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Let f(x) and g(x) are two function which are defined and differentiable for all x \geq x_0. If f\left(x_0\right)=g\left(x_0\right)and f^{\prime}(x)>g^{\prime}(x) for all x>0, then

 

Option: 1

f(x)<g(x) \text{ for some } x>x_0


Option: 2

f(x)=g(x) \text{ for some } x>x_0


Option: 3

f(x)>g(x) \text{ only for some } x>x_0


Option: 4

f(x)>g(x) \text{ for all } x>x_0


Answers (1)

best_answer

Consider \phi(x)=f(x)-g(x)
\Rightarrow \quad \phi^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)>0

\phi(x) is also continuous and derivable in \left[x_0, x\right].

Using LMVT for  \phi(x) \text { in } \left[x_0, x\right]

                    \phi^{\prime}(x)=\frac{\phi(x)-\phi\left(x_0\right)}{x-x_0}

Since, \phi^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x) are f^{\prime}(x)-g^{\prime}(x)>0  for all x>x_0 .

\begin{array}{lc} \therefore & \phi^{\prime}(x)>0 \\ \text { Hence, } & \phi(x)-\phi\left(x_0\right)>0 \\ & \phi(x)>\phi\left(x_0\right) \quad\left\{\phi\left(x_0\right)=f\left(x_0\right)-g\left(x_0\right)=0\right\} \\ f^{\prime}(x)-g^{\prime}(x)>0\end{array}
Hence, (d) is the correct answer.

Posted by

Pankaj

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