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 $0$.

List-$I$	List-$II$
$A$. All the roots are negative	$I$. $(b - 3)^2 = 36 + P^2$ for $P \in R$
$B$. Two roots are complex	$II$. $-3 < b < 9$
$C$. Two roots are positive	$III$. $b \in (-\infty, -3) \cup (9, \infty)$
$D$. All roots are real and distinct	$IV$. $b = 9$
	$V$. $b = -3$

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

Similar Questions

Solve the equation $3x^{2} - 4x + \frac{20}{3} = 0$.

Find the number of real solutions for the equation $\left( \frac{5}{7} \right)^x = -x^2 + 2x - 3$.

Let $t$ be a real number such that $t^2 = at + b$ for some positive integers $a$ and $b$. Then,for any choice of positive integers $a$ and $b$,$t^3$ is never equal to:

If the sum of the two roots of the equation $4x^3 + 16x^2 - 9x - 36 = 0$ is zero,then the roots are

Vedclass Test Series

Exam Paper Generator

Online Exam Module