A) At least one root
B) Exactly one root
C) At most one root
D) No root
Correct Answer: A
Solution :
| [a] Let |
| \[h(x)=\left| \begin{matrix} f(a) & f(x) \\ g(a) & g(x) \\ \end{matrix} \right|=f(a)g(x)-g(a)f(x)\] |
| Then, |
| \[h'(x)=f(a)g'(x)-g(a)f'(x)=\left| \begin{matrix} f(a) & f'(x) \\ g(a) & g'(x) \\ \end{matrix} \right|\] |
| 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, |
| \[h(a)=\left| \begin{matrix} f(a) & f(a) \\ g(a) & g(a) \\ \end{matrix} \right|=0,h(b)=\left| \begin{matrix} f(a) & f(b) \\ g(a) & g(b) \\ \end{matrix} \right|\] |
| \[\therefore \] From Eq. (i), \[\left| \begin{matrix} f(a) & f(b) \\ g(a) & g(b) \\ \end{matrix} \right|=(b-a)h'(c)\] |
| \[=(b-a)\left| \begin{matrix} f(a) & f'(c) \\ g(a) & g'(c) \\ \end{matrix} \right|\] |
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