Question

The interval increases of the function $y=x-e^x+\tan \left(\frac{\mathrm{\pi }}{7}\right)$ is

• A. $\left(-\infty ,0\right)$
• B. $\left(0,\infty \right)$
• C. $\left(-\infty ,\infty \right)$
• D. $\left(1,\infty \right)$

Answer 1

The correct option is A

$\left(-\infty ,0\right)$

Step 1: Differentiate the given function with respect to $x$.

Given function: $y=x-e^x+\tan \left(\frac{\mathrm{\pi }}{7}\right)$.

Differentiate the given function with respect to $x$.

$$\begin{gathered} \frac{\mathrm{d}}{dx}\left(y\right)=\frac{d}{dx}\left[x-e^x+\tan \left(\frac{\mathrm{\pi }}{7}\right)\right] \\ \Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-e^x\right)+\frac{d}{dx}\left(\tan \left(\frac{\mathrm{\pi }}{7}\right)\right)\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{(}\mathbf{u}\mathbf{\pm }\mathbf{v}\mathbf{)}\mathbf{=}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{\pm }\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\right] \\ \Rightarrow \frac{dy}{dx}=1-e^x+0 \\ \Rightarrow \frac{dy}{dx}=1-e^x \end{gathered}$$

Step 2: Find increasing intervals

Now, for increasing intervals $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{>}\mathbf{0}$.

$$\begin{gathered} \Rightarrow 1-e^x>0 \\ \Rightarrow e^x<1 \\ \Rightarrow \ln \left(e^x\right)<\ln \left(1\right) \\ \Rightarrow x<0 \\ \Rightarrow x\in \left(-\infty ,0\right) \end{gathered}$$

Hence, the function $y=x-e^x+\tan \left(\frac{\mathrm{\pi }}{7}\right)$ is increasing for $\left(-\infty ,0\right)$, so the correct answer is option (A).