Assertion (A): coth x = frac{1-k}{1+k} where | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$$\Rightarrow \frac{e^x + e^{-x}}{e^x - e^{-x}} = \frac{1-k}{1+k}$Applying Componendo and Dividendo:$\f

Vedclass · Accepted Answer

$(A)$ is false but $(R)$ is true

Vedclass · Answer

(D) Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$$\Rightarrow \frac{e^x + e^{-x}}{e^x - e^{-x}} = \frac{1-k}{1+k}$Applying Componendo and Dividendo:$\frac{(e^x + e^{-x}) + (e^x - e^{-x})}{(e^x + e^{-x}) - (e^x - e^{-x})} = \frac{(1-k) + (1+k)}{(1-k) - (1+k)}$$\Rightarrow \frac{2e^x}{2e^{-x}} = \frac{2}{-2k}$$\Rightarrow e^{2x} = -\frac{1}{k}$Since $e^{2x} > 0$ for all $x \in \mathbb{R}$ and $-\frac{1}{k} Reason $(R)$: The function $y = 	anh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$. As $x 	o \infty$,$y 	o 1$ and as $x 	o -\infty$,$y 	o -1$. The graph lies strictly between $y = -1$ and $y = 1$. Thus,the reason is true.

Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$ where $0 < k < 2$.
Reason $(R)$: The graph of $y = \tanh x$ always lies between the lines $y = -1$ and $y = 1$.
Choose the correct option:

Similar Questions

The graphs of the polynomial $x^{2}-1$ and $\cos x$ intersect

Let $[t]$ denote the greatest integer $\leq t$. Then the equation in $x$,$[x]^{2}+2[x+2]-7=0$ has

The real valued function $f(x) = \frac{|x-a|}{x-a}$ is

How many positive real numbers $x$ satisfy the equation $x^3-3|x|+2=0$?

Statement $-1$: Any function $f(x)$ is an even function if $f(-x) = f(x)$ for all $x$ in its domain.
Statement $-2$: The function $f(x) = \frac{1}{\sqrt{1 - x^2}} + \left[ \frac{x^2 + x + 1}{4} \right]$,where $[.]$ denotes the greatest integer function,is an even function.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$ where $0 < k < 2$.Reason $(R)$: The graph of $y = \tanh x$ always lies between the lines $y = -1$ and $y = 1$.Choose the correct option: