The number of solutions of equation esin x − 2e– sin x = 2 is
The number of solutions of equation esin x − 2e– sin x = 2 is
Option 1 -
More than 2
Option 2 -
2
Option 3 -
1
Option 4 -
0
-
1 Answer
-
Correct Option - 4
Detailed Solution:Take esinx = t (t > 0)
⇒
⇒
->t2 – 2t – 2 = 0
->t2 – 2t + 1 = 3
⇒ (t −1)2 = 3
⇒ t = 1 ±
⇒ t = 1 ± 1.73
⇒ t = 2.73 or –0.73 (rejected as t > 0)
⇒ esin x = 2.73
->loge esin x = loge 2.73
⇒ sin x = loge 2.73 > 1
So no solution.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 688k Reviews
- 1800k Answers