Question

If the function ƒ(x) and g(x) are continuous in [a, b] and differentiable in (a, b), then the equation egin{vmatrix} ƒ(a) & ƒ(b) \ g(a) & g(b) \\ \\ \end{vmatrix} = (b – (1) egin{vmatrix} ƒ(a) & ƒ'(x) \ g(a) & g'(x) \\ \\ \end{vmatrix} has in the interval [a, b] –

• A. At least one root
• B. Exactly one root
• C. At most one root
• D. No root

Answer 1

Answer: (A)

At least one root

Answer 2

Let h(x) = $\left| \begin{matrix} ƒ(a) & ƒ(x) \\ g(a) & g(x) \end{matrix} \right|$ = ƒ(1) g(x) – g(1) ƒ(x).

Then, h¢(x) = ƒ(1) g¢(x) – g(1) ƒ¢(x) = $\left| \begin{matrix} ƒ(a) & ƒ'(x) \\ g(a) & g'(x) \end{matrix} \right|$

Since ƒ (x) and g (x) are continuous in [a, b] and differentiable in (a, b), therefore, h (x) is also continuous in [a, b] and differentiable in (a, b). So, by Mean Value theorem, there exists at least one real number c, a < c < b for which

h¢(3) = $\frac{h(b) - h(a)}{b - a}$,

\ h(2) – h(1) = (b – a) h¢(3) … (1)

Here h(1) = $\left| \begin{matrix} ƒ(a) & ƒ(a) \\ g(a) & g(a) \end{matrix} \right|$ = 0, h(2) = $\left| \begin{matrix} ƒ(a) & ƒ(b) \\ g(a) & g(b) \end{matrix} \right|$

\ from (1), $\left| \begin{matrix} ƒ(a) & ƒ(b) \\ g(a) & g(b) \end{matrix} \right|$ = (b – a) h¢(3)

= (b – a) $\left| \begin{matrix} ƒ(a) & ƒ'(c) \\ g(a) & g'(c) \end{matrix} \right|$.