MHTCET 15 May 2024 Shift 1 Paper

Concept:The equation involves $e^{\sin x}$, whose value is always positive and bounded because $\sin x$ lies between $-1$ and $1$.Explanation:Let $e^{\sin x} = y$, where $y > 0$.The equation becomes:$$y - \frac{1}{y} = 4$$Multiplying by $y$:$$y^2 - 4y - 1 = 0$$Solving:$$y = 2 \pm \sqrt{5}$$Since $y > 0$, we take only:$$y = 2 + \sqrt{5}$$Thus:$$e^{\sin x} = 2 + \sqrt{5}$$Now, $2 + \sqrt{5} \approx 4.236$, which is greater than $e \approx 2.718$.So:$$e^{\sin x} > e \Rightarrow \sin x > 1$$But $\sin x$ can never exceed $1$.Therefore, no real value of $x$ satisfies the equation.Answer:The equation has no real solutions.Correct option: D. no