Question

Let $f:\left[ {\dfrac{1}{2},1} \right] \to R$ (the set of all real numbers) be a positive, non-constant and differentiable function such that f’(x) < 2 f (x) and $f\left( {\dfrac{1}{2}} \right)$ = 1. Then the value of $\int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} $ lies in the interval
$\left( a \right)\left( {2e - 1,2e} \right)$
$\left( b \right)\left( {e - 1,2e - 1} \right)$
$\left( c \right)\left( {\dfrac{{e - 1}}{2},e - 1} \right)$
$\left( d \right)\left( {0,\dfrac{{e - 1}}{2}} \right)$

Answer 1

Hint: Given data:
$f:\left[ {\dfrac{1}{2},1} \right] \to R$ (The set of all real numbers) be a positive, non-constant and differentiable function.
Therefore, f (x) > 0, (because f (x) is a positive function).
Now it is given that, $f'\left( x \right) < 2f\left( x \right)$
Therefore, $f'\left( x \right) - 2f\left( x \right) < 0$.................. (1)
Now as we know that ${e^x} > 0$ and ${e^{ - x}} > 0$, $\left[ {x \in R} \right]$
So multiply ${e^{ - 2x}}$ in equation (1) we have,
$ \Rightarrow {e^{ - 2x}}f'\left( x \right) - 2f\left( x \right){e^{ - 2x}} < 0$

Complete step-by-step solution:
Now as we know that $\dfrac{d}{{dx}}{e^{ - 2x}}f\left( x \right) = {e^{ - 2x}}f'\left( x \right) + f\left( x \right)\dfrac{d}{{dx}}{e^{ - 2x}} = {e^{ - 2x}}f'\left( x \right) + f\left( x \right){e^{ - 2x}}\left( { - 2} \right) = {e^{ - 2x}}f'\left( x \right) - 2f\left( x \right){e^{ - 2x}}$ so we have,
$ \Rightarrow \dfrac{d}{{dx}}{e^{ - 2x}}f\left( x \right) < 0$
So as the first derivative of ${e^{ - 2x}}f\left( x \right)$ is less than zero, so ${e^{ - 2x}}f\left( x \right)$ is a decreasing function.
So we can say that ${e^{ - 2x}}f\left( x \right) < {\left( {{e^{ - 2x}}f\left( x \right)} \right)_{x = \dfrac{1}{2}}}$ as $f:\left[ {\dfrac{1}{2},1} \right] \to R$
$ \Rightarrow {e^{ - 2x}}f\left( x \right) < {e^{ - 2\left( {\dfrac{1}{2}} \right)}}f\left( {\dfrac{1}{2}} \right)$
$ \Rightarrow {e^{ - 2x}}f\left( x \right) < {e^{ - 1}}f\left( {\dfrac{1}{2}} \right)$
Now it is given that $f\left( {\dfrac{1}{2}} \right)$ = 1, so we have,
$ \Rightarrow {e^{ - 2x}}f\left( x \right) < {e^{ - 1}}$
$ \Rightarrow f\left( x \right) < \dfrac{{{e^{ - 1}}}}{{{e^{ - 2x}}}}$
$ \Rightarrow f\left( x \right) < {e^{ - 1 + 2x}}$
Now as we calculated that f (x) > 0
$ \Rightarrow 0 < f\left( x \right) < {e^{ - 1 + 2x}}$
Now integrate the above equation from ($\dfrac{1}{2}$ to 1) we have,
$ \Rightarrow \int_{\dfrac{1}{2}}^1 {0dx} < \int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} < \int_{\dfrac{1}{2}}^1 {\left( {{e^{ - 1 + 2x}}} \right)dx} $
Now integrate it using the property that integration of zero is zero, and $\int {{e^{ax + b}}dx} = \dfrac{{{e^{ax + b}}}}{a} + c$, where C is some integration constant so we have,
$ \Rightarrow 0 < \int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} < \left[ {\dfrac{{{e^{ - 1 + 2x}}}}{2}} \right]_{\dfrac{1}{2}}^1$
Now apply integrating limits we have,
$ \Rightarrow 0 < \int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} < \left[ {\dfrac{{{e^{ - 1 + 2}}}}{2} - \dfrac{{{e^{ - 1 + 2\left( {\dfrac{1}{2}} \right)}}}}{2}} \right]$
$ \Rightarrow 0 < \int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} < \left[ {\dfrac{e}{2} - \dfrac{{{e^0}}}{2}} \right]$
Now as we know that something to the power zero is always 1, so we have,
$ \Rightarrow 0 < \int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} < \left[ {\dfrac{e}{2} - \dfrac{1}{2}} \right]$
$ \Rightarrow 0 < \int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} < \left[ {\dfrac{{e - 1}}{2}} \right]$
So, $\int_{\dfrac{1}{2}}^1 {f\left( x \right)dx} $ lies in the interval, $\left( {0,\dfrac{{e - 1}}{2}} \right)$.
So this is the required answer.
Hence option (d) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiation property such as, $\dfrac{d}{{dx}}mn = m\dfrac{d}{{dx}}n + n\dfrac{d}{{dx}}m,\dfrac{d}{{dx}}{e^{nx}} = n{e^{nx}}$ and always recall the basic integration property such as $\int {{e^{ax + b}}dx} = \dfrac{{{e^{ax + b}}}}{a} + c$, where C is some integration constant.