The number of real roots of the equation e^{s | vedclass

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

Question

(D) Given equation: $e^{\sin x} - e^{-\sin x} - 4 = 0$Let $y = e^{\sin x}$. Since $e^{-\sin x} = \frac{1}{y}$,the equation becomes:$y - \frac{1}{y} -

Vedclass · Accepted Answer

None

Vedclass · Answer

(D) Given equation: $e^{\sin x} - e^{-\sin x} - 4 = 0$Let $y = e^{\sin x}$. Since $e^{-\sin x} = \frac{1}{y}$,the equation becomes:$y - \frac{1}{y} - 4 = 0$Multiplying by $y$ $(y 
eq 0)$:$y^2 - 4y - 1 = 0$Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$y = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$Since $y = e^{\sin x} > 0$,we must have $y = 2 + \sqrt{5}$ (as $2 - \sqrt{5} Taking the natural logarithm on both sides:$\sin x = \ln(2 + \sqrt{5})$We know that $\sqrt{4} Since $e^1 \approx 2.718$,$\ln(2 + \sqrt{5}) > \ln(e) = 1$.Since $\sin x \leq 1$ for all real $x$,the equation $\sin x = \ln(2 + \sqrt{5})$ has no real solution.Therefore,the number of real roots is $0$ (None).

The number of real roots of the equation $e^{\sin x} - e^{-\sin x} - 4 = 0$ is

Similar Questions

$\sin 12^\circ \sin 24^\circ \sin 48^\circ \sin 84^\circ = $

If $0 \leq \theta \leq 2 \pi$,$0 \leq \alpha \leq 2 \pi$ and $\sec ^{2018} \theta + \operatorname{cosec}^{2018} \alpha = 2$,then the value of $\cos ^{2020} \theta + \sin ^{2022} \alpha =$

If $\cosh \alpha + \sinh \alpha = e^3$ and $\sinh x = \frac{\alpha}{\alpha+1}$,then $\tanh x =$

If $0 \leq x \leq \frac{\pi}{2}$, then the number of values of $x$ for which $\sin x - \sin 2x + \sin 3x = 0$ is

$\cos ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{3 \pi}{8}\right)+\sin ^3\left(\frac{\pi}{8}\right) \sin \left(\frac{3 \pi}{8}\right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module