Question

Let $f:[\frac{1}{2},1]⟶R$ (the set of all real numbers) be a positive, non-constant and differentiable function such that $f^{\prime }\left(X\right)<2f\left(X\right)$ and $f\left(\frac{1}{2}\right)=1$. Then the value of ${\int }_{1/2}^{1}f\left(X\right)dx$ lies in the interval

• A. $(2e-1,2e)$
• B. $(e-1,2e-1)$
• C. $(\frac{e-1}{2},e-1)$
• D. $\left(0\frac{e-1}{2}\right)$

Answer 1

The correct option is D $\left(0\frac{e-1}{2}\right)$

Given : $f^{\prime }\left(x\right)<2f\left(x\right)$

$\int \frac{f^{{}^{\prime }}\left(x\right)dx}{f\left(x\right)}<2\int dx\ln \left|f\left(x\right)\right|<2x+C$

Put : $x=\frac{1}{2}$

$\ln \left|f\left(\frac{1}{2}\right)\right|<1+C0=1+CC=-1f\left(x\right)<e^{(2x-1)}{\int }_{\frac{1}{2}}^{1}f\left(x\right)dx<{\int }_{\frac{1}{2}}^1{e}^{(2x-1)}dx={\left[\frac{1}{2}{e}^{(2x-1)}\right]}_{1/2}^1{\int }_{\frac{1}{2}}^{1}f\left(x\right)dx<\frac{1}{2}(e-1)0<{\int }_{\frac{1}{2}}^{1}f\left(x\right)dx<\frac{(e-1)}{2}$