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Question : 27 of 160
Marks:
+1,
-0
Solution:
‌‌ Given, ‌2tanh−1x= sinh−1(‌)‌tanh−1(x)=‌‌ln(‌)‌⇒‌‌2tanh−1(x)=ln(‌)‌‌ So, ‌ln(‌)= sinh−1(‌)‌=ln(‌+√(‌)2+1)‌[∵ sinh−1(y)=ln(y+√y2+1)]‌=ln(‌+√‌+1)‌=ln(‌+‌)⇒ln‌=ln(3)‌=‌=3⇒1+x=3−3x‌=4x=2⇒x=‌Now,
cosh−1(‌)=cosh−1(4‌=ln(2+√4−1)‌‌‌[∵cosh−1(z)=ln(z+√z2−1)]‌=ln(2+√3)
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