Mathematics · Trigonometry Ratios and Identities

JEE Main 2024 — 31 January, Shift 2 — Question 15

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

  1. Option A:

    2

  2. Option B:

    more than 2

  3. Option C:

    1

  4. Option D:

    0

    Correct

Answer: D

Step-by-step solution

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

⇒t−2t=2\Rightarrow \mathrm{t}-\frac{2}{\mathrm{t}}=2

⇒t2−2t=2\Rightarrow \frac{\mathrm{t}^{2}-2}{\mathrm{t}}=2

⇒t2−2t−2=0\Rightarrow \mathrm{t}^{2}-2 \mathrm{t}-2=0

⇒t2−2t+1=3\Rightarrow \mathrm{t}^{2}-2 \mathrm{t}+1=3

⇒(t−1)2=3\Rightarrow(\mathrm{t}-1)^{2}=3

⇒t=1±3\Rightarrow \mathrm{t}=1 \pm \sqrt{3}

⇒t=1±1.73\Rightarrow \mathrm{t}=1 \pm 1.73

⇒t=2.73\Rightarrow \mathrm{t}=2.73 or −0.73(-0.73( rejected as t>0)\mathrm{t}>0)

⇒esin⁡x=2.73\Rightarrow \mathrm{e}^{\sin \mathrm{x}}=2.73

⇒log⁡eesin⁡x=log⁡e2.73\Rightarrow \log _{\mathrm{e}} \mathrm{e}^{\sin \mathrm{x}}=\log _{\mathrm{e}} 2.73

⇒sin⁡x=log⁡e2.73>1\Rightarrow \sin \mathrm{x}=\log _{\mathrm{e}} 2.73>1 So no solution.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Trigonometry Ratios and Identities
Topic
Periodicity of trigometric functions,Solutions of trigonometric equations
The number of solutions, of the equation e sin x -2 e -sin x =2 is | JEE Main 2024 PYQ with Solution · DhiX AI