Verified Solution • 2025–26 Session moderate

QJEEM24Ma36c4: JEE Main 2024 Mathematics - Class 11 JEEM24Ma36c4 Solution: The number of solutions, of the equation , is :

The number of solutions, of the equation esin⁡x−2e−sin⁡x=2e^{\sin x}-2 e^{-\sin x}=2 , is :

Multiple Choice Options

  • A 0
  • B 1
  • C 2
  • D more than 2

Step-by-Step Verified Solution

Take e^{\sin x}=t(t>0)

amp;⇒t−2t=2amp;⇒t2−2t=2amp;⇒t2−2t−2=0amp;⇒t2−2t+1=3amp;⇒(t−1)2=3amp;⇒t=1±3amp;⇒t=1±1.73amp;⇒t=2.73 or −0.73 (rejected as tgt;0)amp;⇒esin⁡x=2.73amp;⇒log⁡eesin⁡x=log⁡e2.73amp;⇒sin⁡x=log⁡e2.73gt;1\begin{aligned}& \Rightarrow \mathrm{t}-\frac{2}{\mathrm{t}}=2 \\& \Rightarrow \frac{\mathrm{t}^2-2}{\mathrm{t}}=2 \\& \Rightarrow \mathrm{t}^2-2 \mathrm{t}-2=0 \\& \Rightarrow \mathrm{t}^2-2 \mathrm{t}+1=3 \\& \Rightarrow(\mathrm{t}-1)^2=3 \\& \Rightarrow \mathrm{t}=1 \pm \sqrt{3} \\& \Rightarrow \mathrm{t}=1 \pm 1.73 \\& \Rightarrow \mathrm{t}=2.73 \text { or }-0.73 \text { (rejected as } \mathrm{t}>0) \\& \Rightarrow \mathrm{e}^{\sin \mathrm{x}}=2.73 \\& \Rightarrow \log _{\mathrm{e}} \mathrm{e}^{\sin \mathrm{x}}=\log _{\mathrm{e}} 2.73 \\& \Rightarrow \sin \mathrm{x}=\log _{\mathrm{e}} 2.73>1\end{aligned}

So no solution.

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