Question

If f(x) = $\begin{vmatrix} sin^3 x & sin^3 a & sin^3 b \\ xe^x & ae^a & be^b \\ \frac{x}{1+x^2} & \frac{a}{1+a^2} & \frac{b}{1+b^2} \end{vmatrix}$ where 0 < a < b < 2π, then show that the equation f '(x) = 0 has atleast one root in the interval (a, b)

Answer 1

The equation f'(x) = 0 has at least one root in the interval (a, b).

Answer 2

The function is given by $f(x) = \begin{vmatrix} \sin^3 x & \sin^3 a & \sin^3 b \\ xe^x & ae^a & be^b \\ \frac{x}{1+x^2} & \frac{a}{1+a^2} & \frac{b}{1+b^2} \end{vmatrix}$.

We want to show that the equation $f'(x) = 0$ has at least one root in the interval $(a, b)$, where $0 < a < b < 2\pi$. This can be shown using Rolle's Theorem applied to the function $f(x)$ on the interval $[a, b]$.

Rolle's Theorem states that if a function $g(x)$ is:

1. Continuous on the closed interval $[a, b]$.
2. Differentiable on the open interval $(a, b)$.
3. $g(a) = g(b)$.

Then there exists at least one value $c$ in $(a, b)$ such that $g'(c) = 0$.

Let's check if the function $f(x)$ satisfies the conditions of Rolle's Theorem on the interval $[a, b]$.

The entries of the determinant in the first column are $g_1(x) = \sin^3 x$, $g_2(x) = xe^x$, and $g_3(x) = \frac{x}{1+x^2}$. The entries in the second and third columns are constants with respect to $x$.

The functions $g_1(x)$, $g_2(x)$, and $g_3(x)$ are elementary functions.

• $g_1(x) = \sin^3 x$: The sine function is continuous and differentiable everywhere. Thus, $\sin^3 x$ is continuous and differentiable everywhere.
• $g_2(x) = xe^x$: The functions $x$ and $e^x$ are continuous and differentiable everywhere. Their product $xe^x$ is continuous and differentiable everywhere.
• $g_3(x) = \frac{x}{1+x^2}$: The numerator $x$ and the denominator $1+x^2$ are continuous and differentiable everywhere. The denominator $1+x^2 \ge 1$ for all real $x$, so it is never zero. Thus, the quotient $\frac{x}{1+x^2}$ is continuous and differentiable everywhere.

Since $0 < a < b < 2\pi$, the interval $[a, b]$ is a subset of the real numbers. Therefore, $g_1(x)$, $g_2(x)$, and $g_3(x)$ are continuous on $[a, b]$ and differentiable on $(a, b)$.

The function $f(x)$ is a determinant whose first column consists of functions of $x$ that are continuous on $[a, b]$ and differentiable on $(a, b)$, and the other two columns are constants. The determinant can be expanded as a linear combination of $g_1(x), g_2(x), g_3(x)$ with constant coefficients.

$f(x) = C_1 g_1(x) + C_2 g_2(x) + C_3 g_3(x)$, where $C_1, C_2, C_3$ are constants obtained from the minors of the first column.

Since $g_1(x), g_2(x), g_3(x)$ are continuous on $[a, b]$, $f(x)$ is continuous on $[a, b]$. Since $g_1(x), g_2(x), g_3(x)$ are differentiable on $(a, b)$, $f(x)$ is differentiable on $(a, b)$.

Now we check the third condition of Rolle's Theorem: $f(a) = f(b)$.

Let's evaluate $f(a)$:

$f(a) = \begin{vmatrix} \sin^3 a & \sin^3 a & \sin^3 b \\ ae^a & ae^a & be^b \\ \frac{a}{1+a^2} & \frac{a}{1+a^2} & \frac{b}{1+b^2} \end{vmatrix}$

In this determinant, the first column is identical to the second column ($C_1 = C_2$). A property of determinants states that if two columns are identical, the determinant is zero. So, $f(a) = 0$.

Let's evaluate $f(b)$:

$f(b) = \begin{vmatrix} \sin^3 b & \sin^3 a & \sin^3 b \\ be^b & ae^a & be^b \\ \frac{b}{1+b^2} & \frac{a}{1+a^2} & \frac{b}{1+b^2} \end{vmatrix}$

In this determinant, the first column is identical to the third column ($C_1 = C_3$). Therefore, the determinant is zero. So, $f(b) = 0$.

We have $f(a) = 0$ and $f(b) = 0$, which means $f(a) = f(b)$.

All three conditions of Rolle's Theorem are satisfied for the function $f(x)$ on the interval $[a, b]$.

Therefore, by Rolle's Theorem, there exists at least one value $c$ in the open interval $(a, b)$ such that $f'(c) = 0$.

This shows that the equation $f'(x) = 0$ has at least one root in the interval $(a, b)$.

Explanation of the solution:

The function $f(x)$ is a determinant with the first column depending on $x$ and the other two columns being constants based on $a$ and $b$. The entries in the first column are continuous and differentiable functions for all real $x$. Thus, $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$. Evaluating $f(x)$ at $x=a$ makes the first column identical to the second column, resulting in $f(a)=0$. Evaluating $f(x)$ at $x=b$ makes the first column identical to the third column, resulting in $f(b)=0$. Since $f(a)=f(b)=0$, the conditions of Rolle's Theorem are satisfied for $f(x)$ on $[a, b]$. By Rolle's Theorem, there exists at least one point $c \in (a, b)$ such that $f'(c)=0$. This proves that the equation $f'(x)=0$ has at least one root in $(a, b)$.