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B
ex−sinx
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C
ex+sinx
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D
ex−cosx
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Solution
The correct option is Cex+sinx Given, S=ex−sinx d2Sdx2⇒d(dSdx)dx
So, we have dSdx=d(ex−sinx)dx ⇒dexdx−dsinxdx=ex−cosx
Thus, d2Sdx2=d(ex−cosx)dx ⇒dexdx−dcosxdx=ex+sinx