Question
Let $$f:\left[ {\frac{1}{2},\,1} \right] \to R$$ (the set of all real number) be a positive, non-constant and differentiable function such that $$f'\left( x \right) < 2f\left( x \right)$$ and $$f\left( {\frac{1}{2}} \right) = 1.$$ Then the value of $$\int\limits_{\frac{1}{2}}^1 {f\left( x \right)dx} $$ lies
in the interval-
A.
$$\left( {2e - 1,\,2e} \right)$$
B.
$$\left( {e - 1,\,2e - 1} \right)$$
C.
$$\left( {\frac{{e - 1}}{2},\,e - 1} \right)$$
D.
$$\left( {0,\,\frac{{e - 1}}{2}} \right)$$
Answer :
$$\left( {0,\,\frac{{e - 1}}{2}} \right)$$
Solution :
We have,
$$\eqalign{
& f'\left( x \right) - 2f\left( x \right) < 0 \cr
& \Rightarrow {e^{ - 2x}}f'\left( x \right) - 2{e^{ - 2x}}f\left( x \right) < 0 \cr
& \Rightarrow \frac{d}{{dx}}\left( {{e^{ - 2x}}f\left( x \right)} \right) < 0 \cr
& \Rightarrow {e^{ - 2x}}f\left( x \right)\,{\text{is strictly decreasing function on }}\left[ {\frac{1}{2},\,1} \right] \cr
& \therefore {e^{ - 2x}}f\left( x \right) < {e^{ - 1}}f\left( {\frac{1}{2}} \right){\text{ or }}f\left( x \right) < {e^{2x - 1}} \cr} $$
Also given that $$f\left( x \right)$$ is positive function so $$f\left( x \right) > 0$$
$$\eqalign{
& \therefore 0 < f\left( x \right) < {e^{2x - 1}} \cr
& \Rightarrow 0 < \int\limits_{\frac{1}{2}}^1 {f\left( x \right)dx} < \int\limits_{\frac{1}{2}}^1 {{e^{2x - 1}}dx} \cr
& \Rightarrow 0 < \int\limits_{\frac{1}{2}}^1 {f\left( x \right)dx} < \left[ {\frac{{{e^{2x - 1}}}}{2}} \right]_{\frac{1}{2}}^1 \cr
& \Rightarrow \int\limits_{\frac{1}{2}}^1 {f\left( x \right)dx} \, \in \left( {0,\,\frac{{e - 1}}{2}} \right) \cr} $$