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Question : 131 of 150
Marks:
+1,
-0
Solution:
Given,
‌| sin‌(sin‌x)−sin‌x |
| ax3+bx5+c |
=−‌ . . . (i)
⇒‌| sin‌sin‌0−sin‌0 |
| a(0)3+b(0)5+c |
=−‌⇒‌=−‌⇒c=0Applying L' Hospital's rule in Eq. (i), we get
‌‌| cos(sin‌x)‌cos‌x−cos‌x |
| 3ax2+5bx4+0 |
=−‌‌⇒‌| cos‌x(cos(sin‌x)−1) |
| x2(3a+5bx) |
=−‌ ‌⇒‌cos‌x‌=−‌‌⇒‌×‌| 2sin‌2() |
| 4(‌)2×(‌)2 |
‌⇒‌‌‌×‌×‌=−‌‌‌(∵‌=1)‌⇒‌‌‌=−‌⇒‌‌6a=−12‌∴‌‌a=−2Hence,
a=−2,b∈R and
c=0
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