frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2 | vedclass

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

Question

(A) Given expression: $E = \frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2}$.We know that $\sinh x = \frac{e^x - e^{-x}}{2}$,so $(e^x - e^{-x})^2 = 4\s

Vedclass · Accepted Answer

$\sinh^2 x + \coth^2 x$

Vedclass · Answer

(A) Given expression: $E = \frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2}$.We know that $\sinh x = \frac{e^x - e^{-x}}{2}$,so $(e^x - e^{-x})^2 = 4\sinh^2 x$.Thus,the denominator is $4(4\sinh^2 x) = 16\sinh^2 x$.Alternatively,let us evaluate option $A$: $\sinh^2 x + \coth^2 x$.$\sinh^2 x + \coth^2 x = \left(\frac{e^x - e^{-x}}{2}\right)^2 + \left(\frac{e^x + e^{-x}}{e^x - e^{-x}}\right)^2$$= \frac{(e^x - e^{-x})^4 + 4(e^x + e^{-x})^2}{4(e^x - e^{-x})^2}$$= \frac{(e^{2x} + e^{-2x} - 2)^2 + 4(e^{2x} + e^{-2x} + 2)}{4(e^x - e^{-x})^2}$$= \frac{(e^{4x} + e^{-4x} + 2 - 4e^{2x} - 4e^{-2x} + 2) + 4e^{2x} + 4e^{-2x} + 8}{4(e^x - e^{-x})^2}$$= \frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2}$.This matches the given expression. Hence,the correct option is $A$.

$\frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2} = \dots$

Similar Questions

$(\sec A + \tan A - 1)(\sec A - \tan A + 1) - 2\tan A = $

If $|\cos x + \sin x| + |\cos x - \sin x| = 2 \sin x$ for $x \in [0, 2\pi]$,then the maximum integral value of $x$ is:

Evaluate: $\sin 21^{\circ} \cos 9^{\circ}-\cos 84^{\circ} \cos 6^{\circ}$

If $\sin (\theta-\alpha), \sin \theta$ and $\sin (\theta+\alpha)$ are in $H.P.$,then the value of $\cos 2 \theta$ is

The sum $S = \sin \theta + \sin 2\theta + \dots + \sin n\theta$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module