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$e^{\sin x} - e^{- \sin x} = 4$ for

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some $x \in [0, π / 2]$

no real value of $x$

$some x \in (- π / 2, π / 2)$

all real values of $x$

answer is D.

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Detailed Solution

$\begin{array}{l} e^{\sin x} = 4 + \frac{1}{e^{\sin x}} \\ - 1 \leq \sin x \leq 1 and e > 1 \\ e^{- 1} \leq e^{\sin x} \leq e < 3 . \\ \Rightarrow e^{\sin x} < 3, also e^{\sin x} > 0 . \\ ∴ \frac{1}{e^{\sin x}} + 4 > 4 . \end{array}$

So two sides of (1) cannot be equal for any real value of $x .$

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