=0 \\ \Rightarrow \mathrm{x}=0 \end{gathered}$$

Since $c=0\in \left(\frac{-\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.


(iv)

The given function is $f\left(x\right)=e^x\sin x$.

Since $\sin x\&e^x$ are everywhere continuous and differentiable.

Therefore, being a product of these two, $f\left(x\right)$ is continuous on $\left[0,\mathrm{\pi }\right]$ and differentiable on $\left(0,\mathrm{\pi }\right)$.

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=e^x\sin x \\ \Rightarrow f\text{'}\left(x\right)=e^x\left(\sin x+\cos x\right) \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow e^x\left(\sin x+\cos x\right)=0 \\ \Rightarrow \sin x+\cos x=0 \\ \Rightarrow \tan x=-1 \\ \Rightarrow \mathrm{x}=\mathrm{\pi }-\frac{\mathrm{\pi }}{4}=\frac{3\mathrm{\pi }}{4} \end{gathered}$$

Since $c=\frac{3\mathrm{\pi }}{4}\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.


(v)

The given function is $f\left(x\right)=e^x\cos x$.

Since $\cos x\&e^x$ are everywhere continuous and differentiable, $f\left(x\right)$ being a product of these two is continuous on $\left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ and differentiable on $\left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$.

Also,
$f\left(\frac{-\mathrm{\pi }}{2}\right)=f\left(\frac{\mathrm{\pi }}{2}\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=e^x\cos x \\ \Rightarrow f\text{'}\left(x\right)=e^x\left(\cos x-\sin x\right) \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow e^x\left(\cos x-\sin x\right)=0 \\ \Rightarrow \sin x-\cos x=0 \\ \Rightarrow \tan x=1 \\ \Rightarrow \mathrm{x}=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Since $c=\frac{\mathrm{\pi }}{4}\in \left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.


(vi)

The given function is$f\left(x\right)=\cos 2x$.

Since $\cos 2x$ is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$ is continuous on $\left[0,\mathrm{\pi }\right]$ and differentiable on $\left(0,\mathrm{\pi }\right)$.

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=1$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=\cos 2x \\ \Rightarrow f\text{'}\left(x\right)=-2\sin 2x \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow -2\sin 2x=0 \\ \Rightarrow \sin 2x=0 \\ \Rightarrow 2x=\mathrm{\pi } \\ \Rightarrow \mathrm{x}=\frac{\mathrm{\pi }}{2} \end{gathered}$$

Thus, $c=\frac{\mathrm{\pi }}{2}\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.


(vii)

The given function is $f\left(x\right)=\frac{\sin x}{e^x}$.

Since $\cos x\mathrm{and}{e}^x$ are everywhere continuous and differentiable, being the quotient of these two, $f\left(x\right)$ is continuous on $\left[0,\mathrm{\pi }\right]$ and differentiable on $\left(0,\mathrm{\pi }\right)$.

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=\frac{\sin x}{e^x} \\ \Rightarrow f\text{'}\left(x\right)=\frac{\cos x-\sin x}{e^x} \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow \frac{\cos x-\sin x}{e^x}=0 \\ \Rightarrow \cos x-\sin x=0 \\ \Rightarrow \tan x=1 \\ \Rightarrow x=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Thus, $c=\frac{\mathrm{\pi }}{4}\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.



(viii)

The given function is$f\left(x\right)=\sin 3x$.

Since $\sin 3x$ is everywhere continuous and differentiable, $\sin 3x$ is continuous on $\left[0,\mathrm{\pi }\right]$ and differentiable on $\left(0,\mathrm{\pi }\right)$.

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=\sin 3x \\ \Rightarrow f\text{'}\left(x\right)=3\cos 3x \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow 3\cos 3x=0 \\ \Rightarrow \cos 3x=0 \\ \Rightarrow 3x=\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2},.... \\ \Rightarrow x=\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{2},\frac{5\mathrm{\pi }}{6} \end{gathered}$$

Thus, $c=\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{2},\frac{5\mathrm{\pi }}{6}\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.


(ix)

The given function is$f\left(x\right)=e^{1-x^2}$.

Since exponential function is everywhere continuous and differentiable, $e^{1-x^2}$ is continuous on $\left[-1,1\right]$ and differentiable on $\left(-1,1\right)$.

Also,
$f\left(1\right)=f\left(-1\right)=1$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(-1,1\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=e^{1-x^2} \\ \Rightarrow f\text{'}\left(x\right)=-2x{e}^{1-x^2} \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow -2x{e}^{1-x^2}=0 \\ \Rightarrow x=0 \end{gathered}$$

Thus, $c=0\in \left(-1,1\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.


(x)

The given function is $f\left(x\right)=\log \left(x^2+2\right)-\log 3$, which can be rewritten as $f\left(x\right)=\log \left(\frac{x^2+2}{3}\right)$.

Since logarithmic function is differentiable and so continuous in its domain, $f\left(x\right)=\log \left(\frac{x^2+2}{3}\right)$ is continuous on $\left[-1,1\right]$ and differentiable on $\left(-1,1\right)$.

Also,
$f\left(1\right)=f\left(-1\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(-1,1\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=\log \left(\frac{x^2+2}{3}\right) \\ \Rightarrow f\text{'}\left(x\right)=\frac{3\left(2x\right)}{x^2+2}=\frac{6x}{x^2+2} \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow \frac{6x}{x^2+2}=0 \\ \Rightarrow x=0 \end{gathered}$$

Thus, $c=0\in \left(-1,1\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.


(xi)

The given function is$f\left(x\right)=\sin x+\cos x$.
Since $\sin x\mathrm{and}\cos x$ are everywhere continuous and differentiable, $f\left(x\right)=\sin x+\cos x$ is continuous on $\left[0,\frac{\mathrm{\pi }}{2}\right]$ and differentiable on $\left(0,\frac{\mathrm{\pi }}{2}\right)$.

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

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(0,\frac{\mathrm{\pi }}{2}\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=\sin x+\cos x \\ \Rightarrow f\text{'}\left(x\right)=\cos x-\sin x \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow \cos x-\sin x=0 \\ \Rightarrow \tan x=1 \\ \Rightarrow x=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Thus, $c=\frac{\mathrm{\pi }}{4}\in \left(0,\frac{\mathrm{\pi }}{2}\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.


(xii)
The given function is$f\left(x\right)=2\sin x+\sin 2x$.

Since $\sin x\&\sin 2x$ are everywhere continuous and differentiable, $f\left(x\right)$ is continuous on $\left[0,\mathrm{\pi }\right]$ and differentiable on $\left(0,\mathrm{\pi }\right)$.

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=2\sin x+\sin 2x \\ \Rightarrow f\text{'}\left(x\right)=2\cos x+2\cos 2x \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow 2\cos x+2\cos 2x=0 \\ \Rightarrow \cos x+\cos 2x=0 \\ \Rightarrow \cos x+2{\cos }^{2}x-1=0 \\ \Rightarrow 2{\cos }^{2}x+\cos x-1=0 \\ \Rightarrow \left(\cos x+1\right)\left(2\cos x-1\right)=0 \\ \Rightarrow \cos x=-1,\cos x=\frac{1}{2} \\ \Rightarrow \cos x=\mathrm{cos\pi },\cos x=\frac{\mathrm{\pi }}{3} \\ \Rightarrow x=\mathrm{\pi },\frac{\mathrm{\pi }}{3} \end{gathered}$$

Thus, $c=\frac{\mathrm{\pi }}{3}\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.


(xiii)


The given function is$f\left(x\right)=\frac{x}{2}-\sin \frac{\mathrm{\pi x}}{6}$.

Since $\sin x\&\frac{x}{2}$ are everywhere continuous and differentiable, $f\left(x\right)$ is continuous on $\left[-1,0\right]$ and differentiable on $\left(-1,0\right)$.

Also,
$f\left(-1\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(-1,0\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=\frac{x}{2}-\sin \frac{\mathrm{\pi x}}{6} \\ \Rightarrow f\text{'}\left(x\right)=\frac{1}{2}-\frac{\mathrm{\pi }}{6}\cos \frac{\mathrm{\pi x}}{6} \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow \frac{1}{2}-\frac{\mathrm{\pi }}{6}\cos \frac{\mathrm{\pi }x}{6}=0 \\ \Rightarrow \cos \frac{\mathrm{\pi }x}{6}=\frac{3}{\mathrm{\pi }} \\ \Rightarrow x=\frac{-6}{\mathrm{\pi }}{\cos }^{-1}\left(\frac{3}{\mathrm{\pi }}\right) \end{gathered}$$

Thus, $c=\frac{-6}{\mathrm{\pi }}{\cos }^{-1}\left(\frac{3}{\mathrm{\pi }}\right)\in \left(-1,0\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(xiv)


The given function is$f\left(x\right)=\frac{6x}{\mathrm{\pi }}-4{\sin }^{2}x$.

Since ${\sin }^{2}x\&x$ are everywhere continuous and differentiable, $f\left(x\right)$ is continuous on $\left[0,\frac{\mathrm{\pi }}{6}\right]$ and differentiable on $\left(0,\frac{\mathrm{\pi }}{6}\right)$.

Also,
$f\left(\frac{\mathrm{\pi }}{6}\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(0,\frac{\mathrm{\pi }}{6}\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=\frac{6x}{\mathrm{\pi }}-4{\sin }^{2}x \\ \Rightarrow f\text{'}\left(x\right)=\frac{6}{\mathrm{\pi }}-8\sin x\cos x \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow \frac{6}{\mathrm{\pi }}-8\sin x\cos x=0 \\ \Rightarrow \sin 2\mathrm{x}=\frac{3}{2\mathrm{\pi }} \\ \Rightarrow x=\frac{1}{2}{\sin }^{-1}\left(\frac{3}{2\mathrm{\pi }}\right) \end{gathered}$$

Thus, $c=\frac{1}{2}{\sin }^{-1}\left(\frac{3}{2\mathrm{\pi }}\right)\in \left(0,\frac{\mathrm{\pi }}{6}\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.



(xv)

The given function is$f\left(x\right)=4^{\sin x}$.

Since sine function and exponential function are everywhere continuous and differentiable, $f\left(x\right)$ is continuous on $\left[0,\mathrm{\pi }\right]$ and differentiable on $\left(0,\mathrm{\pi }\right)$.

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=1$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=4^{\sin x} \\ \Rightarrow f\text{'}\left(x\right)=4^{\sin x}\left(\cos x\right)\log 4 \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow 4^{\sin x}\left(\cos x\right)\log 4=0 \\ \Rightarrow 4^{\sin x}\cos x=0 \\ \Rightarrow \cos x=0 \\ \Rightarrow x=\frac{\mathrm{\pi }}{2} \end{gathered}$$

Thus, $c=\frac{\mathrm{\pi }}{2}\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.


(xvi)

According to Rolle’s theorem, if f(x) is a real valued function defined on [a, b] such that it is continuous on [a, b], it is differentiable on (a, b) and f(a) = f(b), then there exists a real number c ∈(a, b) such that f(c) = 0.

Now, f(x) is defined for all x ∈[1, 4].
At each point of [1, 4], the limit of f(x) is equal to the value of the function. Therefore, f(x) is continuous on [1, 4].

Also, exists for all x ∈(1, 4).

So, f(x) is differentiable on (1, 4).

Also,
f(1) = f(4) = 0

Thus, all the three conditions of Rolle’s theorem are satisfied.

Now, we have to show that there exists c ∈(1, 4) such that.

We have


$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow 2x-5=0 \\ \Rightarrow x=\frac{5}{2} \end{gathered}$$

[Since ∈(1, 4) such that]

Hence, Rolle’s theorem is verified.


(xvii)

The given function is $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$.
Since $\sin x\mathrm{and}\cos x$ are everywhere continuous and differentiable, $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$ is continuous on $\left[0,\frac{\mathrm{\pi }}{2}\right]$ and differentiable on $\left(0,\frac{\mathrm{\pi }}{2}\right)$.

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

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(0,\frac{\mathrm{\pi }}{2}\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)={\sin }^{4}x+{\cos }^{4}x \\ \Rightarrow f\text{'}\left(x\right)=4{\sin }^{3}x\cos x-4{\cos }^{3}x\sin x \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow 4{\sin }^{3}x\cos x-4{\cos }^{3}x\sin x=0 \\ \Rightarrow {\sin }^{3}x\cos x-{\cos }^{3}x\sin x=0 \\ \Rightarrow {\tan }^{3}x-\tan x=0 \\ \Rightarrow \tan x\left({\tan }^{2}x-1\right)=0 \\ \Rightarrow \tan x=0,{\tan }^{2}x=1 \\ \Rightarrow \tan x=0,\tan x=\pm 1 \\ \Rightarrow x=0,x=\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4} \end{gathered}$$

Thus, $c=\frac{\mathrm{\pi }}{4}\in \left(0,\frac{\mathrm{\pi }}{2}\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.


(xviii)

The given function is $f\left(x\right)=\sin x-\sin 2x$.

Since $\sin x\mathrm{and}\sin 2x$ are everywhere continuous and differentiable, $f\left(x\right)$ is continuous on $\left[0,\mathrm{\pi }\right]$ and differentiable on $\left(0,\mathrm{\pi }\right)$.

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=\sin x-\sin 2x \\ \Rightarrow f\text{'}\left(x\right)=\cos x-2\cos 2x \end{gathered}$$

$$\begin{gathered} \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow \cos x-2\cos 2x=0 \\ \Rightarrow \cos x-2{\cos }^{2}x+1=0 \\ \Rightarrow 2{\cos }^{2}x-\cos x-1=0 \\ \Rightarrow \left(\cos x-1\right)\left(2\cos x+1\right)=0 \\ \Rightarrow \cos x=1,\cos x=\frac{-1}{2} \\ \Rightarrow \cos x=\cos \frac{\mathrm{\pi }}{2},\cos x=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow x=\frac{\mathrm{\pi }}{2},\frac{2\mathrm{\pi }}{3} \end{gathered}$$

Thus, $c=\frac{\mathrm{\pi }}{2},\frac{2\mathrm{\pi }}{3}\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.​