The equation e^{sin x} - e^{sin(-x)} - 4 = 0 | vedclass

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

Question

(B) Given equation is $e^{\sin x} - e^{-\sin x} - 4 = 0$.Let $e^{\sin x} = t$. Since $\sin x \in [-1, 1]$,we have $t \in [e^{-1}, e^1]$,i.e.,$t \in [1

Vedclass · Accepted Answer

no real roots

Vedclass · Answer

(B) Given equation is $e^{\sin x} - e^{-\sin x} - 4 = 0$.Let $e^{\sin x} = t$. Since $\sin x \in [-1, 1]$,we have $t \in [e^{-1}, e^1]$,i.e.,$t \in [1/e, e]$.The equation becomes $t - \frac{1}{t} - 4 = 0$.Multiplying by $t$,we get $t^2 - 4t - 1 = 0$.Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $t = \frac{4 \pm \sqrt{16 + 4}}{2} = 2 \pm \sqrt{5}$.Case $1$: $t = 2 - \sqrt{5}$. Since $\sqrt{5} \approx 2.236$,$2 - \sqrt{5} Case $2$: $t = 2 + \sqrt{5}$. Since $\sqrt{5} \approx 2.236$,$t \approx 4.236$. Since $e \approx 2.718$,$t > e$. However,the maximum value of $e^{\sin x}$ is $e^1 = e \approx 2.718$. Since $4.236 > 2.718$,there is no real value of $x$ such that $e^{\sin x} = 2 + \sqrt{5}$.Therefore,the equation has no real roots.

The equation $e^{\sin x} - e^{\sin(-x)} - 4 = 0$ has

Similar Questions

If $k \in ( - \infty , - 2) \cup (2, \infty ),$ then the roots of the equation $x^2 + 2kx + 4 = 0$ are

The values of $a$ for which $(a^2 - 1)x^2 + 2(a - 1)x + 2$ is positive for any $x$ are

Let $\alpha$ and $\beta$ be the roots of $x^2 - \sqrt{2}x + 1 = 0$. Then the value of $\alpha^{50} + \beta^{50}$ is:

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{3} \times 13 = x^{2} \times 247$
$II.$ $y^{1 / 3} \times 14 = 294 \div y^{2 / 3}$

Solve the given two equations and select the correct option.
$I.$ $x^{3} - 468 = 1729$
$II.$ $y^{2} - 1733 + 1564 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module