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Question

Answers

${\text{A}}{\text{. }}f\left( x \right) > 0$ for all $x > {\text{0}}$

${\text{B}}{\text{. }}f\left( x \right) < 0$ for all $x > {\text{0}}$

${\text{C}}{\text{.}}$ No sign of $f\left( x \right)$ can be ascertained

${\text{D}}{\text{. }}f\left( x \right)$ is a constant function

Answer
Verified

Hint- Here, we will be using the concept of increasing functions.

Let \[g\left( x \right) = {e^{ - x}}f\left( x \right)\]

Differentiating above function with respect to $x$, we get

\[

\Rightarrow \dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] = \dfrac{d}{{dx}}\left[ {{e^{ - x}}f\left( x \right)} \right] = {e^{ - x}}\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] + f\left( x \right)\dfrac{d}{{dx}}\left( {{e^{ - x}}} \right) = {e^{ - x}}f'\left( x \right) - {e^{ - x}}f\left( x \right) = {e^{ - x}}\left[ {f'\left( x \right) - f\left( x \right)} \right] \\

\Rightarrow g'\left( x \right) = \dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] = {e^{ - x}}\left[ {f'\left( x \right) - f\left( x \right)} \right] \\

\]

Given, \[{\text{ }}f'\left( x \right) > f\left( x \right){\text{ }} \Rightarrow f'\left( x \right) - f\left( x \right) > 0\]

As we know that \[{e^{ - x}} > 0\] (always)

\[ \Rightarrow g'\left( x \right) = {e^{ - x}}\left[ {f'\left( x \right) - f\left( x \right)} \right] > 0\]

If the first derivative of any function is greater than zero, then it is an increasing function. From the above statement we can say that \[g\left( x \right)\] is an increasing function.

Now, for \[x > 0\] we can say that the value of the function at \[x\] is greater than the value of the function at 0 i.e., \[g(x) > g\left( 0 \right)\]

As given, \[f\left( 0 \right) = 0 \Rightarrow \]At \[x = 0\], \[g\left( 0 \right) = {e^{ - 0}}f\left( 0 \right) = 0\]

\[ \Rightarrow g(x) > g\left( 0 \right) \Rightarrow {e^{ - x}}f\left( x \right) > g\left( 0 \right) \Rightarrow \dfrac{{f\left( x \right)}}{{{e^x}}} > 0 \Rightarrow f\left( x \right) > 0\] provided \[x > 0\].

Hence, \[f\left( x \right) > 0\] for all \[x > 0\].

Therefore, option A is correct.

Note- In these types of problems we have to make an assumption in order to get to the results such as in this problem we assumed some other function as the given function multiplied by an exponentially decreasing term.

Let \[g\left( x \right) = {e^{ - x}}f\left( x \right)\]

Differentiating above function with respect to $x$, we get

\[

\Rightarrow \dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] = \dfrac{d}{{dx}}\left[ {{e^{ - x}}f\left( x \right)} \right] = {e^{ - x}}\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] + f\left( x \right)\dfrac{d}{{dx}}\left( {{e^{ - x}}} \right) = {e^{ - x}}f'\left( x \right) - {e^{ - x}}f\left( x \right) = {e^{ - x}}\left[ {f'\left( x \right) - f\left( x \right)} \right] \\

\Rightarrow g'\left( x \right) = \dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] = {e^{ - x}}\left[ {f'\left( x \right) - f\left( x \right)} \right] \\

\]

Given, \[{\text{ }}f'\left( x \right) > f\left( x \right){\text{ }} \Rightarrow f'\left( x \right) - f\left( x \right) > 0\]

As we know that \[{e^{ - x}} > 0\] (always)

\[ \Rightarrow g'\left( x \right) = {e^{ - x}}\left[ {f'\left( x \right) - f\left( x \right)} \right] > 0\]

If the first derivative of any function is greater than zero, then it is an increasing function. From the above statement we can say that \[g\left( x \right)\] is an increasing function.

Now, for \[x > 0\] we can say that the value of the function at \[x\] is greater than the value of the function at 0 i.e., \[g(x) > g\left( 0 \right)\]

As given, \[f\left( 0 \right) = 0 \Rightarrow \]At \[x = 0\], \[g\left( 0 \right) = {e^{ - 0}}f\left( 0 \right) = 0\]

\[ \Rightarrow g(x) > g\left( 0 \right) \Rightarrow {e^{ - x}}f\left( x \right) > g\left( 0 \right) \Rightarrow \dfrac{{f\left( x \right)}}{{{e^x}}} > 0 \Rightarrow f\left( x \right) > 0\] provided \[x > 0\].

Hence, \[f\left( x \right) > 0\] for all \[x > 0\].

Therefore, option A is correct.

Note- In these types of problems we have to make an assumption in order to get to the results such as in this problem we assumed some other function as the given function multiplied by an exponentially decreasing term.

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