Q.
If limx→0 aex−bcosx+ce−xxsinx=2, then a+b+c=
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a
2
b
4
c
0
d
6
answer is B.
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Detailed Solution
We observe that as x→0, numerator tends to a -b + c whereas the denominator tends to 0. Therefore, for the limit to exist we must have a−b+c=0 …(i)Now, if a−b+c=0, thenlimx→0 aex−bcosx+ce−yxsinxis in the 00 form.∴limx→0 aex+bsinx−ce−xsinx+xcosx=2 [Using L' Hospital's Rule] Here, the numerator is tending to a−c as x→0. Therefore for the limit to exist, we must have a−c=0 …(ii)limx→0 aex+bsinx−ce−xsinx+xcosx is in the form 00 ∴ limx→0 aex+bcosx+ce−x2cosx−xsinx=2⇒ a+b+c2=2⇒a+b+c=4
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