The number of solutions of the equation e^{si | vedclass

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

Question

(D) Let $e^{\sin x} = t$. Since $\sin x \in [-1, 1]$,$t$ must be in the range $[e^{-1}, e^1]$,i.e.,$[1/e, e] \approx [0.368, 2.718]$.Given equation: $

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $e^{\sin x} = t$. Since $\sin x \in [-1, 1]$,$t$ must be in the range $[e^{-1}, e^1]$,i.e.,$[1/e, e] \approx [0.368, 2.718]$.Given equation: $t - \frac{2}{t} = 2$.Multiplying by $t$: $t^2 - 2t - 2 = 0$.Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$t = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$.Since $t > 0$,we take $t = 1 + \sqrt{3} \approx 1 + 1.732 = 2.732$.We require $e^{\sin x} = 1 + \sqrt{3}$.Since $1 + \sqrt{3} \approx 2.732$ and $e \approx 2.718$,we have $1 + \sqrt{3} > e$.Because the maximum value of $e^{\sin x}$ is $e^1 = e$,the equation $e^{\sin x} = 1 + \sqrt{3}$ has no real solution for $x$.Thus,the number of solutions is $0$.

The number of solutions of the equation $e^{\sin x} - 2e^{-\sin x} = 2$ is

Similar Questions

If the roots of the equation $\frac{\alpha}{x - \alpha} + \frac{\beta}{x - \beta} = 1$ are equal in magnitude but opposite in sign,then $\alpha + \beta =$

The roots $\alpha, \beta$ of the equation $x^2-6(k-1)x+4(k-2)=0$ are equal in magnitude but opposite in sign. If $\alpha > \beta$,then the product of the roots of the equation $2x^2-\alpha x+6\beta(\alpha+1)=0$ is

One root of the given equation $2x^5 - 14x^4 + 31x^3 - 64x^2 + 19x + 130 = 0$ is

The roots of the equation $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$,where $a \in R$,are always:

The number of real values of $x$ for which the equality $|3x^2 + 12x + 6| = 5x + 16$ holds is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module