Question

Let $f:\left[\frac{1}{2},1\right]\to R$ (the set of all real numbers) be a positive non-constant and differentiable function such that $f\text{'}\left(x\right)<2f\left(x\right)$ and $f\left(\frac{1}{2}\right)=1$. Then the value of ${\int }_{0.5}^{1}f\left(x\right)dx$ lies in the interval

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

Answer 1

The correct option is D

$\left(0,\frac{[e-1)}{2}\right)$

Finding the intervals for the value ${\int }_{0.5}^{1}f\left(x\right)dx$

Step 1: Given data.

$f:\left[\frac{1}{2},1\right]\to R$

${\int }_{0.5}^{1}f\left(x\right)dx$

$f\text{'}\left(x\right)<2f\left(x\right)$

$f\left(\frac{1}{2}\right)=1$

Since, if we have inequality in differential equation we calculate that, if the function is increasing or decreasing.

we have,

$f\text{'}\left(x\right)<2f\left(x\right)$

Since, we have used the formula,

$$\begin{gathered} \Rightarrow \frac{dy}{dx}+Py\le 0 \\ \Rightarrow \text{integral factor}=e^{\int Pdx} \end{gathered}$$

For

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)-2f\left(x\right)<0 \\ \Rightarrow \text{integral factor}=e^{-\int 2dx} \end{gathered}$$

Step 2: Determining the function

When we have

$f\text{'}\left(x\right)-2f\left(x\right)<0$

Multiply $e^{-\int 2dx}$on both sides

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)e^{-2x}-2e^{-2x}f\left(x\right)<0 \\ \Rightarrow \frac{d}{dx}\left(f\left(x\right)e^{-2x}\right)<0 \\ \Rightarrow \varphi \left(x\right)=f\left(x\right)e^{-2x} \end{gathered}$$

Hence, we get the function which is decreasing from $x\in \left[\frac{1}{2},1\right]$

Step 3: Determining the interval

When, $x>\frac{1}{2}$

$$\begin{gathered} \Rightarrow \varphi \left(x\right)<\varphi \left(\frac{1}{2}\right) \\ \Rightarrow e^{-2x}f\left(x\right)<e^{-1}f\left(\frac{1}{2}\right) \\ \Rightarrow f\left(x\right)<e^{2x-1}\times 1\left[f\left(\frac{1}{2}\right)=1\right] \end{gathered}$$

Then,

$$\begin{gathered} 0<{\int }_{\frac{1}{2}}^{1}f\left(x\right)dx<{\int }_{\frac{1}{2}}^1{e}^{2x-1}dx \\ 0<{\int }_{\frac{1}{2}}^{1}f\left(x\right)dx<{\left[e^{2x-1}\right]}_{\frac{1}{2}}^1 \\ 0<{\int }_{\frac{1}{2}}^{1}f\left(x\right)dx<\frac{[e-1)}{2} \end{gathered}$$

Hence, we get the interval in which this ${\int }_{0.5}^{1}f\left(x\right)dx$lies $\left(0,\frac{[e-1)}{2}\right)$

Therefore, the correct answer is Option D.