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Question : 27 of 160
Marks:
+1,
-0
Solution:
We know that
‌ sinh−1x=log(x+√x2+1)
‌∴ sinh−12=log(2+√4+1)=log(2+√5)
‌cosh−1x=log(x+√x2−1)
‌∴cosh−1√6=log(√6+√6−1)=log(√6+√5)
‌∴ sinh−12+cosh−1√6
‌=log(2+√5)+log(√6+√5)
‌=log[(2+√5)(√6+√5)]
‌e( sinh−1√2+cosh−1√6)=elog((2+√5)(√6+√5)]
‌=(2+√5)(√6+√5)=2√6+2√5+√30+5
‌=5+2√5+2√6+√30=5+2√5+2√6+√5√6
‌=5+(2+√6)√5+2√6=a+(b+√c)√a+b√c
‌∴a=5,b=2,c=6
‌∴‌‌a+b+c=5+2+6=13
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