Application of Rolle's theorem and determinant

Question

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

My approach First try to differentiate the give determinant and find that $f^{'}(x)=0$ then I apply the Rolle's theorem to solve the question.

I face the difficulty in differentiate the determinant.

Answer 1

I am quoting a rule to differentiate a determinant:

"You can differentiate the first row (or column) and keep the entries of the other row untouched. Then you can get a determinant. Do this for all other rows (or columns). Then find the sum of all such determinants."

Applying this rule here,

f'(x) = \begin{vmatrix} 3 \sin^2x \cos x & \sin^3 a & \sin^3 b \\ e^x + xe^x & ae^a & be^b\\ \frac{1-x^2}{(1+x^2)^2} & \frac{a}{1+a^2} & \frac{b}{1+b^2} \end{vmatrix}

However, this does not solve your problem. Let's look at the statement of Rolle's theorem:

If a real-valued function $f$ is continuous on a proper closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f (a) = f (b)$, then there exists at least one $c$ in the open interval $(a, b)$ such that

${\displaystyle f'(c)=0}$

Notice that in this case, $f(a) = f(b)$, right? [If you put $x = a$ or $x = b$, two columns become identical and the determinant vanishes]. That is all you need!

Answer 2

Let $\mathbf{c}$ be the vector-valued function which takes $x$ to $\left<\sin^3x,xe^x,x/(1+x^2)\right>$. Then $f(x)$ is the scalar triple product of $\mathbf{c}(x)$, $\mathbf{c}(a)$, and $\mathbf{c}(b)$: $$ f(x) = \mathbf{c}(x) \cdot \left(\mathbf{c}(a) \times \mathbf{c}(b)\right) $$ Because the cross product of two vectors in $\mathbb{R}^3$ is orthogonal to each of the individual vectors, $f(a) = f(b) = 0$. So by Rolle's Theorem there exists a point $x$ between $a$ and $b$ where $f'(x) = 0$.