Limits Continuity And Differentiability Question 92
Question: If the functions $ f(x) $ and $ g(x) $ are continuous in [a, b] and differentiable in (a, b), then the equation $ | \begin{matrix} f(a) & f(b) \\ g(a) & g(b) \\ \end{matrix} |=(b-a)| \begin{matrix} f(a) & f’(x) \\ g(a) & g’(x) \\ \end{matrix} | $ has in the interval [a, b]
Options:
A) At least one root
B) Exactly one root
C) At most one root
D) No root
Answer:
Correct Answer: A
Solution:
- [a] Let $ h(x)=| \begin{matrix} f(a) & f(x) \\ g(a) & g(x) \\ \end{matrix} |=f(a)g(x)-g(a)f(x) $ Then, $ h’(x)=f(a)g’(x)-g(a)f’(x)=| \begin{matrix} f(a) & f’(x) \\ g(a) & g’(x) \\ \end{matrix} | $ Since, $ f(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’(c)=\frac{h(b)-h(a)}{b-a}, $
$ \therefore h(b)-h(a)=(b-a)h’(c) $ ?. (i) Here, $ h(a)=| \begin{matrix} f(a) & f(a) \\ g(a) & g(a) \\ \end{matrix} |=0,h(b)=| \begin{matrix} f(a) & f(b) \\ g(a) & g(b) \\ \end{matrix} | $
$ \therefore $ From Eq. (i), $ | \begin{matrix} f(a) & f(b) \\ g(a) & g(b) \\ \end{matrix} |=(b-a)h’(c) $ $ =(b-a)| \begin{matrix} f(a) & f’(c) \\ g(a) & g’(c) \\ \end{matrix} | $
BETA
