AIEEE 2012 (Offline) 29th April Code AQuestion 1
The equation $e^{\sin x}-e^{-\sin x}-4=0$ has :
Trigonometric Ratio & Identities
Solution
Concept. $\sin x$ lies in $[-1,1]$, so $e^{\sin x}-e^{-\sin x}$ can never reach $4$. Formulas used. (1) $g(t)=e^t-e^{-t}$ is increasing, so for $|t|\le1$: $g(t)\le e-\frac1e$Step 1 - Bound. With $t=\sin x\in[-1,1]$, by (1) $$e^{\sin x}-e^{-\sin x}\le e-\frac1e\approx2.72-0.37=2.35.$$ Step 2 - Conclude. $2.35<4$, so the left side minus $4$ is always negative. No real root. Answer: (A)