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Question : 25 of 160
Marks:
+1,
-0
Solution:
Consider the expression,
2‌cos‌h‌2‌x+10‌sin‌h‌2‌x=5 Then
2(‌)+10(‌)‌‌=5 2e2x+2e−2x+10e2x−10e−2x‌‌=10 12e2x−8e−2x−10‌‌=0 6e2x−4−2x−5‌‌=0 Further simplify the above,
6(e2x)2−8e2x+3e2x−4‌‌=0 (2e2x+1)(3e2x−4)‌‌=0 3e2x−4‌‌=0,2e2x+1≠0 e2x‌‌=‌ Further simplify the above,
2x=log‌ x=‌‌log‌
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