Question

If a tangent is drawn to the curve $y=e^{x+e^{x+e^{x^{+...\infty }}}}$ at $P(x,y)$ then the range of its slope is

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

Answer 1

The correct option is D $(-\infty ,-1)\cup (0,\infty )$

$y=e^{x+e^{x+e^{x^{+...\infty }}}}$
$\Rightarrow y=e^{x+y}$
Taking log both side, we have
$\ln y=(x+y)$
Differentiate both side w.r.t. $x$
$\frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{y}{1-y}=y^{\prime }$
$\Rightarrow y=\frac{y^{\prime }}{1+y^{\prime }}$

Since, $y$ is an exponential function so, $y>0$
$\Rightarrow \frac{y^{\prime }}{1+y^{\prime }}>0$

Case-1: $y^{\prime }>0$ and $1+y^{\prime }>0$
$\Rightarrow y^{\prime }>0$ and $y^{\prime }>-1$
$\Rightarrow y^{\prime }\in (0,\infty )$

Case-2: $y^{\prime }<0$ and $1+y^{\prime }<0$
$\Rightarrow y^{\prime }<0$ and $y^{\prime }<-1$
$\Rightarrow y^{\prime }\in (-\infty ,-1)$

$\therefore y^{\prime }\in (-\infty ,-1)\cup (0,\infty )$