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Question : 61 of 160
Marks:
+1,
-0
Solution:
Given, function
f(x)={‌| (eax−1)‌log(1+x) | | sin‌2x |
| ,‌ | x>0 |
‌| cos‌4‌x−cos‌b‌x | | tan2x |
| ,‌ | x=0 |
is contiuous at
x=0Since,
f(x) is continuous at
x=0So,
f(x)=f(x)=f(0) and
f(0)=2Now,
‌| (eax−1)‌log(1+x) |
| sin‌2x |
Applying L'hospital rule, twice, since it is an indeterminate form
‌.
So,
‌| aeax‌log(1+x)+ |
| 2‌cos‌x‌s‌i‌n‌x |
This limit is again
‌ form, so apply L'Hospital rule again.
‌‌a2eax‌log(1+x)+| aeax(1+x)−(eax−1) | | (1+x)2 |
+ |
| 2cos2x−2sin‌2x |
‌=‌=‌=aSince,
f(x)=f(0)⇒a=2Now,
‌| cos‌4‌x−cos‌b‌x |
| tan2x |
‌=‌| −2sin‌()sin‌() |
| tan2x |
‌=‌| −2sin‌()sin‌()⋅cos2x |
| â‹…x2 |
‌‌| ‌⋅()⋅()⋅cos2x |
| â‹…x2 |
‌=‌| −2⋅1⋅()⋅1⋅()⋅cos2x |
| 1â‹…x2 |
‌‌[∵‌=1]‌=‌⋅cos2x‌=‌⋅cos2x=‌⋅1‌‌[∵cos2x=1]‌=‌Also,
f(x)=f(0)⇒‌‌‌=2⇒b2=4+16=20Now,
√b2−a2=√20−22=√20−4=√16=4
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